Financial Algebra 1st Edition Chapter 3 Banking Services
Page 151 Problem 1 Answer
We will use your calculator to find the value of g(x) for each of the increasing values of x in the table.
The values will be:
| x | g(x)=−5x+1 |
| 100 | −499 |
| 1000 | −4999 |
| 90000 | −449,999 |
| 900,000 | −4,499,999 |
| 8,000,000 | −39,999,999 |
| 50,000,000 | −249,999,999 |
As x approaches infinity, the value of g(x) decreases without. Therefore,g(x) has no limit.
As the values of x increase towards infinity, the values of g(x)=−5x+1 keeps on decreasing.
Page 151 Problem 2 Answer
Given: f(x)=1/x
To find: lim f(x)
x→∞
Solution: Set up a table with increasing values of x.
| x | f(x)=1x |
| 100 | 0.01 |
| 1,000 | 0.001 |
| 90,000 | 0.00001111111 |
| 900,000 | 0.00000111111 |
| 8,000,000 | 0.000000125 |
| 50,000,000 | 0.00000002 |
| 2,000,000,000 | 0.0000000005 |
We will first set up a table with increasing values of x
The pattern in the table shows that as x approaches infinity,f(x) approaches 0.
It keeps getting closer to 0; it, never reaches 0.
You can say, “The limit of f(x), as x approaches infinity, is 0,”
If f(x)=1/x, then lim x→∞=0
Page 152 Problem 3 Answer
Given:f(x)=1x
To find: lim
x→∞
Solution: The value of the function will always be 1 for any value of x
So, the limit of f(x), as x approaches infinity is 1
Iff(x)=1x, then lim = 1
x→∞
Page 152 Problem 4 Answer
Given: f(x)=(1+0.05/x)x
To find: lim (1+0.05/x)x
x→∞
Solution: We will create a table to find values of the function for v
The table showing various values of the function is; x f(x) to five decimal places
| x | f(x)
to five decimal places |
| 100 | 1.05126 |
| 1,000 | 1.05127 |
| 90,000 | 1.05127 |
| 900,000 | 1.05127 |
| 8,000,000 | 1.05127 |
| 50,000,000 | 1.05127 |
| 2,000,000,000 | 1.05127 |
The pattern in the table shows that as x approaches infinity,f(x) approaches a number around1.05127…
Using a table we get, Lim (1+0.05/x)x
x→∞
=1.05127 rounded to five decimal places.
Page 153 Problem 5 Answer
Given: e=2.718281828
π=3.141592654
To find: Difference between e π and π e
Solution
e π=23.1406926328
Π e=22.4591577184
Difference=e π−πe
=23.1406926328−22.4591577184
=0.6815349144
=0.681 (rounded to the nearest thousandth)
The difference between eπ and πe rounded to the nearest thousandth is0.681
Page 153 Problem 6 Answer
Given: Craig deposits $5,000 at 5.12% interest compounded continuously for four years
To find: Ending balance to the nearest cent
Solution: Balance= Deposit amount×e years
=5000×(2.7182818)4
=$272,990.7
If Craig deposits$5,000 at 5.12% interest compounded continuously for four years, then the balance in the account at the end would be$272,990.7
(rounded off to nearest cent)
Page 154 Problem 7 Answer
According to question,”Infinite” is a concept that is hard to understand for people, because we cannot observe infinity and thus we need our imagination to be able to take into account the meaning of “infinite”.
This then implies that the question of “infinite” profoundly moves the spirit of a human.
Hence, The required asnwer is: “Infinite” is a concept that is hard to understand for people
Page 154 Problem 8 Answer
Given:p= Principle =$2,000r= Interest rate =4%=0.04t= Time expressed in years =3 years
Find the interest if it is computed using simple interest.
