Alekhya

Sets, Function & Relations

Sets Introduction:

1. A set is a group of elements or objects that share at least one characteristic or property.
2. The objects within one set are similar in some respect.
3. Mathematically, a set can be defined as a group of well-defined objects that are similar in some respect. These objects are called the elements of the set.
4. For example, consider the set of vowels in the English alphabet. Let this set is represented by the letter V. This set contains 5 elements- a, e, i, o, u. this can also be written as:
   V ≡ {a, e, i, o, u}
5. The following two conventions should be adhered to while denoting a set
   1. A set must be denoted by a capital letter.
   2. The elements of the set must be denoted in small letters.
6. A few examples of sets are given below:
   A ≡ {1,2,3,4,5}
   B≡ {a,b,c,d,e}
   In the above examples, A and B are the names of the sets while 1,2,3,4,5, and a,b,c,d,e are the elements belonging to these sets respectively.
7. The symbol ∈ is used to denote that an element belongs to a particular set. Hence, to denote that 1 belongs to set A, write it as: 1 ε A
   ε Is a Greek letter called epsilon.
8. In number systems, numbers can be classified in various ways as per their characteristics. The standard notations used to denote the commonly used sets of numbers are as shown below:

Read and Learn More CA Foundation Maths Solutions

N ≡ set of all natural numbers = {1,2,3,4,5 … )
W ≡ set ofall whole numbers= {0,1,2,3,4, … }
I ≡ set ofall Integers = {…..-3, -2, -1,0,1,2,3,…}
Q ≡ set ofall rational numbers
R ≡ set of all real numbers
P ≡ set ofall prime numbers
C ≡ set ofall complex numbers

Cardinality of Sets

1. Cardinality of a set refers to the total number of elements that are present in the set. For example, V= {a, e, i, o, u}
2. The cardinality of this set of vowels V is 5 as the total number of elements in the set is 5, This can be represented as: |V| = 5 or n(V) = 5

Classification of Sets based on Cardinality

Based on the concept of cardinality, sets can be classified into the following types:

Null Set/ Void Set/ Empty Set

Consider the set B≡ { )

This set has no elements. Hence, the cardinality of this set is zero.

This can be represented as:

|B| =0 or n (B) =0

Such a set is called a null set or a void set or an empty set. A null set can be represented as:

B≡ ∅

Infinite Set

Consider the set of natural numbers.

N= set of all natural numbers= {1, 2, 3, 4….}

This set does not have a final element. The number of elements n this set cannot be counted and is hence, infinite.

Such sets, where the cardinality of the set cannot be exactly determined, are called infinite sets. All the standard sets mentioned above are infinite sets.

Finite Set

Consider the set of the factors of a number, say 12. The factors of 12 are 1, 2, 3, 4, 6, and 12.

If this set is denoted by S, then S = {1,2,3,4,6,12}.

The cardinality of this set is 6. Since this set has a finite number of elements, it is known as a finite set. Hence, sets which have a finite number of elements are called finite sets.

Singleton Set /Singlet

A set which has one and only one element in it is called a singleton set or a singlet.

For example,

1. A≡ {a}
2. B≡ {0}

In the above cases, A and B are both singleton sets.

Note:

In the second example, B is not an empty set. It contains the element 0.

B would have been an empty set ifit did not contain any elements, i.e., B ≡ { }

Equivalent and Equal Sets

Equivalent sets

Two finite sets are said to be equivalent if their cardinality is the same, i.e., n(A) = n (B)

For example, if A ≡ {1,2,3,} and B ≡ {4,5,6}

Then, n {A} = n (B} = 3

Hence, A and B are equivalent sets.

Equal sets

Two sets are said to be equal if and only if they contain the same elements.

This implies that every elements of set B is present in set A and every element of set A is present in set B

For example, if A ≡ {a, b, c} and B ≡{c, b, a} then A and B are said to be equal sets.

Equality of sets is denoted as A = B. Note that the order in which the elements of the two equal sets are written is not important.

Two equal sets are always equivalent, but two equivalent sets may or may not be equivalent.

Representation of a Set

There are two ways of representing a set. They are,

1. Roster notation
2. Set Builder notation (or Rule Method)

Roster Notation

Consider the set of vowels:

V ≡ (a, e, i, o, u}

Here, each vowel is individually listed down in the set V.

This method in which each element of a set is specified within the set is called the Roster method.

Note:

To denote an infinite set (such as the set of natural numbers) using the Roster method, write the first few elements followed by ‘…’ to indicate continuation up to infinity.

N ≡ (1, 2,3 }

Set Builder Notation

In the set Builder notation, instead of writing down every single element of a set, the common property of the elements of the set is described.

For example, in the Roster notation the set of the first five natural numbers is written as follows:

D≡ {1,2,3,4,5}

Now, if one were to describe this set in words, the description would say, “set D is a set of all elements x such that x is a natural number and x lies between 1 and 5 (both inclusive)”.

This is written in set Builder notation as follows:

D≡ {x|x ∈ N and 1 ≤ x ≤ 5}

Note:

The symbol | is read as “such that”.

Consider another example,

V≡{a, e, i, o, u)

This can be written as “the set V is a set of elements x, such that x is a vowel from the English alphabet.”

This can be written in set Builder notation as: V≡ {x|x is a vowel from the English alphabet)

Hence, in general, if all the elements of a set A have a common property B, then the set can be expressed in the set builder form in the following way:

A≡{x|x has property B}

Examples:

1. Convert the following sets from the set Builder notation to the Roster notation: P= {y|y ∈ w and y < 3)

Solution:

Here, P is a set of all numbers y, such that y is a whole number less than 3. The whole numbers less than 3 are 0,1 and 2. Hence, set P can be written in Roster notation as: P ={0,1,2}

2. Convert the following sets from the set Builder notation to the Roster notation: M ={x|x²- 5x + 6 = 0}

Solution:

Here, M is a set of all such numbers x that satisfy x² -5x + 6 = 0. To find the values of x, solve this quadratic equation.

x² – 5x + 6 = 0

x²- 3x- 2x + 6 = 0

x(x- 3)- 2(x- 3) = 0

(x – 3)(x- 2) = 0

X=3 or x=2

Hence, the set M can be written in the Roster form as: M ≡ {2,3}

Subsets and Supersets

Consider the set ofall students in a particular college. Now, the set of first year students studying in the same college is a part of this bigger set of all students.

Such a set, which is entirely present in a bigger set, is called a subset of the bigger set.

The bigger set is called the superset of the smaller set. Each element of the subset is definitely an element of the superset as well.

However, every element of the superset need not be part of the subset.

A is said to be the superset of B.

Consider the following example:

A≡ {1,2,3,4,5,6} and B≡ {1,2,3,}

Note that every element of the set B is also an element of the set A.

Hence, B is a subset of A or B is contained in A. A is said to be the superset of B.

Note:

A Null set is a subset of every set. Eveiy set is a subset of itself.

Mathematically, a subset is defined as: A is a subset of B if and only if every element of A is an element of B.

This is denoted as follows:

A⊆ B

For example,

1. A ≡ {1,2}, B ≡ { 1,2}
2. A ≡ {1,2,3}, B ≡ {1,2,3,4} in both cases, A £ B

In the first example,

B is also a subset of A. Hence, A ⊆ B and B ⊆ A.

∴ Set B = set A

Hence, both the sets are equal. Hence, two equal sets are subsets of each other.

Such a subset is called an Improper subset. Subset A is an improper subset of B if A = B.

In the second example,

B is not a subset of A, which can be denoted by B ⊄ A.

Such a subset, where the superset also contains other elements, is called a proper subset.

Set A is said to be a proper subset of B if A ⊆ B but B ⊆ A.

A proper subset is denoted by the symbol ⊂. Thus, in the second example A⊂B.

Power Set

A Power set is the set of all subsets of a set. Consider a set A≡ {1,2,3}

Now, consider a set {1}. This set is a subset of A. similarly, {1, 2} is also a subset of A similarly, list down all the subsets of A. they would be:

{1}
{2}
{3}
{1,2}
{1,3}
{2,3}
{1,2,3} …(Every set is a subset of itself)
{ } …. (Null set is a subset of every set)

Total number of subsets = 8

Hence, a set of these subsets is called the power set of A.

P≡{{ 1 },{2},{3},{1,2},(1,3),(2,3),{1,2,3},{ } } If is the number of elements (cardinality) in a set is n, then the cardinality of the power set is given by 2ⁿ.

Universal Set

Consider a college, consisting of students from three streams- Arts (set A), Commerce (set C) and Science (set S). All these students together form the set of all the students in the entire
college. Such a set is called a Universal set (set U).

A universal set is a set containing all the elements under consideration. It is represented by the capital letter U.

3. The universal set for the set of all numbers is as follows:

U≡ {-∞…….. ∞}. It includes all natural numbers, integers, rational numbers, irrational numbers, and so on.

Complement of a Set

The concept of complement of a set logically flows from the concept of a universal set.

The complement of a set A, which is a subset of a universal set IJ, includes all the elements present in U which are not present in A. it is denoted by A’.

Suppose,

U ≡ (a,b,c,d,e,f,g,h,i) and

A ≡ {a,b,c,d,e},

Then A’, or complement of the set A ≡ {f,g,h,i}

Union and Intersection of Sets

Consider the following example.

In a room, there are 5 people a, b, c, d, e. Out of them, a, b and c are men while d and e are women. Also, a and e study science while b, c and d study commerce.

∴ The set of males is: M ≡ (a,b, c)

The set of females is: F≡ {d, e)

The set of science students is S ≡ (a, e)

The set of commerce students is C≡(b,c,d}

If the number of female students who have taken science is to be found, find the elements common to the set of females (F) and the set of Science students (S).

This is called an intersection of set F and set S and is denoted by F ∩ S.

Hence, in the above case, F ∩ S ≡ (e)

Thus, an intersection of two sets is formed by the elements that are common to both the sets.

Similarly, consider sets M and F. There is no common element between them,

Hence, M ∩ F =ø

Such sets which have no element in common are called disjoint sets.

Now, consider the females who have not taken science. Here, check the set F and remove all elements ofset S present In this set. This is called the difference between two sets.

F – S ≡ {d}

Thus, difference of set A and set 13 is defined as the elements present in A blit not present in B.

A – B ≡ { x| x ∈ A and x ∉ B}

Now, suppose one wants to represent a set containing “either males or commerce students or both”. This would imply taking all the elements from set M and set C together into one set. This is called the union of set M and set C and is denoted by M∪C.

Thus, M∪C ≡{a,b,c,d}

Thus, the elements common to both M and C are not counted twice. Thus, the elements belonging to the intersection set are subtracted from the union set.

Hence, M∪C = M+C – M ∩ C

Note:

Though b and c exist in both sets, they are written only once while writing the union. This is because no element is ever written twice while writing a set.

Venn Diagrams

1. Sets can also be represented using diagrams known as Venn diagrams.
2. This method of representing sets is especially useful when multiple sets are given.
3. The closed figure corresponding to a set is used to denote that only the elements within this
   closed figure belong to this set.
4. Any element lying outside this figure does not belong to this set. Though sets can be represented in the form of circles, triangles, rectangles and so on, the most commonly used figure in Venn Diagrams is a circle.
5. Consider the following example.
   Let U be the universal set containing all the whole numbers between (and not including) 0 and 11.

Hence, U ≡ (1,2,3,4,5,6,7,8,9,10}

Let P be the set containing all the prime numbers between 0 and 11 (not including the two).

Thus. P ≡ (2,3,5,7)

Let E be the set containing all the even numbers between 0 and 11.

Hence, E = (2,4,6,8,10)

Hence P U E ={2,3,4,5,6,7,8,10}

Also, P ∩ E = {2}

This can be represented using a Venn diagram in the following manner.

In the above diagram, the universal set is represented by the rectangle while all the other sets are represented by circles.

The common portion between the two sets P and E is the intersection of these two sets.

The universal set contains two numbers, 1 and 9, which do not belong to the set of prime or even numbers. Hence, they are written outside both the circles but within the universal set.

Now, consider the cardinality of these sets.

N(P) = 4, n(E) = 5 and n (PUE) = 8

Thus if we observe, n (PUE)≠ n(P)+n(E)

As explained earlier, the union set is obtained by adding the individual sets and subtracting the intersection set from the sum of individual sets.

Hence, n (P U E) = n (P) +n (E) – n [P ∩ E)

i.e., n (P U E) = 4 +5 -1 = 8

To generalize this, for any two sets A and B,

n (A U B) =n (A) +n(B) -n (A∩B)

Solved Examples

3. In a class of 100 students, 60 play foot balls, 40 play hockey while 10 do not play any sport. Find the number of students who play both hockey and football.

Solution:

N (U) = 100

Since 10 students do not play any game. Let F and H be the set of students playing football and hockey respectively,

n(FUH) = 100-10 =90

∴ n (F) =60

∴ n (H) = 40

The set of students who play both the games is n (F ∩ H)

∴ n(FUH) = n(F)+n(H)-n(F∩H)

∴ 90= 60+40-n (F ∩ H)

Hence, n (F ∩ H) = 100-90 = 10

∴10 students play both football and hockey.

Similarly, consider a universal set as shown below:

U≡ {1, 2, 3,4, 5, 6, 7,8, 9,15)

Let A be the set of all even numbers. Hence, A≡ {2,4,6,8}

Let B be the set ofal perfect squares. Hence, B ≡ {1,4,9}

Let C be the set ofall composite numbers. Hence, C ≡{4, 6, 8, 9, 15}

This can be represented using a Venn diagram as follows:

Such a Venn diagram is also known as a three-level diagram.

For a three-level diagram, the formula is

n (A U B U C) =n (A)+ n(B) +n(c)- n(A ∩ B) -n(A ∩ C) – n(B ∩ C)+ n (A ∩ B ∩ C)

So in this case,

n(AUBUC) = 4 + 3 + 5-1- 3- 2 +1 = 7

4. Mow many numbers lying in the range 5000 to 10000 are divisible by 7,11 or 137

Solution:

Let A, B and C be the set of numbers divisible by 7,1 1 and 1 3 respectively.

Lowest number in set A = 5005 and the highest number in A = 9996

∴ Number of Elements in A = \(\frac{9996-5005}{7}+1=714\)

Lowest number in set B = 5005 and the highest number in B = 9999

∴ Number of elements in B= 455

Lowest number in C= 5005 and the highest number in C= 9997

∴ Number of Elements in C = 385

Similarly,

Number of elements divisible by both 7 and 11=Number of elements in the set A ∩ B =65

Number of elements in the set A ∩ B =65

Number of elements in the set A ∩ C =55

Number of elements in the set B ∩ C =35

Number of elements in the set A ∩ B ∩ C = 5

∴Number of elements in the set A U B U C

n(A U B U C) = n(A) + n(B) + n(C)- n(A ∩ B)- n(A ∩ C)- n(B ∩ C) + n(A ∩ B ∩ C)

n(A U B U C) = 714+ 455+ 385 – 65 – 55- 35 +5 = 1404

5. Draw Venn diagrams for:

1.  A – B – C
2. A ∩ B ∩ C’
3. A U B U C’

Solution:

1) A – B – C

The shaded part comprises all the elements belonging to the set A – B – C

2) A ∩ B∩ C’

3) A ∪ B ∪ C’

6. Express the following as Venn diagrams:

1. Every bull is an animal
2. Some bulls are in the stock market
3. No animal is a bull

Solution:

Let A = set of Animals, B = set ofbulls, C= set of things in the stock market

Every bull is an animal

The first figure above is the general form of representing “Every bull is an animal”.

The second figure is true if and only if set B = set A i.e., every bull is an animal and vice versa. In the first figure, some animals may not be bulls.

Some bulls are in the stock market.

This can be represented in 3 different ways as shown

1: only some elements are common to set B and set S.

2: set S is a subset of set B i.e., all the things in the stock market are bulls. Hence, only some bulls will be things in the stock market.

3: set S = set B

No animal is a bull

Here, A and B are disjoint sets.

7. Three of a popular actor’s films, A, B and C, were nominated for the best film award. 60 persons were asked which film deserved the award the most. 16 people thought only C deserved it. The number of people who thought that all the three films equally deserved the award was twice the number of people who thought that B and C deserved the award but A did not. The number of people who thought that B deserved the award was equal to twice the number ofpeople who thought all the three equally deserved it. The number of people who thought that A and B, but not C deserved the award was equal to the number ofpeople who thought that only A or A and C but not B deserved it. Find the maximum number ofpeople who thought that all the three films deserved the award?

Solution:

From the given data, we can draw the following Venn diagram.

Here, x denotes the number of people who thought that all the three films deserved the award and m denotes the number of people who thought that oniy A or both A and C but not B deserved the award.

Hence, in the figure, m indicates the sum of the number of people who think that oniy A should get the award and the number of people who think that both A and but not B should get the award.

Thus,

2m+x -m +2x+x+16=60

∴ m+4x=44

∴ x is maximum when m = 0

∴ x = 11

∴2x = 22

The maximum no. of people who thought that all the three films deserved the award =22

Properties of Sots

1. Union of any sot with a null set is equal to the original set. A ∪ ø = A
2. Intersection ofany set with a null set is equal to a null set. A∩ ø = 0
3. The intersection of any set with its complement is a null set. A ∩A’ = 0
4. The union ofany set with its complement is the universal set. A U A = U
5. (A ∩ B)’ = A‘ U B’
6. (A ∪ B)’ = A’ ∩ B’
   The last two properties are called De Morgan’s laws.
7. Associative Property
   (A U B) U C = A U (B U C)
   (A ∩ B) ∩ C = A ∩ (B ∩ C)
8. Distributive Property
   A U (B ∩ C) = (A U B) ∩ (A U C)
   A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Note:

For any two sets A and B,

n(A ∪ B) = n(A) +n(B)- n (A ∩ B)

For any three sets A, B and C, n(A U B U C) = n(A) + n{B) + n(C)- n(A ∩ B)- n(A ∩ C)-n(B ∩ c) + n(A ∩ B ∩ C)

Exercise – 1: Sets

Choose the most appropriate option or options (1) (2) (3) or (4).

1. The number of subsets of the set {2, 3, 5} is

1. 3
2. 8
3. 6
4. none of these

Answer: (2)

No of subsets of a set with n elements = 2ⁿ

2. The number of subsets of a set containing n elements is

1. 2ⁿ
   
2. 2^-n
3. n
4. none of these

Answer: (1) 2ⁿ

No of subsets of a set with n elements = 2ⁿ

3. The null set is represented by

1. \(\{\Phi\}\)
2. {0}
3. ø
4. none of these

Answer: (3) ø

Null set = ∅

4. A = (2, 3, 5, 7), B { 4, 6, 8, 10} then A ∩ B can be written as

1. { }
2. \(\{\Phi\}\)
3. (AUB)’
4. None of these

Answer: (1) { }

There is no element common to A & B

A ∩ B = { }

5. The set {x|0<x<5} represents the set when x may take integral values only

1. {0, 1, 2, 3, 4, 5}
2. {1,2, 3, 4}
3. {1, 2, 3, 4, 5}
4. none of these

Answer: (2){1, 2,3,4}

{x|0 < x < 5}

X can take 1, 2, 3, 4 but not 0, 5

6. The set {0, 2, 4, 6, 8, 10} can be written as

1. {2x | 0<x <5}
2. (x : 0<x<5}
3. {2x : 0≤x≤5}
4. none of these

Answer: (3} {2x : 0 ≤ x ≤ 5}

{0,2(1}, 2(2), 2(3), 2(4), 2(5)}

(2x : 0 ≤ x ≤ 5}

Using Q.7 to Q.10 If P = (1, 2, 3, 5, 7}, Q = (1, 3, 6, 10, 15},
Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

7. The cardinal number of P ∩ Q is

1. 3
2. 2
3. 0
4. none of these

Answer: (2)2

P ∩ Q = (1, 3} -> 2 elements

8. The cardinal number of P U Q is

1. 10
2. 9
3. 8
4. none of these

Answer: (3) 8

Cardinal no ->n(P U Q}

n(PUQ} = n(P) + n (Q) – n (P ∩ Q}

=5+5-2=8

9. n (P’) is

1. 10
2. 5
3. 6
4. none of these

Answer: (1) 10

n (P’}) = n (Universal set} – n (P)

= 15-5 = 10

10. n(Q’) is

1. 4
2. 10
3. 4
4. none of these

Answer: (2)10

n (Q’) = n (Universal set} – n (P)

= 15-5 = 10

11. The set of cubes of the natural number is

1. a finite set
2. an infinite set
3. a null set
4. none of these

Answer: (2) an infinite set

-> required set = {1, 8, 27,…}

12.The set {2*|x is any positive rational number} is

1. an infinite set
2. a null set
3. a finite set
4. none of these

Answer: (1) an infinite set

-> required set = {2, 4, 8,…… }

13. {1- (-1)^x} for all integral x is the set

1. {0} 00
2. {2}
3. (0, 2}
4. none of these

Answer:(3) {0, 2}

-> {1 – (-1)^x} = 0 when x is even

{1 – (-1)^x}= 2 when x is odd

For x ∈ integral -> set is {0, 2}

14. E is a set of positive even number and 0 is a set of positive odd numbers, then E ∪ O is a

1. set of whole numbers
2. a set of rational number
3. N
4. none of these

Answer: (2) N

E U 0 = all natural no. -> N

15. If R is the set of positive rational number and E is the set of real numbers then

1. R⊆E
2. R⊂E
3. E⊂R
4. none of these

Answer: (2)R⊂ E

as all rational no belong to real no.

16. If N is the set ofnatural numbers and I is the set ofpositive integers, then

1. N=1
2. N⊂1
3. N⊆ 1
4. none of these

Answer: (1) N = 1

->N = {1,2, 3,…….}

I = {1, 2,3…… }

N = 1

17. If I is the set of isosceles triangles and E is the set of equilateral triangles, then

1. I⊂E
2. E⊂l
3. E =I
4. none of these

Answer: (2) E ⊂ 1

-> as all equilateral triangles are isosceles

But all isosceles are not equilateral

18. If R is the set of isosceles right angled triangles and I is set of isosceles triangles, then

1. R=1
2. R⊃I
3. R⊂l
4. none of these

Answer: (3)R⊂1

-> all right angle isosceles triangle are isosceles but all isosceles are not right-angled.

19. {n(n+1)/2: n is a positive integer} is

1. a finite set
2. an infinite set
3. is an empty set
4. none of these

Answer: (2) infinite set

{1,3,6……………….}

20. If A = {1, 2, 3, 5, 7}, and B = {x² : x∈A}

1. n(b) = n(A)
2. n(B) > n(A)
3. n(A)= n(B)
4. n(A)<n(B)

Answer: (3) n (A) = n (B)

A = {1,2, 3, 5, 7} n (A) = 5

B = (1,4, 9, 25, 49) n(B) = 5

N (A) = n (B)

21. AUA is equal to

1. A
2. E
3. ø
4. none of these

Answer: (1) A

A U A = A

22. A ∩ A is equal to

1. ø
2. A
3. E
4. none of these

Answer: (2)A

->A ∩ A = A

23. (AUB)’ is equal to

1. (A’nB)’
2. A’UB’
3. A’nB’
4. none of these

Answer: (3) A’ n B’

(A U B)’ = 1 – (A U B)

= 1 – (A U B)

(A U B)’ = . A’, n B’

24. (A∩B)’ is equal to

1. (A’ u B)’
2. A’ U B’
3. A’ ∩ B’
4. none of these

Answer: (2) A’ U B’

(A ∩ B)’ = A’ U B’

25. A∪ E is equal to (E is a superset of A)

1. A
2. E
3. ø
4. none of these

Answer: (2) E

26. A∩E is equal to (E is a superset of A)

1. A
2. E
3. ø
4. none of these

Answer: (1) A

27. E∪E is equal to (E is a superset of A)

1. E
2. ø
3. 2E
4. none of these

Answer: (1) E

28. A∩ E’ is equal to (E is a superset of A)

1. E
2. ø
3. A
4. none of these

Answer: (2) ø

29. A ∩ ø  is equal to (E is a superset of A)

1. A
2. E
3. ø
4. none of these

Answer: (3) ø

30. AUA’ is equal to (E is a superset of A)

1. E
2. ø
3. A
4. none of these

Answer: (3) E

= E

31. If K = { 1, 2, 3. 4, 5, 6, 7, 8, 9}, the subset of E satisfying 5 + x > 10 is

1. {5, 6, 7, 8, 9}
2. {6, 7,8, 9}
3. {7, 8, 9}
4. none of these

Answer: (2) {6, 7, 8, 9}

->E = 1 to 9

5 + x > 10

X > 5

x = 6, 7, 8, 9

32. If A Δ n = (A-B) U (B-A} and A = { 1. 2, 3, 4). B = {3,5,7} than A ΔB is

1. {1,2, 4, 5, 7}
2. {3}
3. {1,2, 3, 4, 5, 7}
4. none of these

Answer: (c) A Δ B = (A – B) U (B – A)

A = {1,2, 3, 4}

B = {3, 5,7}

A- B = {1, 2, 4}

B – A = (5, 7}

(A – B) U (B – A) = {1, 2, 4, 5, 7}

33. If A has 32 elements, B has 42 elements and A U B has 62 elements, the number of elements in A ∩ B is

1. 12
2. 74
3. io
4. none of these

Answer: (1) 12

n (A ∩ B) = n (A) + n (B) -n (A∩B)

62 = 32 + 42 – n (A ∩ B)

n (A ∩ B) = 74- 62 -> 12

34. In a group of 20 children, 8 drink tea but not coffee and 13 like tea. The number of children drinking coffee but not tea is

1. 6
2. 7
3. 1
4. none of these

Answer: (2) 7

35. The number of subsets of the sets {6, 8, 11} is

1. 9
2. 6
3. 8
4. none of these

Answer: (3) 8 No. of subsects = 2ⁿ= 2³ = 8

36. The sets V = {x / x+2=0), R={x / x²+2x=0} and S = {x : x² + x- 2 = 0} are equal to one another if x is equal to

1. -2
2. 2
3. 1/2
4. none of these

Answer: (1) -2

If we go by choices

V = 0 = -2 + 2

R = 4 – 4 = 0

S=4-2-2=0

V = R = S for- 2

37. If the universal set E = {x |x is a positive integer <25}, A = {2, 6, 8, 14, 22}, B = {4, 8, 10, 14} then

1. (A n B}’=A’ U B’
2. (A n B]’= A’ n B’
3. (A’ n B)’= 0
4. none of these

Answer:(1) (A ∩ B)’ = A’ U B’

-> (A ∩ B) = {8, 14}

(A ∩ B)’ = {1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15 ………}

A’ = {1, 3,4,……24}

B’ = {1, 2, 3 ……….24}    A’ U B’ = {1, 2, 3, 4, 5, 6, 7………. }

(A ∩ By = A’ U B’

38. If the set P has 3 elements, Q four and R two then the set P x Q x R contains

1. 9 elements
2. 20 elements
3. 24 elements
4. none of these

Answer: P = 3 elements, Q = 4 elements R = 2 elements

n (P x Q x R) = 3 x 4 x 2 = 24 -> (3) 24

39. Given A = {2, 3}, B = {4, 5}, C = {5, 6} then A x (B∩C) is

1. {(2, 5), (3,5)}
2. {(5,2),(5,3)}
3. {(2, 3), (5, 5)}
4. none of these

Answer: A = {2,3} B = {4, 5}, C = {5, 6}  A x (B ∩ C)  B ∩ C = {5}

A x (B ∩ C) = {(2, 5), (3,5)}

(1) {(2,5), (3,5)}

40. A town has a total population of 50,000. Out of it 28,000 read the newspaper X and 23,000 read Y while 4,000 read both the papers. The number of persons not reading X and Y both is

1. 2,000
2. 3,000
3. 2,500
4. none of these

Answer: (2) 3000

Total Population 50000

41. If A = {1, 2, 3, 5, 7} and B = {1, 3, 6, 10, 15}. Cardinal number of A-B is

1. 3
2. 4
3. 6
4. none of these

Answer: (1) 3

A – B = {2, 5, 7} ->3

42. At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. The number of women doctors attending the conference is

1. 2
2. 4
3. 1
4. none of these

Answer: (3) 1

Total Indian attendees -> 29 + 23 = 52

Let M = 23 men w = 29 women D represent doctors

We are given

n (M U D) = 24

n (M) + n (D)- n (M ∩ D) = 24

23 + 4 – n (M ∩ D) = 24

n (M ∩ D) = 3

no. of male doctors = 3

Total doctors 4 & all Indian

no. of female doctors = 4-3 = 1

43. Let A = {a, b}. Set of subsets of A is called power set of A denoted by P(A}. Now n(P(A)) is

1. 2
2. 4
3. 3
4. none of these

Answer: (2) 4

A = {a, b)

P (A) power set

n (P(a))

No. of subsets = 2² = n (Power set) = 4

44. Out or 2000 employees In an office 40% preferred Coffee (c), 54% liked (T), 64% used to smoke (S). Out of the total 28% used C and T, 32% used T and S and 30% preferred C and S, only 6% did none of these. The number having all the three Is

1. 360
2. 300
3. 380
4. none of these

Answer: (1) 360

Coffee -> 48% = n (c) = 960

Tea -> 54% = n (T) = 1080

Smoke -> 64% = n(s) = 1280

n (c ∩ T) = 560 b (T ∩ S) = 640 n (C ∩ S) = 600

120 ->none

n(CUSUT) = 2000 -120 = 1880

1880 = n (c) + n (T) + n (s) – n(C∩T) – n (T∩S) – n (C ∩ S) + n (C ∩ S ∩ T)

1880 = 960 + 1080 + 1280 – 560 – 640 – 600 + n (C∩S∩T)

1880- 1520 = n (C∩S∩T) = 360

45. Following set notations represent: A⊂B; x ∉ A; A⊃B; {0}; A⊄ B

1. A is a proper subset of B; x is not an element of A; A contains B; singleton with an only element zero; A is not contained in B.
2. A is a proper subset of B; x is an element of A; A contains B; singleton with an only element zero; A is contained in B.
3. A is a proper subset of B; x is not an element of A; A does not contain B; contains elements other than zero; A is not contained in B.
4. None

Answer: (1)

A ⊂ B = A is proper subset of B
X ∉ A = x does not belong to A
A ⊃ B = A contains B
{0} = singleton with element 0
A ⊄ B = A is not contained in B

46. Represent the following sets in set notation: – Set of all alphabets in English language, set of all odd integers less than 25, set of all odd integers, set of positive integers x satisfying the equation x²+5x+7=0 :

1. A = {x:x is an alphabet in English}, 1 = (x:x is an odd integer>25},I = {2, 4, 6, 8 ….} I = {x: x²+5x+7=0}
2. A= {x:x is an alphabet in English},I = {x:x is an odd integer<25}, 1 = {1, 3, 5, 7 ….} I = {x: x²+5x+7=0}
3. A = {x:x is an alphabet in English},I = {x:x is an odd integer £ 25},I = (1, 3, 5, 7 ….} I = {x: x²+5x+7=0}
4. None

Answer: (2)

—> A = (x : x alphabet in english}

—>I = (x: x is odd integer < 25}

—>I = (x: x is set ofall odd int) = {1, 3.5,….. }

->I = {x: x² + 5x + 7 = 0}

47. Rewrite the following sets in a set builder form: – A = {a, e, i, o, u} B = {1, 2, 3, 4 ….} C is a set of integers between -15 and 15.

1. A = {x:x is a consonant}, B= {x:x is an irrational number}, C = {x:-15<x<15^ x is a fraction}
2. A = (x:x is a vowel}, B= {x:x is a natural number}, C = {x: -15³X³15^ x is a whole number}
3. A = {x:x is a vowel}, B = {x:x is a natural number}, C = {x: -15 <x<15^ X is a whole number}
4. None

Answer: (1)

A = {x: x is a vowel}

B = {1, 2, 3, 4,…..}->B={x| x is a natural no.}

C = {x: – 15 < x < 15 A x is a whole no}

48. If V = (0, 1, 2, …9}, X = {0, 2, 4, 6, 8}, Y = {3, 5, 7} and Z = {3 7} then YUZ, (VUY) ∩ X, (XUZ)UV are respectively: –

1. {3,5, 7},{0, 2,4,6, 8}, {0, 1, 2, …9}
2. {2,4, 6},{0, 2,4, 6, 8}, {0,1, 2, …9}
3. {2, 4, 6}, {0,1, 2, …9},{0, 2, 4, 6, 8}
4. None

Answer: (1) Go by choices

49. In question No. (48) (X U Y) ∩ Z and (ø ∪ V) nø are respectively: –

1. {0, 2, 4, 6, 8},ø
2. {3, 7}, ø
3. {3, 5, 7}, ø
4. None

Answer: (b) (3, 7},ø

-> (ø u V) ∩ ø -> ø

-> (x U y) n z -> x U y = (0, 2, 3,4,5, 6, 7, 8}

Z = (3, 7}

X U y ∩ Z = (3, 7}

50. What is the relationship between the following sets? A = {x:x is a letter in the word flower} B = {x:x is a letter in the word flow} C = {x:x is a letter in the word wolf} D = {x:x is a letter in the word follow}

1. B=C=D and all these are subsets of the set A
2. B=C≠D
3. B≠C≠D
4. None

Answer:(1) B = C = D ⊂ A

51. Comment on the correctness or otherwise or the following statements: –
(1) {a, b, c) = (c, b, a}
(2) {a,b,c,d)) ⊂ (a.c.d)
(3) {b} ∈ {{b}}
(4) {b} ⊂ ((b)) and
(5)ø ⊂{{b}}.

1. Only (4) is incorrect
2. (2) (3) are incorrect
3. (1) (2) are incorrect
4. All are incorrect

Answer: (1) only (4) is incorrect

->(b) ⊄ {{b}} as b is not an element of({b}}

-> ø ⊂ {{b}} as ø is not an element of {b}}

52. If A=(a, b, c). B=(a, b). C = (a, b, d}, D=(c, d) and E=(d) state which of the following statements are correct: –

(1)B⊂A
(2) D ≠ C
(3) C⊃E
(4) D ⊂ E
(5) D ⊂ B
(6) D = A
(7) B ⊄ C
(8) E⊂A
(9) E ⊄B
(10) a∈ A
(11) a⊂A
(12) (a) ∈ A
(13) (a) ⊂ A

1. (1) (2) (3) (9) (10) (13) only are correct
2. (2) (3) (4) (10) (12) (13) only are correct
3. (1) (2) (4) (9) (11) (13) only are correct
4. None

Answer: (1) is correct

53. Let A = (0), B = (0 1), C =ø, D = (0), E = (x|x is a human being 300 years old), F = (x|x ∈ A and x e B} state which of the following statements are true: –

(1) A ⊂ B
(2) B = F
(3) C ⊂D
(4) C= E
(5) A = F
(6) F = 1 and
(7) E = C = D

1. (1) (3) (4) and (5) only are true
2. (1) (2) (3) and (6) only are true
3. (1) (2) (3) and (4) are true
4. None

Answer: (1) is correct

54. If A = (0, 1} state which of the following statements are true: –

(1) (1) ⊂ A
(2) {1} ∈ A
(3) ø ∈ A
(4) 0 ∈ A
(5) 1 ⊂ A
(6) {0} ∈ A
(7) ø ⊂ A

1. (1) (4) and (7) only are true
2. (2) (3) and (6) only are true
3. (1) (4) and (6) only are true
4. None

Answer: (1) is correct

55. State whether the following sets are finite, infinite or empty: –

(1) X = (1, 2, 3,…..500)
(2) Y ={y: y=a²; a is an integer)
(3) A = (x:x is a positive integer multiple of 2)
(4) B = (x:x is an integer which is a perfect root of 26<x<35)

1. finite infinite infinite empty
2. infinite finite infinite empty
3. infinite infinite finite empty
4. None

Answer: (1) is correct

56. If A = {1, 2, 3, 4} B = {2, 3, 7, 9) and C = (1, 4, 7, 9} then

1. A ∩ B ≠ ø B ∩ C≠ ø A ∩ C≠ ø but A ∩ B ∩ C = ø
2. A ∩ B = ø B ∩ C = ø A ∩ C =ø A ∩B ∩C=ø
3. A ∩ B ≠ ø B ∩ C ≠ ø A ∩ C≠ ø A ∩ B ∩ C ≠ ø
4. None

Answer: (1) A ∩ B = (2, 3} B∩C = (7,9) A ∩ C = {1,4}

A∩B∩C=ø

57. If the universal set isX = (x:x ∈ N, 1≤ x ≤ 12} and A = (1, 9, 10), B = (3, 4, 6, 11, 12} and C = (2, 5, 6} are subsets of X then set A U (B ∩ C) is_.

1. {3, 4, 6, 12}
2. {1, 6, 9, 10}
3. {2, 5, 6,11}
4. None

Answer: (2) {1,6, 9, 10}

-> B ∩ C = {6}

A U (B ∩ C) = {1, 6, 9,10}

58. Identify the elements of P if set Q = (1, 2, 3} and P x Q = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5,3), (6,1), (6,2), (6,3)}

1. {3, 4, 5}
2. {4, 5, 6}
3. {5, 6, 7}
4. None

Answer: (2) (4,5.6}

Set P will have the preimages of ordered

pairs of p x Q P= {4,5.6}

59. If A = (2, 3}, B = {4, 5), C = (5, 6} then A x (B U C) is

1. {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
2. {(2,5), (3,5)}
3. {(2, 4), (2, 5), (3, 4), (3, 5), (4, 5), (4, 6), (5, 5), (5, 6)}
4. None

Answer: (1)

B U C -> {4, 5, 6}

A x {B U C} = {(2, 4}, (2, 5), (2, 6), (3, 4), (3,5), (3,6)}

60. Complaints about works of canteen had been about Mess (M) Food (F) and Service (S). Total complaints 173 were received as follows :-n(M)=110,n(F)=55,n(S)=67,n(M∩F∩S’)=20,n(M∩S∩F’)=11 and n(F∩S∩M’)=16.Determine the complaints about all the three

1. 6
2. 53
3. 35
4. None

Answer: (1) 6

n (M∩F∩S’) = 20

n (M∩S∩F’) = 11

n (F(∩S∩M’) = 16

-> n (M U FUS) = 173

173 = n (M) + n (F) + n(S) – n (M∩F) – n

(F∩S) – n (S∩M) + n (M∩S∩F)

173 = 110 + 55 + 67 – (20 + x) – (11 + x) – (16 +x) + x

59 = 20 + x+11+ x+16 + x-x

2x = 12 -> x = 6

61.Out of total 150 students 45 passed in Accounts 50 in Maths. 30 in Costing 30 in both Accounts and Maths. 32 in both Maths and Costing 35 in both Accounts and Costing. 25 students passed in all the three subjects. Find the number who passed at least in any one of the subjects.

1. 63
2. 53
3. 73
4. None

Answer: (2) 53

n (A∩M) = 30

n (M∩C) = 32

n (A∩C) = 35

n (A∩M∩C) = 25

n (A∪M∪C) = 45 + 50 + 30-32-35 + 25

62. After qualifying out of 400 professionals, 112 joined industry, 120 started practice and 160 joined as paid assistants. There were 32, who were in both practice and service 40 in both practice and assistantship and 20 in both industry and assistantship. There were 12 who did all the three. Find how many could not get any of these.

1. 88
2. 244
3. 122
4. None

Answer: (1) 88

n (Industry) 112 n (I n paid) = 20

n (practice) 120 n (Paid n practice) = 40

n (paid) 160

n (I U P U S)

= 112+ 120 + 160-20-40-32 + 12

= 32

no who are unemployed

= 400-312 = 88

63. On a survey of 100 boys it was found that 50 used white shirt 40 red and 30 blue. 20 were habituated in using both white and red shirts 15 both red and blue shirts and 10 blue and white shirts. Find the number ofboys using all the colours.

