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Exercises 5.6 Chapter Test

1.

$1000 is placed in an account for 5 years that pays 9% interest.

  1. Calculate the simple interest earned.

  2. Calculate the interested earned if it compounded daily.

Answer.

  1. $450

  2. $568.23

Solution.

  1. Using the simple interest formula, \(I = P_0rt\text{,}\) we can compute the interest with \(P_0 = 1000\text{,}\) \(r = 0.09\text{,}\) and \(t = 5\text{.}\)

    \begin{equation*} I = 1000(0.09)(5) = 450 \end{equation*}

  2. Using the compound interest formula, \(P_n = P_0 \left( 1 + \frac{r}{k} \right)^{Nk}\text{,}\) we can compute the future value of the account with \(P_0 = 1000\text{,}\) \(r = 0.09\text{,}\) \(k = 365\)and \(N = 5\text{.}\)

    \begin{equation*} P_n = 1000\left(1 + \frac{0.09}{365}\right)^{5 \cdot 365} = 1568.23 \end{equation*}
    To find the interest subtract the initial investment from the future value:
    \begin{equation*} I = 1568.23 - 1000 = 568.23 \end{equation*}

2.

Judy has placed money in an account that pays 10% interest compounded monthly. After 3 years, she has $1617.82 in the account. How much money to she put in the account?

Answer.

$1200

Solution.

Use the compound interest formula, \(P_n = P_0\left(1 + \frac{r}{k}\right)^{Nk}\text{,}\) to solve for \(P_0\text{.}\)

\begin{equation*} 1617.82 = P_0 \left(1 + \frac{0.10}{12}\right)^{3 \cdot 12} \end{equation*}
\begin{equation*} 1617.82 = P_0 \left(1.348181842\right) \end{equation*}
\begin{equation*} P_0 = \frac{1617.82}{1.348181842} \end{equation*}
\begin{equation*} P_0 = 1200 \end{equation*}
3.

You are paying off your large-screen TV that cost $3,000 with 24 monthly payments of $162.50. What is the annual simple interest rate that you are being charged?

Answer.

15%

Solution.

We will need to use the simple interest formula, \(I = P_0rt\text{,}\) to solve for r. But first compute the amount of interest you are paying.

24 payments of $162.50 is: \(25 \cdot 162.50 = 3900\)

You borrowed $3,000 so you paid $900 in interest over 2 years (24 months).

\begin{equation*} 900 = 3000(r)(2) \end{equation*}
\begin{equation*} 900 = 6000r \end{equation*}
\begin{equation*} r = 900 \div 6000 = 0.15 \end{equation*}
\begin{equation*} r = 15\% \end{equation*}
4.

Kate has started a college fund for her daughter by putting money into an annuity that pays 6% interest.

  1. If she puts in $50 each month, how much money will she have in 18 years?

  2. If she wants to have $50,000 after 18 years of making monthly payments, what must her monthly payment be?

Answer.

  1. $19,367.66

  2. $129.08

Solution.

In both parts, we will be using the annuity formula, \(P_n = \frac{d\left(\left(1 + \frac{r}{k}\right)^{Nk}-1\right)}{\left(\frac{r}{k}\right)}\text{.}\)

  1. To find the future value in 18 years:

    \begin{equation*} P_n = \frac{50\left(\left(1 + \frac{0.06}{12}\right)^{18\cdot 12}-1\right)}{\left(\frac{0.06}{12}\right)} \end{equation*}
    \begin{equation*} P_n = 19367.66 \end{equation*}

  2. To solve for d:

    \begin{equation*} 50000 = \frac{d\left(\left(1 + \frac{0.06}{12}\right)^{18 \cdot 12}-1\right)}{\left(\frac{0.06}{12}\right)} \end{equation*}
    \begin{equation*} 50000 = d\left(387.3531944\right) \end{equation*}
    \begin{equation*} d = \frac{50000}{387.3531944}=129.08 \end{equation*}

5.

You want to be able to withdraw $40,000 each year for 20 years from your retirement account. If your account earns 7% interest, how much do you need to save for your retirement?

Answer.
$423,760.57
Solution.

We will use the payout annuity formula, \(P_0 = \frac{d\left(1 - \left(1 + \frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}\text{,}\) since we are taking money out of an account.

\begin{equation*} P_0 = \frac{40000\left(1 - \left(1 + \frac{0.07}{1}\right)^{-20 \cdot 1}\right)}{\left(\frac{0.07}{1}\right)} \end{equation*}
\begin{equation*} P_0 = \frac{40000\left(1 - \left(1.07\right)^{-20}\right)}{0.07} \end{equation*}
\begin{equation*} P_0 = 423,760.57 \end{equation*}
6.

Ben purchases a car for $18,395.

  1. If sales tax is 6.5% of the purchase price, find the amount of the sales tax.

  2. If the car license fee is 1.2% of the purchase price, find the amount of the license fee.

  3. If Ben makes a $2500 down payment, find the amount of the loan that he will need.

  4. Ben gets an amortized loan at 7.5% interest for 4 years. What will his monthly payments be?

Answer.

  1. $1,195.68

  2. $220.74

  3. $17,311.42

  4. $418.57

Solution.