According to question,
I=prt
=2000×0.04×3
=240
Hence, The interest if it is computed using simple interest is:$240
Page 154 Problem 9 Answer
Given:
p= Principle =$2,000
r= Interest rate =4%=0.04
t= Time expressed in years =3 years
Find the interest if it is compounded annually.
p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3years
Then by using the formula,
B=p(1+r/n)n×t
=2000(1+0.04/1)1×3
=2000(1.04)3
≈2,249.73
Then, The interest is the balance decreased by the principle:
I=B−p
=2,249.73−2,000
=249.73
Hence, The interest if it is compounded annually is:$249.73
Page 154 Problem 10 Answer
Given:p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3 years
n=Number of periods per year=2
Find the interest if it is compounded semiannually.
p= Principle =$2,000
r= Interest rate =4%=0.04
t= Time expressed in years =3 years
n= Number of periods per year =2
(Given)
Then, by using the formula:
B=p(1+r/n)n×t
=2000(1+0.04/2)2×3
=2000(1.02)6
≈2,252.32
And then according to question,
I=B−p
=2,252.32−2,000
=252.32
Hence, the interest if it is compounded semiannually is:$252.32
Page 154 Problem 11 Answer
Given:p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3years
n=Number of periods per year=4
Find the interest if it is compounded quarterly.
p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3 years
n=Number of periods per year=4
Then, by using the formula:
B=p(1+r/n)n×t
=2000(1+0.04/4)4×3
=2000(1.01)12
≈2,253.65
And then according to question,The interest is the balance decreased by the principle:
I=B−p
=2,253.65−2,000
=253.65
Hence, The interest if it is compounded quarterly is:$253.65
Page 154 Problem 12 Answer
Given:p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3 years
n=Number of periods per year=12
Find the interest if it is compounded monthly.
p= Principle =$2,000
r= Interest rate =4%=0.04
t= Time expressed in years =3 years
n= Number of periods per year =12
(Given)
Then, by using the formula:
B=p(1+r/n)n×t
=2000(1+0.04/12)12×3
≈2,254.54
And then according to question,The interest is the balance decreased by the principle:
I=B−p
=2,254.54−2,000
=254.54
Hence,the interest if it is compounded monthly is:$254.54
Page 154 Problem 13 Answer
Given:p=Principle=$2,000
r= Interest rate=4%=0.04
t= Time expressed in years=3 years
n= Number of periods per year=365
Find the interest if it is compounded daily.
p= Principle =$2,000
r= Interest rate =4%=0.04
t= Time expressed in years =3 years
n= Number of periods per year =365
(Given)
Then, By using the formula:
B=p(1+r/n)n×t
=2000(1+0.04/365)365×3
≈2254.98
And then according to question,The interest is the balance decreased by the principle:
I=B−p
=2,254.98−2,000
=254.98
Hence, the interest if it is compounded daily is:$254.98
Page 154 Problem 14 Answer
Given:p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3 years
n=Number of periods per year=365×24=8760
Find: the interest if it is compounded hourly..
According to question,
By using the formula:
B=p(1+r/n)n×t
=2000(1+0.04/8760)8760×3
≈2254.99
Then, As we know that:
The interest is the balance decreased by the principle:
I=B−p
=2,254.99−2,000
=254.99
Hence, The interest if it is compounded hourly is:$254.99
Page 154 Problem 15 Answer
Given:p=Principle=$2,000
r=Interest rate=4%=0.04
t=Time expressed in years=3 years
n=Number of periods per year=365×24×60=525,600
Find: the interest if it is compounded every minute.
According to question,
B=p(1+r/n)n×t
=2000(1+0.04/525,600)525,600×3
≈2254.99
Then, As we know that:
I=B−p
=2,254.99−2,000
=254.99
Hence, the interest if it is compounded every minute is:$254.99
Page 154 Problem 16 Answer
Given:p= Principle =$2,000r= Interest rate =4%=0.04t= Time expressed in years =3 years
Find: the interest if it is compounded continuously.
According to question,
B=per×t
=2000e0.04×3
=2000e0.12
≈2254.99
Then, As we know that,
The interest is the balance decreased by the principle:
I=B−p
=2,254.99−2,000
=254.99
Hence, The interest if it is compounded continuously is:$254.99
Page 154 Problem 17 Answer
According to question,
Result of first part:$240
(simple interest)
Result pf second part:$254.99
(compounded continuously)
Then, the difference of interest between compounded continuously and simple interest.
$254.99−$240=$14.99
Hence, The difference in interest between simple interest and interest compounded continuously is:$14.99
Page 154 Exercise 1 Answer
Given:p=Principle=$50,000r=Interestrate=4\dfrac{1}{8}%=4.125%=0.04125t=Timeexpressedinyears=6years
Find: the interest earned .