1. 20
2. 25
3. 30
4. None

Answer: (2) 25

n (w) = 50 n (R) = 40 n (B) = 30

n (W ∩ R) = 20 n (R ∩ B) = 15

n (B ∩ w) = 10.

n (W ∪ R ∪ B) =n (w) + n (R) + n (B) – n (w ∩ R) – n (R ∩ B)- n (B ∩ w) + n (B ∩ R ∩ w)

100 = 50 + 40 + 30 – 20 – 15 – 10 + n (B ∩ R ∩ w)

n (B ∩ R ∩ w) = 25

64. Out of 60 students 25 failed in paper (1), 24 in paper (2), 32 in paper (3), 9 in paper (1) alone, 6 in paper (2) alone, 5 in papers (2) and (3) and 3 in papers (1) and (2). Find how many failed in all the three papers.

1. 10
2. 60
3. 50
4. None

Answer: (1) 10

n (P1 ∩ P2’∩ P3’) = 9

n (P1∩P2∩P3’) = 6

n (P2∩P3∩P1′) = 5 n (P1∩ P2 ∩P3’) = 3

n (P2) = n (P2 alone) + n (P2 ∩ P3 ∩ P1’) + n (P1 ∩ P2 ∩ P2’) + n (P1 ∩ P2 ∩ P3)

24 = 6 + 5 + 3 + n (P1 ∩ P2 ∩ P3)-> 10

65. Out of 1000 students 658 failed in the aggregate, 166 in the aggregate and in group-] 434 in aggregate and in group-II, 372 in group-1, 590 in group-11 and 126 in both the groups. Find out how many failed in all the three.

1. 106
2. 224
3. 206
4. 464

Answer: (1) 106

1000 students,

658 failed in aggregate,

166 aggregate & G1,

434 in aggregate & G2,

372 in G1,

590 in G2,

126 in both G1 & G2

1000 = n (A) + n (G1) + n (G2) – n (G1 ∩ A) – n (G2 ∩A)- n (G1 ∩ G2) + n (G1 ∩ G2 ∩ A)

1000 = 658 + 372 + 590 – 166- 434 – 126 + n (G1 ∩ G2 ∩ A)

n (G1 ∩ G2 ∩ A) = 106

66. For any set A, (A’)’ is equal to

1. A’
2. A
3. none of these

Answer: (2) A

67. Let A and B be two sets in the same universal set. Then, A – B =

1. A ∩ B
2. A’∩ B
3. A ∩ B’
4. none of these

Answer: (3) A ∩ B’

68. For any two sets A and B, A ∩ (A U B) =

1. A
2. B
3. D
4. none of these

Answer: (1) A

69. If A = {1, 3, 5, B} and B = {2, 4), then

1. 4 6 A
2. {4} c A
3. B⊂A
4. none of these

Answer: (4) none of these

70. For any two sets A and B, (A – B) u (B – A) =

1. (A – B) ∪ A
2. (B – A) U B
3. (AUB)-(A∩ B)
4. (A U B) n (A ∩ B)

Answer: (3) (AUB)-(A∩ B)

71. Which of the following statement is false:

1. A – B = A ∩ B’
2. A – B = A – (A ∩ B)
3. A – B = A – B’
5. A – B = (A U B) – B

Answer: (3) A – B = A – B’

72. For any three sets A, B and C

1. A ∩ (B – C) = (A ∩ B)- (A ∩ C)
2. A ∩ (B – C) = (A ∩ B) – C
3. A U (B – C) = (A U B) ∩ (A U C’)
4. All of these

Answer: (4) All of these

73. Let A = {x: x ∈R, x > 4} and B = {x ∈ R: x < 5}. Then, A n B =

1. [4, 5]
2. [4,5]
3. [4,5]
4. [4,5]

Answer: (2) (4.5)

i.e. all real no. between 4 & 5 excluding 4 &5

74. Let u be the universal set containing 700 elements. If A, B are sub-sets ofu such that n (A)- 200, n (B) = 300 and n (A ∩ B) = 100. Then, n (A’ ∩ B’) =

1. 400
2. 600
3. 300
4. none of these

Answer: (3)

= universal – b (A U B)

= 700 – (n (A +n(B)-n(A ∩ B))

= 700 -(200 + 300 -100)

= 300

75. Let A and B be two sets such that n [A] = 16, n [B] = 14, n [A U B] = 25. Then, n [A ∩ B) is equal to

1. 30
2. 50
3. 5
4. none of these

Answer: (3)

= n (A) + n (B) – n (A U B)

= 30-25

= 5

76. If A = {1, 2, 3, 4, 5}, then the number of proper subsets of A is

1. 120
2. 30
3. 31
4. 32

Answer: (3) no. of proper subsets

= 2ⁿ-1

= 32-1

= 31

77. In set-builder method the null set is represented by 

1. {}
2. ø
3. {x:x≠x}
4. {x:x=x}

Answer: (3) (x:x≠x)

In set builder form

78. If A and B are two disjoint sets, then n (A U B) is equal to

1. n (A) + n (B)
2. n (A) + n (B)- n (A ∩ B)
3. n (A) + n (B) + n [A ∩ B]
4. n (A)∩ (B)

Answer: (a) n (A) + n (B)

We know n (A U B) = n (A) + n (B)

– n (A ∩ B)

n (A U B) = n (A) + n (B)

n (A ∩ B)00000 = 0 disjoint sets.

79. For two sets A U B = A iff

1. B⊆A
2. A⊆B
3. A ≠B
4. A = B

Answer: (1) B ⊆ A

A U B = A

Means A is superset of B

80. If A and B are two sets such that n (A) = 70, n (B) = 60, n (A U B) = 110, then n [A ∩ B) is equal to

1. 240
2. 50
3. 40
4. 20

Answer: (4) 20

n (A ∩) = n (A) + n (B) – n (A U B)

= 70 + 60-110

= 20

81. In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both and bus. Then, persons travelling by car or bus is

1. 80%
2. 40%
3. 60%
4. 70%

Answer: (3) 60%

n (C U B) = n (C) + n (B) – n (C∩B)

= 20 + 50-10

= 60%

82. lf A∩B = B, then

1. A⊆ B
2. B⊆A
3. A = ø
4. B = ø

Answer: (2) B ⊆ A

83. An investigator interviewed 100 students to determine their preference of three drinks: milk, coffee and tea. The investigator reported that 10 students take all three drinks milk, coffee and tea; 20 students take milk and coffee; 25 students take milk and tea; 20 students take coffee and tea; 12 students take milk only; 5 students take coffee only and 8 students take tea only. Then the number of students who did not take any of three drinks is

1. 10
2. 20
3. 25
4. 30

Answer: (4) 30

n (M∩C∩T) = 10

n (MUCUT)

= 12+5+10+10

= 70

do not drink

= 100-70 = 30

84. Two finite sets have m and n elements. The number of elements In the power set of first set Is 48 more than the total number of elements in power set of the second set. Then, the values, of m and n are:

1. 7,6
2. 6,3
3. 6,4
4. 7,4

Answer: (3) 6, 4

Let x be no of elements in second set’s power set n (P(S₂)) = x

N (P (S1)) = 48 + x

X = 2ⁿ

48+x=2^m

48 = 2^m – 2ⁿ

2ⁿ (2^m-n– 1) = 48

2ⁿ (2^m-n – 1) = 16 x 3

2ⁿ= 16

2^m-n-1= 3

2ⁿ = 16

2^m-n = 4

n=4

2^m/2ⁿ=4

2^m=64

m = 6

85. In a class of 175 students the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered Mathematics alone?

1. 35
2. 48
3. 60
4. 22

Answer: (3)

86. For any two sets A and B, A ∩ (A U B)’ is equal to

1. A
2. B
3. ø
4. A ∩ B

Answer: (3) ø

-> no interaction

87. Let F1 be the set of all parallelograms, F2 the set of all rectangles, F3 the set of all rhombuses, F4 the set ofall squares and F5 the set oftrapeziums in a plane. Then F1 may be equal to

1. F₂ ∩ F₃
2. F₃ ∩ F₄
3. F₂ U F₃
4. F₂ U F₃ U F₄U F₁

Answer: (4) F₂∪F₃∪F₄∪F₁

As all rectangle, square, rhombus are parallelograms only.

88. Out of 20 members in a family, 11 like to take tea and 14 like coffee. Assume that each one likes atleast one of the two drinks. Find how many like both coffee and tea:

1. 2
2. 3
3. 4
4. 5

Answer: (4)

Given that

n[T] = 11; n(C) = 14

and n(TUC) = 20

n(TUC) = n(T) + n(c) – n[T∪C]

= 11+14-20 = 5

(4) is correct

89. In a group of 70 people, 45 speak hindi, 33 speak English and 10 speak neither Hindi nor English. Find how many can speak both English as well as Hindi:

1. 13
2. 19
3. 18
4. 28

Answer: (3) n(H} = 45; n(E} = 33

n(H ∪ E}’ = 10 => n(H ∪ E) = 70-10 = 60

∴ (H∪E} = n(H) + n(E} – n(H ∪ E)

= 45 + 33-60 = 18

(3} is correct

90. In a survey of 300 companies, the number of companies using different media – newspapers (N), Radio (R) and Television (T) are as follows: n(N) = 200, n(R) = 100, n(T) = 40, n(N∩R] = 50, n(R∩T) = 20, n(N∩T) = 25 and n(N∩R∩T) = 5. Find the number of companies using none of these media:

1. 20 companies
2. 250 companies
3. 30 companies
4. 50 Companies

Answer: (2) n(N∪R∪T} = n(N} + n(R) + n(T) – n(N∩R] – n(N∩T} – n(R∩T} + n(N∩R∩T}

= 200 + 100 + 40-50-20-25 + 5

= 250

No. of companies using no media

= 300 – n(NURUT)

= 300-250 = 50

(4) is correct

91. In a town of 20,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then the number of families which buy A only is

1. 6600
2. 6300
3. 5600
4. 600.

Answer: (1) Given

n(A) = 40% ; n(B) = 20% ; n(C) = 10%

n(A∩B) = 5%; n(B∩C) = 4%

n(C∩A) = 4%; n(A∩B∩C) = 2%

No. of families which buy only A

= n(A) – n(A∩B) – n(A∩C) + n(A∩B∩C) = 40-5-4+2 = 33%

= 20,000 x 33% = 6600

(1) is correct

92. Out of total 150 students, 45 passed in accounts, 30 in economics and 50 in maths, 30 in both accounts and maths, 32 in both maths and economics, 35 in both accounts and economics, 25 students passed in all the three subjects. Find the numbers who passed atleast in anyone of the subjects:

1. 63
2. 53
3. 73
4. None.

Answer: (2) Total students = 150

n(A) = 45 ; n(E) = 30; n(M) = 50;

n(A∩M) = 30; n(M∩E) = 32

n(A∩M) = 35; n(M∩E∩M) = 25

∴ n (A∪E∪M) = 45 + 30 + 50 – 30 – 35 + 25 = 53

(2) is correct.

93. If A = {p,q,r,s}, B = {q,s,t), C = {m,q,n} find C – (A∩B)

1. {m,n}
2. {P<q}
3. {r,s}
4. {p,r}

Answer: (1)

⊂-(A ∩ B) ={m,q,n} – {p,q,r,s} ∩ {q,s,t}

= {m,q,n} – {q,s}

= { m;n }

(1) is correct.

94. If A = {X : X² – 3X + 2 = 0} B = {X:X²-4x + 12 = 0}, B-A =

1. {-6}
2. {1}
3. {1,2}
4. {2,-6}

Answer: x²- 3x + 2 = 0

Or x(x-2)-1(x-2)=0

Or (x-2) (x-1) = 0

x = 1;2

A = {1,2}

And x² + 4x – 12 = 0

Or x² + 6x – 2x – 12 = 0

Or x(x+6)-2(x+6)=0

Or (x+6) (x-2) = 0

x = -6 ; 2

B -A = {-6;2} -{1;2} ={-6}

A (1) is correct

95. For any two sets A and B, A∩(A’UB)=___________ where A’ represent the compliment of the set

1. A∩B
2. AUB
3. AUB
4. None of these

Answer: Tricks: Take an example and then decide the answer

Let U = {0,1,2,3,4,5}

A = {0,1,2,3}

B = {2,3,4,5}

A¹ = U – A = {4;5}

A¹ U B = {4,5} U{2,3,4,5}

= {2,3,4,5}

∴ A ∩ (A¹ U B}

= {0,1,2,3} ∩ {2,3,4,5}

∴ (1) is correct

2nd method = (A∩A’) U (A∩B} = { } U (A∩B} = A ∩ B

96. If A c B, then following is true

1. A ∩ B = B
2. A U B = B
3. A ∩ B = A
4. A ∩ B

Answer: (2)

97. There are 40 students, 30 of them passed in English, 25 of them passed in maths and 15 of them passed in both. Assuming that every student has passed atleast in one subject. How many students passed in English only but not in math’s.

1. 15
2. 20
3. 10
4. 25

Answer: (1) total students = 40

n(E) = 30; n(M) = 25

(E∩M)= 15

No. of stds. Passed in English only

n(E) – n(E∩M)

= 30-15 = 15

(1) is correct

98. If A = (1,2,3,4,5), B = (2,4) and C = (1,3,5) then (A -C) x B is

1. {(2,2),(2,4),(4,2),(4,4),(5,2),(5,4)}
2. {(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)}
3. {(2,2),(4,2),(4,4),(4,5)}
4. {(2,2(,(2,4),(4,2),(4,4)}

Answer: (4) is correct

(A-C)xB = {2,4}x{2,4}=

{(2,2),(2,4),(4,2),(4,4)}

99. For any two sets A and B the set (AUB’)’ is equal to (where denotes compliment of the set)

1. B-A
2. A-B
3. A’-B’
4. B’-A’

Answer: (1) is correct

Tricks: Let U = {1,2,3,4,5}

A = {0,1,2,3}; B = {1,2,3}

B’ = U – B = {0,4,5}

A∪B’ = {0,1,2,4,5}

(AUB’)’ = U- (A U B’)= {3}

Then go by choices

For (1) B – A = {1,2,3} – {0,1,2} ={3}

(A∪B’)’ =U -(AUB’) = {3}

2 nd Method

(A∪B’)’

= A’D(B’)’

= A’ ∩ B

= B – A ∩ B

= B- A

100. The number of proper sub-set of the set {3,4,5,6,7} is

1. 32
2. 31
3. 30
4. 25

Answer: (2) No. of proper sub -sets = 2ⁿ– 1

= 2⁵ -1 = 31

101. For a group of 200 person, 100 are interested in music, 70 in photography and 40 in swimming, furthermore 40 are interested in both music and photography, 30 in both music and swimming, 20 in photography and swimming and 10 in all the three. How many interested in photography but not in music and swimming? 

1. 30
2. 15
3. 25
4. 20

Answer: (4) is correct

Let A = No. of persons interested in Music

B = No. of persons interested in photography

C = No. of persons interested in swimming

n(A) = 100; n(B) = 70; n(C) = 40;

n(A∩B) = 40; n(A∩C) = 30;

n(B∩C) = 20; n(A∩B∩C)= 10.

n(B∩A¹ ∩ C¹) = n(B) – n(B∩A) – n(B∩C)+n(A∩BnC)

= 70-40-20+10

= 20

102. Of the 200 candidates who were interviewed for a position at call center, 100 had a two wheeler, 70 had a credit card and 140 had a mobile phone 40 of them had both a two-wheeler and a credit card, 30 had both a credit card and mobile phone, 60 had both a two-wheeler and a mobile and 10 had all the three. How many candidates had none of them?

1. 0
2. 20
3. 10
4. 18

Answer: (3) is correct

Let n(A) = No. of candidates having two wheeler

n(B) = No. of candidates having credit cards

n(C) = No. of candidates having mobile phone.

Given

n(A) = 100; n(B) = 70; n(c) = 140

n(A∩B) = 40; n(B∩C)= 30; n(C∩A) = 60

n(A ∩ B ∩ C) = 10

n(A U B U C) = 100 + 70 + 140- 40-30-60 + 10 = 190

No. of candidates having none = 200 – 190= 10

103. In a class of 50 students 35 opted for Maths, 37 opted for commerce. The number of such student who opted for both maths and commerce is

1. 13
2. 15
3. 22
4. 28

Answer: (3) is correct

n(M) = No. of students opted for Maths = 35

n(C) = No. of students opted for commerce = 37

So; (MUC) = 50

n(M∩C) = 35 + 37- 50 = 22

104. A = (2,3), B=(4,5), C=(5,6) then Ax(B∩C)

1. [(5,2),(5,3)]
2. [(2.4),(5,3)]
3. [(2,5),(3,5)]
4. [(3,5),(2,6)]

Answer: (2)

B ∩ C = {4,5} ∩ {5,6} = {5}

= {(2,5);(3,5)}

105. In a class of 80 students, 35% play only cricket, 45% only tennis, how many play cricket?

1. 80
2. 54
3. 36
4. 44

Answer: Given n(C – T) = n(C) – n(C ∩ T) = 35%

n(T- C) = n(T) – n(C∩T) = 100

or; 35 + n(C∩T) + 45 + n(C∩T) – n(C∩T) = 100

or 80 + n(C∩T) = 100

n(C∩T) = 20%

n(C) = 35 + n(C∩T) = 35 + 20 = 55%

= 80 x 55% = 44

106. If set A = (x:x/2 ∈ z,0 ≤ 12}

B = {x: x is one digit prime number} and

C = {x:x/3 ∈ N, x ≤ 12} then A ∩ (B ∩ C) =

1. 0
2. Set A
3. Set B
4. Set C

Answer: (1)

A = {2,4,6,8,10}

B = {2,3,5,7}

C = {3,6,9,12}

A ∩ (B ∩ C) = ø

No common element in all 3 sets.

107. Let A be the set of the squares ofnatural numbers and x ∈ A, y ∈ A. Then__________

1. x + y ∈ A
2. x- y ∈ A
3. x/y∈ A
4. xy ∈ A

Answer: (4) is correct

A = {x/x is the squares of natural Nos.} = {1,4,9,16,25……….. }

Tricks: then go by choices

Let x = 1: y = 4 ∈ A

x+y = 1 + 4 = 5 ∉ A.

x – y = 1 -4 = 3 ∉ A.

x/y=1/4 ∉ A.

But xy = 1 x 4 = 4 ∈ A.

(4) is correct.

108. The number of sub-sets formed from the letters of the word “ALLAHABAD”

1. 128
2. 16
3. 32
4. None

Answer: (3) is correct

Let X = {Letters of word ALLHABAD)

= {A, L, H, B, D}

No. ofsub-sets = 2⁵ = 32

109. If f(x) =\( \frac{x-1}{x} \text { and } g(x)=\frac{1}{1-x} \)then fog(x) =

1. x-1
2. x
3. 1-x
4. -x

Answer: \(\mathrm{fog}(\mathrm{x})=\mathrm{f}\{\mathrm{g}(\mathrm{x})\}=\frac{g(x)-1}{g(x)}\)

\(=\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}=\left(\frac{1-1+x}{1-x}\right) \times \frac{(1-x)}{1}\)

= x
Option (2) is correct

110. In a class of 35 students, 16 students play football and 24 students play cricket. Assume that each one play atleast one game, then number of students who play both the games is_______________

1. 5
2. 11
3. 12
4. 17

Answer: N (F ∩ C) =n (F) + n(C)- n(F U C)

= 16 + 24-35 = 5

Option (1) is correct.

111. If A = {ø,{ø}} then the Power Set of A is

1. {ø}, {0}
2. {ø,{ø},{ø}},A}
3. A
4. {A}, {ø}

Answer: (2)

A = {ø;{ø; }}

P(A) = {{}};{ø};{{ø}};{ø;{∅}}

= {ø;{ø};{{ø};A}}

112. If A = {x:x = 3ⁿ -2n – 1, where n∈ N}, B = {x:x = 4(n – 1}, where n∈N}. Then

1. A ⊂B
2. B⊂A
3. A = B
4. None

Answer: (1)

Putting n= 1,2,3……….; we Ret

A = {x/x = 3ⁿ – 2n – 1}

= {0 ; 4 ; 20 ; }

B = {x/x = 4(n – 1)}

= {0;4;8; 12; 16; 20;………}

Clearly ; A ⊂ B

113. The numbers of proper sub-sets of the set {3, 4, 5, 6, 7} is :

1. 32
2. 31
3. 30
4. 25

Answer: (2)

Formula

No. of proper sub-sets = 2ⁿ -1

= 2⁵-1 = 31

114. If A = {1, 2, 3, 4, 5, 6, 7} and B = {2, 4, 6, 8}. Cardinal number of A – B is:

1. 4
2. 3
3. 9
4. 7

Answer: A ∩ B = {1, 2, 3, 4, 5, 6, 7} n {2, 4,6,8}

= {2, 4, 6} => n (A ∩ B) = 3

n (A – B) = n(A) -n(A ∩ B)

=7-3=4

115. If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}; B = {1, 3, 4, 5, 7, 8}; C = {2, 6, 8} Then find (A – B) U C

1. {2, 6}
2. {2,6,8}
3. {2, 6, 8, 9}
4. None

Answer: (3)

A-B = A-(A ∩ B)

= {1,2, 3,4, 5, 6, 7, 8,9}

{1,3, 4, 5, 7, 8}

= {2,6,9}

(A – B) U C = {2 ; 6 ; 9} U {2 ; 6 ; 8}

{2 ; 6 ; 8 ; 9}

(3) is correct.

116. The no. of sub-set of the set (3, 4, 5} is:

1. 4
2. 8
3. 16
4. 32

Answer: (2)

No. of sub-sets = 2ⁿ = 2³ = 8

Function And Relation

Introduction

A Function is a rule that describes the dependence between two quantities and is generally denoted by f(x).

Take for e.g., f (x) = 2x+5

Let y= f(x)

In the above case, x is the input, y is the output and f(x) is the function that processes the input according to a certain rule to give the output.

The function f here multiplies the input value x by 2 and then adds 5 to it to give the corresponding value of the output y.

For this function, different values of x give different values of y. By plotting these points and joining them the following figure is obtained.

Any function in general can be reprinted through the diagram given below.

If x be the input and y the output fora function/, then the function is denoted as y = f(x).

This indicates that when input x is processed by a function /, the output obtained is y.

Thus, x is also called the independent variable, while y is called the dependent variable.

Therefore, a function can be considered to be a machine that transforms a given input into the corresponding output according to a certain rule defined by that function.

Some other examples of functions are

f(x) = x² + 3x-7

f(x) = cosx

f(x) = log ( |x| + 5)

A function can be represented using formulae, tables, graphs, algorithm, arrows etc.

The concepts of an ordered pair, Cartesian product and relation need to be understood to understand functions.

Cartesian Product

Consider two sets A and B

A ≡ {1,2,3}

B≡ {1.2,3.4}

The set of all ordered pairs of elements from A and elements from B, known as the Cartesian product, gives all the ways to possibly relate the elements of set A with those of set B.

The Cartesian product A x B is given by

Ax B ≡ {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)(3,1), (3,2), (3,3), (3,4)}

Ax B ≡ {(a,b)| a ∈ A, b ∈ B}

Also,BxA = {(b, a) b∈ B, a ∈ A)

In a Cartesian product, the order in which a and b are written is important.

∴Ax B ≠ B x A

It is evident that the number of elements in the Cartesian product is the product of the number of elements in the two sets A and B.

Consider the following example.

Let set S= {I,II,III} represent three students and set M≡ {56,29,45}represent their scores out of 100 in maths in some order.

Then the different ways in which the students and marks can be related is represented by the Cartesian product Ax B.

∴ S x M≡ {(I,56), (I,29), (I,45), (II, 56), (II,29), (II, 45), (III, 56), (III, 29), (III, 45)}

Relations

Understand the concept of relation before moving on to functions.

Consider two sets P and Q.

P ≡ {1,2,3,4}

Q≡ {1,2,3,4,5,6,7,8,9,10}

Let R ≡ {(1,1), (2,4), (3,9)}

Observe that R ⊆ PxQ.

But elements of R have been chosen in a specific way i.e. (p, q) ∈ R only if q = p²

Thus R = {{p,q)| q = p², p ∈ P, q ∈ Q}

R is called a relation from P to Q and this is denoted by p R q or p~q

In general, any relation is a set of ordered pairs (p , q) ∈ PxQ Such that there exists some definite relationship R between p and q.

Note:

If R ≡ {{p,q)\q = p²,p∈P,q∈Q],it is not necessary that all elements of P and Q are a part of the relation.

Domain, Co- Domain and Range

The above relation R from p to Q can be expressed in the form of a diagram as follows:

The direction of the arrow indicates the direction of the relation.

In the above figure the relation exists from set P to set Q and not vice versa.

The set of all first elements in the ordered pairs that belong to R is called the domain of R.

In this example, Domain ≡ {1,2,3}

Hence Domain ⊆ p.

The set of all second elements in the ordered pairs that belong to R is called the range of R.

In this example, Range = {1,4,9}.

Hence, Range ⊆ Q.

Q is called the co-domain of R. Thus, range ⊆ co-domain.

In other words, A function can be denoted as y= f{x}, where x is the independent variable and y is the dependent variable as discussed above.

Domain: The domain of a function is set of all the values that an independent variable can take i.e. values of “x” in this case.

Range: The range of a function is set of all the values which a dependent variable will take corresponding to the values taken by independent variables i.e. Values of “y” obtained corresponding to the value of x.

Types of Relations

Many-Many Relation

Consider the following relation.

A ≡ {2,6,5,7},B ≡ {14,30,35,42,59}

Relation R₁ has ordered pairs (a,b)such that be is multiple of a

R₁ = {(2,14), (2,30), (2,42), (6,30), (6,42), (5,30), (5,35), (7,14), (7,35), (7,42)}

In this relation, more than one element in the domain has the relation with the same element in the range. Also, one element in the domain has the relation with more than one element in the range. Such a relation is called a many-many correspondence.

One-Many Relation

If A ≡ {1,2,3,4}, B ≡ {-2, -1,0,1,2}

Relation R₂ has ordered pairss (a, b)such that b is the square root of a.

R₂ ≡ {(1,-1), (1,1), (4, -2), (4,2)}

In this relation, one element in the domain has the relation with more than one element in the range. Such a relation is called a one-many correspondence.

One-One Relation

If A≡ {1,2,3}, B ≡{-1,0,1}

If A ≡ {1,2,3), B ≡ {-1,0,1}

Relation R₃ has ordered pairs (a, b) such that b = (a- 1)

R₃ ≡ {(2,1), (2,1)}

In this relation, one element in the domain has the relation with exactly one element in the range. Such a relation is called a one – one correspondence.

Many-One Relation

A≡ {-2,-1,0,1,2), B ≡ {0,1,2,3,4)

Relation R₄ has ordered pairs [a,b] such that b is the square of a

R₄ = {(-2,4), (-1,1), (1,1), (2.4)}

In this relation, more than one element in the domain has the relation with the same element in the range. Such a relation is called a many-one correspondence.

Equivalence Relation

Let S be a set and R be a relation between S and itself.

R is called an equivalence relation on S if the relation R has the following three properties:

Reflexivity: Every element of S is related to itself

Symmetry: If an element s is related to t then t is related to s

Transitivity: If an element s is related to t and t is related to u, then s is related to u.

Consider the set S ≡ {1,2,3}

Now a relation R from S ~S is said to be an equivalence relation if:

S~S = {(1,1), (2,2), (3,3), (1,3), (3,1), (1,2), (2,1), (2,3), (3,2)}

Here the reflexivity relation exists as every element i.e., 1, 2, 3 is related to itself.

In addition, the symmetry relationship exists as every element is related to all other elements and vice versa i.e.,1 is related to 2 and 2 is related to 1.

Also, transitive relationship is satisfied since reflexive and symmetry relationships are satisfied i.e.,1 is related to 2,2 is related to 3 and simultaneously 1 is related to 3.

Congruence and equality are two common equivalence relations.

Functions

Any relation between sets A and B, is a function ifit satisfies the following conditions.

• It should be a one- one or many -one relation.
• Every element in A should be a part of the domain.

Thus, every function is a relation. However, not every relation need be a function.

Thus, if A and B are two sets and if there exists a relation so that each element of A is related with one and only one element of B, then a function from A to B exists.

Example 1:

Is the following relation a function?

Solution:

This relation is a many – many relations because c has a relation with 5 as well as 15. Also, b and c both have a relation with 15. Therefore, this is not a function.

Example 2:

Find whether the relation shown above represents a function or not.

Solution:

This is a one- one relation. However, every element of A is not in the domain.

Hence, this is not a function.

Example 3:

Find the value of f(x) = \(\frac{x^6}{2 x-3}+\mathrm{x} \text { at } \mathrm{x}=\frac{3}{2} \text { and at } \mathrm{x}=1\)

Solution:

To find the value of a function at some value of the independent variable, substitute this value of the variable in the function.

\(\text { Put } \mathrm{x}=\frac{3}{2} \text { in } f(x)=\frac{x^6}{2 x-3}+x\)

Thus, \(f\left(\frac{3}{2}\right)=\left(\frac{1.5^6}{0}\right) \frac{3}{2}=\infty\) ,which is not defined.

Hence the function f(x)is not defined at x= 3/2

At x =1, \(f(1)=\frac{1^6}{2-3}+1=0\)

Types of Function

Into and on to Functions

A Function f : A → B is called a function from A onto B if the range of f = co-domain.

Thus, if every element of B has a pre-image in A, then f: A→B is a function from A onto B.

If there is at least one element of B that has no pre-image in A, then f : A → B is a function from A into B.

Thus functions can be of four kinds – one-one into, one-one onto, many-one into and manyone onto.

Example 4:

Is the given relation an onto function?

Find the co-domain and range for the function.

Solution:

In this function,

Co-domain ≡ {5,10,15,20}

Range≡ {5,10,15}

This is not an onto function because the element 20 in B does not have a pre-image in A.

Injective, Surjective and Bijective Functions

An injective function, also called an injection is a one- one function.

Thus, a function f is an injective if and only if whenever f (x) = f (y),x = y

A many-one function is not an injection.

A subjective function, also called a surjection, is an onto function.

Thus a function f: X → Y is a surjective if and only if its range f{x} is equal to its co-domain Y.

A function f: X → Y is a bijective function or a bijection if for every y in Y, there is exactly one x in X such that f (x)=y.

Alternatively, f is bijective ifit is both one- one {injective} and onto {surjective}.

Even and Odd Functions

Even Functions

Let y=f (x) be a function such that x, y ∈ R.

If the values of f (x) and f(- x) are the same for all values of x in the domain, then the function is said to be an even function. For example, consider f(x) = x²

Here, f(1) = 1. also/(-1) = 1.

Similarly,f(-1) = -1² =1² = f (x)

Thus, f (x) is an even function. Other examples of even functions are cos x,|x|, x²ⁿ ; n ∈ N

An even function of the form y=f(x) is symmetrical about they-axis. Similarly, an even function ofthe form x=f (y) is symmetrical about the x-axis.

The following graph shows the even function f (x) = x²- 3x² + 2

Thus, the graph of an even function appears the same on both sides of the y- axis. The left half of the plane appears as the mirror image of the right half of the plane.

Properties of Even Functions:

The sum, difference, product, or division of any two even functions is also an even function.

Odd Functions

Let y = f (x)be a function such that x, y ∈ R.

If f (x) = -f(-x) for all values of x in the domain, then the function is said to be an odd function.

Consider the function f(x) = x³

Here,f(1) = 1 while f(-1) = -1

Similarly, for any x, f(-x) = (-x)³ = -x³ = -f(x)

Thus f (-x) = -f (x) and f (x) = -f (-x)

Hence, this is an odd function.

Other examples of odd functions are sin x, tan x, x ²ⁿ⁺¹; n ∈ N

For odd functions the inverted left half of the graph looks like the mirror image of the right half of the graph.

In other words, an odd function is symmetrical about the origin.

The following graph shows the odd function f(x) = x³ + sin x

Properties of Odd Functions:

The sum or difference of any two odd functions is an odd function.

Division or product of an even number of odd functions is an even function whereas that of an odd number of odd functions is an odd function. i.e„ odd function raised to
an even power is an even function whereas and odd function raised to an odd power is an odd function.

Neither Even nor Odd Functions

There are functions that satisfy the conditions neither of an even nor of an odd function. Such functions are said to be neither even nor odd functions.

For example: f (x) = x³ + 1

f(-x) = (-x)³ + 1 = -x³ + 1

∴ f (x) ≠ ± f(-x).

Thus, it is neither an even nor an odd function.

Note:

The only function that is both even and odd is the function defined by f (x) = 0 for all real x.

Every function can be expressed as a sum of an odd and an even function.

Even function x Odd function ≡ Odd function

Even function/odd function ≡ Odd function

Even function ± Odd function ≡ Neither even nor odd function

Composite Functions

Let two functions f and g be defined as f : A → B and g: B → C, such that f(x) = 2x + 1 and g (y) = y + 5

Then a function from A to C, g°f: A A → C, can be defined as

g° f (x) = g [ f (x)]

= f(x) + 5

= 2x + 1 + 5 = 2x + 6

If A = {1,3,5 }, then the function g° f (x)can be shown as follows:

Such a function is called a Composite function.

Example 5:

f(x) = x⁴

g(x) =√x,x > 0

Find g°f (2).

Solution:

g°f(x) = g(f(x))

f(2) = 2⁴ = 16

g(f(2)) = g(16)

g(f(2)) = g(16) = √16 = 4

Note that g°f(x) = g[f(x)] = f[x⁴]

= √x⁴

= x²

Example 6:

f(x) = 2x² – 5x + 9

g(x) = 3x + 7

Find f°g(x)and g°f(x). is fºg(x) = g°f(x)?

Solution:

f°g(x) = f[g(x)] = f(33x + 7)

Substitute g(x)for x in f (x).

f[g(x)] = 2(3x + 7)²- 5(3x + 7) + 9

= 2(9x² + 42x + 49)- 15x- 35 + 9

= 18x² + 69x + 72

g°f(x) = g[f(x)] = g(2x²- 5x + 9)

= 3(2x² – 5x + 9) + 7

= 6x² – 15x + 34

Thus ,f° g (x) ≠ g° f (x)

Piecewise Functions

A function, which has different definitions depending on the value of the independent variable, is a piecewise Function. Thus, the definition of function changes according to the domain.

For example:

y = x/2 for x ≤ 0,

= x for 0 < x < 2 and

= 2 for x ≥ 2

The modulus function is a piecewise function.

Example 7:

Find the value of f(2), f(-2)and f(0) for the function

f(x) = 2x + 3 for- ∞ ≤ x ≤ 0

= 2x- 3 for 0 < x ≤ ∞

Solution:

Here, two functions are defined for two different domains.

So, to find f(-2), use f(x) = 2x + 3

Thus, f(-2) =-1

However, for f(2) use the function f(x) = 2x- 3

Thus, f (2) = 2(2) -3 = 1

At x=0,we use f(x) = 2x + 3

∴ f(0) = 3

Inverse Functions

If f: A -> B Is a one one onto function, then for each x ∈ B, there exists one and only one y ∈ B such that f(x) = y

If one defines a relation from B to A such that for each £ B, there exists one and only one x ∈ A such that f(x) = y, then the relation is a function from B to A denoted by

f¹:B → A, and defined as f¹ (y) = x.

such a function is called an inverse function.

Consider the two functions given below:

f(x) = 2x + 3, where x ∈ I

g(x) =x-3/2 where x∈ I

For x = 3,

f(3) = 2(3) + 3 = 9

g(9) = (9-3)/2,= 3 = x

We can see that g(f(x)) = x for all x∈ l.

In such a case,g(x) is called the inverse of f(x). This is denoted by g (x) = f¹(x)

Note:

g(x) = f¹ {x) does not necessarily mean that f(x)) = g¹(x)

In the above example,

f(x) = 2x + 3 where x∈ I

g(x) = x-3/2, where x∈ I

Algebraically,f(g(x)) = 2(x- 3/2 + 3 = x)

But if x is even, say x=4, g(4) = 0.5 ∉ I

f(g(4)) is not defined as g(4) does not belong to the domain of f.

f(x) ≠ g^-1(x)

However, if the domains of f(x)and g(x)were x ∈ R, then the two functions would be inverses of each other.

Note that not all functions have an inverse.

Consider the function y = x², x ∈ I

The range of this function is y ∈ I, y≥ 0

For y = 1,x can be -1 or 1.

Thus, each value of y does not give a unique value of x.

If the inverse of the function y=x² was g(x), then g (1) would be both 1 and -1 which would make g(x) a one-many relation, which cannot be a function.

Condition for A Function to Have an Inverse:

The function should be a one-one relation. This ensures that the inverse relation is also one-one. If the function is many-one, the inverse relation is one- many and hence is not a function.

The co-domain of the original function becomes the domain of its inverse and vice-versa. Thus, the inverse relation will be a function if and only if the co-domain = range; of the function is a one-one onto function.

In other words, for a function to have an inverse it must be a bijection. Graphically a function y= f(x) has an inverse if no two points on the graph have the same y coordinate

Example 8:

Find out if the inverse of the following functions exists.
1. f(x) = x², where the domain and the co- domain are the set of integers;
2. f(x) = x- 3, where the domain and the co- domain are the set of integers.
3.f(x) = x², where the domain is {x|x ∈ R,x ≥ 0 }

Solution:

f(x) = X²

1. Here, we have f(-1) = f(1) = 1. Hence, this is a many-one relationship. Therefore, the inverse of this function does not exist.

2. f (x) = x- 3

Here, for each value of x, we have only one value of x-3.

Hence,it is a one- one relationship.

Also, f(x) = x- 3, so x = f(x) + 3

Thus, every value of f (x) corresponds to a unique integer value of x. So the function f(x) is an onto function.

Hence, this function is both one-one and onto. Hence, it has an inverse function.

f(x) =x²,{x|x∈R,x ≥ 0}

In this function, the domain is restricted to the set of positive real numbers and so is the co-domain.

The function is a one-one onto function.

∴ f(x)has an inverse.

Steps to Find the Inverse of a Function:

The inverse of a function can be found easily using three algebraic steps. However, this method should be used only after verifying that the inverse exists.

The steps are:

1. Write down the function in the form y = f(x).
2. Solve for x, Replace x with f^-1 (y)
3. Replace y with x to get the inverse function.

For example, consider/(x) = x- 3

Step 1: write down y = f(x).

Hence y = x- 3

Step 2: solve for x.

Thus, x- y + 3

Since y = /(x),x = f^-1{y)

Thus f^-1 (y) = y + 3

Step 3: Replace y by x in the above equation to get f^-1 (x) = x + 3

Thus, if we follow the four steps given above, we can find the inverse of any function.

Example 9:

Find the inverse function of:

1. \(f(x)=\frac{2 x+3}{7 x+5}\)

2. f(x)=e^x

Solution:

\(\text { 1) } f(x)=\frac{2 x+3}{7 x+5}\)

Step 1:

\(\mathrm{Y}=f(x)=\frac{2 x+3}{7 x+5}\)

Step 2:

y(7x + 5) = 2x + 3

7xy + 5y = 2x + 3

7xy- 2x = 3- 5y

x{7y- 2) = 3- 5y

\(x=\frac{3-5 y}{(7 y-2)}\)

Step 3:

Writing x as f^-1(y),

\(x=f^{-1}(y)=\frac{3-5 x}{(7 y-2)}\)

Step 4: Replacing y in the above equation by x,

\(f^{-1}(x)=\frac{3-5 x}{(7 x-2)}\)

2) f(x) = e^x

Step 1: y = f(x) = e^x

Step 2: y = e^x

Take log on both sides

log y = log e^x

log y = x log e [ log a^b = b log (a)]

∴ log y = x [log e = 1]

∴ x = log y

Step 3: x = f¹ (y) = log y

Step 4: Replacing y in the above equation by x,

f¹ (x) = log x

Some Standard Functions

Constant Function

A Constant function is f(x) = c, where c ∈ R. The range of the function is {c} The graph of a constant function is as shown.

