## 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