According to question,
By using the formula:
B=per×t
=50000e0.04125×6
=50000e0.2475
≈64,040.97
Then,The interest is the balance decreased by the principle:
I=B−p
=64,040.97−50,000
=14,040.97
Hence, The interest is:$14,040.97
Page 154 Exercise 2 Answer
Given: p= Principle=$9,000
r=Interest rate=4.1%=0.041
t=Time expressed in years=2 years
n=Number of periods per year=12
Find: How much interest does State Bank pay.
According to question,
By using the formula:
B=p(1+r/n)n×t
=9000(1+0.041/12)12×2
≈9,767.74
Then,The interest is the balance decreased by the principle:
I=B−p
=9,767.74−9,000
=767.74
Hence, interest does State Bank pay is:$767.74
Page 154 Exercise 3 Answer
Given:p= Principle =$9,000r= Interest rate =4.01%=0.0401t= Time expressed in years =2 years
Find: How much interest does Kings Savings pay.
According to question,
By using the formula:
B=per×t
=9000e0.0401×2
=9000e0.0802
≈9,751.53
Then,The interest is the balance decreased by the principle:
I=B−p
=9,751.53−9,000
=751.53
Hence, The interest does Kings Savings pay is:$751.53
Page 154 Exercise 4 Answer
According to question,
Result of first part: $767.74
Result of second part: $751.53
Then the result of part first is higher and thus State Bank earns more interest.
The difference of the interests:
$767.74−$751.53=$16.21
Therefore, The State Bank will result in$16.21 more interest.
Hence, The State Bank will result in$16.21 । more interest.
Page 154 Exercise 5 Answer
As we know that,There are many factors that could affect Whitney’s choice, such as:
Location bank Opening hours of the bank Other services offered by the bank
Hence, The answer is above explanation.
Page 155 Exercise 6 Answer
Given:p= Principle =$1,000r= Interest rate =16%=0.16t= Time expressed in years =5 years
Find: what is the ending balance.
According to question,
By using the formula,
B=per×t
=1000e0.16×5
=1000e0.8
≈2,225.54
Hence, The ending balance is:$2,225.54
Page 155 Exercise 7 Answer
Given:p= Principle =$1,000r= Interest rate =5%=0.05t= Time expressed in years =5 years
Find: What is the ending balance,
According to question,
B=per×t
=1000e0.05×5
=1000e0.25
≈1,284.03
Hence, The ending balance is: $1,284.03
Page 155 Exercise 8 Answer
According to question,
Firstly2225.54 is the ending bal. At a 16% interest rate for a 5-year period.
And then 1284.02 is the ending bal. At a 5% interest rate for a 5-year period.
2225.54−1284.02=941.52
Hence, The difference between the two ending balances is: 941.52
Page 155 Exercise 9 Answer
Given:p= Principle =$30,000r= Interest rate =4\ddrac{1}{2}%=4.5%=0.045t= Time expressed in years =0.5 years (6 months)
Find: interest earned.
According to question, By using the formula:
B=per×t
=30,000e0.045×0.5
=30,000e0.0225
≈30,682.65
Then, The interest is the balance decreased by the principle:
I=B−p
=30,682.65−30,000
=682.65
Hence, The interest is: $682.65
Page 155 Exercise 10 Answer
Given:p= Principle=$4,000
r=Interest rate=3.8%=0.038
t=Time expressed in years=3years
n=Number of periods per year=4
Find: How much interest does Option 1 pay.
According to question,
B=p(1+r/n)n×t
=4000(1+0.038/4)4×3
≈4,480.60
Then, The interest is the balance decreased by the principle:
I=B−p
=4,480.60−4,000
=480.60
Hence, The interest is: $480.60
Page 155 Exercise 11 Answer
Given:p= Principle =$4,000r= Interest rate =3.5%=0.035t= Time expressed in years =3 years
Find: interest does Option 2 pay
According to question,
B=per×t
=4000e0.035×3
=4000e0.105
≈4442.84
Then, The interest is the balance decreased by the principle:
I=B−p
=4,442.84−4,000
=442.84
Hence, The interest is:$442.84