Linear Function

A linear function is of the form y = ax + b, where, a and b are real numbers. The graph of a linear function is a straight line. It intersects the x- axis at-b/a and it intersects the y-axis at be. The function y = x, passes through the origin.

Modulus Function

The modulus runction is used to find the positive value of an expression. It is also called the absolute value function. The function is denoted by

f(x) = |x|.

Thus.

f(x) = x if x ≥ 0

= -x if x < 0

The modulus function is an even function. Graphically it can be shown as,

Exercise – 2: Relation

Choose the most appropriate option/options (1) (2) (3) or (4).

1. “Is smaller than” over the set of eggs in a box is

1. Transitive (T)
2. Symmetric (S)
3. Reflexive (R)
4. Equivalence (E)

Answer: (1) Transitive (T)

Let a be S₁ & b be in S₂: a < b

∴ (a, b) are a part of relation

Let c be in S₃: b < c

∴ a < b < c

-> a < c

(a, c) are part of relation not reflexive because it is smaller than not equal to & not symmetric as a < b but b < a then.

2. “Is equal to” over the set of all rational numbers is

1. (T)
2. (S)
3. (R)
4. E

[By using R = Reflexive; T = Transitive, S = Symmetric and E = Equivalence from Q. No. 2 to 8]

Answer: (4) Equivalence

a, a∈R->a = a

-> (a, a) ∈ R reflexive

a,b∈R->a=b

-> b = a -> (b, a) ∈ R symmetric

a,b∈R->a = b

b, c∈R->b = c

∴ a = c

∴ (a, c)∈ R

∴ transitive

∴ equivalence

3. “has the same father as”……over the set of children

1. R
2. S
3. T
4. All of these

Answer: (4) All of these

-> a has same father as himself reflexive

-> a & b have same father as b & a symmetric

-> a & b have same father & b & c have same father

Implies a & c have same father -> transitive

4. “is perpendicular to” over the set ofstraight lines in a given plane is

1. R
2. S
3. T
4. E

Answer: (2) S

a line cannot be perpendicular to itself

∴ not reflexive

a line perpendicular to another line

L₁ ⊥ L₂ -> L₂ ⊥ L_{1      ∴}  symmetric

A line perpendicular to L₂ & L₂ ⊥ L₃

Implies L₁ || L₃

∴ not transitive

5. “is the reciprocal of’…… over the set of non-zero real numbers is

1. S
2. R
3. T
4. none of these

Answer: (1)S

a number is not a reciprocal of itself

∴ not reflexive

a number is reciprocal ofanother say

a =1/b -> b =1/a        ∴ symmetric

let a, b e R & let b, c ∈ R

a=1/b      b=1/c    → 1/a=1/c->a=c

a,c ∉ R not transitive

6. {(x, y)/x ∈ x, y ∈ y, y = x } is

1. R
2. S
3. T
4. All of these

Answer: (4) All of these

7. {(x,y) / x + y = 2x where x and y are positive integers}, is

1. R
2. S
3. T
4. E

Answer: (4)E

{(x,y):y=2x-x}

->{(x,y):x = y}

It is reflexive, Transitive, symmetric

∴ equivalence

8. “Is the square of over n set of real numbers is

1. R
2. S
3. T
4. none of these

Answer: (4) none of these

[a, a] ∉ R as a²          ∴  not reflexive

(a, b) ∈ R ≠>(b, a) ∈ R as a² = b ≠>b² = a

∴ not symmetric

(a, b) ∈ R ⇔ (b, c) ∈ R a² = b, b² = c => a⁴ =c not transitive

9. If A = {1,2,4}, B = {2,4,5} C = {2,5}, then (A-B) x (B – C) is

1. {(1,2),(1,5),(2,5}}
2. {(1,4}}
3. (1,4)
4. none of these

Answer: (2) {[1,4)}

A = {1, 2, 4}

B = {2, 4, 5}

C = {2, 5}

A- B = {1}

B – C = {4}

A – B x B – C

=>{(1,4)}

10. If R is a relation on the set A = { 1,2,3,4,5,6,7,8,9} given by x R y ⇔ y = 3x, then R =

1. {(3,1), (6,2), (8,2), (9,3)
2. {(3,1),(6,2),(9,3)
3. {(3,1), (2,6), (3,9)}
4. None of these.

Answer: (4) none of these

R – {(1, 3], (2,6),………….. (9,27)}

11. Let A = {1,2,3} B = {1,3,5}. If relation R from A to B is given by R = {(1,3),(2,5),(3,3) then R^-1  is

1. {(3,3),(3,1),(5,2)}
2. {(1,3),(2,5),(3,3)}
3. { (1,3),(5,2)}
4. none of these

Answer: (1) {(3,3), (3, 1), (5,2)}

R = ((a,b) (a, c)}

R^-1 = {(b, a), (b, c)}

12. If A = {1,2,3}, B = {1,4,6,9} and R is relation from A to B defined by ‘x is greater than y. The range of R is

1. {1,4,6,9}
2. {4,6,9}
3. { 1 }
4. none of these

Answer: (3) 1

R : {(x, y) : x > y, x ∈ A, y ∈ B)

Range => 1

As x ∈ A will be greater than only 1 from

y ∈ B

13. If R = {(x,y) : x,y ∈ Z, x² + y² ≤ 4} is a relation on Z, then domain of R is

1. {0,1,2}
2. {0.-1.-2}
3. {-2,-1,0,1,2}
4. none of these

Answer:(3) domain (-2,-1,0,1,2}

14. A relation R Is defined from (2,3,4,5) to {3,6,7,10} by: x R y ⇔ x is relatively prime to y. then, domain of R is

1. {2,3,5}
2. {3,5}
3. {2,3,4}
4. {2,3,4,5}.

Answer: (4) (2,3,4,5)

Domain of R : A→ B = A

15. Let R be a relation on N defined by x + 2y = 8. The domain of R is

1. {2,4,8}
2. {2,4,6,8}
3. {2,4,6}
4. {1,2,3,4}.

Answer: (3) {2.4,6)

For y to be a N no

X should be even < 8 & > 0

{2,4,6}

16. If the set A has p elements, B has q elements, then the number of elements in A x B is

1. p + q
2. p + q +1
3. pq
4. P2

Answer: (3)

n (A) = P    n(B) = q          (3) pq

n (AxB) = pq

17. Let R be a relation from a set A to a set B, then

1. R = A U B
2. R = A ∩ B
3. R⊆ AxB
4. R ⊆ B x A.

Answer: (3)R⊆AxB

All relations are a subset of the Cartesian product

18. R is a relation from a finite set A having m elements to a finite set B having n elements then the number of relations from A to B is

1. 2^mn
2. 2^mn-1
3. 2mn
4. mⁿ

Answer: (1) 2^mn

If a relations is from A→ B

A has in elements B has n elements

Then no. of relations is 2^mn

19. If R is a relation on a finite set having n elements, then the number of relations on A is

1. 2ⁿ
2. 2^n²
3. n²
4. nⁿ.

Answer: (2) 2^n²

Relation → A → A

A has n elements no. of relations is

2^mn = 2^n²

20. \(\text { If } f(x)=\frac{2+x}{2-x} \text {, then } f^{-1}(x) \text { : }\)

1. \(\frac{2(x-1)}{x+1}\)
2. \(\frac{2(x+1)}{x-1}\)
3. \(\frac{x+1}{x-1}\)
4. \(\frac{x-1}{x+1}\)

Answer: (1)\( \mathrm{f}(\mathrm{x})=\frac{2+x}{2-x}=y(\text { let })\)

2 + x = 2y- xy
Or x + xy = 2y-2
Or x(1 + y) = 2 (y-1)

\(\text { Or } \mathrm{x}=\frac{2(y-1)}{1+y}\) \(f-1(x)=\frac{2(x+1)}{x-1}\)

(1) is correct

21. In the set N of all natural numbers the relation R defined by a R b “if and only if a divide b’, then the relation R is:

1. Partial order relation
2. Equivalence relation
3. Symmetric relation
4. None of these.

Answer: (1) it is transitive relation, i.e. partial order relation

22. On the set of lines, being perpendicular is a__________

1. Reflexive
2. Symmetric
3. Transitive
4. None of these

Answer: (2) is correct

It is symmetric relation because it x is perpendicular to y

Then y is also perpendicular to x

23. If A = {1,2,3} then the relation R = {{1,1},(2,3),(2,2),{3,3),(1,2)} on A is:

1. Reflexive
2. Symmetric
3. Transitive
4. Equivalence

Answer: (1) is correct

Reflexive relation.

x R x; (x;x)∈R

here, (1,1),(2,2),(3,3)∈ R

So; it is Reflexiv

24. The range of the reltation {(1,0)(2,0),(3,0)(4,0)(0,0) is

1. {1,2,3,4,0}
2. {0}
3. {1,2,3,4}
4. None

Answer: (2) is correct

Range = {0}

25. If a relation S = {1,2,3} then {(1,1),(2,2),(1,2),(2,1)} is symmetric and

1. Reflexive but not transitive
2. Reflexive as well as transitive
3. Transitive but not reflexive
4. Neither transitive nor reflexive

Answer: If S = {1,2,3} then

Then relation {{1,1); (1,2); (2,1)} is symmetric and transitive but not reflexive.

(3) is correct

26. If A = {1, 2} and B = {3,4}. Determine the number of relations from A and B:

1. 3
2. 16
3. 5
4. 6

Answer: (2)

No. of Relations = 2^n(AxB)

=2^(2×2) = 16.

27. A = {1,2, 3, 4 .. 10} a relation on A, R = {x,y):x + y = 10, x ∈ A, Y ∈ A, x ≥ Y} then domain of R^-1 is

1. {1,2, 3,4,5}
2. {0, 3, 5, 7, 9}
3. {1,2, 4, 5, 6, 7}
4. None

Answer: (1)

Given ; A = {1, 2, 3,…….; 10}

R = {x;y) / x + y= 10;X ∈A;Y ∈A;x≥y}

=> R = (5 ; 5) ; (6 ; 4) ; (7 ; 3) ; (8 : 2) ; (9 :1)

=> R^-1 = (5 ; 5) ; (4; 6) ; (3 ; 7) ; (2 ; 8) ; (1 ;9)

Domain of R^-1 = (5 ; 4; 3 ; 2 ;1)

Exercise – 3: Function

Choose the most appropriate option/options (1) (2) (3) or (4}.

1. If A = {x,y,z}, B = {p, q, r, s) which of the relation on A to B are function.

1. {n, p),(x,q), (y, r), (z, s}},
2. {(x, s}, (y, s), (z, s)}
3. {(y, p), (y, q), (y, r), (z, s),
4. {(x, p), (y, r), (z, s}}

Answer:(2) in a function, element from domain can have only one range in range and every element has a image

2. {(x, y)|x+y = 5} where x, y ∈ R is

1. not a function
2. a composite function
3. one-one mapping
4. none of these

Answer: (3) one-one mapping

3. {(x, y) | x = 4} where x, y ∈ R is a

1. not a function
2. function
3. one-one mapping
4. none of these

Answer: (1) not a function

as only one element has an image in the rang

4. {(x, y}, y=x²} where x, y ∈ R is

1. not a function
2. a function
3. inverse mapping
4. none of these

Answer: (2) a function

As every element in domain has only one image

5. {(x, y}|x<y} where x, y ∈ R is

1. not a function
2. one-one mapping
3. a function
4. none of these

Answer: (1) not a function

6. The domain of {(1, 7), (2, 6)} is

1. {1, 6}
2. {7, 6}
3. {1, 2}
4. {6,7}

Answer: {(1,7), (2, 6)}

Domain (3) {1, 2}

Set of first elements of a pair of a function

7. The range of {(3, 0), (2, 0), (1, 0), (0, 0}} is

1. {0,0}
2. {0}
3. {0, 0, 0, 0}
4. none of these

Answer: (2) {0}

Set of second elements of a pair of function

8. The domain and range of {{x,y} : y = x²} where x, y ∈ R is

1. (reals, natural numbers)
2. (reals, positive reals)
3. (reals, reals)
4. none of these

Answer: (2) (reals, positive reals)

Domain can take any real value & y = x² -> y will only take positive real no.

9. Let the domain of x be the set {1}. Which of the following functions are equal to 1

1. f(x) = x², g(x) = x
2. f(a) = x, g(x) = 1-x
3. f(x) = x² + x + 2, g(x) = (x+ l)²
4. none of these

Answer: (1) f(x) = x², g (x) = x

f1) = 1² = 1 & g(1) =1

∴ 1

10. If f(x) = 1/1-x, f(-1} is

1. 0
2. 1/2
3. 0
4. none of these

Answer: (2) 1/2

\(f(x)=\frac{1}{1-x}\) \(f(-1)=\frac{1}{1-(-1)}=\frac{1}{2}\)

-> (2) 1/2

11. If g(x} = (x-1)/x, g(-1/2) is

1. 1
2. 2
3. 3/2
4. 3

Answer: (4) 3

\(g(x)=\frac{x-1}{x}\) \(g(-1 / 2)=\frac{-3 / 2}{-1 / 2} \rightarrow 3\)

12. If f(x} = 1/1-x and g(x) = (x-1}/x, than fog(x} is

1. x
2. 1/x
3. -x
4. none of these

Answer: (1) x

\(f(x)=\frac{1}{1-x} g(x)=\frac{x-1}{x}\) \(f o g(x)=f(g(x))=\frac{1}{1-g(x)}\) \(=\frac{1}{1-\frac{(x-1)}{x}}\) \(=\frac{1(x)}{x-x+1} \rightarrow \frac{x}{1} \rightarrow \mathrm{X}\)

13. If f(x) = 1/1-x and g(x) = (x-1)/x, then gof(x) is

1. x-1
2. x
3. 1/x
4. none of these

Answer: (2) x

\(f(x)=\frac{1}{1-x}\)

\(g(x)=\frac{(x-1)}{x}\) gof(x)

\(g(\mathrm{f}(\mathrm{x}))=\frac{f(x)-1}{f(x)}\) \(=\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}\) \(\rightarrow \frac{\frac{1-1+x}{1-x}}{\frac{1}{1-x}} \rightarrow \frac{x}{1} \rightarrow \mathrm{x}\)

14. The function f(x) = 2^x is

1. one-one mapping
2. one-many
3. many-one
4. none of these

Answer:(1) one-one mapping

15. The range of the function f(x) = log₁₀(1 + x) for the domain of real values of x when is 0≤ x ≤ 9 is

1. [0,1]
2. [0,1,2]
3. [0, -1]
4. none of these

Answer: (1) [0, 1]

16. The Inverse function f^-1 of f(x) = 2x is 

1. 1/2x
2. x/2
3. 1/x
4. none of these

Answer: (2) x/2

f(x) = 2x

y = 2x->x =y/2

∴ Inverse -> f^-1 =x/2

17. If f(x)= x+3, g(x) = x², then fog(x) is

1. x² + 3
2. x² + x + 3
3. (x + 3)²
4. none of these

Answer: (1) x² + 3

Fog x = f(g(x)) = g(x) + 3 = x² + 3

18. If f(x) = x+3, g[x] = x², then f(x).g(x] is

1. (x + 3)²
2. x² + 3
3. x³ + 3x²
4. none of these

Answer:(3) x³ + 3x²

f(x).y(x) = (x + 3) (x²) = x³ + 3x²

19. The Inverse h^-1 when h(x) = log₁₀  x is

1. log₁₀x
2. 10x
3. log₁₀(1/x)
4. none of these

Answer: (2) 10x

h(x) = log₁₀ x

y = log₁₀x -> 10y = x

h^-1 = 10x

20. For the function h(x) = 10^1+x the domain of real values of x where 0 ≤ x ≤ 9, the range is

1. 10 ≤ h(x) ≤ 10¹⁰
2. 0 ≤ h(x) ≤ 10¹⁰
3. 0<h(x)<10
4. none of these

Answer:(1) 10 ≤ h(x) ≤10¹⁰

h(x).10x+1, 0 ≤ x ≤ 9

at 0 smallest→ 10¹

At 9 biggest value→ 10¹⁰

10 ≤ h(x) ≤ 10¹⁰

21. Let A = (1,2,3), B = (2,3,4), then which of the following is a function from A to B?

1. {(1,2),(1,3),(2,3),(3,3)}
2. {(1,3), (2,4)}
3. {(1,3),(2,2), (3,3)}
4. {(1,2),(2,3),(3,2),(3,4)}.

Answer: (3) {(1,3), (2,2), (3,3)}

for a function every element in domain should have an image & not more than one

22. If f: Q —» Q is defined as f (x) = x², then f^-1 (9) is equal to

1. 3
2. -3
3. {-3,3}
4. 0

Answer:(3) {-3.3}

y = x²

x=y

f^-1(x)=y

f^-1=9

=3

or -3->{-3,3}

23. Let f(x) = | x-1|. Then,

1. f(x²) = [f(x)]²
2. f(x+y) = f (x) f(y)
3. f(|x|) = |f(x)|
4. none of these

Answer: (4) none of these

24. If A = {1,2,3}, B = {x, y}, then the number of functions that can be defined from A into B is

1. 12
2. 8
3. 3f(x)
4. -f(x)

Answer: (2) 8

25. Let A = {x ∈ R : x ≠ 0, – 4 ≤ x ≤ 4} and f: A —> R be defined by f(x) = |x|/x for x ∈ A. Then A is

1. {1,-1,}
2. {x:0≤x≤4}
3. {1}
4. {x:-4≤ x ≤ 0}

Answer: (1) {1,-1}

\(\frac{|x|}{x}=\left\{\begin{array}{cc}
1 & 4 \geq x>0 \\
-1 & -4 \leq x<0
\end{array}\right.\)

26. If f:R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x² + 7, then the values of x such that g (f(x)) = 8 are

1. 1,2
2. -1,2
3. -1,-2
4. 1,-2

Answer: (3) -1, -2

g(f(x)) = 8

f(x)² + 7 = 8

(2x + 3)² =1

2x + 3 = 1 or 2x + 3 = 1

2x = -2 X = -1

x = -2

27. Let R is the set of real numbers, such that the function f: R→ R and g : R → R are define by f(x) = x² + 3x + 1 and g(x) = 2x- 3. Find (fog):

1. 4x² + 6x + 1
2. x² + 6x +1
3. 4x²- 6x + 1
4. x² – 6x + 1.

Answer: (3) f(x) = x² + 3x + 1

g(x) = 2x – 3

fog = f{g(x)}

= f (2x – 3)

= (2x- 3)² + 3(2x – 3) +1

4x²- 2.2X.3 + 9 + 6x – 9 + 1

= 4x² – 6x + 1

(3) is correct

28. If R is the set ofreal numbers such that the function f: R →R is defined by f(x) = (x+1)², then find (fof)

1. (x+1)²+ 1
2. x²+1
3. {(x+1)²+1}²
4. None

Answer:(3) f(x) = (x + 1)²

fof=f{f(x)} = f{(x+1)²)

= { (X + 1)² + 1)²

(3) is correct

29. If f: R → R. f(x) = 2x + 7, then the inverse of f is:

1. f^-1 (x) = (x-7)/2
2. f^-1 (x) = (x + 7)/2
3. f^-1(x) = (x – 3)/2
4. None.

Answer: (1) Let y = f(x) = 2x+ 7

Or 2x = y-7

Or\(x=\frac{y-7}{2}\)

\(f^{-1}(x)=\frac{x-7}{2}\)

30. Let f: R → R be such that f(x) = 2^x, then f(x+y) equals:

1. f(x) + f(y)
2. f(x). f(y)
3. f(x) ÷ f(y)
4. None of these

Answer: (2) f(x) = 2X

f(x+y) = 2x+y = 2x2y = f(x).f(y)

(2) is correct

31. If A = {1,2,3,4,}; B = {2,4,6,8,} f(1) = 2, f(2) = 4, f{3) = 6 and f(4) = 8, and f: A->B then f^-1 f: A->B then f-1 is:

1. {(2,1),(4.2),(6,3),(8,4)}
2. {(1,2),(2,4),(3,6),(4,8)}
3. {(1.4),(2,2),(3,6),(4,8)}
4. None of these

Answer:(1) f^-1 = {(2,1),(4,2),(6,3)1(8,4)}

(1) is correct

32. If f(x) = x² + x -1 and 4f(x) = f(2x) then find ‘x’

1. 4/3
2. 3/2
3. -3/4
4. None of these

Answer:(2) f(x) = x² + x-1

And 4f(x) = f(2x)

Or 4(x² + x- 1) = (2x)²+ 2x-1

Or 4x² + 4x – 4 = 4x² + 2x-1

Or4x-2x = 4-1

Or 2x = 3 x = 3/2

(2) is correct

33. X = {x,y,w,z} y ={1,2,3,4); H={(x,1);(y,2);(y,3);(z,4);(x,4)}

1. H is a function from x to y
2. H is not a function from x to y
3. H is a relation from y to x
4. None of these

Answer:(2) H is not a function from x to y because x has 2 images 1 & 4

(2) is correct

34. Given the function f(x) = (2x+3), then the value of f(2x) – 2f(x) + 3 will be:

1. 3
2. 2
3. 1
4. 0

Answer: (4) f(x) = 2x + 3

f(2x) – 2 f(x) + 3

= 2(2x) + 3 – 2(2x+3)+3

= 4x + 3 – 4x – 6 + 3 =0

(4) is correct

35. If f(x) = 2x + h then find f(x+h)-2f(x)

1. h-2x
2. 2x-h
3. 2x+h
4. None of these

Answer:(1) f(x) = 2x + h

f(x + h) – 2f(x)

= 2x + 2h + h – 4x – 2h

= h – 2x

(1) is correct

36. If F: A→ R a real valued function defined by f(x) =1/x then

1. R
2. R -{1}
3. R-{0}
4. R-N

Answer: (2)

f(x) =1/x is defined at all x ∈ R except x = 0

A = R – {0}

(3) is correct

37. If f: R — R, f(x) = x + 1, g: R → R g[x) = x² +1 then fog(-2) equals to

1. 6
2. 5
3. -2
4. None

Answer: (1) f(x) = x + 1

g(x) = x² + 1. => g(-2) = (-2)² +1 = 5

fog (-2) = f{g(-2)) = f(5)

=5+1=6

(1) is correct

38. If f(x-1) = x²- 4x + 8, then f(x+1) =

1. x² + 8
2. x² + 7
3. x² + 4
4. x² – 4x

Answer: (3) f(x-1)=x²-4x + 8

=(x-1-1)2-4 (x-1+1) + 8

f(x+1) = (x+ 1 + 1)² – 4(x+ 1 + 1)+8

= (x+2)²- 4(x+2) + 8

= x² + 4x + 4-4x-8 + 8

= x² + 4

(3) is correct

39. If A – {±2, +3}B = {1,4,9)ANS F = {(2,4),(-2,4),(3,9){-3,4)} Then F is defined as:

1. One to one function from A into B
2. One to one function from A onto B
3. Many to one function from A onto B
4. Many to one function from A into B.

Answer:(3)

(3) is correct

40. \(\text { If } f(x)=\frac{x}{\sqrt{1+x^2}} \text { and } g(x)=\frac{x}{\sqrt{1-x^2}} \text { find fog? }\)

1. x
2. 1/x
3. \(\frac{x}{\sqrt{1+x^2}}\)
4. \(x \sqrt{1+x^2}\)

Answer:

(1) fog = f{g(x)}\(=f\left(\frac{x}{\sqrt{1-x^2}}\right)\)

\(=\frac{\frac{x}{\sqrt{1-x^2}}}{\sqrt{1+\frac{x^2}{1-x^2}}}\) \(=\frac{\frac{x}{\sqrt{1-x^2}}}{\sqrt{\frac{1-x^2+x^2}{1-x^2}}}\) \(=\frac{x}{\sqrt{1-x^2}} \times \frac{\sqrt{1-x^2}}{\sqrt{1}}=x\)

(1) is correct

41. E(x) = 3 +x, for -3 < x < 0 and 3 – 2x for 0<x<3, then value of f(2) will be

1. -1
2. 1
3. 3
4. 5

Answer: (1) is correct

f(x) = 3 + 2x; when -3 < x < 0

= 3- 2x ; when 0 < x < 3

f(x=2) = 3- 2x² = -1

2 lies in 2nd condition

42. The range of the function f: N —> N; = (-1)^x-1 , is

1. {0,-1}
2. {1,-1}
3. {1,0}
4. {1,0,-1}

Answer: (2) is correct

f(x)-(-1)x-1

If x= odd no.

f(x) = odd no.

f(x) = 0; even N.

Range = {1 ; -1)

Domain = {Any real No.}

43. If f: R —> R is a function, defined by f(x) = 10x – 7, if g(x) = f^-1 {x), then the value of g(x) is
equal to

1. \(\frac{1}{10 x-7}\)
2. \(\frac{1}{10 x+7}\)
3. \(\frac{x+7}{10}\)
4. \(\frac{x-7}{10}\)

Answer: (3) is correct

Let y= f(x) = 10x-7

Or 10x = y + 7

\(x=\frac{y+7}{10}\) \(f-1(x)=\frac{x+7}{10}\) \(g(x)=\frac{x+7}{10}\)

44. If f(x) = x+2, g(x} = 7^x then go f(x} =

1. 7^x.x+2.7^x
2. 7^x+2
3. (7^x)+2
4. None

Answer: f(x) = x+2 ; g(x) = 7x

g,pf(x) = g{f(x)} = g(x+2) = 7^x+2

(2) is correct

45. If f(x)=\(\log \left(\frac{1+x}{1-x}\right) \text { then } f\left(\frac{2 x}{1+x^2}\right)\)

1. f(x)
2. 2f(x)
3. 3 f(x}
4. -f(x)

Answer: (2)

\(f\left(\frac{2 x}{1+x^2}\right)=\log \left(\frac{1+\frac{2 x}{1+2 x^2}}{1-\frac{2 x}{1+x^2}}\right)\) \(=\log \left(\frac{1+x^2+2 x}{1+x^2-2 x}\right)\) \(=\log \frac{(1+x)^2}{(1-x)^2}=\log \left(\frac{1+x}{1-x}\right)^2\) \(=2 \log \left(\frac{1+x}{1-x}\right)=2 f(x)\)

(2) is correct

46.\(\text { If } f(x)=\frac{x^2-25}{x-5} \text { then } f(5) \)

1. 0
2. 1
3. 10
4. Undefined

Answer: (4) is correct

\(f(5)=\frac{x^2-25}{x-5}=\frac{5^2-25}{5-5}=\frac{0}{0}\)

Undefined

47. \(\mathrm{f}(\mathrm{x})=\left(a-x^n\right)^{\frac{1}{n}}\),a > 0 and n is positive integer then f [f(x)] =

1. x
2. a
3. x^1/n
4. a ^1/n

Answer: (1) is correct

\(\mathrm{f}\{\mathrm{f}(\mathrm{x})\}=\mathrm{f}\left\{\left(\mathrm{a}-\mathrm{x}^{\mathrm{n}}\right)^{1 / n}\right\}\) \(=\left[a-\left\{\left(a-x^n\right)^{\frac{1}{n}}\right\}^n\right]^{\frac{1}{n}}\) \(=\left[a-\left(a-x^n\right)\right]^{\frac{1}{n}}=\left[x^n\right]^{\frac{1}{n}}=x\)

48. If A = {1,2,3} and B = {4,6,7} then the relation R = {(2,4)(3,6)} is

1. A function
2. A function from A to B
3. Both {a} and (b}
4. Not a function

Answer: (4) is correct.

Note: -1 has no image

49. \(\text { If } f(x)=\frac{x}{x-1} \text {, then } \frac{f(x / y)}{f(y / x)}=\) ___________

1. x/y
2. y/x
3. -x/y
4. -y/x

Answer:

\(f(x)=\frac{x}{x-1}\) \(f(x / y)=\frac{x / y}{x / y-1}=\frac{x / y}{\frac{x-y}{y}}=\frac{x}{x-y}\) \(f(y / x)=\frac{y / x}{y / x-1}=\frac{y / x}{\frac{y-x}{x}}=\frac{y}{-(x-y)}\) \(\frac{f(x / y)}{f(y / x)}=\frac{\frac{x}{x-y}}{y}=-\frac{x}{y}\)

50. Let N be the set of all natural number; E be the set of all even natural numbers then the function f:N→ E defined as f{x) = 2x; V x ∈ N is:

1. One-one into
2. One-one onto
3. Many-one into
4. Many-one onto

Answer:(2) is correct

N = {1,2, 3,…;n}

E = {2,4, 6……;2n}

Clearly it is one-one onto mapping

51. Which of these is a function from A →B; A = {x,y,z}; B {a,b,c,d}

1. {(x,a)(x,b)(y,c)}
2. {(x,a}(x,b)(y,c)(z,d)}
3. {9x,a} (y,b) (z.d)}
4. {(a,x)(b,z)(c,y)}

Answer: (3)

52. If (x) = 2x + 2, g(x) = x², fog(4) = ?

1. 100
2. 10
3. 34
4. 55

Answer: Fog (x) = f{g(x)}

= f(x²) = 2.x² + 2

Fog (4) = 2 x 4² + 2 = 34

(3) is correct

53. The domain (D) and range (R) of the function f(x) = 2-|x+1|is

1. D = Real numbers, R = (2,∞)
2. D= Integers, R= (0,2)
3. D = Integers, R = (-∞, ∞)
4. D = Real numbers, R = (-∞,2]

Answer: (4)

Lety = f(x) = 2-|x+1|

For any real values of x: f(x) is defined

Domain = D Real number

Minimum value of |x+1| is zero

Maximum value of range

=2-0=2

Range = – ∞ < y ≤ 2

= (-∞; 2]

54. If f (x) = 100 x then f^-1 (x) =

1. x/100
2. 1/100x
3. 1/100
4. None of these

Answer: (1) is correct

Let y = f (x) = 100x

\(X=\frac{y}{100} ; \text { So, } f^{-1}(x)=\frac{x}{100}\)

55. f: R —> R is defined by f (x) = 2^x then f is

1. One- one and onto
2. Many to one
3. One-one and into
4. One to many

Answer: (2) is correct

56.\(\text { If } f(x)=\frac{x-1}{x} \text { and } g(x)=\frac{1}{1-x} \text { then fog }(x)=\)

1. x-1
2. x
3. 1-x
4. -x

Answer: fog(x) = f{g(x)}\(=\frac{g(x)-1}{g(x)}\)

\(=\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}=\left(\frac{1-1+x}{1-x}\right) \times \frac{(1-x)}{1}\)

= X

Option (2) is correct

57. \(\text { If } \mathrm{f}(\mathrm{x})=\frac{x+1}{x+2}=\text { then } \mathrm{f}\left[\mathrm{f}\left(\frac{1}{x}\right)\right]=\)____________

1. \(\frac{2 x+3}{3 x+5}\)
2. \(\frac{2 x+5}{3 x+2}\)
3. \(\frac{3 x+2}{5 x+3}\)
4. \(\frac{5 x+2}{2 x+3}\)

Answer: (1)

\(f\left(\frac{1}{x}\right)=\frac{1 / x+1}{1 / x+2}=\frac{1+x}{x} \times \frac{x}{1+2 x}\) \(=\frac{1+x}{1+2 x}\)

Now, \(\mathrm{f}\left[f\left(\frac{1}{x}\right)\right]=\mathrm{f}\left(\frac{1+x}{1+2 x}\right)\)

\(=\frac{\frac{1+x}{1+2 x}+1}{\frac{(1+x)}{1+2 x}+2}=\frac{\frac{3 x+2}{1+2 x}}{\frac{5 x+3}{1+2 x}}\) \(=\frac{3 x+2}{5 x+3} .\)

58. If f(x) = x² and g(x) = √x then

1. go f(3) = 3
2. go f(-3) = 9
3. go f(9) = 3
4. go f(-9) = 3

Answer:(1)

f(x) = x² ; g(x) = √x

go f(x) = g {f(x)}

√f(x) = √x²

= X

gof(3) = 3

(1) is correct.

 

Sequence & Series Introduction

Terms arranged in n definite order from ;i sequence. The terms in a sequence may lie numbers, letters, symbols or even words.

Here, we consider general sequences, before moving on to three specific mathematical sequences known as progressions i.e., arithmetic, geometric and harmonic before considering
number series.

Sequences

A Sequence is a logically ordered list of elements related to each other by some relationship.

Identify the pattern followed by the terms in a sequence and use the pattern to find the terms of the sequence, sum of the terms in the sequence or to identify properties of the sequence.

Terms of a sequence are generally denoted by T₁,T₂, T₃,……. T_n

A lot of sequences either display a difference-based or a multiplicative pattern.

However, there are infinite ways in which sequences can be formed. Only a couple of patterns are discussed below to show how to analyze a sequence and mathematically represent it.

Read and Learn More CA Foundation Maths Solutions

Difference

In these sequences, consecutive terms are related to each other in terms of the difference between the two.

This difference can be constant or may follow a logical pattern itself.

For example, consider the sequence 1, 3,7,13, 21, 31

Observe that the difference between successive terms is 2,4,6,8, and 10.

Hence, the pattern followed by the difference is that they are all multiples of 2. Hence, the term after 31 should be 31+12=43. Mathematically, each term of the sequence can be expressed as:

T₁ =1+2×0=1
T₂=1+2×1=3
T₃ =3+2X2=7
T₄ =7+2X3=13
T₅ =13+2×4=21
T₆ =21+2X5=31

Thus, the n,h term of the sequence can be written as T_n = T_n 4+ 2(n — 1)

We can see that the nth term depends on the previous term as well as its position in the sequence.

Hence, T₇= 31+2×6=43

The advantage of representing a sequence mathematically is that later terms of the sequence can be found easily without the need to write the entire sequence.

For instance, in the above case, if T₃₈ is known and the value of T₄₀ is to be found, there is no need to write the entire sequence up to 40 terms. Instead, using the value of T₃₈, the value of T₃₉ and consequently, that of T₄₀ can be found in just 2 steps.

Solved Examples

Find the 7th term of the sequence 1,2,4,7,11, 16

Solution:

T₁ =1
T2=1+1=2
T3=2+2=4
T4=4+3=7
T5=7+4=11
T₆ =11+5=16

The nth term or this sequence can he expressed as

T_n =T_n-1+(n- 1)
T₇ =T₆ + (7- 1)
= 16+6
=22

Here the 7th term of this sequence would be 16+6=22

Cumulative Sequence

Consider the sequence 1,1,2,3,5,8,13, 21…. From the sequence observes that after the second term, the next term is the sum of the previous two terms. Hence the sequence mathematically represented in the following manner

T₁= 1
T₂=1
T₃ = 1+1=2
T₄ =2+1=3
T₅ =3+2=5
T₆ =5+3= 8
T₇ = 8+5=13
T₈ =13+8=21
and so on.

As can be seen, the next term is the sum of its previous two terms, hence

T_n = T_n-1 + T_n-2

∴T₉= T₈+ T₇

∴ T₉ =21+13=34

In these types of sequences, the pattern is formed with the help of its previous terms.

Solved Examples:

Find the next term of the series 3,4,11,24, 43

Solution:

Difference between the 1st and 2nd term=l

Difference between the 2ml and 3rd term=7=l+6

Difference between the 3rd and 4lh term=13=7+6

Difference between the 4th and 5lh term =19=13+6

Thus, the difference between the 5th and 6th terms=19+6=25.

Hence the next term is 43+25=68.

Progressions

There are 3 specific types of sequences which show a specific mathematical relationship among their terms. These 3 types, also known as progressions, are named Arithmetic
Progression, Geometric Progression and Harmonic Progression

Arithmetic Progression

The terms of a sequence are said to be in Arithmetic Progression (A.P.) when they differ by a constant value known as their Common difference, dented by d. In other words, the
difference between any two consecutive terms in an A.P. is constant. The first term of an A.P. is generally denoted by a.

If d > 0, the A.P. is said to be an increasing A.P.

If d < 0, the A.P. is said to be a decreasing A.P.

If d = 0, each term of the sequence is equal.

Some A.P.s are given below.

1)1,2,3,4……where a= 1 and d= 1
2) 3.7, 11. 15…where a= 1 and d= 1
3) 8,2,-4,-10 … where a = 8 and d =-6

THE nth TERM OF AN A.P.

The first, second, third…nth terms of an A.P. are denoted by T₁, T₂ T₃,….T_n

For any arithmetic progression having the first term a and common difference d,

T₁ = a = a + [1-1) d
T₂ = a + d = a + (2-1) d
T₃ = a + 2d = a + (3-1) d and so on.

Continuing thus, the n,h term of an A.P. is, Tn=a+(n-1) d

Solved Examples

1. Find the fifteenth term of term A.P. -3,-9,-15,…….

Solution:

T_n=a+(n-1) d

Here, a = -3, d= -9-(-3) = -6 and n =15

T₁₅= -3 + (14)(-6)

T₁₅ = -87

2. The ninth term exceeds the fifth term of an A.P. by 32. The sum of the ninth and fifth terms is 114. Find the eighth term of the A.P.

Solution:

T₉ = a + (9 – 1)d = a + 8d

T₅ = a + (5- 1)d = a + 4d

T₉ + T₅ = 32

∴ (a+8d) – (a+4d)=32

∴ 4d=32

∴ d=8

T₉ + T₅ =(a +8d) + (a+4d) =2a+12d

T₉ + T₅ =114

114 = 2(a+6d)

a + 6d = 57

T8= a + 7d = a + 6d + d = 57 + 8 = 65

3. The 54th and the 4th terms of an A.P. are – 61 and 64 respectively. Find the 23rd term.

Solution:

The 54th and 4th term of the given A.P. can be represented as shown below.

a +53d= – 61 ……………………(i)

a + 3d = 64 ……………..(ii)

Subtracting we get

50d = -125

d= -5/2,

a= 64-3d=143/2

Hence the 23rd term = a+22d= 143/2+22(-5/2)=33/2

Sum of N Terms of An A.P.

Let the first term and common difference of an A.P. containing n terms he a and d respectively. Let Tn be the last term of the A.P. Then, the sum of n terms of the A.P. is
S_n= n/2[2a+ (n- 1)d] =n/2[a + T_n]

Solved Examples

1. The sixth and eighth terms of an A.P. are 38 and 52 respectively. Find the sum of the first twelve terms of the A.P.

Solution:

T₆ = a + 5d = 38

T₈ = a + 7d = 52

Solving the two equations, d= 7 and a=3

The sum of12 terms of the A.P. is

S₁₂=12/2[2×3 + (11)7] = 498

2. How many terms of the series -12, -9, -6, … must be taken so that the sum may be become 78?

Solution:

The given sequence is an A.P. with a = -12 and d =3

For an A.P. the sum of the first n terms is

s_n =n/2[2a + (n-l)d]

The sum ofn terms of this sequence is 78

n/2 x [2a + (n- 1)d] = 78

n/2 x [—24 + (n- 1)3] = 78

nx (n- 9) = 52

n=13

Average of the Terms of an A.P.