  1. \(\displaystyle 0.065 \cdot 18395 = 1195.68\)

  2. \(\displaystyle 0.012 \cdot 18395 = 220.74\)

  3. \(\displaystyle 18395 + 1195.68 + 220.74 - 2500 = 17311.42\)

  4. Use the loans formula, \(P_0 = \frac{d\left(1 - \left(1 + \frac{r}{k}\right)^{-Nk}\right)}{\left(\frac{r}{k}\right)}\text{,}\) to solve for the amount the payment (d).

    \begin{equation*} 17311.42 = \frac{d\left(1 - \left(1 + \frac{0.075}{12}\right)^{-4\cdot 12}\right)}{\left(\frac{0.075}{12}\right)} \end{equation*}
    \begin{equation*} 17311.42 = d \left(41.35837114\right) \end{equation*}
    \begin{equation*} d = 17311.42 \div 41.35837114 = 418.57 \end{equation*}

7.

Complete the first two lines of the amortization table for a $250,000 mortgage at 4.5% interest. Assume it is a 30 year mortgage.

Payment Number Monthly Payment Interest Paid on Principal Loan Balance
1
2

Answer.

Payment Number Monthly Payment Interest Paid on Principal Loan Balance
1 $1,266.71 $937.50 $329.21 $249,670.79
2 $1,266.71 $936.27 $330.44 $249,340.35

Solution.

Use the loans formula, \(P_0 = \frac{d\left(1 - \left(1+\frac{r}{k}\right)^{-Nk}\right)} {\left(\frac{r}{k}\right)}\) to solve for the monthly payment, d.

\begin{equation*} 250000 = \frac{d\left(1 - \left(1+\frac{0.045}{12}\right)^{-360}\right)} {\left(\frac{0.045}{12}\right)} \end{equation*}
\begin{equation*} 250000 = d(197.361159) \end{equation*}

\

\begin{equation*} d = 250000 \div 197.361159 = 1266.71 \end{equation*}

To find the interest owed during the first month, compute \(I = P_0 rt\) for one month (or \(\frac{1}{12}\) year).

\begin{equation*} I = 250000 \left(0.045\right)\left(\frac{1}{12}\right) = 937.50 \end{equation*}

If $937.50 of the first payment is interest, then \(1266.71 - 937.50 = 329.21\) will go toward the balance of the loan.

After the first payment, the balance will be reduced by $329.21. The new balance is \(250,000 - 329.21 = 249,670.79.\)

The interest owed during the second month is computed using that new balance of the loan:

\begin{equation*} I = 249670.79 \left(0.045\right)\left(\frac{1}{12}\right) = 936.27 \end{equation*}

Then the amount that will go toward the balance of the loan is \(1266.71 - 936.27 = 330.44\text{.}\) The balance of the loan is now \(249,670.79 - 330.44 = 249,340.35\text{.}\)

8.

Julia puts $100 in an account that pays 4% interest (compounded yearly) every year for 10 years. Then she lets her money sit in the account, without making payments, for another 15 years.

  1. How much money will she have in the account?

  2. How much interest did she earn?

Answer.

  1. $2,162.23

  2. $1,162.23

Solution.

First, use the annuity formula to compute the amount of money she will have after 10 years of making deposits.

\begin{equation*} P_n = \frac{100\left(\left(1+\frac{0.04}{1}\right)^{10 \cdot 1}-1\right)}{\left(\frac{0.04}{1}\right)} \end{equation*}
\begin{equation*} P_n = \frac{100\left(\left(1.04\right)^{10}-1\right)}{0.04}=1200.61 \end{equation*}

Then use the compound interest formula to find the future value of $1200.61 is after sitting in the account for 15 years.

\begin{equation*} P_n = 1200.61\left(1+\frac{0.04}{1}\right)^{15 \cdot 1} \end{equation*}
\begin{equation*} P_n = 1200.61\left(1.04\right)^{15}=2162.23 \end{equation*}

The total amount in the account will be $2,162.23. Julia made a total of 10 payments of $100 into the account.

\begin{equation*} 10 \cdot 100 = 1000 \end{equation*}
\begin{equation*} I = 2162.23 - 1000 = 1162.23 \end{equation*}

She earned $1,162.23 in interest.

9.

Geoff borrow $2,000 at 8% simple interest.

  1. If he takes out a 4-year loan, what will his monthly payments be?

  2. If he takes a 3-year loan, what will his monthly payments be?

  3. How much money will he save (over the life of the loan) if he takes the 3-year loan instead of the 4-year loan?

Answer.

  1. $55

  2. $68.89

  3. $160

Solution.

  1. First compute the interest charged:

    \begin{equation*} I = P_0rt \end{equation*}
    \begin{equation*} I = 2000 \left(0.08\right) \left(4\right) = 640 \end{equation*}

    Geoff will have to pay back 2000 + 640 = $2,640 in 48 monthly payments.

    \begin{equation*} 2640 \div 48 = 55 \end{equation*}

    His monthly payments will be $55.

  2. If it is a 3-year loan, the interest charged will be:

    \begin{equation*} I = 2000 \left(0.08\right)\left(3\right)=480 \end{equation*}

    He will have to pay back 2000 + 480 = 2480 in 36 monthly payments.

    \begin{equation*} 2480 \div 36 = 68.89 \end{equation*}

    His monthly payments will be $68.89

  3. He is paying a total of $2,480 with the 3-year loan and $2,640 with the 4-year loan. If he takes the 3-year loan, he will save \(2640-2480 = \$160\text{.}\)