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2. [0/4 Points] DETAILS PREVIOUS ANSWERS MY NOTES James wants to take out a loan. He can afford to make monthly payments of 1

James wants to take out a loan. He can afford to make monthly payments of 100 dollars and wants to pay the loan off after exactly 30 years.

What is the maximum amount that James can afford to borrow if the bank charges interest at an annual rate of 8 percent, compounded monthly?

(Give your answer, in dollars, correct to the nearest dollar.)

Nicola borrows 60000 dollars from a bank that charges interest at an annual rate of 10 percent, compounded monthly.

Calculate the monthly payment that Nicola would have to make in order for the loan to be paid off after exactly 20 years.

(Give your answer, in dollars, to the nearest cent. You should not include the dollar sign or any commas in your answer.)

Wanda wants to take out a loan. Suppose she can afford to make monthly payments of 600 dollars and the bank charges interest at an annual rate of 8 percent, compounded monthly.

What is the maximum amount that Wanda could afford to borrow if the loan is to be paid off eventually?

(Give your answer, in dollars, correct to the nearest dollar.)

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Answer #1

Question-2

.

L=PMT\left[\frac{1-\left(1+\frac{r}{n}\right)^{-nt}}{\frac{r}{n}}\right]

PMT=100

r=8% = 0.08

t=30 years

n=12 for monthly payment

L=100\left[\frac{1-\left(1+\frac{0.08}{12}\right)^{-12\cdot 30}}{\frac{0.08}{12}}\right]

L=100\left[\frac{1-\left(1+0.0067\right)^{-360}}{0.0067}\right]

L=100\left[\frac{1-\left(1.0067\right)^{-360}}{0.0067}\right]

L=100\left[\frac{1-0.09035}{0.0067}\right]

L=100\left[\frac{0.90964}{0.0067}\right]

L=100\left[135.76719\right]

L=13628.34941

L \approx {\color{Red} 13628 }

.

Question-3

.

L=PMT\left[\frac{1-\left(1+\frac{r}{n}\right)^{-nt}}{\frac{r}{n}}\right]

L=60000

r=10% = 0.1

t=20 years

n=12 for monthly payment

60000=PMT\left[\frac{1-\left(1+\frac{0.1}{12}\right)^{-12\cdot 20}}{\frac{0.1}{12}}\right]

60000=PMT\left[\frac{1-\left(1+0.0083\right)^{-240}}{0.0083}\right]

60000=PMT\left[\frac{1-\left(1.0083\right)^{-240}}{0.0083}\right]

60000=PMT\left[\frac{1-0.13754}{0.0083}\right]

60000=PMT\left[\frac{0.86245}{0.0083}\right]

60000=PMT\left[103.90981\right]

PMT=\frac{60000}{103.90981}

PMT=579.01298

PMT \approx {\color{Red} 579.01 }

.

.

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Question-4

.

L=PMT\left[\frac{1-\left(1+\frac{r}{n}\right)^{-nt}}{\frac{r}{n}}\right]

PMT=600

r=8% = 0.08

t=infinite

n=12 for monthly payment

L=600\left[\frac{1-\left(1+\frac{0.08}{12}\right)^{-12\cdot \infty }}{\frac{0.08}{12}}\right]

L=600\left[\frac{1-\left(1+\frac{0.08}{12}\right)^{-\infty }}{\frac{0.08}{12}}\right]

L=600\left[\frac{1-0}{\frac{0.08}{12}}\right]

L=600\left[\frac{1}{\frac{0.08}{12}}\right]

L=600\left[\frac{12}{0.08}\right]

L={\color{Red} 90000}

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