The average or the arithmetic mean ofn terms of an A.P

\(=\frac{S_n}{n}=\frac{n}{2} \times \frac{\left|a+T_n\right|}{n}\)

The average of n terms of an A.P

\(=\frac{a+T_n}{2}\)

The average of n terms of an A.P.

\(=\frac{(a+d)+\left(T_n-d\right)}{2}\)

The average of n terms of an A.P.

\(=\frac{(a+2 d)+\left(T_n-2 d\right)}{2}\)

Continuing thus, we see that the average of the terms of an A.P. is equal to the average of its first and nth terms, second and f(n-l),th terms, third and (n-2),th terms and so on.

In general, the average of the terms of the A.P. is equal to the average of the kth term from the beginning and the kth term from the end, or it is equal to average of any two terms
of the A.P. that are equidistant from the beginning and the end.

Also, if n is even the average of the A.P.is equal to the average of its

(n/2) th and (1+n/2)th terms.

if n is odd, the average of the terms of the A.P. is equal to the (n+1/2) th term of the A.P.

Solved Examples

1. The sum of the first nine terms of an A.P. is 387. Pind the fifth term,

Solution:

Since n is odd, the average of the first n terms of the A.P, is equal to the (n = 1)/2th

term = 5″‘ term.

Hence, the average of the first 9 terms of the A.P. = the (9 + 1)/2th term = 5,th term.

∴the 5′” term of the A.P. = 387/9= 43

If any two consecutive terms of an arithmetic progression are known, the series can be completely determined.

2. The fourteenth and fifteenth terms of an A.P. are 25 and 32 respectively. Find the 30th term, sum of the first 30 terms and the first term of the A.P.

Solution:

T₁₄ = a + 13d = 25

T₁₅ = a + 14d = 32

d = 7 and a = -s66

T30-66+29X 7 = 137

The sum of the first n terms = n/2 (a+T_n)

The sum of the first 30 terms

=30/2(-66+137)=1065

3. A teacher observes that the marks that the students in her class have scored are all different. She arranges her students in a line in increasing order of their marks such
that difference in marks scored by any two students next to each other is 4. The lowest marks that any student has scored are 11. The sum of the marks that all her students
have scored is 585. Find the marks scored by the student standing in the middle of the line.

Solution:

Since the difference in the marks of any two adjacent students is 4, the marks of the students standing in the line form an A.P. with a = 11 and d=4

Let there be n students in the class.

s_n = n/2 [2a + (n- l)d]

∴585 = n/2[22 + (n- 1)4]

∴(n) (2n+9) = 585

On solving this equation,

∴n = 15

The 8th student stands in the middle of the line. Since n = 15, the average of the first 15 terms is equal to the value of the (15+1)/2,h term = 8th term.

Hence, the value of the 8lh term of this A.P.

= 585/15 = 39

Hence, the marks of the 8lh student are 39.

Note:

When three terms are in Arithmetic progression, the middle term is the arithmetic mean of the other two. It is always convenient to take three terms in an A.P. as (a-d), a and (a + d).
Similarly, four terms in an A.P. could be taken as a-3d, a-d, a + d and a + 3d; five terms could be taken as a -2d, a – d, a, a+ d, a + 2d.

The advantage of representing terms in tin’s way is that the sum of terms is then obtained in only one unknown, i. e. when the sum of these terms is written mathematically, d cancels out and the sum is expressed only in terms of a.

Some Properties of an A.P.

If each term of an A.P. is increased, decreased, multiplied or divided by the same non -zero number, then the resulting sequence is also an A.P. In case the terms are increased or decreased by some quantity, the common difference of the new A.P. remains equal to that of the original A.P.

In case the terms are multiplied or divided by a constant c (c≠0), the common difference d accordingly changes to d x c or d/c.

The number of elements in an arithmetic series from n₁ to n₂ with a step size (or common difference) ofm is1 = (n₁-n₂)/m

Solved Examples

1. How many multiples of 13 lie between 1000 and 5000? What is the sum of all these multiples?

Solution:

The lowest multiple of 13 that is greater than 1000 is 1001. The greatest multiple of 13 that is lesser than 5000 is 4992.

Here, m=13, n₁=1001 and n₂=4992

\(\text { There are } \frac{4992-1001}{13}+1\)

=308 multiples of 13 between 1000 and 5000. The sum ofall these multiples is

S=n/2(a+T_n)

S=308/2(1001+4992)

∴ S=9,22,922

If the sum of the first p terms of an A.P. is equal to the sum of the first q terms of the A.P. such that p and q are different, then the sum of (p + q) terms of the A.P. is zero.

2. The sum of the first 16 terms of an A.P. is equal to the sum of the first 24 terms of the A.P. Find the sum of the first 40 terms of the A.P.
Solution:

∴S₁₆=S24

∴S_(16+24)=0

∴S₄₀=0

Geometric Progression

The terms of a sequence are said to be in Geometric progression (G.P.) when they increase or decrease by a constant factor. This constant factor is called the common ratio, denoted by r, and can be found by dividing any term of the sequence by the preceding term.

If the first term is positive and common ratio is greater than 1 (or if the first term is negative and the common ratio is less than1 and positive), the G.P. is an increasing G.P.

If the first term is positive and the common ratio is less than 1 and positive (or if the first term is negative and the common ratio is greater than 1), the G.P. is and decreasing G.P.

In other words, if all terms are greater than the preceding terms, the G.P. is an increasing G.P. else it is a decreasing G.P.

If the common ratio is equal to 1, all terms of the G.P. are equal.

A sequence with all terms equal is both, an A.P. and a G.P.

If the first term is a, the terms of the progression are a, ar, ar², ar³……………

THE nth TERM OF A.G.P

If T₁, T₂, T₃ ….Tn denote consecutive terms of a G.P. then

T₁= a= ar^1-1

T₂ – ar= ar^2-1

T₁= ar2= ar^3-1

Continuing thus, the n,h term of the geometric progression is given by T_n= ar^n-1

Solved Examples

1. Find the fifth tern if the G.P. whose first terms is 3 and the common ratio is 1/3.

Solution:

a=3 and r= (1/3)

The 5th term =ar^5-1 = ar⁴ =3(1/3)^5-1

=3/81

=1/27

2. The product of the first five terms of an A.P. is 28. Find the third term.

Solution:

ax arx ar²xar³xar⁴=28

a⁵r¹⁰ =28

(ar²)⁵= 28

ar² =⁵√28

∴the third term = ⁵√28

Alternatively, In such a case, assume the central term to be a and find the other terms from this point onwards, In this case, there are 5 terms in all. Hence, let the third term of this G.P. be a.

Hence, the 1st, 2nd 4th, and 5″‘ terms will be a/r², a/r, ar, and ar².

a/r²xa/rxaxarxar²=28

a⁵=28

The third term =a= ⁵√28

In case the G.P. has an even number of terms, one can take the two central terms to be a/r and ar and proceed from there.

Sum of N Terms of a G.P.

The sum of n terms or a G.P. with

r< 1 is a(1-rⁿ)/1-r The sum of n terms of a G.P. with

r <1 is a(rⁿ-1)/r-1  The sum of an infinite number of terms of a decreasing G.P.

=a/1-r

Geometric Mean

If terms

a₁,a₂,a_n are in G.P. then the Geometric mean G of these n terms is given by

G=ⁿ√a₁xa₂xa₃x…………..xa_n

If three terms are in G.P. then the middle term is the Geometric mean of the other two terms. If a, b and c are in G.P. (a, c > 0 or a, c < 0), then b is the geometric mean of a and c, and is given by b = ac or b² = ac

If n is even, the geometric mean of the terms of the G.P. is equal to the geometric mean of its (n/2) th and 1+(n/2)th terms.

If n is odd, the geometric mean of the terms of the G.P. is equal to the

(n+1/2)th term of the A.P.

Note:

While solving problems, three terms in G.P. can be assumed to be a/r, a and ar. Similarly, four terms in G.P. can be assumed to be a/r³, a/r, ar and ar³.

The advantage of representing terms in this way is that the product of terms is then obtained in only one unknown, i.e. in the product, r cancels out and the product is then expressed only in terms of a.

Some Properties of a G.P.

If each term of a G.P.is multiplied or divided by the some non-zero quantity, then the resulting sequence is also a G.P. with the common ratio remaining the

The reciprocals of the terms of a given G.P. also form a G.P. where the ratio is the reciprocal of that of the earlier G.P.

In a finite G.P. the product of two terms equidistant from the first and the last terms is same as the product of the first and the last term.

Exercise – 1: Arithmetic Progression (AP)

Choose the most appropriate option ( 1 ), ( 2 ), ( 3) or (4).

1. The nth element of the squence 1,3,5,7………………is

1. n
2. 2n – 1
3. 2n +1
4. none of these

Answer: (2) 2n – 1

Seq =1,3,5,7…

a = 1 d = 2

a_n = 1 + (n-l)d= 1 + [n-1)2

= 1 + 2n – 2 = 2n – 1

2. The nth element of the sequence -1, 2, -4, 8 is

1. (-1 )ⁿ2^n-1
2. 2^n-1
3. 2ⁿ
4. none of these

Answer: (1)(-1)ⁿ2^n-1

Seq = -1,2, -4, 8

a = 1

n = 1 (-1)1 21-1 =-l

n = 2 (-1)2 22-1 = 2

n = 3 (-1)3 23-1 = -4

a_n= (-1)ⁿ 2^n-1

3. \(\sum_{i=4}^7 \sqrt{2 i-1}\)

1. \(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\)
2. \(\sqrt[2]{7}+2 \sqrt{9}+2 \sqrt{11}+2 \sqrt{13}\)
3. \(\sqrt[2]{7}+2 \sqrt{9}+2 \sqrt{11}+2 \sqrt{13}\)
4. None of these

Answer: (1) \(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\)

\(=\mathrm{i}=4 \sum_{i=4}^7 \sqrt{8-1}=\sqrt{7}\) \(\sqrt{7}+\sqrt{9}+\sqrt{11}+\sqrt{13}\) \(=\mathrm{i}=5 \sqrt{2 i-1}=\sqrt{9}\)

4. The sum to 00 of the series -5,25,-125,625…..can be written are

1. \(\sum_{k=1}^{\infty}(-5)^k\)
2. \(\sum_{k=1}^{\infty} 5^k\)
3. \(\sum_{k=1}^{\infty}-5^k\)
4. none of these

Answer: (1)

\(\sum_{k=1}^{\infty}(-5)^k\)

= -5,25,-125,625

= (-5)¹, (-5)²,( 5)³

= (-5)^k

5. The first three terms of sequence when nth term tn is n2- 2n are

1. -1,0,3
2. 1, 0, 2
3. -1,0,-3
4. none of these

Answer: (1) -1,0,3

= nth terms = tn = n² – 2n

=1st term = t₁ = 1 – 2 = -1

= 2nd term = t₂ = 4 – 4 = 0

= 3rd term = t₃ = 9 – 6 = 3

6. Which term of the progression -1,-3,-5,………Is -39

1. 21st
2. 20 th
3. 19th
4. none of these

Answer:(2) 20 th

= a = -1 d = -2

a_n= -39

-1 + (n-1) (-2) = -39

(n-1) (-2) = -38

n – 1 = 19. Therefore, n = 20

7. The value of x such that 8x + 4, 6x- 2, 2x + 7 will form an AP is

1. 15
2. 2
3. 15/2
4. none of the these

Answer: (3) 15/2

= d will be same if in AP

= d = a_n+1+an

= 6x – 2 – 8x- 4 = 2x + 7 – 6x + 2

-2x – 6 = -4x + 9

2x=15 Therefore, x =15/2

8. The mth term of an A. P. is n and nth term is m. The rth term ofit is

1. m + n +r
2. n + m- 2r
3. m + n + r/2
4. m + n – r

Answer: (4) m + n – r

1) = a_m = a+(m-1)d

a + dm- d = n

2) = a_n= a + (n-1)d

a + dn – d = m

Substracting 2) from 1]

d + dn – d = m

a + dm – d = n
–      –       +

a (n-m) = m-n

d = -1

a = m + n – 1

a_r = m + n -1 (r-1) (-1)

=m+n-1-r+1

= m + n – r

9. The number of the terms of the series………….will amount to 155 is

1. 30
2. 31
3. 32
4. none of these

Answer: (1) 30 and (2) 31

\(=10+9 \frac{2}{3}+9 \frac{1}{3}+9 \ldots \ldots\) \(=10+\frac{29}{3}+\frac{28}{3}+9+\ldots \ldots\)

= a= 10 d= -1/3

S_n = n/2 (2a + (n- l)d)

155 =n/2(20 + (n-1)-1/3

310 = 20n -1/3n(n-1

310 = 20n-1/3(n2-n)

930 = 60n – n² + n

n² – 61n + 930 =

n(n-30) -31(n-30) = 0

n = 31 n = 30

10. The nth term of the series whose sum to n terms is 5n² + 2n is

1. 3n – 10
2. 10n – 2
3. 10n – 3
4. none of these

Answer: (3)

= S_n = 5n² + 2n

S₁ = a₁ = 5+2 = 7

S₁ = a₁ + a₂ = 5(2)² + 2(2) = 24

S₂-S₁ = a₂

24-7 = a₂

a₂= 17

d = a₂-a₂= 10

a_n = 7 + (n-1) 10

= 7 + 10n – 10

= 10n – 3

11. The 20th term of the progression 1, 4, 7, 10,……is

1. 58
2. 52
3. 50
4. none of these

Answer: (1) 58

= 1,4,7,10

a = 1, d = 3

a_n = 1 + (n-1)3

= 1 + 3n – 3

= 3n – 2

a₂₀ = 3(20)-2 = 58

12. The last term of the series 5, 7, 9,…..to 21 terms is

1. 44
2. 43
3. 45
4. none of these

Answer: (3) 45.

= a₂₁ = 5 + (n-1)2

= 5 + (20)2

= 45

13. The last term of the A.P. 0.6, 1.2, 1.8,… to 13 terms is

1. 8.7
2. 7.8
3. 7.7
4. none of these

Answer: (2) 7.8

= a₁₃ = 0.6 + (12) 0.6

= 0.6 (1+12)

= 0.6 x 13 = 7.8

14. The sum of the series 9, 5, 1,…. to 100 terms is

1. -18,900
2. 18,900
3. 19,900
4. none of these

Answer: (1) -18,900

S₁₀₀ = n/2 (2a + (n – 1)d) = 50(18 + 99(-4))

= 18-396

= -18,900.

15. The two arithmetic means between -6 ami 1 4 is

1. 2/3,1/2
2. \(2 / 3,7 \frac{1}{3}\)
3. \(-2 / 3,-7 \frac{1}{3}\)
4. none of these

Answer: (2) 2/3, 7 1/3

Two A-M between – 6 & 14

a=-6 a4=14

a₄=-6+3d=14

3d=20 d=20/3

\(\mathrm{a}_2=-6+\frac{20}{3}=\frac{2}{3} \mathrm{a}_3=-6+\frac{40}{3}=+7 \frac{1}{3}\)

16. The sum of three integers in AP is 15 and their product is 80. The integers are

1. 2, 8, 5
2. 8, 2, 5
3. 2, 5, 8
4. 8, 5, 2

Answer: (3) & (4)

= let 3 integers be a – d, a, a + d

Sum = a-d + a + a + d = 3a=15->a = 5.

(a – d) (a + d) a = 80

(a² – d²)a = 80

a² – d² = 16

a² – 16 = d²

25 – 16 = d² = 9 -7 d =±3.

AP when a = 5, d = 3

2,5,8

AP when a = 5, d = -3

8,5,2.

17. The sum of n terms of an AP is 3n² + 5n. The series is

1. 8,14,20,26
2. 8, 22, 42, 68
3. 22, 68, 114,
4. none of these

Answer: (1)

S₁ = 3 + 5 = 8 = a,

S₂ = 12 + 10 = 22

a₂ = S2 – S1 = 14, d = 6

Series -> 8, 14, 20,……..

18. The number of numbers between 74 and 25,556 divisible by 5 is

1. 5,090
2. 5,097
3. 5,095
4. none of these

Answer: (2) 5097

-7 75.80,……….25,555

It forms an AP with a = 75 d = 5

25,555 = Last term = a_n

A_n = 75 + (n-1)5

25,555 = 75 +(n-1)5

25480 = (n-1)5

5096 = n-1

5097 = n.

19. The pth term of an AP is (3p- 1)/6. The sum of the first n terms of the AP is

1. n(3n + 1)
2. n/12 (3n + 1)
3. n/12 (3n – 1)
4. none of these

Answer: (2) n/2(3n + l)

\(\rightarrow \mathrm{a}_1=\frac{3(2)-1}{6}=\frac{2}{6}=\frac{1}{3}\) \(a_2=\frac{3(2)-1}{6}=\frac{5}{6}\) \(d=\frac{5}{6} \frac{-2}{6}=\frac{3}{6}=\frac{1}{2}\) \(3_n=\frac{n}{2}(2 a+(n-1) d)=\frac{n}{2}\left(\frac{2}{3}+\right.\) \(\left.(n-1) \frac{1}{2}\right) \text {. }\) \(\frac{n}{2}\left(\frac{2}{3}+\frac{1}{2} n-\frac{1}{2}\right)=\frac{n}{2}\left(\frac{1}{6}+\frac{1}{2} n\right)\) \(=\frac{n}{12}(1+3 n)\)

20. The arithmetic mean between 33 and 77 is

1. 50
2. 45
3. 55
4. none of these

Answer: (3) 55.

AM = a + C

\(A M=\frac{a+c}{2}=\frac{33+77}{22}=\frac{110}{2}=55\)

21. The 4 arithmetic means between -2 and 23 are

1. 3, 13, 8, 18
2. 18, 3, 8, 13
3. 3, 8, 13, 18
4. none of these

Answer: (3) 3,8,13,18

->-2,_,_,_,_,_,23

a = 2

a₆= 23 = a + 5d = 23

-2 + 5d = 23

d = 5.

a₂=-2+5=3

a₃=3+5=8

a₄=8+5=13,

a₅=13+5=18.

Given = -2,3,8,13,18,23.

22. The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign. The 3rd term of the AP is

1. 6×4/11
2. 6
3. 4/11
4. none of these

Answer: \(6 \frac{4}{11}.\)

-> S₅ = -S₁₀

\(\frac{5}{2}(28+4 d)=-\left(\frac{10}{2}\right)(28+9 d)\)

28 + 4d = -56 -18d

22d = -84

d=-84/22=42/11

a₃=14+2(-42/11)

=70/11

= 6×4/11

23. The sum of a certain number of terms of an AP series -8, -6, -4,……is 52. The number of terms is

1. 12
2. 13
3. 11
4. none of these

Answer: (2) 13.

S_n = 52

a = -8, d==2

S_n =n/2 (-16 + (n-1)2)

52 x 2 = n(-16+2n-2) = 104

= -18n + 2n²

2n²- 18n-104 = 0

n²- 9n – 52 = 0

n²- 13n + 4n – 52 = 0

n(n-13)+4(n-13)=0

n = 13

24. The first and the last term of an AP are -4 and 146. The sum of the terms is 7171. The number of terms is

1. 101
2. 100
3. 99
4. none of these

Answer: (1) 101

S_n = n/2(a + 1)

7171=n/2(-4 + 146)

14342 = n(142)

n = 101

25. The sum of the series 3 x1/2 + 7 + 10 x1/2 + 14 + …. to 17 terms is

1. 530
2. 535
3. 535×1/2
4. none of these

Answer: (3) 535×1/2

a = 3.5 d=3.5

S₁₇ = 17/2(3.5×2 + (17 -1)3.5)

= 17/2(7 + 56) = 535×1/2

26. Three numbers are in AP and their sum is 21. If 1, 5, 15 are added to them respectively, they form a G. P. The numbers are

1. 5, 7,9
2. 9,5,7
3. 7, 5,9
4. none of these

Answer: (1) 5,7,9

= S = 3a = 24 = a = 7

If 1,5,15 are added terms are

a-d + 1, a + 5, a + d + 15

8 – d, 12, 22+d -> GP

∴12/8-d=22+d/12

144 = (22+d) (8-d)

144=176-22d+8d-d²

d² + 14d – 32 = 0

d² + 16d – 2d – 32 = 0

d(d+16)-2(d+16) =0

d = 2, d = -16

If a = 7, d = 2

= AP = 5,7,9

If a= 7, d= -16

AP = 23,7,-9

27. If p, q and r are in A.P. and x, y, z are in G.P. then xq-r. yr-q. Zp-q is equal to

1. 0
2. -1
3. 1
4. none of these

Answer: (3) 1

q-p = r-q

p-q = q-r

2q = p+r

y² = xz

x^q-r.y^r-p.z^p-q

x^p-q.z^p-q.y^r-p

(x·z)^p-q.y^r-p

(y)^2p-2q.y^r-p

y^2p-2Q+r-P

y^p+r-2q →y^p+r-2q →y^2q-2q→y⁰= 1 (p +r = 2q)

28. The sum of 3 numbers in A.P. is 15. If 1, 4 and 19 be added to them respectively, the results are is G. P. The numbers are

1. -26, 5, -16
2. 2, 5, 8
3. 5,8,2
4. none of these

Answer: (1),(2)

a-d, a, a+d

be in AP

a-d+a+d+a=15

a=5

if 14,19 added

terms 6-d, 9, 24+d = GP

81 = (6-d)(24+d)

81 = 144 + 6d – 24d – d²

d² + 18d – 63 = 0

d² + 21d -3d – 63 = 0

d(d+21) -3 (d+21) = 0

d=21,d=3, -> d=3, and d=-21

Series = 2,5,8 if a = 5, d=3

Series = -26,5,-16 if a = 5, d = -21

29. If the terms 2x, (x+10) and (3x+2) be in A.P., the value of x is

1. 7
2. 10
3. 6
4. none of these

Answer: (3) 6.

same common difference = d

x + 10 – 2x = 3x + 2 – x- 10

10-x = 2x-8

18 = 3x

x = 6

30. If A be the A.M. of two positive unequal quantities x and y and G be their G. M, then

1. A < G
2. A>G
3. A ≥ G
4. A ≤ G

Answer: (b) A > G.

If two unequal quantities present

AM > GM

A > G

31 The A.M. of two positive numbers is 40 and their G. M. is 24. The numbers are

1. (72, 8)
2. (70,10)
3. (60,20)
4. none of these

Answer: (1) 72,8

AM=a+c/2         ac=GM²

80 = a + c             ac = 576

\(80=\frac{576}{c}+c \quad a=\frac{576}{c}\)

80c = 576 + c²

C²- 80c + 576 = 0

C²- 72c- 8c + 576 = 0

C(c-72)-8 (c-72) = 0

C = 72, C = 8

If C = 72,      a = 8

If C = 8,        a= 72,8

= 72,8

32. Three numbers are in A.P. and their sum is 15. If 8, 6, 4 be added to them respectively, the

1. 2, 6, 7
2. 4,6,5
3. 3,5,7
4. none of these

Answer: (3) 3,5,7

Sum = 15

13a = 15s = a= 5

If 8,6,4 is added respectively

13-d, 11,9 + d -> GP

121 = (13-d) (9+d)

121 = 117 + 13d -9d -d²

d² – 4d + 4 = 0

d²- 2d -2d + 4 = 0

d(d-2)-2(d-2) = 0

d = 2.

Series -> 3,5,7.

33. A sum of is paid off in 30 instalments such that each instalment is *?10 more than the proceeding instalment. The value of the 1st instalment is

1. 36
2. 30
3. 60
4. none of these

Answer: (4) none of the above

S₃₀ = 6240

6240 =30/2(2a + 290)

416 = 2a + 290

2a =126 a = 63

34. If x, y, z are in A.P. and x, y, (z + 1) are in G.P. then

1. (x- z)² = 4x
2. z² = (x- y]
3. z = x- y
4. none of these

Answer: (1)

x,y, (z+1) = GP

y² = (z + 1)x

2y = x +z

y = x + z

\(y^2=\left(\frac{x+2}{4}\right)^2 \rightarrow 4(z+1) x=(x+z)^2\)

4xz + 4x = x² + z²+ 2zx

2xz + 4x = x² + z²

4x = x² + z² – 2xz

(x- z)²= 4x

35. The numbers x, 8, y are in G.P. and the numbers x, y, -8 are in A.P. The value of x and y are

1. (-8, -8)
2. (16,4)
3. (8,8)
4. none of these

Answer: (1) (2)

X,8,y = GP

X,y,-8 = AP

2y-64/2 = —8 -> 2y²- 64 = -8y

2y² + 8y- 64 = 0

Y² + 4y- 32 = 0

Y² + 8y- 4y- 32 = 0

y(y+8) – 4y- 32 = 0

y(y+8) – 4 (y+8) = 0

y = 4 y=-8

x= 16 x=-8

36. The sum of all odd numbers between 200 and 300 is

1. 11,600
2. 12,490
3. 12,500
4. 24,750

Answer: (4) 12,500

201,203,…….299

AP with a = 201,       d = 2

I = 299, a = 201, d = 2

299 = 201 + (n – 1) 2

98= (n- 1)2

49 = n-1= n = 50

S₅₀= n/2 (a + 1) = 25 (201 + 299)

= 12,500

37. The sum of all natural numbers between 500 and 1000 which are divisible by 13, is

1. 28,405
2. 24,805
3. 28,540
4. none of these

Answer: (1) 28,405

= 507,520,……. 988

AP, a = 507      d = 13      I = 988

988 = 507 + (n – 1) 13

481 = (n – 1) 13

37 = n-1 = n = 38

S₃₈ = 38/2 (507 + 988) = 28,405

38. If unity is added to the sum of any number of terms of the A.P. 3, 5, 7, 9…….the resulting sum is

1. ‘a’ perfect cube
2. ‘a’ perfect square
3. ‘a’ number
4. none of these

Answer: (2) Perfect square

If 1 is added to any sum

It is same as adding1 as

A term of AP, So AP becomes

AP: 1,3, 5, 7……….

These are n natural odd number

Now sum of first n natural odd no. is n2 i.e. perfect square

39. The sum of all natural numbers from 100 to 300 which are exactly divisible by 4 or 5 is

1. 10,200
2. 15,200
3. 16,200
4. none of these

Answer: (3) 16,200

Natural no. divisible by 4 or 5

= natural no divisible by 4 + natural no divisible by 5

= natural no divisible by 4 & 5 i.e. 20

100, 104………..300

AP -> a = 100 d = 41 = 300

N = 51 terms

= S₅₁ = 51/2 (100+300)

Sum of natural no divisible by 4 = 10200

For divisible by 5

100,105,…..300

AP, a = 100     I = 300     d = 5     n=41

S₄₁ = 41/2 (100 + 300) = 8200

Sum of natural no divisible by 5 = 8200

For4&5

100,120,…..300

AP a = 100 d = 20, I = 300, n = 11

S₁₁ = 11/2 (100 + 300) = 2200

∴ sum of natural no divisible by 4 or 5

= 10200 + 8200-2200= 16200

40. The sum of all natural numbers from 100 to 300 which exactly divisible by 4 and 5 is

1. 2,200
2. 2,000
3. 2,220
4. none of these

Answer: (1) 2200

Divisible by 4 & 5

i.e. divisible by 4 x 5 = 20

100,120……300

AP -> a = 100    d = 20     I = 300     n = 11

S₁₁ =11/2 (100 + 300)

= 2200

41. A person pays 975 by monthly instalment each less then the former by 5. the first instalment is 100.The time by which the entire amountwill be paid is

1. 10 months
2. 15 months
3. 14 months
4. none of these

Answer: (2) 15 months

a = 100

d = -5

100,95,…….

S_n = 975, n =?

975 =n/2 (200 + (n – 1) (-5))

1950 = n (200 – 5n + 5)

1950 = 205 – 5n²

5n²- 205n+ 1950 = 0

n² – 41n + 390 = 0

n²-26n- 15n + 390 = 0

n (n – 26) – 15 (n-26) = 0

n = 15 or n = 26

15 months

42. A person saved 16,500 in ten years. In each year after the first year he saved 100 more than he did in the precedinh year. The amount of money he saved in the 1st year was

1. 1000
2. 1500
3. 1200
4. none of these

Answer: (3)

d = 100

S₁₀ = 16500

16500 = 10/2 (2a + 9 (100))

33000 = 10 (2a + 900)

3300 -900 = 2a

a = 1200

Exercise – 2: Geometric Mean

Choose the most appropriate option (1), (2), (3) or (4)

1. The 7th term of the series 6, 12, 24,……… is

1. 384
2. 834
3. 438
4. none of these

Answer: (1) 384

GP: 6, 12, 24,

A = 6 r=12/6=2

a₇ = ar^7-1 = ar⁶

= 6(2)⁶ =384

2. t₈ of the series 6, 12, 24,…is

1. 786
2. 768
3. 867
4. none of these

Answer: (2) 768

GP: 6, 12, 24….

a = 6, r = 2

a₈= ar^8-1 = ar⁷

= 6 (2)⁷ =768

3. t₁₂ of the series -128, 64, -32, ….is

1. – 1/16
2. 16
3. 1/16
4. none of these

Answer: (3) 1/16

t₁₂ = a¹¹ =- 128 (-1/2 )11 =1/16

4. The 4th term of the series 0.04, 0.2, 1, … is

1. 0.5
2. 1/2
3. 5
4. none of these

Answer: (3) 5

0.04, 0.2, 1,………….

a = 0.04 r= 5

a₄=ar³

=0.04(5)³ = 5

5. The last term of the series 1, 2, 4,…. to 10 terms is

1. 512
2. 256
3. 1024
4. none of these

Answer: (1) 512

a = 1, r=2

a₁₀ = ar⁹ = 1(2)⁹ = 512

6. The last term of the series 1, -3, 9, -27 up to 7 terms is

1. 297
2. 729
3. 927
4. none of these

Answer: (1) 729
a = 1, r = -3

a₇ = ar⁶= 1(-3)⁹ =729

7. The last term of the series x2, x, 1, …. to 31 terms is

1. x²⁸
2. 1/x
3. 1/x²⁸
4. none of these

Answer: (3) 1/x²⁸

a = x², r = 1/x

a₃₁=ar₃₀=x²(1/x)³⁰

=x²x1/x³⁰=x^-28=1/x²⁸

8. The sum of the series -2, 6, -18, …. to 7 terms is

1. -1094
2. 1094
3. – 1049
4. none of these

Answer: (1) -1094

a = -2, r = -3

a₇= ar⁶ = -2(-3)⁶ = -1458

GP = -2, 6,-18,……-1458

S_n =Ir-1/r-1 =-1094

9. The sum of the series 243, 81, 27, …. to 8 terms is

1. 36
2. \(\left(36 \frac{13}{30}\right)\)
3. \(36 \frac{1}{9}\)
4. none of these

Answer: (4) none of these

a = 243 r = -1/3

\(S_8=\frac{a\left(1-r^n\right)}{1-r}=\frac{243\left(1-\frac{1}{3}_3^8\right)}{1-\frac{1}{3}}\) \(=\frac{243(0.999847) .3}{2}\)

= 364.4445 = 364.45

10. The sum of the series \(\frac{1}{\sqrt{3}}+1+\frac{3}{\sqrt{3}}+\ldots . . \text { to } 18 \text { terms is }\)

1. \(9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\)
2. 9841
3. \(\frac{9841}{\sqrt{3}}\)
4. none of these

Answer:

\(9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\) \(a=\frac{1}{\sqrt{3}}, \quad r=\sqrt{3}\) \(S_{18}=\frac{a\left(r^n-1\right)}{r-1}=\frac{1}{\sqrt{3}} \frac{\left(\left(\sqrt{3)}^{18}-1\right)\right.}{\sqrt{3}-1}\) \(\frac{1}{\sqrt{3}} \frac{\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{1}{\sqrt{3}} \frac{\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)}{3-1}\) \(=\frac{1}{2 \sqrt{3}}\left(\sqrt{3}^{18}-1\right)(\sqrt{3}+1)\) \(=9841 \frac{(1+\sqrt{3})}{\sqrt{3}}\)

11. The second term of a G P is 24 and the fifth term is 81. The series is

1. 16,36,24, 54,..
2. 16, 24,36,54,..
3. 24,36, 53,…
4. none of these

Answer: (3)

16, 24, 36, 54

ar = 24 ar⁴ = 81

\(\frac{a r^4}{a r}=\frac{81}{27}=r^3=\frac{81}{27}\) \(=r^3=\left(\frac{3}{2}\right)^3 \quad r=\frac{3}{2}\) \(\mathrm{ar}=24=\mathrm{a}=\frac{24.2}{3}=16\)

12. The sum of 3 numbers of a G P is 39 and their product is 729. The numbers are

1. 3, 27, 9
2. 9,3, 27
3. 3, 9. 27
4. none of these

Answer: (3)

Sum = 39 product =729

a/r, a, ar

a/r.a.ar=729

a³=729

a/r+a+ar=39

9/r+9+9r=39

9(1/r+1+r)=39

\(9\left(\frac{1+r+r^2}{r}\right)=39\)

9 (1 + r + r²) = 39x

9 + 9x + 9x² = 39r

9r²-30r + 9 = 0

9r²- 27r- 3r + 9 = 0

9r(r-3)-3 (r-3) = 0

r=3/9 r=3

when, r = 3, GP = 3, 9,27

When r= 1/3 GP = 27, 9,3

13. In a G. P, the product of the first three terms 27/8. The middle term is

1. 3/2
2. 2/3
3. 2/5
4. none of these

Answer: (1) 3/2

GP= a/r, a, ar

a/r.a.ar=27/8=a³=27/8=a=3/2

14. If you save 1 paise today, 2 paise the next day 4 paise the succeeding day and so on, then your total savings in two weeks will be

1. 163
2. 183
3. 163.83
4. none of these

Answer:(3) 163.83 Rs

a₁₄ = ar¹³, a = 1, r = 2

= 1(2)¹³ =8192

\(S_{14}=\frac{I r-a}{r-1}=\frac{16384-1}{1}=16383 \text { paise }\)

= 163.83 Rs.

15. Sum of n terms of the series 4 + 44 + 444 + …………. is

1. 4/9 {10/9 (10ⁿ -1) -n}
2. 10/9 ( 10ⁿ -1 ) -n
3. 4/9(10ⁿ-!) -n
4. none of these

Answer:(1) 4/9 {10/9 (10ⁿ -1) -n}

4 + 44 + 444 +………

4/9(9 + 99 + 999 +…..)

=4/9[(10-1)+(10²-1)+(10³-1)+…..(10ⁿ-1)]

=4/9(10+10²+10³……10ⁿ-n)

=4/9(10+10²+10³……10ⁿ-1)-n)

\(=\frac{4}{9}\left(10\left(\frac{10^n-1}{9}\right)-n\right)\) \(=\frac{4}{9}\left(\frac{10}{9}\left(10^n-1\right)-n\right)\)

16. Sum of n terms of the series 0.1 + 0.11 + 0.111 + … is

1. 1/9 {n – ( 1-(0.1)ⁿ)}
2. 1/9 {n- (1-(0.1)ⁿ)/9)
3. n-1 – (0.1)ⁿ/9
4. none of these

Answer:(2) 1/9 {n- (l-(0.1)ⁿ)/9)

= 0.1 +0.11 + 0.111 +……….

= 1/9(0.9 + 0.99 +……)

= 1/9[(1-1/10)+(1-1/10²+…..)

= 1/9[n-(1/10+1/10²+……)]

= 1/9[n-1/10(1+1/10+1/10²+…….)]

\(=\frac{1}{9}\left[n-\frac{1}{10}\left(\frac{1-\frac{1}{10}}{1-\frac{1}{10}}\right)\right]\) \(=\frac{1}{9}\left[n-\frac{1}{10}\left(\frac{1-10^{-n}}{9}\right) 10\right]\) \(=\frac{1}{9}\left[n-\left(\frac{1-10^{-n}}{9}\right)\right]\)

=1/9 {n- (l-(0.1)ⁿ)/9)

17. The sum of the first 20 terms of a G. P is 244 times the sum of its first 10 terms. The common ratio is

1. ±√3
2. ±3
3. √3
4. none of these

Answer:(1) ±√3

\(\frac{a\left(r^{20}-1\right)}{r-1}=244 \frac{(a)\left(r^{10}-1\right)}{r-1}\)

r²⁰ – 1 = 244 (r¹⁰– 1)

y² – 1 = 244 (y- 1)

y² – 1 = 244y – 244

y² – 244y + 243 = 0

y² – 243y – y + 243 = 0

y (y – 243) – 1 (y – 243) = 0

y = 243 y=1

y = r¹⁰

r¹⁰ = 243 r¹⁰ = 1

(±√3)¹⁰=243 r=±1

r= ±√3

18. Sum of the series 1 + 3 + 9 + 27 +….is 364. The number of terms is

1. 5
2. 6
3. 11
4. none of these

Answer: (2) 6

\(S n=\frac{a\left(r^n-1\right)}{r-1}\) \(=364=\frac{1\left(3^n-1\right)}{2}\)

19. The product of 3 numbers in G P is 729 and the sum of squares is 819. The numbers are

1. 9, 3, 27
2. 27, 3, 9
3. 3, 9, 27
4. none of these

Answer: (3) 3, 9, 27

Product 729 ; sum of squares = 819

a/r. a . ar = 729

a³ = 729

a = 9

a/r,9,9r = 9²/r² + 81 + 81r² = 819

81(1/r²+1+ r²) = 819

81 (1 + r² + r⁴) = 819r²

81 + 81r- + 81r⁴ = 819r²

81r⁴ – 819r² + 81r² + 81 = 0 = 81 r⁴

738r² + 81 = 0

9r⁴ – 82r²+ 9 = 0

9r⁴ – 81r² — 1 r² + 9 = 0

\(9 r^2\left(r^2-9\right)-1\left(r^2-9\right)=r^2 \frac{1}{9}, r^2=9=r \frac{ \pm 1}{3}\)

r = ±3

When ± 3, a = 9

GP = 3,9,27 & -3,9,-27

When ± 1/3 a=9

GP= 27,9,3 & -27,9,-3

20. The sum of the series 1 + 2 + 4 + 8 + .. ton term

1. 2ⁿ-1
2. 2n – 1
3. 1/2ⁿ -1
4. none of these

Answer: 2ⁿ – 1

\(S_{11}=\frac{1\left(2^n-1\right)}{2-i}=2^n-1\)

21. The sum of the infinite GP 14, – 2, + 2/7,- 2/49, + … is

1. \(4 \frac{1}{12}\)
2. \(12 \frac{1}{4}\)
3. 12
4. none of these

Answer: \(12 \frac{1}{4}\)

\(S_n=\frac{a}{1-r}=\frac{14}{1+\frac{1}{7}}=\frac{7 \times 14}{8}=\frac{49}{4}=12 \frac{1}{4}\)

22. The sum of the infinite G. P. 1 – 1/3 + 1/9 – 1/27 +… is

1. 0.33
2. 0.57
3. 0.75
4. none of these

Answer: (3) 0.75

\(\mathrm{S}_{\mathrm{n}}=\frac{a}{1-r}=\frac{1}{1+\frac{1}{3}}=\frac{3}{4}=0.75\)

23. The number of terms to be taken so that 1 + 2 + 4 + 8 + will be 8191 is

1. 10
2. 13
3. 12
4. none of these

Answer: (2) 13

s_n= 8191

a = 1 r = 2

\(S_n=\frac{1\left(2^n-1\right)}{1}=8191\)

2ⁿ = 8192 n=B

24. Four geometric means between 4 and 972 are

1. 12,36, 108,324
2. 10,36, 108,320
3. 12, 24,108,320
4. none of these

Answer: (1)

a = 4 I = 972

ar^n-1 = 972

ar⁵ = 972

4r⁵ = 972

r⁵ = 243 = r = 3

4,12,36,108,324, 972

25. The sum of 1 + 1/3 + 1/32 + 1/33 + … + 1/3n-1 is

1. 2/3
2. 3/2
3. 4/5
4. none of these

Answer: (b)3/2

a = 1 x =1/3

\(S=\frac{a}{1-x}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}\)

26. The sum of the infinite series 1 + 2/3 + 4/9 + .. is

1. 1/3
2. 3
3. 2/3
4. none of these

Answer: (2) 3.

a = 1 x = 2/3

\(S=\frac{a}{1-x}=\frac{1}{1-\frac{2}{3}}=\frac{3}{1}=3\)

27. The sum of the first two terms of a G.P. is 5/3 and the sum to infinity of the series is 3. The common ratio is

1. 1/3
2. 2/3
3. – 2/3
4. none of these

Answer: (2) & (3)

\(\frac{a}{1-x}=3\)

a = 3(l-x) = 3 -3x

a(l+x)=5/3

(3-3X)(1+X)=5/3

3(1-X)(1+x)=5/3==1-X²=5/9

\(1-\frac{5}{9}=x^2=\frac{4}{9}=x^2=x=\frac{2}{3}\)

28. The sum of three numbers in G.P. is 70. If the two extremes by multiplied each by 4 and the mean by 5, the products are in AP. The numbers are

1. 12, 18, 40
2. 10, 20, 40
3. 40, 20, 10
4. none of these

Answer: (2) & (3)

a + ax + ax² = 70

A.P = 4a, 5ax, 4ax²

10ax = 4a + 4ax²

10ax = 4a(1+x²)

5/2 x=1+x²

2x²- 5x + 2 = 0

2x²- 4a- x + 2 = 0

2x(x- 2) — 1(x — 2) = 0

x= 1/2,X = 2.

29. Given x,y,z. are in G.P. and Xp = Yq = Zσ then 1/p,1/q.1/a are In

1. A.P
2. G.P
3. Both A.P & G.P
4. None of these

Answer: (1)AP

X^p = Y^q = Z^σ

 y² = xz

X^p = Y^q = Z^σ = k

\(x=k^{\frac{1}{p}}, y=k^{\frac{1}{q}}, z=k^{\frac{1}{\sigma}}\) \(k^{\frac{2}{q}}=k^{\frac{1}{p}} \cdot k^\sigma\) \(k^{\frac{2}{q}}=k^{\frac{1}{p}}+\sigma\) \(\frac{2}{9}=\frac{1}{p}+\frac{1}{\sigma}=2\left(\frac{1}{q}\right)=\frac{1}{p}+\frac{1}{\sigma}\) \(\frac{1}{p}, \frac{1}{q}, \frac{1}{\sigma} \text { are in AP. }\)

30. The sum of 1.03 + ( 1.03 )²+ ( 1.03 )³+ …. to n terms is

1. 103 {(1.03)ⁿ– 1}
2. 103/3 {(1.03)ⁿ– 1}
3. (1.03)n -1
4. none of these

Answer: (2)

1.03(1 + (1.03) +(1.03)² +……(1.03)ⁿ-1)

\(1.03\left(\frac{(1.03)^n-1}{1.03-1}\right)\) \(\frac{1.03}{0.03}\left((1.03)^n-1\right)\) \(\left.=\frac{103}{3}\right)\left((1.03)^n-1\right)\)

31. The nth term of the series 16, 8, 4…..is 1/217. The value of n is

1. 20
2. 21
3. 22
4. none of these

Answer: (3) 22

a = 16 r = 1/2.GP

ar^n-1=a_n=1/2¹⁷

\(\frac{16}{2^{n-1}}=\frac{1}{2^{17}}=\frac{1}{2^{n-1}}=\frac{1}{2^{17} \cdot 2^4}\) \(=\frac{1}{2^{n-1}}=\frac{1}{2^{21}}\)

n- 1 = 21 -> n = 22

32. The sum of n terms of a G.P. whose first terms 1 and the common ratio is 1/2 , is equal to 1×127/128. The value of n is

1. 7
2. 8
3. 6
4. none of these

Answer: (2)8

\(S_{n}=1 \frac{127}{128}=\frac{255}{128}\) \(\frac{\left(1-0.5^n\right)}{1-0.5}=\frac{255}{128}\) \(\frac{\left(1-0.5^n\right)}{0.5}=\frac{255}{128}\) \(1-0.5^n=\frac{127.5}{128}\)

33. t₄ of a G.P. in x, t₁₀ = y and t₁₆ = z. Then

1. x² = yz
2. z² = xy
3. y² = zx
4. none of these

Answer: (3)

T₄ = X, t₁₀ = y,t₁₆ = z.

ax³ = x ax⁹ = y ax¹⁵ = z

y² = a²x¹⁹

y² = xz

34. 1 f x, y, z are in G.P., then

1. y²= xz
2. y (z² + x²) = x(z² + y²)
3. 2y = x+z
4. none of these

Answer: (1)

x,y,z in GP

y² = xz

35. At 10% C.I. p.a., a sum of money accumulate to 9625 in 5 years. The sum invested initially is

1. 5976.37
2. 5970
3. 5975
4. 5370.96

Answer: (1)

Sum of money at end 9625

Time series

Initial investment =? = a

Amt = a

Amount at end first yr= a + 0.1a = a(1.1)

Amount at end second yr = a(1.1) + a(0.1)(1.1)

= a(1.1+0.11)

= a(1.21) = a(1.1)²

Amount at end of third = a(1.1)³

Amount at end of n = a(1.1)ⁿ

9625 = a(1.1)ⁿ

9625 = a(1.1)⁵

a = 5976.37

36. The population of a country was 55 crores in 2005 and is growing at 2% p.a C.I. the population is the year 2015 is estimated as

1. 5705
2. 6005
3. 6700
4. none of these

Answer:

(3) 6700

Population initial = 55

Time 10 years

Population at end ofyr= 55 + 55(0.2) = 55(1.02)

At end ofyr 2 = 55(1.02) + 55(1.02)

(0.02) = 55(1.02)²

At end ofyrs = 55(1.02)ⁿ

S₁₀ = a(l + r)ⁿ = 55(1.02)¹⁰

S₁₀ = 6700.

Exercise – 3 (Mix- AP and GP)

1. What is the arithmetic mean of the arithmetic progression 6,8,10,12,14,16?

1. 22
2. 11
3. 24
4. 12

Answer: Arithmetic mean = sum of the terms/number of terms

\(=\frac{6+8+10+12+14+16}{6}\)

= 66/6

= 11

Choice (2)

2. What is the geometric mean of the geometric progression 2,4,8,16?

1. 32
2. √32
3. 64
4. 8

Answer:

Geometric mean= [(2)(4)(8)(16)]^1/4

=[(21)(22)(23)(24)]^1/4=2^10/4 = √32

Alternative solution:

Geometric mean of a geometric progression

\(=\sqrt{\text { first term } \times \text { last term }}\) \(=\sqrt{(2)(16)}\)

=32

Choice (2)

3. What is the fourth term of the arithmetic progression in which the first term is 4 and the seventh term is 28?

1. 16
2. 8
3. 12
4. 2

Answer: The fourth term is equidistant (3 teams away) from the first and the seventh term in the arithmetic progression.

the fourth term is the arithmetic mean of the first term and the seventh term in an A.P.

fourth term \(=\frac{4+28}{2}\)

32/2=16

choice (1)

4. What is the sum to 7 terms of the arithmetic progression in which the first term is 2 and the common difference is 4?

1. 49
2. 98
3. 126
4. 77

Answer: Sum of the first n terms of an A.P

\(=s_n=\left(\frac{n}{2}\right)[2 a+(n-1) d]\) \(s_n=\left(\frac{7}{2}\right)[2(2)+(7-1) 4]\) \(=\left(\frac{7}{2}\right)[4+24]=\frac{7}{2}(28)=98\)

choice (2)

5. What is the sum to 8j terms of an arithmetic progression in which the first term is 3 and the last term is 31?

1. 136
2. 58.5
3. 132
4. Cannot be determined

Answer: Given first term=3 and the last term=31

As we don’t know the total number of terms, we cannot find the common difference.

we cannot find the sum of first 8 terms of the Progression, choice (4)

6. What is the arithmetic mean of an arithmetic progression with 13 terms, in which the 7lh term is 9?

1. 9
2. 91/7
3. 95/7
4. Cannot be determined

Answer: Given total number of terms is 13, i.e. n=13 in an A.P. if n is odd

then \(\frac{(n+1)^{t h}}{2}\) term is the arithmetic mean of that A.P.

as \(\frac{(13+1)}{2}=7^{t h}\) ,term is the arithmetic mean

Given 7th term is 9

9 is the arithmetic mean

Choice (1)

7. What is the sum to 15 terms of an arithmetic progression’ whose 8th term is 4?

1. 30
2. 60
3. 40
4. Cannot be determined

Answer: The 8th term is equidistant from first term and 15 terms

8th term is the arithmetic of first 15 terms sum of first 15 terms = 15x (eight term)=15(4)

=60

Choice (2).

8. What is the sum of all the terms in an arithmetic progression in which the first term is 5, the last term is 15 and the number of terms is 11?

1. 55
2. 110
3. 115
4. Cannot be determined

Answer: in A.P

S_n=(n/2)[first term+last term]

S_n=(11/2)[5+15]=110

Choice (2)

9. What is the seventh term of an arithmetic progression whose first term is 9 and the common difference is 3?

1. 27
2. 36
3. 33
4. 30

Answer: in an A.P. n 1,1 term (t_n) = a+ (n-1) d

t₇ = 9 + (7- 1)3 = 9 + 6(3) = 27

choice (1)

10. What is the fourth term so a ‘geometric progression in which the second term is 4 and the sixth term is 64?

1. 8
2. 32
3. 64
4. 16

Answer: fourth term is equidistant from the second term and the sixth term.

in a G.P. the fourth term is the geometric mean of the second term and the sixth terms

Fourth term = \(\sqrt{(\text { second term)(sixth term) }}=\sqrt{(4)(64)}=16\)

choice (4).

11. What is the sum to 4 terms of a geometric progression whose first term is 6 and the common ratio is 3?

1. 300
2. 360
3. 270
4. 240

Answer: In a G.P.

\(S_n=\frac{a\left(1-r^n\right)}{1-r}\) where is the common ration, a the first term and sn in the sum to n term.

\(S_4=\frac{6\left(1-3^4\right)}{1-3}=\frac{6(1-81)}{2}\)

=(3)(80)= 240

Choice (4)

12. In an arithmetic progression having 100 terms, the m th term form the beginning and the mthTerm from the end is 10 and 20 respectively what is the sum of all the terms?

1. 3000
2. 1500
3. 3200
4. Cannot be determined

Answer:

As the average of the kth term from the beginning and the kth term from the end is equal to the arithmetic mean the average of 10 and 20 is the arithmetic mean i.e. the mean is

\(\frac{10+20}{2}=15\)

in an A.P. as S_n = n (AM.)

S₁₀₀ = 100(15) = 1500

Choice (2)

13. If m n and p are arithmetic progression then the m th term and pth term of arithmetic progression are in

1. Arithmetic progression
2. Geometric progression
3. Not neccessarly in arithmetic progression or geometric progression

Answer: Given m,n, and pare in A.P.

Let n-m =p- n = k

n = m + k,p = m + 2k

consider an A.P.

a₁,a₂,a₃,a₄,…….whose connon difference is d.

a_m + d = a_m + 1[a_m is mth term]

a_m+ kd = a_m+k = a_n m+k

a_m + kd = a_m+2k = a_p m+2k

a_p – a_n = [a_m + 2kd]- (a_m + kd] = kd

a_n – a_m = a_p – a_n = kd        ∴ a_m ,a_n and a_p are in A.P

m term, nth term and pth term are in A. P

Choice (1)

14. If m . n and p are in arithmetic progression then m th term, n th term and pth term of a geometric progression are in

1. Arithmetic progression
2. Geometric progression
3. Not neccessarly in arithmetic progression or geometric progression

Answer: Given m, n and p are in A.P.

Let n-m=p-n=k

n = m + k,p = m + 2k

Consider a G.P

g1,g2,g3——— whose common ratio is r.

gm(r) = gm+1 [gm is mth term]

gm(rk)gm+k = gn [m+k=n]

gm(r2) = gm+2k = gp [m+2k=p]

\(\frac{g_p}{g_m}=\frac{g_m r^{2 k}}{g_m r^k}=r^{-k}\) \(\frac{g_n}{g_m}=\frac{g_p}{g_{11}}=r^k\)

gm,gn and gp are in G. P.

Choice (2)

15. If sum of first 51 terms of an arithmetic progression is zero , then which of the following terms is zero?

1. 13th
2. 26th
3. 17th
4. Cannot be determined

Answer: In an A.P. if the sum of the first n terms is Zero, and dn is odd, the

\(\frac{(n+1)^{t h}}{2}\) term is zero.

\(\left(\frac{51+1}{2}\right)^{t h}\)

26th term is zero

Choice (2)

16. If the sum of first 20 terms of an arithmetic progression is 30 and the sum of first 50 terms is also 30 then what is the sum of 21st term and the 50th term?

1. 0
2. 30
3. 15
4. Cannot be determined

Answer: Given S₂₀=30 and S₅₀ = 30

S₅₀ = S₂₀ + t₂₁+——-+t₅₀

30 = 30 + t₂₁ +————-t₅₀

t₂₁+t₂₂——–+t₅₀=0

the mean of t₂₁ and t₅₀ is zero.

choice (1)

17. If the positive numbers m, n and p are in geometric progression then Ion gm, logn and log p are in

1. Arithmetic progression
2. Geometric progression
3. Cannot be determined

Answer: Given m, n and p are in G. P.

\(\text { Let } \frac{n}{m}=\frac{p}{n}=r\)

n = mr and p = mr²

Let log m=a…..(1)

Log n == log (mr)

= log m + log r[ log(xy) = logx + logy]

Let log r=b    logn = a + b…..(2)

Log p = log (mr)²log m+ log r² =log m +2 log r [ log r² = 2logr]

∴ log p = a + 2b…..(3)

From 1,2,3

Log m, log n and log p are in A.P.

Choice (1)

18. If the geometric mean of two distinct positive numbers is 4, then the arithmetic mean of these two numbers is

1. <4
2. =4
3. >4
4. Cannot be determined

Answer: For any two unequal positive numbers a and b

\(\frac{a+b}{2}>\sqrt{a b}\)

Given √ab =4

Arithmetic mean, \(\frac{a+b}{2}>4\)

Choice (3)

19. What is the seventh term of a geometric progression whose first term is 3 and common ratio is 2?

1. 96
2. 384
3. 192
4. 288

Answer: In a G.P.

T_n = ar^n-1

T₇ = 3(2)^7-1

= 3(64) = 192

Choice (3)

20. What Is the product of first 9 terms of a geometric progression’ having a total of 13 terms given that 51th term is 2?

1. 512
2. 32
3. 16
4. Cannot be determined

Answer: The 5th term is equidistant from the first term and the 9th term

5th term is the geometric mean of first and ninth term and moreover, it is the geometric mean of first 9 terms the product of the first n terms in a G.P = (Geometric mean of first n
terms)ⁿ

Product = 2⁹ = 512

Choice (1)

21. What is the geometric mean of the geometric progression having a total of 13 terms given the 7lh term is 4?

1. 2
2. 4
3. 16
4. Cannot be determined

Answer: In G.P. if the number of terms is n which is odd, then the middle term i.e. \(\left(\frac{n+1}{2}\right)^{t h}\) term is the geometric mean.

\(\left(\frac{13+1}{2}\right)^{\text {th }}\) term i. e. the 7th term is the geometric mean i. e. the geomeric mean is 4.

Choice (2)

22. What is the sum of infinite geometric series 1,1/2,1/4,1/8,1/16,….?

1. \(1 \frac{255}{256}\)
2. 2
3. 3
4. 4

Answer: Common ratio = \(\frac{1 / 2}{1}=\frac{1}{2}\)

\(S_a=\frac{a}{1-1}=\frac{1}{1-1 / 2}=2\)

Choice (2)

23. Every number of an infinite geometric progression of positive terms is equal to m times the sum of the numbers is equal to m times the sum of the numbers that follow it. What is the common ratio of the progression?

1. \(\frac{m}{m+1}\)
2. \(\frac{1}{m+1}\)
3. \(\frac{2}{m+1}\)
4. Cannot be determined

Answer:

Let the geometric progression be a, ar, ar², …. given a = m \(\left(\mathrm{ar}+\mathrm{ar}^2+\cdots=\frac{\mathrm{ar}}{1-\mathrm{r}}\right.\)

=>mr = 1-r ⇒ mr + r = 1

\(\mathrm{r}=\frac{1}{m+1}\)

Choice (2)

24. What is the sum to 7 terms of a geometric progression whose first term is 1 and the 4th  term is 27?

1. 1093
2. 2186
3. 3279
4. 4372

Answer: Given a= 1,t4= 27

t₄ = ar³

27 = (1)r³->r = 3

\(S_7=\frac{1\left(1-3^7\right)}{1-3}=\frac{1-2187}{2}=\frac{-2186}{-2}=1093\)

Choice (1)

25. What is the sum of the cubes of first (9 natural numbers?

1. 45
2. 2025
3. 91125
4. Cannot be determined

Answer: Sum of the first 9 natural numbers = \(\frac{9(9+1)}{2}=45\)

Sum of the cubes of first n natural numbers = (sum of first n natural numbers)²

∴ Sum of the cubes of first 9 natural numbers =45²=2025

=2025

Choice (2)

26. In an arithmetic progression, the first and the last terms are 9 and 96 respectively if the sum of all the terms of the progression is 4200, what is the common difference ?

1. 80
2. \(\frac{76}{79}\)
3. 70
4. \(\frac{87}{79}\)
5. \(\frac{83}{79}\)

Answer:

Let the number of terms in the arithmetic progression be n.

\(\frac{n}{2}[9+96]=4200 \Rightarrow n=80 \text {. }\)

21th term = a+20d where a=9.

96= 9+(n-1)d = 9 + 79d

\(\mathrm{d}=\frac{87}{79}\)

Choice (4)

27. The sum of the first 30 terms of an arithmetic progression is 4500 the ratio of the first 20 terms and the last 10 terms in 4:5 find the first term.

1. 6.5
2. 7.5
3. 6
4. 5
5. 7

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

\(\frac{30}{2}[2 a+29 d]=4500\) 2a + 29d = 300

Sum of the first 20 terms= \(\frac{4}{9}(4500)=2000\)

\(\frac{20}{2}[2 a+19 d] 2000\)

2a = 19d = 200

Subtraction the second equation from the first 10d = 100=>d = 10

First term = \(\frac{300-29 d}{2}=5\)

Choice (4)

28. The sum of all the three digit numbers which leave a remainder of 3 when divided by 7 is

1. 64215
2. 70720
3. 64320
4. 70821
5. 64160

Answer: The first three-digit number when divided by 7 leaving a remainder of 3 is 101 and the last three-digit number when divided by 7 leaving a remainder of 3 is 997.

101 = 7 X 14 + 3

997 = 7 X 142 + 3

Number of 3 digit multiples of 7= \(\frac{997-101}{7}+1\) = 129 the sum of all these numbers =

\(\frac{129}{2}[101+997]=129(549)=70821\)

Choice (4)

29. The sum of all the natural numbers from 200 to 600 (both inclusive ) which are neither divisible by 8 nor by 12 is

1. 1,23,968
2. 1,33,068
3. 1,33,268
4. 1,87,332
5. 1,34,168

Answer: Let the sum of all the natural numbers from 200 to 600 which are divisible neither by 8 nor by 12 be P.

P= sum ofall natural numbers from 200 to 600 – (sum ofall natural numbers divisible by either 8 or 12)

=sum of the first 600 natural numbers- sum of the first 200 natural numbers – (sum of all 3 digit natural numbers from 200 to 600 divisible by 8= sum of all 3 digit
natural numbers from 200 to 600 divisible by k12 – sum of all 3 digit numbers from 200 to 600 numbers divisible by both 8 and 12)

\(=\frac{(600)(601)}{2}-\frac{(199)(200)}{2}\)

– (200+208+…..+600)

+ (204+212+…. +600)

– (216+240+….+600) =(180300-19900)

\(-\left[\frac{51}{2}(200+600)+\frac{34}{2}(204+600)-\frac{17}{2}(216+600)\right]=133268\)

Choice (3)

30. The sum of four terms in arithmetic progression is 64. The sum of the squares of the first and the last term is 64 more than the sum of the squares of the second and the third terms. Find the first term, if the second term is less than third term.

1. 6
2. 8
3. 12
4. 14
5. 10

Answer: Let the four numbers be a-3d, a-d, a + d and a-3d. a -3d+a-d+a+d+a+3d=64

4a = 64

a=16

(a- 3d)² + (a + 3d)² = (a- d)² + (a + d)² + 64

=> 18d² = 2d² + 64

d=√4 = 2 as d > 0

The first term = a- 3d = 10.

Choice (5)

31. There are four distinct numbers in a sequence such that the first and the last terms are equal, the first three numbers progression and the last three are in geometric progression, find the common ratio of the last three numbers.

1. 1
2. -1
3. -1/2
4. -2
5. 2

Answer: Let the second, third and fourth numbers which are in geometric progression be a, ar and ar² respectively the first number = the fourth number = ar²

Since the first three are in arithmetic progression, ar², a and ar are in arithmetic progression.

=>2a =ar²+ ar

As the numbers are distinct, a≠0 and r≠1

=> 2=r²+r=>r²+r-2=0=>r=1 or -2 since r≠1,r=-2

Since r≠1,r=-2

Choice (4)

32. Four times the ninety first term of an arithmetic progression is equal to 5 times its eighty first terms. If the thirty first term of the progression is -630, find its fifty first term.

1. 810
2. 729
3. 630
4. 567
5. 600

Answer: Let the first term and the common difference of the arithmetic progression be a and d.

4(a+90d) = 5 (a+80d)=>-40d=a

Thirty-first term of the at arithmetic progression =a +30d= -630 d=+63.

Fifty first term of that arithmetic progression =a+50d=-40d+50d=+10d=630

Choice (3)

33. The ratio of the sum of the first n terms of two arithmetic progressions is given by 5n+4:8n-15. Find the ratio of the 12th terms of the two arithmetic progressions.

1. 125:161
2. 119:169
3. 123:166
4. 115:169
5. 119:238

Answer: Let the two arithmetic progressions’ have first terms and common differences as a₁,d₁ and a₂,d₂ respectively. Ratio of the sum of their first n terms

\(=\frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]} \quad \frac{a_1+\frac{n-1}{2} d_1}{a_2+\frac{n-1}{2} d_2}=\frac{5 n+4}{8 n-15} \ldots \ldots\)

Ratio of their 12th terms = \(\frac{a_1+11 d_1}{a_2+11 d_2}\)

Substituting \(\frac{n-1}{2}=11\) in the L.H.S. we get ratio of their 12th terms.

\(=\frac{5(23)+4}{8(23)-15}=\frac{119}{169}\)

Choice (2)

34. The number of terms of the series 40,37, 34,… for which the sum is 282 is _________

1. 13
2. 11
3. 16
4. 18
5. 12

Answer: The first term of the arithmetic progression series is 40 and the common difference is -3 let the number of terms in the series for which the sum is 282 be n

\(\frac{n}{2}[2(40)+(n-1)(-3)]=282\) \(\frac{n}{2}[83-3 n]=282\)

3n²- 83n + 564 = 0

3n(n-12)-47(n-12)=0

(n-12)(3n-47)=0

n=12 or 47/3 as n is an integer, n = 12.

Choice (5)

35. The sum of the first, sixth, eighth, ninth and thirteenth terms of an arithmetic progression is 243. Find the sum of the first 13 terms of the arithmetic progression.

1. 1134
2. 1296
3. 1458
4. 526.5
5. 1588

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

a + (a + 4d) + (a + 5d) + (a + 7d) + (a + 8d) + a + 12d) = 243

6a = 36d = 243……. (1)

Sum of the first 13 terms = [2a + 12d] = 13[a = 6d] \(=\frac{(13)(243)}{6}(\text { from }(1))=526.5[latex]

Choice (4)

36. If 5x+y,3x+2y and 3x+y are in arithmetic progression and 5x+1, 3(x+1) and 3x are in geometric progression, find (x,y).

1. (4,3)
2. (5,2)
3. (3/2,3/2)
4. (3,3)
5. (3/4,3/4)

Answer: 2(3x + 2y)=5x + y + 3x + y

6x + 4y = 8x + 2y=>x = y

[3(x + 1)]² = (5x + 1)3x

9x² + 18x + 9 = 15×2 + 3x

6x²- 15x- 9 = 0

Dividing both sides by 3

2x²- 5x- 3 = 0

(2x =1)(x-3) = 0

X=-1/2 or 3; going by the choices, x=3,

Choice (4)

37. The maximum value of the sum of the series series 50,46,42,…is

1. 356
2. 492
3. 368
4. 650
5. 338

Answer: Let the number of terms of the series which give the maximum sum be n

Sum =n/2[2(50)+(n-1)(-4)]

=n/2[104-4n]=n[52-2n]=2n[26-n]

When the sum of two quantities is constant, their product is maximum when the quantities are equal as n and 26-n have a constant sum of 26 their product is maximum when n = 26-n

26/2=13=n

Maximum sum = (2)(13)(13)=338

Choice (5)

38. In a geometric progression the first three terms are 10n +36,6ct + 12 and 4a respectively, find the sixth term of the geometric progression if

1. a > 0
2. 5
3. 2
4. 3
5. 9
6. 6

Answer: (6n + 12)² =(10a+36) 4a

36a² + 144a + 144 = 40a² + 144a

144 = 4a², a = 6 as a > 0

sixth term of the G. P = [latex](10 a+36)\left(\frac{6 a+12}{10 a+36}\right)^5=96\left(\frac{48}{96}\right)^5=3\)

Choice (3)

39. A ball dropped from 36 m above the ground rebounds to 1/3rd of the height it falls from. If it continues to rebound in this manner, find the total distance the ball can cover.

1. 96m
2. 72m
3. 54m
4. 76m
5. 108m

Answer:

the height to which the ball rebounds after the first fall = 36 (1/3)=12m the 2nd time it falls from 12m height. The distance the ball rebounds after the second time it falls =

36(1/3)(1/3)= 4m, it falls from 4 m height

Likewise the distance rebounded by the ball after every time can be found to be (1/3) of distance rebounded after the previous time it fell

Total distance it covers, 36+12+12+4+4+…a= (36+12+4+…a)+(12+4+…a)

=(36+12+4+…..a)+(12+4+…..a)= \(\left(\frac{36}{1-1 / 3}\right)+\left(\frac{12}{1-1 / 3}\right)\)

54 + 18=72m.

Choice (2)

40. Three distinct numbers in geometric progression have a product of 1728 the sum of the products taking two numbers at a time is 456. Find the least number.

1. 8
2. 9
3. 12
4. 16
5. 6

Answer: Let the numbers in geometric progression be a/r,a and ar(a/r)(a)(ar)=1728

\(a=\sqrt[3]{1728}=12\) \(\frac{a}{r}(a r)+a(a r)+\frac{a}{r}(a)=456\) \(a^2\left(1+r+\frac{1}{r}\right)=456\) \(\frac{r+r^2+1}{r}=\frac{456}{12^2}=\frac{19}{6}\)

6r+6r² + 6 = 19r

6r² -13r + 6 = 0

3r(2r- 3)- 2(2r- 3) = 0

(3r-2) (2r-3)=0, r= 2/3 or 3/2

The least number = \(\frac{12}{3 / 2}\)

=8

Choice (1)

41. The value of sum of first 75 terms of the sequence 150 x 2 + 148 x 4 + 6146 x 6 + is

1. 252850
2. 292600
3. 273950
4. 284050
5. 231300

Answer: Given series is 150 x 2 + 148 x 4 + 146 x 6 +….75 terms

The numbers 150, 148, 146 … from an arithmetic progression, with a=150, d=-2.

The numbers 2, 4, 6…… from an arithmetic progression with a=2, d=2

Hence, tn then nth term of the series is:

[150 + (n- 1)(—2)][2 + (n- 1)2] = (152- 2n)2n.

t_n = 304n – 4n²

Hence sum to 75 terms is s75 = 304(1 + 2 + 3 …+ 75)- 4(1² + 2² + 3² + 75²)

\(=304 \times \frac{75 \times 76}{2}-4 \times \frac{75 \times 76 \times 151}{6}=292600\)

Choice (2)

42. The sum of the first 20 terms of the series \(\frac{3}{2}+\frac{15}{4}+\frac{47}{8}+\cdots \text { is }\)

1. \(\frac{\left(2^{20}\right)(420)-2^{20}+1}{2^{20}}\)
2. \(\frac{\left(2^{20}\right)(210)-2^{20}+1}{2^{20}}\)
3. \(\frac{\left(2^{20}\right)(210)+2^{20}+1}{2^{20}}\)
4. \(\frac{\left(2^{20}\right)(420)+2^{20}+1}{2^{20}}\)
5. \(\frac{\left(2^{20}\right)(420)-2^{20}-1}{2^{20}}\)

Answer:

\(\frac{3}{2}+\frac{15}{4}+\frac{47}{8}+\cdots \text { for } 20 \text { terms }\) \(=2-\frac{1}{2}+4-\frac{1}{4}+6-\frac{1}{8}+\cdots \text { for } 20 \text { terms }\)

=2+4+6+…for 20 terms— \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\) for 20 terms = 2(1 + 2 + 3+…. +20)

\(\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^{20}\right)}{1-\frac{1}{2}}=\left[\frac{(420)(2)^{20}-2^{20}+1}{2^{20}}\right]\)

Choice (1)

43. The maximum number of terms common to the arithmetic progression 3,7, 11, 15, 19,23,…403 and 5,11,17,23,29,35,… 505 is

1. 30
2. 33
3. 35
4. 40
5. 37

Answer: The first common term to both arithmetic progressions is 11 and the common differences of the first and second arithmetic progressions are 4 and 6 respectively the next term common to both arithmetic progressions is 23 = 11+ L.C.M, of common differences of the two progressions= 11+12.

It can be confirmed that the term common to both series have a common difference of 12 the last term common to both progressions is less than minimum of last terms of both
progressions i.e. 403. Let n terms be common to both progressions.

The last term common to both arithmetic progressions is 11 + 12(n – 1). This is less than 403.

\(n-1<\frac{(403-11)}{12}=32 \frac{2}{3}\)

\(n<33 \frac{2}{3}\) Hence n must be a maximum of 33.

44. Find the value of \(\frac{1}{3^2-2^2}+\frac{1}{7^2-2^2}+\frac{1}{11^2-2^2}+\) \(\cdots+\frac{1}{39^2-2^2}\)

1. 5/41
2. 31/41
3. 18/37
4. 19/37
5. 10/41

Answer:

\(\frac{1}{3^2-2^2}=\frac{1}{(3-2(3+2)}=\frac{1}{4}\left[1-\frac{1}{5}\right]\) \(\frac{1}{7^2-2^2}=\frac{1}{(7-2)(7+2)}=\frac{1}{4}\left[\frac{1}{5}-\frac{1}{9}\right]\) \(\frac{1}{39^2-2^2}=\frac{1}{(39-2)(39+2)}=\frac{1}{4}\left[\frac{1}{37}-\frac{1}{41}\right]\)

the required sum is 1/4[1-1/41}=10/41

Choice (5)

45. If an is defined as follows, find the value of a50

1. a_n = 1,if n = 0

2. a_n = a_n – 1,if n = 3k, k being a positive integer

3. a_n = 3 a_n-1 If n = 3k + 1, where k is a whole number

4. a_n = 2 a_n-1 if n = 3k + 2, where k is a whole number

1. 6¹⁷
2. 3¹²2¹⁷
3. 3¹⁷2¹⁸
4. 6¹⁸
5. none of these

Answer: Tabulating the values of a, we have a₀ = 1

a₁ = 3

a₂ = 2(3)

a₃ = 2(3)

a₄ = 2(3)²

a₅ = 22(3)²

a₆ = 22(3)²

a₇ = 22(3)³

a₈ = 23(3)³

46. the first 15 terms of the series 32.1 + 42. 2 + 52. 3 + 62. 4 + 72. 5 +…..is _____________

1. 16120
2. 24420
3. 21840
4. 20840
5. 19840

Answer: Each of the terms being added is of the form (a + 2)² a where 1 <a <15

Their sum =\( \sum_{a=1}^{15}(a+2)^2 a=\sum_{n=1}^{15} a^3+4 a^2+4 a\)

\(=\left(\frac{a(a+1)}{2}\right)^2+4\left(\frac{1}{6}\right) a(a+1)(2 a+1)+\frac{4 a(a+1)}{2}\) \(=\left(\frac{15(16)}{2}\right)^2+4\left(\frac{1}{6}\right)(15)(16)(31)+\frac{4(15)(16)}{2}\)

=(120)² + (4)(40)(31) + 480 = 19840

Choice (5)

47. 3100 – 2(396 + 397 + 398 + 399) =

1. 3⁹⁶
2. 2(3⁹⁶)
3. 3⁹⁷
4. 4(3⁹⁶)
5. 5(3⁹⁶)

Answer:

3¹⁰⁰ – 2(3⁹⁶+ 3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 3¹⁰⁰– (2)3⁹⁶(3⁴ – 1)/(2) = 3⁹⁶

Choice (1)

48. A Eugenics research worker is working with two types of bacteria. One type triples in number every ten minutes while the other one becomes 5 times in the same time, if the total number of bacteria after half an hour is 1118, what was the total numberat the beginning?

1. 62
2. 246
3. 16
4. 32
5. 64

Answer: Let there be x bacteria of the first kind and y of the second.

27x + 125y = 1118

The remainder of 1118/27 is 11

Taking successive multiplies of 125, the remainders are 17,7,24,14,4,21,11 (Add 17 or subtract 10)

27(9)+125(7)=1118

Hence 27(9+125)+125(7-27) and other such expression are also equal to 1118, but x=9, y=8 is the only solution in positive integers.

At the beginning there were 9+7=16 bacteria.

Choice (3)

49. The sum to infinity of the series a² (a² + b²) + a⁴(a⁴ + b⁴) + a⁶(a⁶ + b⁶) +…. where |a|≤1 and |b| ≤1 is

1. \(\frac{a^2 b^2+a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
2. \(\frac{a^2 b^2-a^4-2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
3. \(\frac{a^2 b^2-a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
4. \(\frac{a^2 b^2+a^4+2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)
5. \(\frac{a^2 b^2-a^4-2 a^6 b^2}{\left(1-a^2 b^2\right)\left(1-a^4\right)}\)

Answer:

The series when expended becomes

a⁴ + a²b² + a⁸ + a⁴b⁴ + a¹² + a⁶b⁶ + …..∞

= a⁴ + a⁸ + a¹² +……∞

\(=\frac{a^4}{1-a^4}+\frac{a^2 b^2}{1-a^2 b^2}=\frac{a^4\left(1-a^2 b^2\right)+a^2 b^2\left(1-a^4\right)}{\left(1-a^4\right)\left(1-a^2 b^2\right)}=\frac{a^4-2 a^6 b^2+a^2 b^2}{\left(1-a^4\right)\left(1-a^2 b^2\right)}\)

Choice (1)

50. If S=3+7x+llx2 + 15×3 + …..(3 +4n)xn…and |x| < 1,S =

1. \(\frac{3+x}{(1-x)^2}\)
2. \(\frac{3+x}{1+x^2}\)
3. \(\frac{3-x}{(1-x)^2}\)
4. \(\frac{3-x}{(1+x)^2}\)
5. none of these

Answer:

S=3+7X+11x² + 15x³ +……-> (1)

Sx = 3x + 7x² +11x³+…..-> (2)

(1)- (2)

5(1- x) = 3 + 4x + 4x² + 4x³ + ……..

= -1+4/1-x = x + 3/1-x

S = (x + 3)/(1- x)²

Choice (1)

51. Chidanand and Dhamesh joined a company on 1 jan 1991. Chidanand was offered a starting salary of Rs. 3000 p.m. with an kannual increment of Rs.500 p.m. Dhamesh was put on a starting salary of RS.2000 p.m. with a six monthly increment of Rs. 350 p.m. if they continued working till 31 December 2000, how much more or less did Dharmesh receive compared to chidanand?

1. Rs.9000 more
2. Rs.9000 less
3. Rs.7500 more
4. Rs.7500 less
5. Rs. 4500 more

Answer: Chidanand’s first monthly salary is Rs.3000

His last monthly salary is Rs.7500

His average monthly salary is Rs.5250

Dharmesh’s first monthly salary is Rs.2000

His last monthly salary is is Rs 8650

His average monthly salary is Rs 5325

Dharkmesh gets 10(12) (5325-5250) = Rs. 9000 more than chidanand.

choice (1)

52. The sum of the first n terms of an arithmetic progression is p. the sum of the first 2n terms is Q. if Q= (4/P, find the sum of the first 3n terms of the arithmetic progression

1. P+Q
2. P-Q
3. Q
4. P
5. Q-P

Answer:

The sum of the first n terms is p₁ (say),that of the next n terms is P₂(say) and that of the third set of n terms is P₃ (say)

P₁, P₂, P₃ Have to be in arithmetic progression.

As P₃ are in arithmetic progression.

P₃ are in aritmetic progression. P₃ = 2(Q- P)- P = 2Q- 3P …. (1)

The sum of the first 3n terms is P₁ + P₂ + p₃ = Q+P₃ = Q + (2Q- 3P)from (1) = 3Q- 3P.

As Q =4/3p,P₁ + P₂ + P₃ = 4P-3P = P

Choice (4)

53. The ratio of the sum of first 20 terms of an arithmetic progression and the sum of its first 30 terms is 10:31 if the common difference of the arithmetic progression is 2, find the first term.

1. \(-6 \frac{3}{8}\)
2. \(-9 \frac{7}{8}\)
3. \(-8 \frac{3}{8}\)
4. \(-10 \frac{5}{8}\)
5. \(-9 \frac{5}{8}\)

Answer: Let the first term and the common difference of the terms in arithmetic progression be a and d respectively

\(=\frac{\frac{20}{2}[2 a+19 d]}{\frac{30}{2}[2 a+29 d]}=\frac{10}{31} \Rightarrow \frac{2(2 a+19 d}{3(2 a+29 d)}=\frac{10}{31}\)

124a + 117d = 60a + 870d

64a = -308d

a=-308/64(2)

\(-9 \frac{5}{8}\)

Choice (5)

54. Find the sum (1)(3)+ (3)(5)+ (5)(7) + (7)(9) + …………………………………. up to 50 terms.

1. 171700
2. 171850
3. 1717750
4. 1701650
5. 171650

Answer: (1)(3)+(3)(5)+(5)(7)+(7)(9)+……

The first parts of each term 1.3,5,7,9,… are in arithmetic progression and its nth term is (2n-1) the second parts of each term 3,5,7,9,1 1,… are also in arithmetic progression and its nth term is (2n+1)

the n term of the series is (2n-1) (2n+1) =4n² – 1

the sum ofn term of the series is (4n2-1)

= Σ(4n²- 1) =4 ∑ n² – ∑1 =\(\frac{4 n(n+1)(2 n+1)}{6}-n\)

The sum of the first 50 terms of the series \(=\frac{4 \times 50 \times 51 \times 101}{6}-50\)

=100 x 17 x 101- 50

= 171700-50=171650

Choice (5)

55. Find the sum \(\frac{1}{2}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\)

\(\frac{5^2}{2^5}+\frac{6^2}{2^6}+\cdots\)

1. 4
2. 5
3. 6
4. 7
5. 8

Answer:

\(\text { let } S=\frac{1^2}{2}+\frac{2^2}{2^2}+\frac{3^2}{2^3}+\frac{4^2}{2^4}+\frac{5^2}{2^5}+\frac{6^2}{2^6}+\cdots\) \(\frac{S}{2}=\frac{1}{2^2}+\frac{2^2}{2^3}+\frac{3^2}{2^4}+\frac{4^2}{2^5}+\frac{5^2}{2^6}+\cdots\) \(S-\frac{S}{2}=\frac{S}{2}=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{1}{2^4}+\frac{9}{2^5}+\frac{11}{2^6}+\cdots\) \(\frac{S}{4}=\frac{1}{2^2}+\frac{3}{2^3}+\frac{5}{2^4}+\frac{7}{2^5}+\frac{9}{2^6}+\cdots\) \(=\frac{1}{2}+\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots\right]\) \(=\frac{1}{2}+\frac{1 / 2}{1-1 / 2}=\frac{1}{2}+1=\frac{3}{2} \Rightarrow S=\frac{3}{2} \times 4=6\)

56. A total of 810 balls are arranged in N layers. The ith layer from the top where 2 < i < N has 4 more balls than the layer immediately above it. Which of the following is not a possible value of N?

1. 27
2. 18
3. 15
4. 10
5. 20

Answer: The number of balls in the 1st, 2nd … nth layer from the top be f let the top are in arithmetic progression. Let the number of balls in the 1st layer from the top be f
total number of balls = N/2(2f+ (N – 1)4)

810 = N(f+ (n- 1)2)

To satisfy the above equation, N must be a factor of 810 only choice (5) violates this condition

Choice (5)

57. X is the set of the first 1990 natural numbers. A is an arithmetic progression having at least 3 elements. First element is the least element of x and its last element is the greatest element of x How many possibilities does a have?

1. 10
2. 11
3. 12
4. 9
5. 13

Answer: The least element of A is 1 and its greatest element is 1990 suppose A has N elements and a common difference of D. then 1990= 1+(N-1) D (N-1)D=1989 =3²(17) (13)——(1)
N ≥ 3

1989 has (3) (2) (2) or 12 factors or 6 pairs of factors.

(N-1)D=1(1989)= 3(663)=9(221)=13(153)=17(117)=39(51)

As N ≥ 3, ,N-1 ≥ 2

N-1 can be 3,9,13,17,39,51,117,153,221,663 or 1989 i.e. N can have 11 values.

Choice (2)

58. Find the number of common terms to the sequences 7,11.15,19,23…..411 and 8,13,18,23,…..463

1. 88
2. 87
3. 21
4. 22
5. 20

Answer: The two equerries are Aps. For any two Aps, their common terms also from on AP.

Moreover, the common difference of that AP is the LCM of the common differences of the Aps the common difference of the first and second Aps are 4 and 5 respectively the AP of common terms has a common difference of 20.

First common term to the two Aps = 23 suppose the AP of common terms has n terms then the nth cannot exceed min (last terms of the Aps) i.e. min (411,463)=411. Also N must be the largest value satisfying this condition.

∴ 23 + 20(N- 1) ≤ 411

20N = 3 ≤ 411

N ≤ 20.4

N = 20

Choice (5)

59. The nth term of the series 6,13,20,Is 636. Find n.

1. 90
2. 91
3. 100
4. 110
5. 101

Answer: The nth term of an arithmetic progression with the first term a and common difference d is given by a +(n-1) d. for the given series.

a=6 and d=7 its nth term =>6+(n-1)7=636⇒n=91

Choice (2)

60. The sum of sixty numbers in arithmetic progression is 7,800. The largest of the numbers is 12 times the smallest. Find the smallest number.

1. 30
2. 25
3. 15
4. 10
5. 20

Answer: Let the smallest numberbe a and common difference of the arithmetic progression be largest number = 12a

Sum of the 60 numbers =60/2[a+12a]

390a=7800

a=20

Choice (5)

61. The sum of three numbers in arithmetic progression is 36.the sum of the squares of the three numbers is 464. Find the smallest number.

1. 8
2. 10
3. 12
4. 16
5. 6

Answer: Let the three numbers be a-d a and a+d

a-d + a + a + d = 36.

3a=36               a=12

(12- d)² + 12² + (12 + d)² = 464

d² = 16

d=√16=±4

If d is either +4 or -4 the numbers are 8, 12,16 the smallest is 8.

Choice (1)

62. Find the sum of all the numbers divisible by 8 and lying between 300 and 950.

1. 90,960
2. 1,00,480
3. 96,480
4. 50,544
5. 52,648

Answer: The first number divisible by 8 above 300 is 304= (8x 38) the last number before 950 divisible by 8 is 944 i.e.8 x 118

Total number ofmultiples between 304 and 944

\(=\frac{9 \cdot 4-30 \cdot}{8}+1=81, n=81 .\)

Sum of all these numbers 81/2(304- 944) = 50544.

Choice (4)

63. The seventh term of a geometric progression is 3645. If the product of its first 3 terms is 3375, find the first term of the geometric progression.

1. 5
2. 10
3. 15
4. 3
5. 6

Answer: Let the first, second and third terms of the geometric progression be a/r, a and ar respectively

a/r a ar = a³ =3375 =>a=√3375 = 15

Seventh term of the geometric progression =\(\frac{a}{r} r^6\)= ar⁵(since first term is a/r) 15r⁵=3645

r⁵ = 243

r=⁵√243 = 3

a/r=5

Choice (1)

64. A geometric progression has a sum to infinity. The value of the cube of any term is 1 /8th of the cube of the sum of the terms which follow it . find the sixth term of the geometric progression if the first term is 4.

1. \(\frac{2}{3}\)
2. \(\frac{64}{81}\)
3. \(\frac{128}{243}\)
4. \(\frac{256}{729}\)
5. \(\frac{512}{2187}\)

Answer: Let the series be 4, 4r, 4r², 4r³

Given that,(4)³ = \(\frac{1}{8}\left\{\frac{4r}{1-r}\right\}^3\)

\(\Rightarrow 4=\frac{1}{2}\left(\frac{4 r}{1-r}\right)\)

r=2/3

the sixth term = ar⁵ = (4)(2/3)⁵=128/243

Choice (3)

65. There are three numbers in arithmetic progression having a sum of 51. If the first is increased by 2, the second is increased by 1 and the third is increased by 3, the numbers would be in geometric progression. Find the smallest number in the geometric progression.

1. 10
2. 12
3. 14
4. 11
5. 13

Answer: Let the first second and third numbers in arithmetic progression be a- d a and a + d.

a-d + a = a = d = 3a = 51=>a = 17

a-d + 2,a + 1 and a + d + 3 i.e. (19- d) (20 + d) = 324

=> d² + d- 56 = 0=>d = -8 or 7

So, the numbers in geometric progression are either 27, 18 and 12 or 12, 18 and 27, so whatever be the case the smallest is 12.

Choice (2)

66. The product of three numbers in geometric progression is 1728. If their sum is 37, find the smallest number.

1. 9
2. 10
3. 12
4. 8
5. 6

Answer:

Let (a/r) a and ar be the numbers in geometric progression. Given that (a/r) (a)(ar) = 37

=>12/r+ 12 + 12r=37=>12r² – 25r +12 = 0

=>(3r- 4)(4r- 3) = 0=>r = 3/4 or 4/3

the numbers are 9,12,16 or 16, 12, 9, in either case, the smallest is 9.

Choice (1)

67. A geometric progression has the sum to infinity 4. The sum to infinity of the sum of the squares of the terms of the geometric progression is 6. Find the first term of the geometric progression.

1. 24/11
2. 32/11
3. 96/11
4. 40/11
5. 16/11

Answer:

Let the first term be a and the common ratio be r.

\(\frac{a}{1-r}=4 \ldots .\) \(\frac{a^2}{1-r^2}=6 \ldots \ldots(2)\) \(\Rightarrow \frac{a}{1-r}\left(\frac{a}{1+r}\right)=6\)

from (1)and (3), We have,

\(a=\frac{6}{4}(a+r)=(1-r) \Rightarrow \frac{11}{2} r=\frac{5}{2}\)

r=5/11 first term = 4(1-r)=24/11

Choice (1)

68. An infinite geometric progression having a finite sum has its first term as 12. The difference of the third and the fifth terms of the geometric progression is 9/4 if the common ratio is a rational number; find the sum to infinity of the terms of the geometric progression.

1. 12 or 15
2. 16 or 10
3. 20 or 12
4. 24 or 8
5. 12 or 8

Answer:

Let the common ratio of the geometric progression be r. the third term of the geometric progression =12r²

The fifth term of the geometric progression 12r⁴

As the geometric progression has a sum to infinity, -1 < r < 1

Hence r² < r⁴ = 9/4

Let r = a

12(a- a²) =9/4

48a- 48a² -9 = 0

16a² – 16a + 3 = 0, dividing both sides by 3

16a² – 12a -4a + 3

4a (4a- 3)- 1(4a- 3) = 0

(4a- 3)(4a- 1) = 0

a = 3/4 or 1/4

r=±1/2 as r is rational sum to infnity= \(\frac{12}{1 \pm \frac{1}{2}}\)

= 24 or 8

Choice (4)

69. Find the sum of the squares of the first 12 terms of the arithmetic progression for which the sum of the first n terms is 3n² + 6n

1. 12450
2. 13170
3. 14100
4. 26316
5. 24216

Answer:

Given s_n = n(3n + 6)

a = T₁ = S₁ = 1(3 + 6) = 9

T₂ = S₂– S₁ = 2(6 + 6)- 9 = 15

d=T₂– T₁ = 15- 9 = 6

the series is 9,15,21,….

Sum of the squares of the first 12 terms of the arithmetic progression

=9² + 15² + 21²+ …….+ 75²

=3²(3²+5²+7²+…..25²)

=3²(1²+ 2²+ 3² +4² +…..+ 25²)+- 3²(1²)

=3²[(1² + 2² + 3² + 4² +….. 26²)- (2² + 4² +……+ 26²)- 1²]

\(=9\left[\frac{(26)(27)(53)}{8}-\frac{2^2(13)(14)(27)}{6}-1\right]\)

=9[117(53)-4(63)(13)-1]

=9(6201-3276-1]

=9[2924]=26316

Choice (4)

70. The sum of the first 12 terms of log31/2X + log31/4 x log31/6X + … 312 find x.

1. 9
2. 3
3. 27
4. 81
5. 243

Answer:

\(\log \frac{1}{3^2} x+\log \frac{1}{3^4} x+\log \frac{1}{3^6} x+\cdots\) \(\frac{\log _3 x}{\log _3 3^{1 / 2}}+\frac{\log _3 x}{\log _3 3^{1 / 4}}+\frac{\log _3 x}{\log _3 3^{1 / 6}}+\cdots \text { for } 12 \text { terms }\)

(Rewriting each term being added which is in the form

\(\log _{\frac{1}{b^2}} a a s \frac{\log a}{\log b^{\frac{1}{2}}}\)

and taking the base for both numerator and denominator as 3)

\(\frac{\log _3 x}{\frac{1}{2} \log _3{ }^3}+\frac{\log _3 x}{\frac{1}{4} \log _3{ }^3}+\frac{\log _3 x}{\frac{1}{6} \log _3 3}+\cdots \text { for } 12 \text { term }\)

= 2log₃ x +4log₃ x +6log₃ x +8log₃ x + + 24log₃ x

=(2+4+6+….+24) log₃ x= 156(log₃ x) = 132

=> log₃ x= 2=> x= 3² = 9

Choice (1)

71. If p² = q² -q⁴ + q⁶ – q⁸ +….. ∞, find P given |q| < 1

1. \(\frac{ \pm q}{\sqrt{1-q^2}}\)
2. \(\frac{ \pm(q+1)}{\sqrt{q^2+1}}\)
3. \(\frac{ \pm q}{\sqrt{q^2-1}}\)
4. \(\frac{ \pm q}{\sqrt{q^2+1}}\)
5. \(\frac{ \pm(q-1)}{\sqrt{q^2+1}}\)

Answer:

Given p² = q²- q⁴ + q⁶– q⁸ +….is an infinite geometric progression with |q| <1 =>|q²| < 1. the common ratio of the geometric progression is -q²

\(p^2=\frac{q^2}{1-\left(q^2\right)}=\frac{q^2}{1+q^2} \Rightarrow p=\frac{ \pm q}{\sqrt{1+q^2}}\)

Choice (4)

72. There are three numbers in geometric progression. When the middle number is tripled. The numbers will be in arithmetic progression. If the common ratio is greater than 1, find its value.

1. 4+3√2
2. 3+4√2
3. 3√2-4
4. 3+2√2
5. 6+3√2

Answer: Let the three numbers be a, 3ar and ar² therefore a, 3ar and ar2 are in arithmetic progression

2(3ar) = a + ar²

r² – 6r + 1 = 0

\(r=\frac{6 \pm \sqrt{32}}{2}\)

3±2√2

as r > 1,r = 3 + 2√2 Choice (4)

73. The 10th term of an arithmetic progression is 15. II the sum of the squares of the 7th, 10th and the 13th terms is minimum, the common difference is_.

1. 0
2. 1
3. 1/2
4. 2
5. 3/2

Answer: Let the first term and common difference of the arithmetic progression be a and d respectively given that a+9d=15.

(a+6d)²+(a+9d)²+(a+12d)²

(a+9d-3d)²+(a+9d)²+(a+9d+3d)²

(15-3d)²+15²+(15+3d)²

=3(15)²+18d² this is minimum, when d=0.

Choice (1)

74. The sum to infinity of a geometric progression is 39/2. The sum to infinity of squares of the terms of the geometric progression is 253.5. Find the sum to infinity of cubes of the terms of the geometric progression.

1. 252.5
2. 240.5
3. 253.5
4. 260.5
5. None of these

Answer:

Let the first term and common ratio of the geometric progression be a and r respectively

\(\frac{a}{1-r}=\frac{30}{2} \ldots \ldots\) \(\frac{a^2}{1-r^2}=253.5=\frac{507}{2} \ldots . .2\) \(\left(\frac{a}{1-r}\right)^2=\left(\frac{39}{2}\right)^2\)

Squaring (1 ) on both sides… (3)

Dividing (2)by (3)and simplifying

\(\frac{\frac{a^2}{(1-r)(1+r)}}{\frac{a^2}{(1-r)^2}}=\frac{\frac{507}{2}}{\frac{1521}{4}} \Rightarrow \frac{1-r}{1+r}=\frac{2}{3} \Rightarrow r=\frac{1}{5}\) \(\text { from }(1), a=\frac{39}{2}(1-r)=\frac{78}{5}\)

sum of the cubes of the terms of the GP

\(=\frac{a^3}{(1-r)^3}=\frac{\left(\frac{78}{5}\right)^3}{\left(\frac{1}{5}\right)^3}=\frac{(78)(78)(78)}{124}=\frac{78(39)(39)}{31} \text { which is }>1000\)

Choice (5)

75. Find the 50th term of the following series. 1 + 3 + 7 + 13 + 21 +….

1. 2451
2. 2561
3. 2781
4. 2341
5. 2671

Answer:

S₅₀ = 1 + 3 + 7 + 13 +…… T₄₉ + T₅₀ … ( 1)

Also s₅₀ = 1 + 3 + 7 + 13 + …..T₄₉ + T₅₀ … (2)

(1)- (2) =>0 = (1 + 2 + 4 + 6 +…..50 terms)- T50

=> T₅₀ = 1 + (2 + 4 + 6 + ……49 term)

=> T₅₀ = 1(49)(50)/2=1 + 2450 = 2451

Choice (1)

76.The terms a₁, a₂……..a₁₇ in arithmetic progression. If a₃ +a₈ + a₁₁ + a₁₄ = 100, then a₁ + a₂+ +a₃……… a₁₇ =__________

1. 780
2. 850
3. 390
4. 425
5. 455

Answer:

if1st term T₁ = a, then nth Term T_n = a + (n- 1)d.

Given a₃ + a₈ + a₁₁ + a₁₄= 100

a₁+ 2d + a₁ + 7d + a₁10d + a₁+ 13d = 100

=>4a₁+ 32d = 100=>2a₁+ 16d = 50……(1)

To find: a₁+ a₂ + a₃ +…… a₁₇

= n/2 [first term + last term | =>17/2 |a₁ + a₁₇| = 17/2 |a₁ + a₁₇ + 16d]

=17/2[2a₁ + 16d] = 17(50)/2 = 425

Choice (4)

77. If five distinct numbers a, b, c, d, e are in arithmetic progression with common difference of ‘D’, then a- 3b+5c-5d+2e is equal to D

1. 2D
2. 3D
3. 0
4. -2D

Answer:

Let x be the common difference

=>b = a + x, c = a + 2x, d = a + 3x and e = a + 4x.

a- 3b + 5c- 5d + 2e

= a- (3a + 3x) + (5a + 10x) – (5a + 15x) + 2a + 8x)

= (a — 3a + 5a- 5a + 2a) + (-3x + 10x- 1 5x + 8x) =0 + 0 = 0

Choice (4)

78. If in a geometric progression the fifth term is 8 times the second term and the sum of the first, third and sixth terms Is 111, what is the 7th term?

1. 384
2. 208
3. 224
4. 416
5. 192

Answer: Let the first term of GP= A and common ratio = R

=>AR^N-1 = Nth term =>AR⁴= 8(AR)

=>R³ = 8=>R = 2 ……..(1)

A+AR² + AR⁵ = 111

=>A{1 + R² + R⁵) = 111…..(2)

Put R =2 in (2) => A( 1 + 4 + 32) = 11 1 => A =111/37 = 3

7th Term = AR⁶=3x (2⁶)=192

Choice (5)

79. If the sum of \(1+\frac{2}{x}+\frac{4}{x^2}+\frac{8}{x^3}+\cdots \infty\) is finite and x>0, then which of the following is true?

1. X> 2
2. X < 2
3. X > 1/2, x < 2
4. X < 1/2
5. 1/2 < x < 2

Answer:

Since the sum \(1+\frac{2}{x}+\frac{4}{x^2}+\cdots \infty\) is finite, its common ratio should be less than 1.

\(S_{\infty}=\frac{a}{1-r}\)

common ratio = \(\frac{2}{\frac{x}{1}}=\frac{2}{x}\)

But r < 1

\(\Rightarrow \frac{2}{x}<1 \Rightarrow \frac{x}{2}>1 \Rightarrow x>2\)

Choice (1)

80. In an arithmetic progression, the 10th term is 11 and the 11th term is 10, How many consecutive terms (starting from the first term) o f the arithmetic progression should be considered so as to make their sum equal to zero?

1. 33
2. 41
3. 37
4. 39
5. 35

Answer:

T_n=a+(n-1)d

(a= first term,d= common difference, Tn = n th term)

T₁₀ = a+9d=11……………(1)

T₁₁= a+10d=10…………….(2)

solving (1) and (2)

d= – 1, a = 20

Say ‘n’ terms are needed so as to make sum =0

=>n/2[2a+(n-1)d]=0

=>n/2[40+(n-1)(-1)]=0

=>40 – n + 1 = 0=>n =41

first 41 terms of the arithmetic progression when added give the sum as zero

Choice (2)

81. The first, second and third terms of a geometric progression with distinct terms, are equal to the first, ninth and thirty-first terms, respectively of an arithmetic progression. If the first term of either progression is 11. Find the common ratio of the geometric progression.

Answer: Let the GP be 11,11r, 11r²,r=1

The 1st 9th and 31st terms or the arithmetic progression are 11, 11+8d,11+30d

11r = 11 + 8d=>11 (r- 1) = 8d …. (1)

11r² =11 + 30d =>11(r²-1 )= 30d….(2)

(2) + (1)=>r+= 1 15/4 =>r =11/4

choice (2)

82. The first term of an arithmetic progression consisting of integers, is the common ratio of a geometric progression. The first term of the geometric progression is the common difference of the arithmetic progression. The sum of the first 11 terms of the arithmetic progression is 341 and the sum of the first three terms of the geometric progression is 18. Find the sum of the common difference and the common ratio.

1. 8
2. 10
3. 12
4. 14
5. 7

Answer: If r is the common ratio of the geometric progression and d is the common difference of arithmetic progression first term of arithmetic progression = r and that of geometric
progression =d.

11/2[2r +(11-1)d] = 341

=>r+ 5d = 31…. (1)

=>d+ dr +dr² = 18

As arithmetic progression consists of only integers,both r and d are integers, so, Geometric progression also consists of only integers, By trial and error we get r=1and d=6
r + d = 7

Choice (5)

83. The three terms a + 2b, b+2c are c +2a are in arithmetic progression. The three terms 16a, 2(b + c), and c are in geometric progression. The ratio of the common difference of the arithmetic progression to the common ratio of the geometric progression could be

1. -2
2. 4
3. 8
4. -16
5. 0

Answer:

b+2c- (a+2b) =(c+2a)-(b+2c)

=>2b + 4c = c + 3a + 2b => a = c … (1)

(2b + 2c)² = 16ac=>4b² + 8bc + 4c² = 16ac …. (2)

From (1) and d(2) ,we get

4(b + c)² = 16c²=>b² + 2bc- 3c = 0

=>(b + 3c) (b- c) = 0=>b = c or – 3c

When b =c, the common difference

= (b + 2c)- (a + 2b) = 3c- 3c =0

the common ratio = \(\frac{2(b+c)}{16 a}=\frac{2(2 c)}{16 c}=\frac{1}{4}\)

their ratio = \(\frac{0}{1 / 4}=0\)

When b =-3c the common difference = (b + 2c)- (a + 2b) = (-3c + 2c)- (c- 6c) = 4c

the common ratio = \(\frac{2(b+c)}{16 a}=\frac{2(c-3 c)}{16 c}=\frac{-1}{4}\)

their ratio = \(\frac{4 c}{1 / 4}=-16 c\)

0 is given is the choices

Choice (5)

84. A number of saplings were lying at a point P, by the side of a straight road. Raju planted these saplings in a straight line with a distance of 20 meters between consecutive saplings. The first sapling was planted 20 m from P and all the saplings were planted on the same side of P. Raju carried only one sapling at a time and returned to P after planting each sapling, including the last one. If he travelled a total distance of 2.2 km, find the number of saplings that he planted.

1. 10
2. 12
3. 14
4. 16
5. 11

Answer: To carry the first sapling to plant it and return, Raju has to travel 40 m. to carry the second one to plant it and return, Raju has to travel 80 m. Hence until he plants the final sapling (nth sapling) he will have to travel

40(1 + 2 + ….n) = 2200(given)

n(n+1)/2 = 55=>n² + n – 110 = 0=>n = 10 or- 11

Hence n = 10, as n > 0 Choice (1)

85. In an arithmetic progression, the sum of the squares of the seventh term and the eleventh term is 482. The product of its second term and the sixteenth term is 29. Find the product of its first and the seventeenth terms.

1. 31
2. -31
3. -33
4. 33
5. None of these

Answer: Let the first term and common difference of the arithmetic progression be a and d respectively.

Given that t_7² + t_11² = 482

=>2a² + 32ad + 136d² = 482 …. (1)

And (a + d) (a + 15d) = 29

=a² + 16ad+15d²=29………..(2)

By (1) – 2 x (2), we get

106d² = 482- 58 = 424

=>d² = 4………(3)

Consider, t₁ x t₁₇ = a(a + 16d) = a² + 16 ad

= (2)- 15 x (3) = 29 -15X4 = -31

Choice (2)

86. The sum of the squares of the first and fifth term of an arithmetic progression is 218. If the third term is less than the first term, and the sum of these two terms is 16, find the first term.

1. 10
2. 11
3. 12
4. 13
5. 14

Answer: Let the first term of the arithmetic progression be a and the common difference be d.

Sum of the squares of the first and the fifth term

=a²+(a+4d)²=218

a²+4ad+8d²=109……………. (1)

a+(a+2d)=16

=>a+d=8=>a=8-d……………..(2)

Substituting a=8-d in (1) (8-d)²+4(8-d)d+8d²=104

5d²+16d-45=0 D=-5 or 1.8 As t₃<t₁,d<0,d=5

Substituting=-5 in (2). A=13. Hence, the first term of the arithmetic progression is 13

Choice (4)

87. The sum of the first n terms of an arithmetic progression is given by 3n²+4n. Find the 6th term of the arithmetic progression.

1. 124
2. 128
3. 37
4. 132
5. 41

Answer: Let the first term and the common difference of the arithmetic progression be a and d respectively.

n/2(2a+(n-1)d)=3n²+4n=n(3n+4).

Comparing both sides d/2 = 3 (Whatever multiplies of n2)

In L.H.S. equals that in R.H.S.)

d=6.      2a-d/2=4,    a=7.

The 6lh term of the arithmetic progression = a+5d=37.

Choice (3)

88. The sum of all the perfect squares from 50 to 2500 is

39825
40975
41895
44785
42785

Answer: The Sum of all the perfect squares from 50 to 2500

=the (sum of all perfect squares from 1 to 2500) – (Sum of all squares from 1 to 50).

Sum of the squares first n natural numbers

\(\frac{n(n+1)(2 n+1)}{6}\)

We have 50 perfect squares from 1 to 2500 whose sum is 1/6(50)(51)(101)

=25x17x101 =171700/4 = 42925…..(1)

Sum of all perfect squares from 1 to 7

= 1² + 2² + 3² + 4² + 5² + 6² + 7²

= 1/6(7)(8)(15) = 140……(2)

Required sum = 42925- 140 = 42785.

Choice(5)

89. a₁ = 1.5 and for n > l,a_n = 3a_n-1 + 4.Find a₃₀

1. 3³⁰ -1
2. 3²⁹(7/2)-2
3. 3²⁹(7/2) + 2
4. 3³⁰ +1
5. 3³⁰(7/2) + 2

Answer:

a₁ = 1.5

a₂ = 3a₁ + 4

a₃ = 9a₁ + 4(1 + 3)

a₄ = 27a₁ + 4(1 + 3 + 3²) … and

a₃₀ = 3²⁹a₁ + 4 + 4(3) + 4(3²) + (3²⁸)

\(=3^{29}(3 / 2)+\frac{4\left(3^{29}-1\right)}{2}=2\left(\frac{7}{2}\right) 3^{29}-2\)

Choice (2)

90. How many terms, at the maximum, of the progression 2, 5, 8……can be considered, if the sum of the terms is to be less than 3000?

1. 60
2. 44
3. 68
4. 70
5. 54

Answer: Let the maximum number of terms for which the sum of the terms of the A.P is less than 3000 be n.

The A.P has a first term of 2 and a common difference of 3. Sum of first n terms

\(\left.=\frac{n}{2}[2(2)+n-1) 3\right]<3000\)

n[1+3n] < 6000.

going by the choices only when n = 44,

we have the value ofthe expression n(3n + 1)being less than 6000 .

Hence n = 44.

Choice (2)

Permutations and Combinations Introduction

The concept of permutations and combinations is used to select or arrange some items out of a group according to certain predetermined conditions. This concept helps find the number of ways to arrange, select or reject these items.

The concepts of permutations and combinations derive from the principles of Factorials and the Fundamental principle of counting.

Factorial

The factorial of a natural number n, denoted by n! or |n_is the product of all-natural numbers from 1 up to n.

n! = 1 x 2 x 3 x … x (n- 1) x n

The factorials of the first few natural numbers are as given below.

Remember:

n! = nx [(n- 1) x (n- 2) x (n- 3) x … x 1]
n! = n x (n- 1)!
n! = n x (n- 1) x (n- 2)!

Read and Learn More CA Foundation Maths Solutions

The factorial of 0 is defined to be one.

∴ 0! = 1

The factorial of negative numbers is not defined.

Permutations and Combinations

The task of selecting a number of objects from a given set of objects involves a number of choices the number of choices available depends on the purpose of selection.

If the order of selection is unimportant, the selection is called a combination, if the order of selection is important, the selection is called a permutation.

Consider the case ofselecting two out of three students A, B and C.

Case 1:

Suppose the task is to select two students out of A, B and C and rank them according to the marks they have scored. The different possibilities that exist are as follows:

Note that the selection (A, is different from the selection (B, A] here.

Total number of selections = 6

Here, the students are first selected and then arranged for a purpose. This kind of selection is called a permutation.

Case II:

Now suppose that the task is just to select two students out of three. In such a case, the order in which the students are selected is not important.

Hence, the possibilities in this case are (A, B), (B, C] and (A, C]

Note that (A, and (B, in the above case mean the same thing as both groups imply that students A and B have been selected.

Here, they do not need to be ranked.

total number of selections = 3

Here, the important thing is to just select the students and not to arrange them.

This kind of selection is called a combination.

Remember:

Permutations involve selection and then, arrangement of objects while combinations involve only selection of objects.

Example:

Suppose 4 people out of 7 are to be arranged for a photograph. This involves first selecting the 4 people and then arranging them. This is a permutation.

On the other hand, if 4 people out of 7 are to be chosen for a team, it just involves selecting the 4 people without any arrangement. This is a combination.

Solved Example:

Find out whether the following tasks involve permutations or combinations.

1. Arranging five girls on five chairs for a photograph
2. Forming a committee of five people from then boys and five girls.
3. Electing a captain and a vice -captain from a team of11.
4. Selecting a king or a queen from a pack of cards.

Solution:

1. Here, the girls need to be arranged such that there are five girls on five chairs. Since, it involves arranging the five girls; it is a case of permutations.
2. Here, just select five people out of ten boys and five girls. There is no arrangement involved. So, it is a case of combinations.
3. This task involvers selecting two people, and then arranging them in order, the first one being the captain and the second being the vice-captain. Hence, this is a case of permutations.
4. Here, just select two cards. In the previous task, a player selected was to be ‘made’ the captain or vice -captain. Here, just select a king or a queen out of the kings and queens in the pack of cards. Hence, this is a case of combinations.

To find the number of ways in which the permutations and combinations can be counted, first understand the fundamental principle of counting.

Fundamental Principle of Counting

Suppose one needs to travel from X to Z via Y. There are three routes a, b, c, available to go from X to Y and two routes p, q available to go from Y to Z Ifone travels from X to Y taking the route a, then one can travel from Y to Z in two ways.

Thus, for every choice of route that one makes for travelling form X to Y, there are two ways to travel from Y to Z. As there are three routes to travel from X to Y, there are 2+2+2+=3x 2 ways to travel form X to z via Y.

The routes are (a,p),(a,q), b,p), (b,q),(c,p) and (c,q). If one considers travelling from X to Y as task 1, and travelling from Y to Z as another task 2, task 1 can be permed in 3 ways, task 2 can be performed in 2 ways, and task 1 and task 2 together can be performed in 3x 2 = 6 ways.

In general, the fundamental principle of counting says that, if we can accomplish one takes in xx ways, another task in x2 ways, and so on, until the n”‘ task which can be done in x„ ways
then all these tasks can be accomplished in X1×X2×X3× …xn ways.

Remember:

Consider two tasks; task 1 that can be done in m ways & task 2 that can be done in n ways.
Task 1 and task 2 can be done in m x n ways
Task 1 OR task 2 can be done in m + n ways.

Solved Examples:

1. Josephine likes to have passwords that have eight characters, the first four being different vowels and the last four being different numbers. How many different passwords can she have?

Solution:

The first four characters of Josephine’s password are to be chosen form the five vowels the first vowel can be chosen in 5 different ways. As all of them have to be different, the second
can be chosen only in 4 ways, the third in 3 ways and the last in 2 ways.

Similarly, the four numbers have to be different. Hence, the four numbers can be chosen form the 10 digits from  0 to 9 in 10, 9, 8 and 7 ways respectively.

Thus, the number of different passwords that Josephine can have.

= 5x4x3x2x 10 x9x8x7 = 604800

2. How many numbers greater than 5000 can be formed by arranging the digits 4, 5, 7, 9 using each digit only once?

Solution:

Since the number has to be greater than 5000, the first digit can be chosen in 3 ways only, I.e., it can be 5,7 or 9

Once this diÿit is chosen, the next, digit can be chosen in 3 ways, the next in 2 and the next in 1.

Thus, 3x3x2x1 = 18 such numbers can be formed.

Hence, 18 numbers greater than 5000 can be formed by arranging the digits 4,5,7, 9 using each digit only once.

3. How many numbers can be formed using the digits 4, 5, 7,9, if repetition ofdigits is allowed?

Solution:

In this case, there is no repetition of digits, also, there is no condition as to whether should be greater than or less than something, for each place one can choose any of the 4 digits

Thus, 4x4x4x4 = 256 such numbers can be formed.

Hence, 256 numbers can be formed using any of the digits 4, 5, 7, 9 if repetition of digits is allowed.

4. Four medical representatives Rl, R2, R3 and R4 are to visit one doctor each on Thursday. The doctors are Dl, D2, D3, and D4, Rl Will not visit only D1 and R2 will visit only D2 or D3. In how many ways can the visits be done such that no two representatives visit the same doctor?
Solution:

R2 can visit ldoctor (D2 or D3) in 2 ways. Then Rl is left with 2 choices (D2 /D3 or D4)

After this, R3 and R4 have 2 and 1 choices.

∴ the total number of ways in which the vistis can be done = 2x2x2x1= 8

5. Eight buildings in a residential complex are to be painted on their outsides in any one colour form among cream, blue, green, pink and white. The buildings are in a row. Two buildings can have the same colour but two adjacent buildings cannot. In how many ways can the buildings
be painted?

Solution:

The first building can be painted using any one of the five colours in 5 ways. The second building can be painted using any of the remaining four colours in 4 ways. The third building
can have any colour except the one with which the second building is painted.

Thus it can be painted in 4 ways. Thus, each building can be painted in 4 different ways using a colour except the one used to colour the previous building. Hence, the fourth, fifth, sixth, seventh and eighth buildings can be painted using 4 colours each.

Thus, the total number of ways in which the buildings can be painted = 5 x 4⁷

Permutations

If there are n objects ami one is supposed to arrange r (r < n) objects out of these, then the number of ways in which this can be done is written as nPr and is read as “the number of
permutations of n objects taken rata time”

The number of permutations is given as:

\({ }^3 P_2=\frac{n!}{(n-r)!}\)

The case where 2 out of 3 students were to be ranked can be written as \({ }^3 \mathrm{P}_2\).

\({ }^3 P_2=\frac{n!}{(3-2)!}=6\)

Remember:

If r= 0,

\({ }^n P_2=\frac{n!}{(n-0)!}=1\)

If r= 1,

\(n P_1=\frac{n!}{(n-1)!}=n\)

if r = n ,

\({ }^n \mathrm{P}_{\mathrm{n}}=\frac{n!}{(n-n)!}=\frac{n!}{0!}=\frac{n!}{1}=n!\)

if r = n-1,

\({ }_n P_{n-1}=\frac{n!}{(n-n+1)!}=\frac{n!}{1!}=\frac{n!}{1}=n!\)

Permutations with Repetitions

If one has to form words using the letters A, B and C, one can make 3P.j= 6 words i.e., ABC, ACB, BAC, BCA, CAB AND CBA.

But, if one has to from words using the letters A, B and A, then one can make only 3 different words- AAB, ABA and BAA.

Thus, repetition of letters reduces the number of words that can be formed.

Therefore, in case the n elements that are to be arranged among themselves have some repeated elements among them, the number of permutations is not equal to MP„ or not equal
to n!

If n objects are to be arranged among themselves and these contain p identical objects of one kind q identical objects of another kind, r identical objects of still another kind and so on, the total number of ways in which they can be arranged is given by \(\frac{n!}{p!q!r!\ldots}\)

Solved Examples:

1. How many words can be formed by arranging the letters of the word WEEDED?

Solution:

Total number of letters in the word WEEDED=6.

D occurs 2 times and E occurs 3 times.

Hence, the total number of words that can be formed by arranging these letters is:

\(\frac{6!}{2!\times 3!}=60\)

Thus 60 words can be formed by arranging the letters of the word WEEDED.

2. Akash has is DVDs out of which 3 are of a movie A, 5 are of a movie B and the rest are all different in how many ways can he arrange the DVDs on a shell?
Solution:

There are 15 objects out of which 3 are of one kind and 5 are ofanother kind.

The total number ofarrangements possible \(=\frac{15!}{3!\times 5!}\)

Circular Arrangements

The number of ways in which n objects can be arranged in a circle= (n-l)!

Explanation:

Consider four people A, B, C and D sitting on a circular table.

All the four arrangements shown above are one and the same as the beginning and end of a circle cannot be determined. However, had the four arrangements been linear, had the four arrangements been linear, they would be ABCD, DABC, CDAB and BCDA, which would be four different linear arrangements.

Thus, 4 linear arrangements. Would give rise to 1 circular arrangement.

But 4 people would give 4! Linear arrangements

4 people would give 4!/4 = 3! Circular arrangements.

Similarly, if there were n objects, n linear arrangements would give 1 circular arrangement.

n! Linear arrangements would give n! /n = (n-l)! Circular arrangements.

The number of ways in which n objects can be arranged in a circle = (n-l)!

Solved Examples:

1. In how many ways can 5 people be seated around a circular table?

Solution:

n objects can be arranged in a circle in (n-l)! Ways.

Similarly, 5 people can be seated around a circular table in (5-1)! = 4! =24 ways.

2. Seven girls and five boys are to be seated around a circular table such that two of the girls want to sit next to each other, in how many ways can they be seated?

Solution:

There are 12 people to be seated in a circle, as two girls want to sit next to each other; we consider them to be one entity.

Hence there are 11 people to be arranged in a circle, this can be done in 10! Ways

The two girls can be arranged among themselves in 2! Ways.

Thus the total number of ways in which the seven girls and five boys can be arranged = 10!x 2!

3. Five boys and five girls are to be arranged in a circle such that no two girls are next to each other, in how many ways can they be arranged?

Solution:

Since no two girls are to be next to each other, the boys and girls need to be arranged alternately.

Hence, the five boys can be arranged in a circle in 4! Ways.

There are five places in between the boys, in which the girls can be arranged in 5! Ways.

The total number of ways in which the five boys and five girls can be arranged= 4!x 5!

Note that once the boys are seated, the positions of the girls are relative to the positions of the boys and hence their arrangement is treated as a linear arrangement.

Some circulate arrangements are such that clockwise or anticlockwise arrangements are the same. A necklace is a typical example ofsuch a circular arrangement.

The number of arrangements of n objects in such a fashion is \(\frac{(n-1)!}{2}\)

4. In how many ways can 6 different beads be arranged to from a necklace?

Solution:

In a necklace, clokwise and anticlockwise arrangements are same.

n objects canbe arranged in a necklace in [ltex]1 / 2 \times(n-1)!\text { ways. }[/latex]

Thus, 6 beabs can form a necklace in \( 1 / 2 \times(6-1)!=\frac{5!}{2}=60 \text { ways. }\)

Combinations

If there are in objects and out of them (r < n) are to be selected, them the number of ways in which this can be done is ” C r and is read as “the number of combinations of objects taken r at a time”.
The number of combinations is given as:

\({ }^n C_r=\frac{n!}{(n-r)!r!}\)

If 2 students out of 3 are to be selected, it can be written as \({ }^3 C_2\)

\({ }^3 C_2=\frac{3!}{(3-2)!2!}=3\)

Thus, there, are 3 different ways in which 2 objects out of 3 can be selected. These 3 different possible cases are as given in case II.

Explanation:

Consider n objects out of which r are to be selected. Any combination of r objects out of these n objects can be arranged in \(r P_r\) ways, thus one combination of r objects gives r! Permutations.

As the number of combinations of n objects taken r at a time \({ }^n \mathrm{C}_{\mathrm{r}}\) these combinations in all give

\({ }^{\mathrm{n}} \mathrm{C}_r \times r \text { ! Permutations. }\)

This is nothing but the number of permutations of n objects taken r at a time.

\(\begin{aligned}
& { }^n \mathrm{P}_{\mathrm{r}}={ }^n \mathrm{C}_{\mathrm{r}} \times r! \\
& { }^n \mathrm{C}_{\mathrm{r}}=\frac{{ }^n \mathrm{Cr}}{r!}=\frac{n!}{(n-r)!r!}
\end{aligned}\)

Remember:

If r=0

\({ }^n C_0=\frac{n!}{(n-0)!\times 0!}=1\)

If r=1,

\({ }^n C_1=\frac{n!}{(n-1)!\times 1!}=n\)

If r=n,

\({ }^n C_n=\frac{n!}{(n-n)!\times n!}=1\)

If r =n-1

\({ }^n C_{n-1}=\frac{n!}{(n-n+1)!\times(n-1)!}=\frac{n!}{1!\times(n-1)!}=n\)

Some more useful rules involving Combinations

1. The number of ways in n which identical things can be divided in r groups when each person may get any number of things, including zero=\( (n+r-1) C_{(r-1)}\)

2. The number of ways in n which identical things can be divided in r groups when each person may get at least one thing = \({ }_{(n+r-1)} C_{(r-1)}\)

Solved Examples:

1. In how many ways can 24 chocolates be distributed among 4 children such that each one of them gets some or none?

Solution:

The number of distributions is \((21+1-1) C_{(4-1)}={ }^{27} C_3=2925\)

2. How many solution does the equation x + y +z= 10 have, if x, y, and z are natural numbers?

Solution:

Since x, y and z are all natural numbers; they cannot have a value 0. Hence, their value has to beadiest 1. Hence; this is similar to a problem of distributing 10 chocolates (each chocolate
corresponds to the number 1) among 3 people so that each person has at least one chocolate.

Thus, the number of solutions of this equation is given by \( { }^{10 \cdot 1} \mathrm{C}_{3 \cdot 1}={ }^9 \mathrm{C}_2=36\)

Remember:
The fundamental principle of counting says that, if we can accomplish one task in xx x x2 x x3 x ,…x„ ways.

If task 1 can be done in m ways and task 2 can be done in n ways then:

1. Task 1 and task 2 can be done in m xn ways.

2. Task 1 or task 2 can be done in m + n ways

The number of permutations of n objects taken rat a time is given by: \({ }^n P_r=\frac{n!}{(n-r)!}\)

If n objects are to be arranged among themselves and these contain p identical objects of one kind, q identical objects of another kind, r identical objects of still another kind and
so on, the total number of ways in which they can be arranged is given by \(\frac{n!}{p!q!r!_m}\)

The number of permutations of n different things taken r at a time when repetition is allowed = n x n x n x…..r times = nr

The number of ways in which n objects can be arranged in a circle when clockwise and anticlockwise arrangements are different = (n-l)l

The number of ways in which n objects can be arranged in a circle when clockwise and anticlockwise arrangements are same = (n-l)l / 2

Number ofselections of k consecutive things out of n things in a row =n-k+l

The number of combinations of n objects taken r at a time is given by

\({ }^n C_r=\frac{n!}{(n-r)!r!}\)

 

\(\begin{aligned}
& { }^n P_r={ }^n C_r x r! \\
& { }^n C_r={ }^n C_{n-r} \\
& \text { if }{ }^n C_x={ }^n C_y \text { then } x=y \text { or } x+y=n \\
& { }^n C_r={ }^{n \cdot 1} C_r+{ }^{n \cdot 1} C_{r-1}
\end{aligned}\)

the number of ways in which none or some elements from a set of n elements is

\({ }^n \mathrm{C}_{10}+{ }^n \mathrm{C}_1+\) \({ }^n C_2+\ldots .+{ }^n C_n=2 n\)

the number of ways in which n identical things can be divided in r groups when each person may get any number of things, including zero =\(n+r-1) C_{(r-1)}\)

the number of ways in which n identical things can be divided in r groups when each person may get at least one thing \(=11+r-1) C_{(r-1)}\)

Exercise – 1 Permutations

Choose the most appropriate option (1) (2) (3) or (4)

l. \({ }^4 P_3\) i is evaluated as

1. 43
2. 34
3. 24
4. None of these

Answer: (3) 24 \(\text { We know } \mathrm{nP}_{\mathrm{r}}=\frac{n!}{(n-r)!}\)

\({ }^4 P_3=\frac{4!}{(4-3)!}=\frac{4!}{1!}=24 \)

2.\({ }^4 P_4\) is equal to[/latex]

1. 1
2. 24
3. 0
4. none of these

Answer: (2) 24 \(\text { We know }{ }^n \mathrm{P}_{\mathrm{r}}=\frac{n!}{(n-r)!} \quad{ }^4 \mathrm{P}_4=\frac{4!}{0!}=\frac{4!}{1}=24\)

3. 7! is equal to

1. 5040
2. 4050
3. 5050
4. none of these

Answer: ( 5040 71 = 7x6x5x4x3x2x1 = 5040

4. 0! is a symbol equal to

1. 0
2. 1
3. Infinity
4. none of these

Answer: (1 01 = 1

5. In \(n P_{r^{\prime}}\) n is always

1. an integer
2. a fraction
3. a positive integer
4. none of these

Answer: Integer

6. In \(n P_{r^{\prime}}\) the restriction is

1. n > r
2. n ≥ r
3. n ≤ r
4. none of these

Answer: n ≥ r.

7. \(\text { In nP } P_r=n(n-1)(n-2) \ldots \ldots \ldots \ldots \ldots \ldots \ldots(n-r+1) \text {, }\) the number of factors is

1. n
2. r-1
3. n-r
4. r

Answer: r

8. \(n P_r\) can also written as

1. \(\frac{n!}{(n-r)!}\)
2. \(\frac{n!}{r!(n-r)!}\)
3. \(\frac{r!}{(n-r)!}\)
4. None of these

Answer: \(\text { ( } \frac{n!}{(n-r)!}\)

9.\(m+n P_2=56, m-n P_2=30 \text { then }\)

1. m =6, n = 2
2. m = 7, n= 1
3. m=4, n=4
4. none of these

Answer:

\(=\frac{(m+n)!}{(m+n-2)!}=56\)

= (m+n) (m+n-1) = 56

= m + n = 8

\(\frac{(m-n)!}{(m-n-2)!}=30\)

(m-n) (m-n-1) = 30

m – n = 6

10.\(\text { if } 5 \mathrm{P}_{\mathrm{r}}=60 \text {, }\) then the value of r is

1. 3
2. 2
3. 4
4. none of these

Answer: (1) 3 \(\frac{5!}{(5-r)!}=60\)

2 = (5 – r)! r = 3

11. \(\text { If } n_1+n_2 \mathbf{P}_2=132, n_1-n_2 \mathbf{P}_2=30\) then,

1. n 1 =6, n2=6
2. n1=10,n2=2
3. n1 = 9, n2 = 3
4. none of these

Answer: (3)

n₁=9  n₂=3

=(n₁+n₂)(n₁+n₂-1)=132

n₁+n₂=12

n₁=9

(n₁-n₂)(n₁-n₂-1)=30

n₁-n₂=6

n₂=3

12. The number of ways the letters of the word ‘COMPUTER’ can be rearranged is

1. 40,320
2. 40,319
3. 40,318
4. none of these

Answer:

40319

= Computer can be arranged in 8! Ways 8! = 40320

But we have to rearranged i.e., computer will not be counted 40320- 1 = 40319 ways.

13. The number of arrangements of the letters in the word ‘FAILURE’, so that vowels are always coming together is

1. 576
2. 575
3. 570
4. none of these

Answer: ( 576

= FAILURE = 7 letters

If vowels are clubbed as x

Total 4 letters FLR x

These can be arranged in 41 ways. Also, the vowels can arrange in 41 ways

No. of arrangement = 41 x 41= 576

14. 1 0 examination papers are arranged in such a way that the best and worst papers never come together. The number ofarrangements is

1. 9(8)!
2. 10!
3. 8(9)!
4. none of these

Answer: 89!

= Best fit worst paper be clubbed as 1 total papers =10-2 + 1 =9

9 Papers can be arranged in 91 ways

The 2 papers can be internally arranged in 2 ways No of arrangement’s = 2×91

When worst paper & best paper can be arranged

Now total arrangements = 101

Arrangement when both papers do not come together 10×91-2×91 = 9! [10-2)

= 101-2×9! = 9! x 8 = 89!

15. n articles are arranged in such a way that 2 particular articles never come together. The number of such arrangements is

1. (n-2) (n – 1]!
2. (n-1) (n-2)l
3. n!
4. none of these

Answer: ( (n – 2) (n-1)!

16. If 12 school teams are participating in a quiz contest, then the number of ways the first, second and third positions may be won is

1. 1,230
2. 1,320
3. 3,210
4. none of these

Answer:

(1320

= 12 school terms, 3 positions

Can be given & arranged in \({ }^{12} \mathrm{P}_3 \quad=\frac{12!}{9!}=12 \times 11 \times 10=1320\)

17. The sum of all 4-digit number containing the digits 2, 4, 6, 8, without repetitions is

1. 1,33,330
2. 1,22,220
3. 2,13,330
4. 1,33,320

Answer: 1,33,320.
= 4 digit number from 2,4,6,8

Without repetition can be made in 41 ways = 24 ways

24 numbers will be formed.
6 no. will end with 2
6 no. will end will 4
6 no. will end will 6
6 no. will end will 8

= 1,33,320

18. The number of 4-digit numbers greater than 5,000 can be formed out of the digits 3,4,5,6 and 7(No. digit is repeated]. The number ofsuch is

1. 72
2. 27
3. 70
4. none of these

Answer:  72

= 4 digit no. > 5000 first digit cannot be 3,4 it will be 5,6,7

3x4x3x2 =72

19. 4-digit numbers to be formed out of the figures 0, 1, 2, 3, 4 [no digit is repeated] then number of such numbers is

1. 120
2. 20
3. 96.
4. none of these

Answer: 96

= 4 digit number of 0,1,2,3,4          0 cannot occupy first digit’s place

4x4x3x2 = 96

20. The number of ways the letters of the word ‘TRIANGLE’ to be arranged so that the word ‘angle’ will be always present is

1. 20
2. 60
3. 24
4. 32

Answer: 24
Triangle has 0 letters.
If we chib angle as 1 No. of letters = 8- 5 + 1 = 4
4 letters can be arranged in 41 ways. There is no internal arrangement = 24 ways

21. If the letters word ‘DAUGHTER’ are to be arranged so that vowels occupy the odd places, then number of different words are

1. 2,880
2. 676
3. 625
4. 576

Answer: 2,880

= Daughter’s 5 consonants 3 vowels

Daughters has 4 odd places

3 vowels can occupy in 4P3 ways

First 5 consonants can be arranged in 51 ways

Total = 51 x4p3 = 2880 ways

22. The number of ways in which 7 girls form a ring is

1. 700
2. 710
3. 720
4. none of these

Answer: 720
= 7 girls form a ring = i.e. circular permutation

\(=\frac{1}{0}(n-1)!\quad=\frac{1}{0} \times 6!\quad=6 \times 5 \times 4 \times 3 \times 2=360 \times 2 \quad=720\)

23. The number of ways in which 7 boys sit in a round table so that two particular boys may sit together is

1. 240
2. 200
3. 120
4. none of these

Answer:

240
= Two come togethers <— club = total 7 – 2 + 1 = 6 boys to be arranged
= m 2 (6-1) ! = 240 ways

24. If 50 different jewels can be set to form a necklace then the number of ways is

1. 1/2×50!
2. 1/2×49!
3. 49!
4. none of these

Answer: \(\text { (} 1 / 249 \text { ! }\)

= so different vowels form a necklace in \(\frac{1}{2}(n-1)!\text { Ways }\)

\(\text { i.e. } \frac{1}{2}(50-1)!=\frac{1}{2} 49 \text { ! }\)

25. 3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together. The number of ways is

1. 70
2. 27
3. 72
4. none of these

Answer: (72
= 3 ladies 3 gents
Constraint —> only 2 & only 2 sit together 2 ladies can let selected & arranged in \({ }^3 P_2\) ways

The remaining ladies out of 4 cannot Occupy places adjustment to the 2 only in 2 ways

The remaining 3 men can occupy places in 31 ways

\(\text { Total } \rightarrow{ }^3 \mathrm{P}_2 \times 31 \times 2=72 \text { ways. }\)

26. The number of ways in which the letters of the word ‘DOGMATIC’ can be arranged is

1. 40,319
2. 40,320
3. 40,321
4. none of these

Answer: 40,320
DOGMATIC 8 letters can be arranged in 8!

27. The number of arrangements of10 different things taken 4 at a time in which one particular thing always occurs is

1. 2015
2.  2016
3. 2014
4. none of these

Answer: 2016 \(=4 \times{ }^9 \mathrm{P}_3 \text { ways to arrange }=2016\)

28. The number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is

1. 3.020
2. 3,025
3. 3,024
4. none of these

Answer: 3024 \(={ }^9 \mathrm{P}_4=3024\)

29. Mr. X and Mr. Y enter into a railway compartment having six vacant seats. The number of ways in which they can occupy the seats is

1. 25
2. 31
3. 32
4. 30

Answer: (30. = Mr x can occupy & vacant seats in 6 ways
Mr Y will occupy remaining 5 seat in 5 ways. Total 30 ways

30. The number of numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5, 6, 7 is

1. 210
2. 200
3. 110
4. none of these

Answer: 210

= 3 digit numbers > 100
From 1.2,3,4,5,6,7
first digit can be 1,2,3,4,5,6,7
Second digit can be repeated therefore ways
Similarly third digit with remaining 5 no. in 5 ways =7x6x5 = 210 ways

31. The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 is

1. 124
2. 120
3. 125
4. none of these

Answer: 120

= two digit & three digit between 10 – 1000

0 can occupy first place The no. of ways = 6x5x4 = 120 ways

32. In a group of boys, the number of arrangements of 4 boys is 12 times the number of arrangements of 2 boys. The number of boys in the group is

1. 10
2. 8
3. 6
4. none of these

Answer: 6 \(\frac{(n)(n-1)(n-2)(n-3)}{n(n-1)}=12 \quad n=6 .\)

33. The total number of 9-digit numbers of different digits is

1. 10(9)!
2. 8(9)!
3. 9(9)!
4. none of these

Answer:  99!
= first digit cannot be 0. 9 ways

= total ways = 9x9x8x7x6x5x4x3x2xl = 99!

34. The number of ways in which 6 men can be arranged in a row so that the particular 3 men sit together, is

\((1) ${ }^4 P_4$
(2) ${ }^1 P_4 \times{ }^3 P_3$
(3) $(3)^2$\)
(none of these

Answer:\(\text { (}{ }^4 \mathrm{P}_4 \times{ }^3 \mathrm{P}_3\) = let 3 particulars men be 1

Total men 6 – 3 +1 = 4. They can sit in 41 ways

Internal arrangement among 3 particular Men can be made in 31 ways.

= Total = 41 x 31 ways

35. There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B is

1. 24
2. 4 x 2
3. 5
4. none of these

Answer: (a] 24.
= S speakers

Total = 5 — 2 + 1 = 4 speakers They can speak in 41 ways.

Now there will be no internal arrangement i.e. A is before B-‘- 41 ways

36. There are 10 trains plying between Calcutta and Delhi. The number of ways in which a person can go from Calcutta to Delhi and return by a different train is

1. 99
2. 90
3. 80
4. none of these

Answer:  90.
= Let us say he takes an one train for going = 10 ways are possible

While coming he can take any train from the removing 9 trains in = 9 ways

Total = 10 x 9 = 90

37. The number of ways in which 8 sweats of different sizes can be distributed among 8 persons of different ages so that the largest sweat always goes to be younger assuming that each one of then gets a sweat is

1. 8
2. 5040
3. 5039
4. none of these

Answer: (5040
Sweats = 8 person (diffjv younger one gets largest rest 7 will get 7 sweats in 7! Ways

38. The number of arrangements in which the letters of the word ‘MONDAY’ be arranged so that the words thus formed begin with M and do not end with N is

1. 720
2. 120
3. 96
4. none of these

Answer:96.
= MONDAY 6 letters

Word should begin with M = 1 Way.

It cannot end with N, N can occupy any expect last position = 5 ways

Rest 4 letters can arrange in 4! Ways

1x1x2x3x4x4 =96 ways.

39. The total number of ways in which six V and four ’-’signs can be arranged in a line such that no two ’-’ signs occur together is

1. (7)!./ (3)!
2. (6)! x (7)!/(3)!
3. 35
4. none of these

Answer: 35 = Among 6 + here are 7 intervals

_+_+_+_+_+_+

No. of filling 7 gaps by 4’-‘signs

\(={ }^7 \mathrm{C}_4=35\)

40. The number of ways in which the letters of the word ‘MOBILE’ lie arranged so that consonants always occupy the odd places is

1. 36
2. 63
3. 30
4. none of these.

Answer: 36.

= M O B I L E = 3 consonants 3 vowels

3 consonants can be occupy places in \({ }^4 P_3 \text { ways }=31 \text { ways }\)

Vowels can occupy 3 places in \({ }^3 P_3 \text { ways }\)

= 3! = 6 total 3! x 3! = 36

41. 5 persons are sitting in a round table in such way that Tallest Person is always on the rightside of the shortest person; the number ofsuch arrangements is

1. 6
2. 8
3. 24
4. none of these

Answer: 6.

Let tallest, shortest come together

Total persons = 5- 2 + l = 4

4 People can arrange in (n-1] ways = 31 ways = 6 ways there is no internal arrangement 6 ways.

Exercise – 2 – Combinations

Choose the most appropriate option (1, 2, 3 or 4)

1. The value of \({ }^{12} \mathrm{C}_4+1{ }^{12} \mathrm{C}_3 \text { is }\)

1. 715
2. 710
3. 716
4. none of these

Answer:

715

\(\begin{aligned}
& \frac{12!}{4!8!}+\frac{12!}{3!9!} \\
& \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}+\frac{12 \times 11 \times 10}{3 \times 2 \times 1}
\end{aligned}\)

495+220=715

2. \(\text { If }{ }^n p_r=336 \text { and } n C_r=56 \text {, }\) , then n and r will be

1. (3.2)
2. (8, 3)
3. (7,4)
4. none of these

Answer:

(2) 8, 3)

\(\frac{{ }^n P_r}{{ }^n P_r}=\frac{336}{56}=\frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}}=6\)

r ! = 6 r = 3, n = 8

3. \(\text { If }{ }^{18} \mathrm{C}_{\mathrm{r}}={ }^{18} \mathrm{C}_{\mathrm{r}+2}\) the value of \({ }^r \mathrm{C}_5\) is

1. 55
2. 50
3. 56
4. none of these

Answer:

(3) 56

\({ }^{18} C_r={ }^{18} C_{r+2}\) \(=\frac{\frac{18!}{r!(18-r)!}}{\frac{18!}{(r+2)!(18-r-2)!}}=1\) \(=\frac{\frac{18!}{r!(18-r)(17-r)(16-r)!}}{\frac{18!}{(x+2)(r+1) r!(16-r)!}}=1\) \(=\frac{(r+2)(r+1)}{(18-r)(17-r)}=1\)

r² + 3r + 2 = 306 – 35 r + r2

38r= 304 = r = 8

\(8 C 5=\frac{8!}{5!3!}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56\)

4. \(\text { If }{ }^n C_{r-1}=56,{ }^n C_r=28 \text { and }{ }^n C_{r+1}=8 \text {, }\) then r is equal to

1. 8
2. 6
3. 5
4. none of these

Answer:

\(\frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{28}{8}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)^{!}(n-r-1)!}}=\frac{7}{2}\) \(\frac{\frac{n!}{r!(n-r)(n-r-1)!}}{\frac{n!}{(r+1) r!(n-r-1)!}}=\frac{7}{2}\) \(\frac{r+1}{n-r}=\frac{7}{2} \quad=2 r+r=7 n-7 r\)

9 r- 7n + 2 = 0 (1)

\(\frac{{ }^n C_r-1}{{ }^n C_r}=\frac{56}{28}\) \(\frac{\frac{n!}{(r-1)!(n-r-1)!}}{\frac{n!}{r!(n-r)!}}=2 \frac{\frac{n!}{(r-1)!(n-r-1)(n-r)!}}{\frac{n!}{r(r-1)!(n-r)!}}=\) \(\frac{r}{n-r+1}=2=r=2 n-2 r+2\)

3r- 2n = 2 (2)

Multiply (2) by 3 = 9r- 6n = 6
Subtracting (2) from (1)

9r- 7n = -2

9r – 6n = 6

-n = – 8 = n = 8, r = 6

5. A person has 8 friends. The number of ways in which he may invite one or more of them to a dinner is.

1. 250
2. 255
3. 200
4. none of these

Answer: 255

= 8 friends, lie invites one or more

Total ways

\(={ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_2+{ }^8 \mathrm{C}_3+{ }^8 \mathrm{C}_4+{ }^8 \mathrm{C}_5+{ }^8 \mathrm{C}_6+{ }^8 \mathrm{C}_7+{ }^8 \mathrm{C}_8\)

= 8 + 28 + 56 + 70 + 56 + 28 + 8 +1

= 255

6. The number of ways in which a person can chose one or more of the four electrical appliances: T.V, Refrigerator, Washing Machine and a cooler is

1. 15
2. 25
3. 24
4. none of these

Answer:  15

\(={ }^4 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^4 \mathrm{C}_3+{ }^4 \mathrm{C}_4\)

He may select 1 or more in above way =4+6+4+1=15

7. \(\text { If }{ }^n C_{10}={ }^n C_{14} \text { then }{ }^{25} C_n \text { is }\)

1. 24
2. 25
3. 1
4. none of these

Answer: 25

n = 10 + 14 = 24

25C24 \(=\frac{25!}{24!1!}=25\)

8. Out of 7 gents and 4 ladies a committee of 5 is to be formed. The number of committees such that each committee includes at least one lady is

1. 400
2. 440
3. 441
4. none of these

Answer: 441

7 gents 4 ladies committee = 5 members

Committee includes atleast 1 lady

Different cases will be 1 lady, 2 ladies, 3 ladies or 4 ladies

When 1 ladies selected

\(\rightarrow{ }^1 \mathrm{C}_1 \times{ }^7 \mathrm{C}_4=4 \times 35=140\)

When 2 ladies selected

\(\rightarrow{ }^1 \mathrm{C}_2 \times{ }^7 \mathrm{C}_3=6 \times 35=210\)

When 3 ladies selected

\(\rightarrow{ }^1 \mathrm{C}_3 \times{ }^7 \mathrm{C}_2=4 \times 21=84\)

When 4 ladies selected

\(\rightarrow{ }^4 \mathrm{C}_4 \times{ }^7 \mathrm{C}_1=1 \times 7=7\)

Total when atleast 1 or more ladies
selected = 441

9. \(\text { If }{ }^{28} C_{2 r}:{ }^{2 \cdot 1} C_{2 r-1}=225: 11 \text {, }\) then the value of r is

1. 7
2. 5
3. 6
4. none of these

Answer: (1)7

\(\frac{{ }^{28} C_{2 r}}{{ }^{24} C_{2 r-4}}=\frac{\frac{28!}{(2 r)!(28-2 r)!}}{\frac{24!}{(2 r-4)!(28-2 r)!}}=\frac{225}{11}\) \(\frac{\frac{28!}{(2 r)!}}{\frac{24!}{(2 r-4)!}}=\frac{225}{11}\) \(=\frac{28 \times 27 \times 26 \times 25}{(2 r)(2 r-1)(2 r-2)(2 r-3)}=\frac{225}{11}\)

= 24024 = (2r) (2r – l)(2r – 2)(2r – 3)

= 14 x 13 x 12 x ll = 2r (2r – 1) (2r – 2) (2r- 3)

2r = 14 -> r = 7

10. The number of diagonals in a decagon is

1. 30
2. 35
3. 45
4. none of these

Answer: (b)35

A decagon has 10 points
diagonal will join 2 points
No of ways 2 points can be selected

\(={ }^{10 \prime} \mathrm{C}_2=45 \text { lines }\)

Total lines joining 2 points = 45 (-) lines making the sides of decagon (10)

Total diagonals 35

11. There are 12 points in a plane of which 5 are collinear. The number of triangles is

1. 200
2. 211
3. 210
4. none of these

Answer: (210
-» 12 points 5 collinear

For a triangle 3 non collinear points are required

Note: if are selected any 3 points from 5 collinear they will form a line & not a trianle

Total ways = 12C3-5C3

\(=\frac{12!}{3!9!}-\frac{5!}{3!2!}\)

= 220-10 = 210

12. The number of straight lines obtained by joining 16 points on a plane, no three of them being on the same line is

1. 120
2. 110
3. 210
4. none of these

Answer: 120

-» 16 points; no collinear points for a straight line 2 points are required Total ways of obtaining a straight line

\(={ }^{16} \mathrm{C}_2=\frac{16!}{2!14!}=\frac{16 \times 15}{2}=120\)

13. At an election there are 5 candidates and 3 members are to be elected. A voter is entitled to vote for any number of candidates not greater than the number to be elected. The number of ways a voter choose to vote is

1. 20
2. 22
3. 25
4. none of these

Answer: 25

election 5 candidates, 3 are elected when voter elects 1 candidate

\(\rightarrow{ }^5 \mathrm{C}_1=5\)

when voter elects 2 candidates

\(\rightarrow{ }^5 \mathrm{C}_2=10\)

when voter elects 3 candidates

\(\rightarrow{ }^5 \mathrm{C}_3=10\)

Total ways a voter chooses to votes = 25

14. Every two persons shakes hands with each other in a party anil the total number of handshakes is 66. The number of guests in the party is

1. 11
2. 12
3. 13
4. 14

Answer: 12
Total hand shake = 66
For a hand shake 2 hands are
selected

\(\begin{aligned}
& { }^n C_2=66 \\
& \frac{n!}{2!(n-2)!}=66 \quad \frac{n(n-1)(n-2)!}{(n-2)!}=132
\end{aligned}\)

n(n-l)= 132
n (n – 1) = 12 x 11 n = 12

15. The number of parallelograms that can be formed from a set of four parallel lines intersecting anotherset of three parallel lines is

1. 6
2. 18
3. 12
4. 9

Answer: 18
Set 1 = 4 parallel lines set 2 -* 3 parallel lines are intersecting for a parallelogram 2 lines from set 1 & 2 from set 2 need to be selected

Total ways = \( { }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=6 \times 3=18\)

16. The number of ways in which 1 2 students can be equally divided into three groups is

1. 5775
2. 7575
3. 7755
4. none of these

Answer: 5775

Total ways of selecting

\(={ }^{12} C_4 \times{ }^8 C_4 \times 4 C 4\)

= 495 x 70 x 1 = 34650

So 3 groups are formed & order of selection is not important

So Total no. of ways \(=\frac{34650}{3!}=5775\)

17. 8 points are marked on the circumference of a circle. The number of chords obtained by joining these in pairs is

1. 25
2. 27
3. 28
4. none of these

Answer: 28

for a cords 2 points are required total ways = \({ }^8 \mathrm{C}_2\) = 28 ways

18. A committee of 3 ladies and 4 gents is to be formed out of 8 ladies and 7 gents. Mrs. X refuses to serve in a committee in which Mr. Y is a member. The number ofsuch committees is

1. 1530
2. 1500
3. 1520
4. 1540

Answer:  1540

Committee of 3L4G out of 8L&7G

Mrs x refuses to serve in which Mr y is member

Total committees that can be formed without taking constraint

\(={ }^8 \mathrm{C}_3 \times{ }^7 \mathrm{C}_4=56 \times 35=1960\)

Case: When x & y come together, no. of committees

2 ladies out of 7 & 3G out of 6 gents

\(={ }^7 \mathrm{C}_2 \times{ }^6 \mathrm{C}_3=21 \times 20=420\)

Therefore no. of committees when x y don’t come together

= 1960-420 = 1540

19. \(\text { If }{ }^{5000} \mathrm{C}_{92}={ }^{199} \mathrm{C}_{92}+{ }^{1{ }^{99} \mathrm{C}_91}\) then n is

1. 501
2. 500
3. 502
4. 499

Answer: (499

\({ }^{n+1} C_r={ }^n C_r+{ }^n C_{r-1}\)
n = 499, r = 92

\({ }^{500} \mathrm{C}_{92}={ }^{499} \mathrm{C}_{92}+\mathrm{n}^{\mathrm{n}} \mathrm{C}_{92-1} \quad \mathrm{n}=499\)

20. The Supreme Court lias given a 6 to 3 decision upholding a lower court; the number of ways it can give a majority decision reversing the lower court is

1. 256
2. 276
3. 245
4. 226

Answer:  256
6 to 3 decision -» upholding lower court

No. of judges = 9

To lower a decision, we need 50% T i.e. min 5 judges

No. of ways = \({ }^9 \mathrm{C}_9+{ }^9 \mathrm{C}_8+{ }^9 \mathrm{C}_7+{ }^9 \mathrm{C}_6+{ }^9 \mathrm{C}_5\)

= 1 + 9 + 36 + 84 + 126 = 256

21. Five bulbs of which three are defective are to be tried in two bulb points in a dark room. Number of trials the room shall be lighted is

1. 6
2. 8
3. 5
4. 7.

Answer: 7

5 bulbs. 3 defective

Total way of selecting

22. The ways ofselecting 4 letters from the word ‘EXAMINATION’ is

1. 136
2. 130
3. 125
4. none of these

Answer: 136

All letters distinct EXAMINATION

\(={ }^8 C_4 \text { ways }=70 \text { ways }\)

2 letters same, 2 distinct
\(\rightarrow{ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_2=63 \text { ways }\)

2 1etters alike & 2 letters alike
\(\rightarrow{ }^3 \mathrm{C}_2=3 \text { ways }\)

Total ways = 136 ways

23. \({ }^{51} \mathrm{C}_{31} \text { is equal to }\)

1. ⁵¹C₂₀
2. 2.⁵⁰C₂₀
3. 2.⁴⁵C₁₅
4. none of these

Answer: (1)

⁵¹C₂₀

—> ⁿC_r=ⁿCn-r

⁵¹C₃₁=⁵¹C_51-3

=⁵¹C₂₀

24. A candidate is required to answer 6 out of 12 questions which are divided into two groups containing 6 questions in each group. He is not permitted to attempt not more than four from any group. The number of choices are.

1. 750
2. 850
3. 800
4. none of these

Answer: (850

Total cases

1) 3 from G1 & 3 from G2

\(={ }^6 \mathrm{C}_3 \times{ }^6 \mathrm{C}_3=400\)

2) 2 from G1 & 4 from G2

\(={ }^6 \mathrm{C}_2 \times{ }^6 \mathrm{C}_4=225\)

3) 4 from G1 & 2 from G2

\(={ }^6 \mathrm{C}_4 \times{ }^6 \mathrm{C}_2=225\)

Total ways of choosing = 850

25. The results of 8 matches (Win, Loss or Draw) are to be predicted. The number of different forecasts containing exactly 6 correct results is

1. 316
2. 214
3. 112
4. none of these

Answer: 112
6 correct results can be predicted
\(\text { in }{ }^8 C_6 \text { ways }=\frac{56}{2} 28 \text { ways }\)

The remain 2 matches result are incorrect either it is lose or draw

Ml -> 2 ways M2 -» 2 ways Total 4 ways

Total ways of 6 correct exactly out of 8 matches =28×4=112

26. The number of 4 digit numbers formed with the digits 1, 1, 2, 2, 3, 4 is

1. 100
2. 101
3. 201
4. none of these

Answer: none of these

4 digit numbers 1,1,2,2,3,4

When nil (Hulls art’ different \(4 P_4\) ways – 4 ways = 24

When one digit repeated, remaining 2 different

\({ }^2 \mathrm{C}_1 \times{ }^3 \mathrm{C}_2 \times \frac{11}{2!} \text { (arrangement) }\) = 2 x 3 x 12 = 72

When double repetition occurs

\({ }^2 C_2 \times \frac{4!}{2!2!}=6 \text { ways }\)

Total = 102 ways

27. The number of ways a person can contribute to a fund out of 1 ten-rupee note, 1 five rupee note, 1 two-rupee and 1 one rupee note is

1. 15
2. 25
3. 10
4. none of these

Answer: 15

He can give any1 note in \({ }^4 \mathrm{C}_1\) ways = 4

He may contribute 2 notes in \({ }^4 \mathrm{C}_2\) = 6

He may contribute 3 noes in \({ }^4 \mathrm{C}_3\) = 4

He may contribute all four notes in \({ }^4 \mathrm{C}_4\) ways =1

Total ways = 15

28. The number of ways in which 9 things can be divided into twice groups containing 2,3, and 4 things respectively is

1. 1250
2. 1260
3. 1200
4. none of these

Answer: 1260

9 things into twice group of 2,3 & 4

\(\text { Total ways }={ }^9 \mathrm{C}_2 \times{ }^7 \mathrm{C}_3 \times{ }^4 \mathrm{C}_4\) = 1260

29. \({ }^n C_1+{ }^n C_2+{ }^n C_3+{ }^n C_1+\ldots . .+{ }^n C_n \text { equals }\)

1. 2n -1
2. 2n
3. 2n +1
4. none of these

Answer:

2n- 1

\({ }^n C_1+{ }^n C^2 \ldots \ldots \ldots \ldots \ldots{ }^n C_{n-1}+{ }^n C_n\) \(\text { we know }(1+x)^n={ }^n C_0 x^0+{ }^n C_1 x^1 \ldots \ldots \ldots\) \({ }^n C_n x^n\)

x = 1

\(\begin{aligned}
& (1+1)^n=1+{ }^n C_1+{ }^n C_2 \ldots \ldots \ldots{ }^n C_n \\
& 2^n-1={ }^n C_1+{ }^n C_2 \ldots \ldots{ }^n C_n
\end{aligned}\)

Exercise-3

(Mix Exercise – Perimilnllons & Combinations)

1. In how many ways can the letters of the word ’LEADER’ be arranged?

1. 72
2. 144
3. 360
4. 720
5. None of those

Answer: The word ‘LEADER’ contains 6 letters, namely 1 L, 2E, 1A, 1D) and IR.

Required number of ways =\(\frac{6!}{(1!)(2!)(1!)(1!))}=360\) .

2. In how many different ways ran the loiters of the word ‘RUMOUR’ ho nrmilled?

1. 80
2. 90
3. 30
4. 720
5. None of these

Answer: The word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1U.

Required number of ways =\(\frac{6!}{(2!)(2!)(1!)(1!)}=180\)

3. How many words can he formed from the letters of the word ‘SIGNATURE’ so that the vowels always come together?

1. 720
2. 1440
3. 2880
4. 3600
5. 17280

Answer:

The word ‘SIGNATURE’ contains 9 different letters.

When the vowels IAUE are taken together, they can be supposed to form an entity, treated as one letter.

The, the letters to be arranged are SNGTR (IAUE)

These 6 letters can be arranged in \({ }^6 P_6=6!=720 \text { ways. }\)

The vowels in the group (IAUE) can be arranged amongst themselves in \({ }^4 P_4=4!=24 \text { ways. }\)

Required number of words = (720 * 24) = 17280.

4. In how many different ways can the letters of the word ‘OPTICAL’ he arranged so that the vowels always together?

1. 120
2. 720
3. 4320
4. 2160
5. None of these

Answer:

The word OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5 ! = 120 ways.

The vowels (OLA) can be arranged among themselves in 3 ! = 6 ways.

Required number of ways = (120 x 6) = 720.

5. In how many different ways can the letters of the word ’SOFTWARE’ he arranged in such a way that the vowels come together?

1. 120
2. 300
3. 1440
4. 13440
5. 720

Answer:

The word ‘SOFTWARE’ contains 8 different letters.

When the vowels OAE are always together, they can be supposed to form one letter.

Thus, we have to arrange the letters SFTWR (OAE).

Now, 5 letters can be arranged in 6 ! = 720 ways

The vowels (OAE) can be arranged among themselves in 3 ! = 6 ways.

Required number of ways = (720 x 6) = 4320.

6. In how many different ways can the letters of the word ‘LEADING’ he arranged in such a way that the vowels always come together?

1. 360
2. 480
3. 720
4. 5040
5. None of these

Answer: The word LEADING’ has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LDNG (EAD.

Now, 5 letters can be arranged in 5 ! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3 ! = 6 ways.

Required number of ways = (120 x 6) = 720.

7. In how many different ways can the letters of the word ‘JUDGE’ be arranged in such a way that the vowels come together?

1. 48
2. 124
3. 160
4. None of these

Answer:

The word JUDGE’ has 5 different letters.

When the vowels UE are always together, they can be supposed to form one letter.

Then, we have to arrange the letters JDG (UE).

Now, 4 letters can be arranged in 4 ! = 24 ways.

The vowels (UE) can be arranged among themselves in 2 ! = 2 ways.

Required number of ways = (24 x 2) = 48.

8. In how many different ways can the letters of the word ‘AUCTION’ be arranged in such a way that the vowels always come together?

1. 30
2. 240
3. 144
4. 576
5. None of these

Answer: The word ‘AUCTION’ has 7 different letters.

When the vowels AUIO are always together, they can be supposed to form one letter.

Then, we have to arrange the letters CTN (AUIO).

Now, 4 letters can ho an.ingcd in 4 1 =ÿ 24 ways.

The vowols (AUIO) ran be nrmngi’d among themselves In 4 I = 24 ways,

Required number of ways (24 x 24} – 576.

9. In how many different ways can the letters of the ‘BANKING’ he arranged so that the vowels always come together?

1. 120
2. 240
3. 360
4. 540
5. 720

Answer: In The word ’HANKING’, wo Heat the two vowels Al as one letter. Thus, we have UNKNG (Al).

This has 6 letters of which N occurs 2 times and the rest are different.

Number of ways of arranging these letters =\(\frac{6!}{(2!)(1!)(1!)(1!)(1!)}=360\)

Now, 2 vowels Al can be arranged in 2 ! = 2 ways.

Required number of ways = (360 »2) = 720.

10. In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

1. 810
2. 1440
3. 2880
4. 50400
5. 5760

Answer:

In the word ’CORPORATION’, we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO).

This has 7 letters of which R occurs 2 times and the rest are different.

Number of ways of arranging these letters =

\(=\frac{7!}{2!}=2520 /\)

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in \(\frac{5!}{3!}\) = 20 ways.

Required number of ways = (2520 * 20) = 50400.

11. In how many different ways can the letters of the word ‘DETAIL’ be arranged in such a way that the vowels occupy only the odd positions?

1. 32
2. 48
3. 36
4. 60
5. 120

Answer: There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6)

Now, 3 vowels can be placed at any of the three places out of 4, marked 1, 3, 5.

Number of ways of arranging the vowels =

\({ }^3 P_3=3!=6 .\)

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways f these arrangements =

\({ }^3 P_3=3!=6 .\)

Total number of ways = (6 * 6) = 36.

12. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

1. 63
2. 90
3. 126
4. 45
5. 135

Answer:

Required number of ways = \(\left({ }^7 C_5 \times{ }^3 C_2\right)=\left({ }^7 C_2 \times{ }^3 C_1\right)=\left(\frac{7 \times 6}{2 \times 1} \times 3\right)=63\)

13. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

1. 266
2. 5040
3. 11760
4. 86400
5. None of these

Answer:

Required number of ways=

(⁸C₅x¹⁰C₆)= (8C₅x¹⁰C₄)= \(\left(\frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}\right)=11760 .\)

14. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways car it do done?

1. 564
2. 645
3. 735
4. 756
5. None of these

Answer:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only)

Required number of ways = \(\left({ }^7 \mathrm{C}_3 \times{ }^6 \mathrm{C}_2\right)+\left({ }^7 \mathrm{C}_4 \times{ }^6 \mathrm{C}_1\right)+\left({ }^7 \mathrm{C}_5\right)\)

\(=\left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{6 \times 5}{2 \times 1}\right)+\left({ }^7 C_3 \times{ }^6 C_1\right)+\left({ }^7 C_2\right) \quad=525+\left(\frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 6\right)+\left(\frac{7 \times 6}{2 \times 1}\right)\)

= (525 + 210 + 21) = 756

15. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

1. 159
2. 194
3. 205
4. 209
5. None of these

Answer: We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways =\(\left({ }^6 \mathrm{C}_1 \times{ }^4 \mathrm{C}_3\right)+\left({ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2\right)+\left({ }^6 \mathrm{C}_3 \times{ }^4 \mathrm{C}_1\right)+\left({ }^6 \mathrm{C}_4\right)\)

\(=\left({ }^6 \mathrm{C}_1 \times{ }^1 \mathrm{C}_1\right)+\left({ }^6 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2\right)+\left({ }^6 \mathrm{C}_3 \times{ }^1 \mathrm{C}_1\right)+\left({ }^6 \mathrm{C}_2\right)\) \(=(6 \times 4)+\left(\frac{6 \times 5}{2 \times 1} \times \frac{4 \times 3}{2 \times 1}\right)+\left(\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4\right)+\left(\frac{6 \times 5}{2 \times 1}\right)\)

= (24 + 90 + 80 + 15) = 209.

16. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways-can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

1. 32
2. 48
3. 64
4. 96
5. None of these

Answer:

We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = \(\left({ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2\right)+\left({ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1\right)+\left({ }^3 \mathrm{C}_3\right)\)

\(=\left(3 \times \frac{6 \times 5}{2 \times 1}\right)+\left(\frac{3 \times 2}{2 \times 1} \times 6\right)+1\) =45 + 18 + 1 = 64

17. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

1. 5
2. 10
3. 15
4. 20

Answer:

Since each desired number Is divisible by 5, so we must have 5 at the unit place, So, there is 1 way of doing it.

Tens place can be filled by any of the remaining fi numbers.

So, there are S ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 × 5 × 4) = 20.

18. Evaluate:\(\frac{30!}{28!}\)

Answer: \(\text { We have, } \frac{30!}{28!}=\frac{30 \times 29 \times(28!)}{28!}=(30 \times 29)=870\)

19. Find the value of \(\text { (1) }{ }_{60} P_3 \text { (2) }{ }_4 P_4\)

Answer:

\(\text { (1) }{ }^{60} P_3=\frac{60!}{(60-3)!}=, \frac{60!}{57!}=\frac{60 \times 59 \times 58 \times(57!)}{57!}=(60 \times 59 \times 58)=205320 \text {. }\) \(\text { (b) }{ }^4 P_4=4 \text { ! }=(4 \times 3 \times 2 \times 1)=24 \text {. }\)

20. Find the value of

\((1) { }^{10} \mathrm{C}_3
(2) { }^{100} \mathrm{C}_{98}
(3) { }^{50} \mathrm{C}_{50}\)

Answer:

\(\text { (1) }{ }^{10} \mathrm{C}_3=\frac{10 \times 9 \times 8}{31}=\frac{10 \times 9 \times 8}{3 \times 2 \times 1}=120 \text {. }\) \(\text { (2) }{ }^{100} \mathrm{C}_{98}={ }^{100} \mathrm{C}_{(100-98)}={ }^{100} \mathrm{C}_2=\frac{100 \times 99}{2 \times 1}=4950 \text {. }\) \(\text { (3) }{ }^{50} C_{50}=1 .\left[{ }^n C_n=1\right]\)

21. How many words can be formed by using all letters of the word ‘BIHAR?

Answer: The word BIHAR contains 5 different letters.

Required number of words = 5P5 – 5! = (5x4x3x2xl) = 120.

22. How many words can be formed by using all the letters of the word ‘DAUGHTER’ so that the vowels always come together?

Answer:

Given word contains 8 different letters when the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.

Then, the letters to be arranged are DGHTR (AUE).

These 6 letters to be arranged in 6Pfl = 6! = 720 ways.

The vowels in the group (AUE) may be arranged in 3! = 6 ways.

Required number of words = (720 x6) = 4320.

23. How many words can be formed from the letters of the word ‘EXTRA’ so that the vowels are never together?

Answer: The given word contains 5 different letters.

Taking the vowels EA together, we treat them as one letter.

Then, the letters to be arranged are XTR (EA)

These letters can be arranged in 4! = 24 ways.

The vowels EA maybe arranged amongst themselves in 2! =2 ways.

Number ofwords, each having vowels together = (24 x 2) = 48.

Total number ofwords formed by using all the letters of the given words = 5!= (5x4x3x2xl) = 120.

Number of words, each having vowels never together = (120 – 48) =72.

24. How many words can be formed from the letters of the word “DIRECTOR’ so that the vowels are always together?

Answer:

In the given word, we treat the vowels 1E0 as one letter

Thus, we have DRCTR (IEO).

This group has 6 letters of which R dccurs 2 times and others are different.

Number of ways of arranging these letters=\(\frac{6!}{2!}\) =360.

Now 3 vowels can be arranged among themselves in 3! = 6 ways.

Required number of ways = (360 x 6) = 2160.

25. In how many ways can a cricket eleven be chosen out of a batch of 15 players?

Answer:

Required number of ways = \({ }^{15} \mathrm{C}_{11}={ }^{15} \mathrm{C}_{(15.11)}={ }^{15} \mathrm{C}_4\)

\(=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365 \text {. }\)

26. In how many ways. A committee of 5 members can be selected from 6 men and 5 Ladies, consisting of 3 men and 2 ladies?

Answer:

(3 men out 6) and (2 ladies out of 5) are to be chosen.

Required number of ways =

\(\left({ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2\right)=\left(\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)=200 .\)

27. The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line, is:

1. 185
2. 175
3. 115
4. 105

Answer:

(1) No. of triangles =\(12 c_3-7 c_3\)

\(=\frac{12!}{3!9!}-\frac{7!}{314!}=220-35 \quad=185\)

28. A code word is to consists of two English alphabets followed by two distinct number from 1 to 9. How many such code words are there?

1. 6,15,800
2. 46,800
3. 7,19,500
4. 4,10,800

Answer:

(2) No. of code words = \(26 p_{p_2} \times{ }^9 \mathrm{P}_2\)

\(=\frac{26.25 \cdot 24!}{24!} \times \frac{9!}{7!}=650 \times 72=46,800\)

29. A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow mathematics part-II unless mathematics part-I is also borrowed? In how many ways can he choose the three books to be borrowed?

1. 41
2. 51
3. 61
4. 71

Answer:

30. A supreme court bench consists of 5 judges. In how many ways, the bench can give a majority division?

1. 10
2. 5
3. 15
4. 16

Answer:

\(\text { (d) } \quad=5 c_3+5 c_4+5 c_5=10+5+1=16\)

31. Given: P (7,k) = 60 P (7, k- 3], Then:

1. k=9
2. k=8
3. k=5
4. k=0

Answer: (3]

P(7,K] = 60.p(7,k-3]

\(\frac{7!}{(7-k)!}=60 \times \frac{7!}{\mid 7-(k-3)!} \quad \text { or } \frac{1}{(7-k)!}=\frac{60 \times 1}{|7-(k-3)|!}\) \(\text { or } \frac{1}{(7-k)!}=\frac{60}{(10-k)(9-k)(8-k)(7-k)!}\)

Or(l0-k) (9-k) (8-k] = 60

(Make 3 factors of 60 so that their differences are 1.)

Comparing on both sides: we get

Or(10-k) (0-k) (8-k] = 5.4.3

10-k=5 Or k= 10-5 = 5 k = 5

32. The number of ways in which n books can be arranged on a shelfso that two particular books are not together is:

1. (n-2) x (n-l)l
2. (n-1) x (m+1)!
3. (n-2) x (n-l)l
4. (n-2) x (n+1)!

Answer:

Total no. of ways to arrange book as Qts. = n! – (n-2+l]!.2!

= n! – (n-2)! 2! n(n-l)! -2(n-l)! = (n-l)!. (n-2)!

33. In how many ways can the letters of the word FAILURE be arranged so that the consonants may occupy only old positions?

1. 576
2. 476
3. 376
4. 276

Answer: (1)

Consonants = F, L, R

Vowels = A, I, U, E

Total letters = 7

There will be 7 positions of letters

Odd positions are = 1,3,5,7

Even positions are = 2,4,6

From Questions,

No. of words = \(4 p_3 \cdot 4!=4!\times 4!\)

= 24 x 24 = 576

34. Five bulbs of which three are defective are to be tried in two lights-points in a dark-room. In how many trials the room shall be lighted?

1. 10
2. 7
3. 3
4. None of these

Answer: (2)

No. of bulbs = 5
Defective bulbs = 3
Non-defective bulbs = 2
No. of point in the room = 2

No. of ways to light the darkroom = \(3 c_1 \cdot 2 c_1+3 c_0 \cdot 2 c_2\)

= 3 x 2 + 1 x l = 6 + 1 = 7

35. In how many ways can a parly of 4 men and 4 women be seated at a circular table, so that no two women are adjacent?

1. 164
2. 174
3. 144
4. 154

Answer: (3)

No. of ways to sit 4 women = 4! = 24

Total ways to arrange = 6×24 = 144

36. \(\text { If }{ }^6 P_r=24{ }^6 C_r \text {, then find } r \text { : }\)

1. 4
2. 6
3. 2
4. 1

Answer: 1

\(6 p_r=24.6 c_r\) \(O r \frac{6!}{(6-r)!}=24 \cdot \frac{6!}{r!(6-r)!}\)

Or r!=24 or r!=4!

r = 4

37. Find the number of combinations of the letters of the word COLLEGE taken four together:

1. 18
2. 16
3. 20
4. 26

Answer: (1)

Letters are C, 0, L, L, E, E, G

Case i: 2 same & 2 same letters

= LIFE = Only 1 combination

Case II: LL and 2 out of C, 0, E, G \(=1 \times 4 c_2=6\)

EE and any 2 out of C, 0, L, G \(=1 \times 4 c_2=6\)

Case III: All different letters = Any 4 out of C, 0, L, E, G \(=5 c_4=5\)

Total no. of combinations = l+(6+6) +5 = 18

38. How many words can be formed with the letters of the word ‘ORIENTAL’ so that A and E always occupy odd places:

1. 540
2. 8640
3. 8460
4. 8450

Answer: (2)

As Qts. No. of words = \(4 p_2 \times 6!\) [because there are 4 Positions 1,3,5,7 for A & E)

\(=\frac{4!}{(4-2)} \times 720=12 \times 720=8640\)

39. \(\text { If }{ }^{1000} \mathrm{C}_{98}={ }_{999} \mathrm{C}_{97}+\mathrm{x}_{998} \text {, find } \mathrm{x} \text { : }\)

1. 999
2. 998
3. 997
4. 1000

Answer: (1) Tricks: go by choices

For (1) RHS = \(\left.{ }_{999} C_{97}+999 C_{98} ; \text { if } x=999\right)\)

\(=999{ }_{C_{97}}+999{ }_{C_{98}}=1000 c_{98}=L H S\)

40. How many number greater than a million can be formed with digits 4,5,5,0,4,5,3?

1. 260
2. 360
3. 280
4. 380

Answer: (2)

No. of Numbers \(=\frac{7!}{3!\cdot 2!}-1 \times \frac{6!}{3!.2!}\)

[Zero is fixed at 1st position with permutation 1]

=420-60 = 360

41. A building contractor needs three helpers and ten men apply. In how many ways can these selections take place?

1. 36
2. 15
3. 150
4. 120

Answer: (4)

Selection of 3 helpers out of10

\(10_{C_3}=\frac{10!}{3!(10-3)!}=\frac{10.9 .8 .7!}{3.2 \cdot 1.7!}=120\)

42. There are three blue balls, four red balls and five green balls. In how many ways can they be arranged in a row?

1. 26,720
2. 27,720
3. 27,820
4. 26,620

Answer: [2] No. ofarrangements = \(\frac{12!}{3!.4!5!}=27,720\)

43. If C(n,r): C(n, r+1) =1:2 and C(n, r+1) C(n, r + 2) = 2:3, determine the value of n and r:

1. (14,4)
2. (12,4)
3. (14,6)
4. None

Answer: (1)

Tricks: Go by choices

n = 14; r=4

44. Six seats of articled clerks are vacant in a ‘chartered accountant firm’ how many different batches of candidates can be chosen out of ten candidates?

1. 216
2. 210
3. 220
4. None

Answer: (2)

No. of selections = \({ }^{10} \mathrm{C}_6=\frac{10!}{6!4!}=210\)

45. Six person A, B, C, D, E and F are to be seated at circular table. In how many ways can this be done, if A must always have either B or C on his right and B must always have either C or D on his right?

1. 3
2. 6
3. 12
4. 18

Answer: (4)

B or C is right side of A

Arrangements are AB, AC,

C or Dis right side of B.

Their arrangement are BC or BD

Arrangement can be made as

Case I ABC, D,E,F can be arranged as [4-1]! = 6

Case II ABD C,E,F can be arranged as = [4-1]! = 6

Case III AC, BD, E,F can be arranged as = (4-1)! = 6

∴ total ways to arranged 6 persons = 6+6+6= 18

46. How many six-digit telephone numbers can be formed by using 10 distinct digits?

1. 106
2. 610
3. \({ }^{10} \mathrm{C}_6\)
4. \({ }_{10}^{10} \mathrm{P}_6\)

Answer: (4) Is correct

Total no. of (Units = 10 i.c. they are 0,1,2,3…………..;9

NO of six (I if,it distinct telephone no.

Being distinct dibits = \({ }^{10} P_6\)

47. In how many ways a committee of 6 members can be formed from a group of 7 boys and 4 girls having at least 2 girls in the committee.

1. 731
2. 137
3. 371
4. 351

Answer: 3

48. Number of ways of painting a face of a cube by 6 colours is

1. 36
2. 6
3. 24
4. 1

Answer: (2)

No. of faces of a cube = 6

No. of colours = 6

No, of ways to point 1 face =\({ }^6 \mathrm{C}_1=6\)

49. \(\text { If }{ }^{18} \mathrm{C}_{\mathrm{r}}={ }^{18} \mathrm{C}_{\mathrm{r}+2} \text { find the value of } \mathrm{r}_5 \text {. }\)

55
50
56
None of these

Answer:

\(\text { (3) If }{ }^{18} \mathrm{C}_{\mathrm{r}}={ }^{18} \mathrm{C}_{\mathrm{r}+2}\)

r+r+2 = 18

Or 2r= 16 r= 8

\({ }^r C_5={ }^8 C_5=56\)

50. 7. Books are to be arranged in such a way so that two particular books are always at first and last place. Final the number of arrangments.

1. 60
2. 120
3. 240
4. 480

Answer: (3)

Arrangement of 2 particular books = 2! = 2

i.e. either at lsl place or at last place

Arrangement of rest 5 books = 5! = 120

Total arrangements = 2×120 = 240

51. Find the number of arrangements in which the letters of the word ‘MONDAY’ be arranged so that the words thus formed begin with ‘M1 and do not end with ‘N’

1. 720
2. 120
3. 96
4. None

Answer: (3)

MONDAY

Total no. of words which begins with M = (6-1)! = 120

Total no. of words which begins M and ends With N

Total no. of words which begins with M but do not end with N

=1×4!x1 = 24

= 120-24 = 96

52. In how many ways can 17 billiard balls be arranged if 7 of them are black, 6 red and 4 white?

1. 4084080
2. 1
3. 8048040
4. None of these

Answer:

(1) Total no. of arrangements

\(=\frac{17!}{7!.6!\cdot 4!}=\frac{17.1615 .14 .13 .12 .11 .10 .9 .8 .7!}{7!.6!.4!}=4084080\)

=4084080

53. (n+1)! = 20(n-l)!, Find n

1. 6
2. 5
3. 4
4. 10

Answer: (3)

(n+1) I = 20(n-l)l Or (n+1) n.(n-l)! = 20(n-l)! Or (n+l).n = 5 x 4

n = 4

54. Out of 4 gents and 6 ladies, a committee is to be formed. Find the number of ways the committee can be formed such that it comprises of at least 2 gents and at least the number of ladies should be double of gents.

1. 94
2. 132
3. 136
4. 104

Answer:

55. In how many ways can the letters of ‘REGULATION’ be arranged so that the vowels come at odd places?

1. 14,400
2. 1,144
3. 1,44,252
4. None of these

Answer: (1)

REGULATION

Consonants = R,G,L,T,N

Vowels = E,U,A,l,0

Odd places are 1,3,5,7,9

Arrangement of vowels at these Places = ⁵P₅ = 5! = 120

Consonants will be arranged at remaining places = 5! = 120

Total No. ofarrangments = 120 x 120 = 14,400

56. Six points are on a circle. The number of quadrilaterals that can be formed are:

1. 30
2. 360
3. 15
4. None

Answer: (3)

No. of Quadrilaterals

⁶C₄ =\(\frac{6!}{4!2!}=\frac{6.5 .4!}{4!2!}=15\)

57. The number of ways of arranging 6 boy and 4 girls in a row so that all girls are together is:

1. 61.4!
2. 2(71.4!)
3. 7!.4!
4. 2.(6!.4!)

Answer: (3)

No of arrangement of 6 boys and 4 girls so that all 4 girls are together (6+1)1.41 =71.41 = 120960.

58. How many numbers not exceeding 1000 can be made from the digits 1,2,3,4,5,6,7,8,9 if repetition is not allowed

1. 364
2. 585
3. 728
4. 819

Answer: (2)

Given digits = 1,2,3,………9

Number less than 1000 will be of1 digit 2 digits and of 3 digits.

Total no. of numbers = \({ }^9 \mathrm{P}_1+{ }^9 \mathrm{P}_2+{ }^9 \mathrm{P}_3\)

= 9x9x8+9x8x7 = 585

59. A garden having 6 tall trees in a row. In how many ways 5 children stand, one in a gap between the trees in order to pose for a photograph?

1. 24
2. 120
3. 720
4. 30

Answer: (2)

1*2*3*4*5*6

Clearly there will be 5 positions for children = ⁵P₅ = 120

60. \({ }^{15} \mathrm{C}_3+{ }^{15} \mathrm{C}_3 \text { is equal to: }\)

1. \(16_{c_3}\)
2. \(30_{c_{16}}\)
3. \(15_{c_{16}}\)
4. \(15_{c_{15}}\)

Answer: (1)

\({ }^{15} \mathrm{C}_3+{ }^{15} \mathrm{C}_2={ }^{16} \mathrm{C}_3\)

61. How many ways a team of11 players can be made out of15 players if one particular player is not to be selected in the team.

1. 364
2. 728
3. 1,001
4. 1,234

Answer:

(1) No. of ways to make a 11 – member teams

\(={ }^{15-1} C_{11}={ }^1 C_{11}=\frac{14!}{11!3!}=364\)

62. Find the number of arrangements of 5 things taken out of 12 things, in which one particular thing must always be included.

39,000
37,600
39,600
36,000

Answer: (3)

No. of arrangement of 5 things =

\({ }^{12-1} C_{5-1} \cdot 5!={ }^{11} C_4 \cdot 5!\) \(=\frac{11!}{4!.7!} \times 120=330 \times 120=39600\)

63. In how many ways 3 Prizes out of 5 can be distributed amongst 3 brother equally?

1. 10
2. 45
3. 60
4. 120

Answer: (3)

No. of ways = \({ }^5 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_1=5.4 .3=60\)

64. There are 12 question are to be answered to be Yes or No. how many ways can these be answered?

1. 1024
2. 2048
3. 4096
4. None

Answer: (3)

No. of ways = 2¹² = 4096

65. The letters the word VIOLENT are arranged so that the vowels occupy even place only. The number of permutations is

1. 144
2. 120
3. 24
4. 72

Answer: (1) Vowels = I, 0, E Consonants = V, L, N, T

No. of perms of vowels = ³P₃ = 3! = 6

Total no. of words = 4! x 3! = 24 x 6 = 144

66. \(\text { If } n P_4=20\left(n P_2\right)\) then the value of ‘n’ is___________

1. -2
2. 7
3. -2 and 7 both
4. None of these

Answer: (2) Note: n is always positive GO by choices

67. A man lias 3 sons and 6 schools within his roach, in how many ways, ho can send them to school, if no two of his sons aro to road in the some school?

1. \({ }_6 P_2\)
2. \({ }_6 P_3\)
3. 6³
4. 3⁶

Answer: (2)

No. of ways = ⁶P₃

68. How many permutations can be formed the letters of the word “DRAUGHT”, if both vowels may not be separated?

1. 720
2. 1,440
3. 140
4. 1,000

Answer: (2)

Total Perms. Of DRAUGHT = 61.2! = 720 x 2 = 1440.

69. \(\text { If }{ }^{13} \mathrm{C}_6+2{ }^{13} \mathrm{C}_5+{ }^{13} \mathrm{C}_4={ }^{15} \mathrm{C}_{\text {}} \text { then, } \mathrm{x}=\)

1. 6
2. 7
3. 8
4. 9

Answer:

\({ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_5+{ }^{13} \mathrm{C}_5+{ }^{13} \mathrm{C}_4={ }^{15} \mathrm{C}_{\mathrm{x}}\) \(\text { Or }{ }^{13} \mathrm{C}_6+{ }^{14} \mathrm{C}_5={ }^{15} \mathrm{C}_{\mathrm{x}}\) \(\text { Or }{ }^{15} \mathrm{C}_6+{ }^{14} \mathrm{C}_5={ }^{15} \mathrm{C}_{\mathrm{x}}\) \(\text { Or }{ }^{15} \mathrm{C}_6={ }^{15} \mathrm{C}_x\)

x=6

70. The total number ofshake hands in a group of10 persons to each other are.

1. 45
2. 54
3. 90
4. 10

Answer: (1)

Total no. of handshakes = ¹⁰C₂ = 45

71. A regular polygon has 44 diagonals then the No. of sides are

1. 8
2. 9
3. 10
4. 11

Answer: (4)No. of Diagonals in a polygon of “n” sides = ⁿC₂– n = 44

‘Tricks: Go by choices.

\(\text { For (a) }{ }^8 C_2-8=28-8 \neq 44\) \(\text { For (b) }{ }^9 \mathrm{C}_2-9=36-9 \neq 44\) \(\text { For }(\mathrm{c}){ }^{10} \mathrm{C}_2-10=45-10 \neq 44\) \(\text { (d) }{ }^{11} \mathrm{C}_2-11=55-11=44\)

(4) is correct

72. In how many ways to word “ARTICLE” can be arranged in a row so that vowels occupy even places?

1. 132
2. 144
3. 72
4. 160

Answer: (2)

In word ARTICLE

Vowels = A,I,E

Positions lor vowels = 2,4,6

Vowels can hr arranged In 3! = 6 ways

Rest letters can lie arranged in 4! = 24 ways

Total No. ofsuch arrangements = 6×24 = 144

73. How many different words can be formed with the letters of the word “LIBERTY”

1. 4050
2. 5040
3. 5400
4. 4500

Answer: (2)

LIBERTY

No. of words = 7! = 5040

74. In how ways can a family consist of 3 children have different birthday in a leap year.

1. 366x365x364
2. ³⁶⁶C₆
   
3. ³⁶⁵C₃
4. ³⁶⁶C₃-3

Answer:

(1) 1 Leap year = 366

No. of ways = \({ }^{366} \mathrm{C}_1 .{ }^{365} \mathrm{C}_1 .{ }^{36.1} \mathrm{C}_1\)

= 366x365x364

75. \(\text { If }{ }^{15} \mathrm{C}_{3 \mathrm{r}}={ }^{15} \mathrm{C}_{\mathrm{r}+3} \text { then } \mathrm{r}=\)

1. 2
2. 3
3. 4
4. 5

Answer:

\(

2r =3; so, r = 3/2(In fraction; so invalid)

0r3r+ r+ 3 = 15 0r 4r = 12; so, r=3

76. If 6 times the no. of permutations of n items taken 3 at a times is equal to 7 times the no. of permutations of (n-1) items taken 3 at a time then the value of n will be

1. 7
2. 9
3. 13
4. 21

Answer:

[latex]6 .{ }^n P_3=7 .(\mathrm{n}-1) \mathrm{P}_3 \text { (given) }\) \(\text { Or;6 } \frac{n!}{(n-3)!}=7 \cdot \frac{(n-1)!}{(n-1-3)!}\) \(\text { Or; } \frac{6 . n \cdot(n-1)!}{(n-3)(n-4)!}=\frac{7(n-1)!}{(n-4)!}\) \(\text { Or } \frac{6 n}{n-3}=7\)

Or 6n = 7n – 21 Or n = 21

77. \(\text { If }{ }^{1000} \mathrm{C}_{98}={ }^{999} \mathrm{C}_{97}+{ }^x \mathrm{C}_{98}\) then value of x will be

1. 999
2. 998
3. 997
4. None

Answer:

(1) Tricks: Go by choices

For (1) RHS = \({ }_{1000} \mathrm{C}_{97}+{ }^{999} \mathrm{C}_{901}\)

\(={ }^{999} \mathrm{C}_{97}+{ }_{999} \mathrm{C}_{98}\left[\mathrm{n}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]\) \(={ }^{1000} \mathrm{C}_{98} \text { (L.H.S) }\) \({ }^n C_r={ }^n C_{r-1}={ }^{n+1} C_r\)

78. \({ }^{{ }^G P_r}=360 \text { then find } r\)

1. 4
2. 5
3. 6
4. None

Answer: (1) Tricks: go by choices:

⁶P_r =360 LHS at r = 4 = ⁶P₄ = 6.5.4.3 = 360 R.H.S.

79. If 5 books of English 4 books of Tamil and 3 books of hindi are to be arranged in a single row so that books of same language come together

1. 1,80,630
2. 1,60,830
3. 1,03,680
4. 1,30,680

Answer: (3) Total no. of ways so that same language Books remain together

= 5 !. 4 !. 3 !. 3 !. = 120 x 24 x 6 x 6 = 1,03,680

80. 5 boys and 4 girls are to be seated in row. If the girls occupy even places then the no. of such arrangements

1. 288
2. 2808
3. 2008
4. 2880

Answer: (4)

Total no. ofstudents = 9

For girls’ position may be 2,4,6,8,

Total no. of arrangements of girls ⁴P₄ = 4! = 24

For boys = 5! 4! = 120×24 = 2880

81. A Person has 10 friends of which 6 of them are relatives, He wishes to Invite 5 persons so that 3 of them are relatives. In how many ways he can Invites?

1. 450
2. 600
3. 120
4. 810

Answer:

Friends = 10

Relatives = 6;

so, rest = 4 friends

Selection of 5 in which 3 are relatives = ⁶C₃. 4C₂ = 20×6 = 120 (c) is correct

82. A student has 3 books on computer, 3 books on economics, 5 on commerce. If these books are to be arranged subject wise then these can be placed on a sbelf in the_ number, of ways.

1. 25,290
2. 25,920
3. 4,230
4. 4,320

Answer: (1)

Total ways = 3!x3!x5!x3! = 6 x 6 x 120 x 6 = 25920

83. The number of 4-digit members that can be formed from seven digits 1,2,3,5,7,8,9 such that no digit being repeated in any number, which are greater than 3000 are

1. 120
2. 480
3. 600
4. 840

Answer: For detail see Quicker BMLRS

At 1st place 3,5,7,8,9 these 5 digit are suitable.

So, permutation of 1st place = 5

One of them will be used at that place.

Remaining 6 digits are suitable for next place.

Similarly doing as above, Total ways = 5.6,5.4 = 600

84. A question paper consists 10 questions, 6 in math and 4 in stats. Find out number of ways to solve question paper ifat least one question is to be attempted from each section.

1. 1024
2. 950
3. 945
4. 1022

Answer: (3)

No. of ways to attempt at least one from each (2⁶-1 ) 2⁴-1) = 945.

85. \({ }^{n P_r}=720 \text { and }{ }^n C_r=120 \text { find } r \text { ? }\)

1. 6
2. 4
3. 3
4. 2

Answer: (3)

ⁿP_r=ⁿC_r!=720

120. r! = 720 r!= 6 = 3! r= 3

86. There are 10 students in a class, including 3 girls. The number of ways to arrange them in a row, when any two’ girls out of them never come together

1. \(8_{p_3} \times 7!\)
2. \(3_{p_3} \times 7!\)
3. \(8_{p_3} \times 10!\)
4. None

Answer: (1)

Rost students = 7

Total no. of permutations of rest 7 stds. = 7!

Total no. of place for girls = 7 + 1 = 8. Total ways to arrange girls = ⁸P₃ .(7!)

87. In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include at least two teachers?

1. 220
2. 672
3. 896
4. 968

Answer:

88. The maximum number of points of inter section of10 circles will be

1. 2
2. 20
3. 90
4. 90

Answer: (3)

To intersect 2 circles out of 10 are required

= \({ }^{10} \mathrm{C}_2\) x 2 = 45 x 2 = 90 Nos. of ways

(Each circle will cut at 2 points)

89. How many numbers between 1,000 and 1 0,000 can be formed with the digits 1, 2, 3, 4, 5, 6

1. 720
2. 360
3. 120
4. 60

Answer:

(2) is correct.

Total no. of numbers = \({ }^{(} p_4=6.5 .4 .3=360\)

90. \(\text { If }{ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=8: 3: 1 \text {; }\) then find the value of n.

1. 14
2. 15
3. 16
4. 17

Answer: (2) is correct

Formula,\({ }^n C_r:{ }^{n-1} C_{r-1}=\frac{n}{r}=\frac{3}{1} ; \text { So, } n=3 r\)

\(\&{ }^{n+1} C_{r+1}:{ }^n C_r=\frac{n+1}{r+1}=\frac{8}{3} ;\)

So, 3n + 3 = 8r + 8

Or, 3 x 3r+ 3 = 8r + 8

Or, r= 8- 3 = 5. Hence, n = 3r = 3 x 5 = 15

91. In how many ways 4 members can occupy 9 vacant seats in a row

1. 3204
2. 3024
3. 49
4. 94

Answer: (2) is correct. Total ways =  ⁹P₄

= 9.8.7.6 = 3024

92. The number of arrangements that can be formed from the letters of the word ‘ALLAHABAD”

1. 7560
2. 3780
3. 30240
4. 15320

Answer:

\(\frac{9!}{(4!) \cdot(2!)}=\frac{9.8 \cdot 7 \cdot 6 \cdot 5(4!)}{(4!) \cdot 2 \cdot 1}=7560\)

Option (1) is correct

93. \(\text { If }{ }^{10} \mathrm{C}_3+2 \cdot{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5={ }^{11} \mathrm{C}_5\) then the value of n = _______

1. 10
2. 11
3. 12
4. 13

Answer:

\({ }^{10} \mathrm{C}_3+2 \cdot{ }^{10} \mathrm{C}_4+{ }^{10} \mathrm{C}_5={ }^{11} \mathrm{C}_5\) \(\text { Or; }{ }^{10} \mathrm{C}_3+{ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_5={ }^1 \mathrm{C}_5\) \(\text { Or; }{ }^{11} C_4+{ }^{11} C_5={ }^1 C_5\) \(\mathrm{Or} ;{ }^{12} \mathrm{C}_5={ }^n \mathrm{C}_5 \Rightarrow \mathrm{n}=12\)

Option (3) is correct

94. The number of parallelograms that can be formed by a set of 6 parallel lines intersected by the another set of 4 parallel lines is

1. 360
2. 90
3. 180
4. 45

Answer:

No. of parallelograms = \({ }^6 \mathrm{C}_2 \cdot{ }^4 \mathrm{C}_2\) =15X6 = 90 Option (2) is correct

95. \(^{12} C_3+2,^{12} C_1+{ }^{12} C_5={ }^{14} C_x\) The value of x

1. 3 or 5
2. 5 or 9
3. 7 or 1
4. 9 or 12

Answer:

\({ }^{12} \mathrm{C}_3+2{ }^{12} \mathrm{C}_4+{ }^{12} \mathrm{C}_5={ }^{14} \mathrm{C}_x\) \(\Rightarrow{ }^{12} C_3+{ }^{12} C_4+{ }^{12} C_4+{ }^{12} C_5={ }^{14} C_x\) \(\Rightarrow{ }^{13} \mathrm{C}_4+{ }^{13} \mathrm{C}_5={ }^{11} \mathrm{C}_x\) \(\Rightarrow{ }^{14} C_5={ }^{11} C_x \Rightarrow x=5\) \(\text { And }{ }^{14} C_5={ }^{14} C_{14 \cdot 5}={ }^{14} C_x \Rightarrow x=14-5=9\)

x = 5 or 9

96. The number of ways in which a man can invite one or more of his 7 friends to dinner is

1. 64
2. 128
3. 127
4. 63

Answer:  (3)

No. of ways to invite at least one friend = 2⁷ -1 = 128-1 = 127.

97. The number of words from the letters of the letters of the word BHARAT, in which B and H will never come together, is

1. 120
2. 360
3. 240
4. None

Answer: (3) Use Gap rule.

*A*R*A*T*

Total words = \( \frac{4!}{2!} \times 5 p_2=\frac{24}{2} \times 5 \times 4=240\)

98. The value of N in \(\frac{1}{7!}+\frac{1}{8!}=\frac{N}{9!} \operatorname{ls}\)

1. 81
2. 64
3. 78
4. 89

Answer: (1)

\(\frac{1}{7!}+\frac{1}{8!}=\frac{N}{9!}\) \(\mathrm{Or}^{\prime} ; \frac{1}{7!}+\frac{1}{8 \times 7!}=\frac{N}{9 \times 8 \times 7!}\) \(\operatorname{Or} ; \frac{1}{7!}\left(1+\frac{1}{8}=\right)=\frac{N}{9 \times 8 \times 7!}\) \(\Rightarrow \frac{9}{8}=\frac{N}{9 \times 8}\)

99. If \({ }^n P_r\)= 270 and \({ }^n C_r\)= 120 then r is

1. 4
2. 5
3. 3
4. 6

Answer: (3)

\({ }^n p_r={ }^n C_r, r!\)

Or; 720 = 120 (r!)

Or 6 = r! => r! = 3!

r = 3

100. A bag contains 4 red, 3 black and 2 white balls. In how many ways 3 balls can be drawn from this bag so that so that they include at least one black ball?

1. 46
2. 64
3. 86
4. None

Answer:

101. The letters of the words ‘CALCUTTA’ and ‘AMERICA’ are arranged in all possible ways. The ratio of the number of these arrangements is

1. 1:2
2. 2:1
3. 2:2
4. none of these

Answer: (2) 2:1

Words:

CALCUTTA

N0.of ways= \(\frac{8!}{2!2!2!}=5040\)

(C, A, T are repeated)

AMERICA

No. of ways = \( \frac{7!}{2!} \text { (A is repeated } \times 2 \text { ) } 2520\)

Ratio = 5040 : 2520 = 2:1

102. The number of different words that can be formed with 12 consonants and 5 vowels by taking 4 consonants and 3 vowels in each word is

1. \({ }^{12} \mathrm{C}_4 \times{ }^5 \mathrm{C}_3\)
2. \({ }^{17} \mathrm{C}_7\)
3. 4950 x 7!
4. none of these

Answer: (3) 4950x 7!

-» 12 consonants & 5 vowels

Selection of 4 consonants & 3 vowels = \({ }^{12} \mathrm{C}_1 \times{ }^5 \mathrm{C}_3\)

Also since words need to be formed arrangement is important

7 letters can arrange themselves in 7 ! ways

Total ways =

\({ }^{12} \mathrm{C}_4 \times{ }^5 \mathrm{C}_3 \times 7!=4950 \times 7!\text { ways }\)

103. Eight guests have to be seated 4 on each side of a long rectangular table.2 particular guests desire to sit on one side of the table and 3 on the other side. The number of ways in which the sitting arrangements can be made is

1. 1732
2. 1728
3. 1730
4. 1278.

Answer: (2) 1728

Total guests -> 8

If 2 are seated on one side & 3 on other remaining 3 have to be seated

No. of ways = \({ }^3 \mathrm{C}_2 \times{ }^1 \mathrm{C}_1=3 \text { ways }\)

Now the four on one side can arrange in 4 ! ways & other side also 4 ! ways

Total = 3×4!x4! = 1728

104. A question paper contains 6 questions, each having an alternative. The number of ways an examine can answer one or more questions is

1. 720
2. 728
3. 729
4. none of these

Answer: (2) 728

Any question can be solved in 3 ways

(1) 1st alternative (2) 2nd alternative (3) No answer

6 questions will have 36 ways = 729

But there will be one way which will have no answers to all question which is not required

729 – 1 = 728

105. The number of words that can be made by rearranging the letters of the word APURNA so that vowels and consonants appear alternate is

1. 18
2. 35
3. 36
4. none of these

Answer: (c) 36

A P U R N A 6 letter

3 vowels (A A U)

3 consonants (P R N)

The vowels can arrange in \(\frac{3!}{2!}\) ways = 3 ways (A is repeated)

The consonants can arrange in 3! ways = 6 ways

word can be arranged in 18 ways

We will have 2 cases

VCVCVC or CVCVC i.e., 2 ways Total = 36 ways

106. The number of arrangements of the letters of the word ‘COMMERCE’ is

1. 8
2. 8!/2!2!2!
3. 7 !
4. none of these

Answer: \(\text { (2) } \frac{8!}{2!2!2!}\)

COMMERCE, 8 letters (M-2, E=2, C=2)

Total arrangements = \(\frac{8!}{2!2!2!}\)

107. The number of ways in which 8 different beads be strung on a necklace is

1. 2500
2. 2520
3. 2250
4. none of these

Answer: (2) 2520

8 beads in a necklace

Formula for n beads in necklace is

\(\frac{1}{2}(n-1)!=\frac{1}{2}(8-1)!\quad=\frac{7!}{2}=2520\)

108. The number of different factors the number 75,600 has Is

1. 120
2. 121
3. 119
4. none of these

Answer: (1) 120

75600 no. of factors

-» 756 x 100 -» 756 (52) (22)

-»22. 33. 7. 52. 22-24. 33. 52.7

no. of factors (5+1) (4+1) (3+1) (2+1) = 120

109. (n-1) P_r+r ;(n-1) P_{(r-1)} is equal to

1. \({ }^n \mathrm{C}_r\)
2. \( n/(r n-r)\)
3. \({ }^n \mathrm{P}_r\)
4. none of these

Answer:

\({ }^n P_r\) \({ }^{n-1} P_r+r,(n-1) P_{(r-1)}={ }^n P_r\)

110. (2ⁿ)! can be written as

1. 2ⁿ {1.3.5…. (2n-l)}n
2. 2ⁿ n
3. {1.3.5…..(2n -1)}
4. none of these

Answer: (1) 2n (1. 3. 5…….(2n – 1) n !

(2n) ! = 2n (2n – 1)……… 3x2x1

= 2” (2n – 2)…….(2n-1)…3×1

= 2″ n ! . (1 . 3 . 5…….(2n – 1))