6.6 Applications

Robert Foth

6.6 Applications

Learning Objectives

Upon completion of this section, you should be able to

Solve various financial applications involving simple interest, compounding interest, and annuities.

Which equation to use?

When presented with a finance problem (on an exam or in real life), you’re usually not told what type of
problem it is or which equation to use. Here are some hints on deciding which equation to use based on the
wording of the problem.

The easiest types of problem to identify are loans. Loan problems almost always include words like:
“loan”, “amortize” (the fancy word for loans), “finance (a car)”, or
“mortgage” (a home loan). Look for these words. If they’re there, you’re probably looking
at a loan problem. To make sure, see if you’re given what your monthly (or annual) payment is, or if
you’re trying to find a monthly payment.

If the problem is not a loan, the next question you want to ask is: “Am I putting money in an account and
letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?” If
you’re letting the money sit in the account with nothing but interest changing the balance, then you’re looking
at a compound interest problem. The exception would be bonds and other investments where the interest is
not reinvested; in those cases you’re looking at simple interest.

If you’re making regular payments or withdrawals, the next questions is: “Am I putting money into the
account, or am I pulling money out?” If you’re putting money into the account on a regular basis
(monthly/annually/quarterly) then you’re looking at a basic Annuity problem. Basic annuities are when
you are saving money. Usually in an annuity problem, your account starts empty, and has money in the
future.

If you’re pulling money out of the account on a regular basis, then you’re looking at a Payout Annuity
problem. Payout annuities are used for things like retirement income, where you start with money in your
account, pull money out on a regular basis, and your account ends up empty in the future.

Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly
identify what the question is really asking, and to determine what approach will best allow you to solve the
problem.

Try it Now

For each of the following scenarios, determine if it is a compound interest problem, a savings annuity
problem, a payout annuity problem, or a loans problem. Then solve each problem.

If $2,000 is invested at 7% compounded quarterly, what will the final amount be in 5 years?
How much should be invested each month at 9% for it to accumulate to $8,000 in three years?
Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it
for college, withdrawing money for tuition and expenses each quarter. How much can she take out each
quarter if she has 3 years of school left?
Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an
account earning 3% interest compounded monthly. How much will he have saved up after 3 years?
Keisha is managing investments for a non-profit company. They want to invest some money in an account
earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much
should Keisha deposit into the account?
Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly
payments of $100, how much new equipment can she buy?
How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in
two years?

Answer

Answer Content

This is a compound interest problem. There would be $2829.56 in five years.
This is a savings annuity problem. Deposit $194.40 each month.
This is a payout annuity problem. She can pull out $1833.60 a quarter.
This is a savings annuity problem. He will have saved up $7,524.11/
This is compound interest problem. She would need to deposit $22,386.46.
This is a loans problem. She can buy $4,609.33 of new equipment.
This is a savings annuity problem. You would need to save $200.46 each month

Try It Now 1 Video Solution (1 min 43 secs – CC) Not all
problems in the video.

In addition to identifying the type of problem to be solved you may be required to combine multiple finance
formulas to reach a solution.

Example 1

Sarah started investing into an account that earns an annual rate of 5.5% compounded monthly. Initially she
invested $25 a month for five years. She then boosted her contribution to $100 a month for an additional five
years. How much is in the account after those ten years?

Solution

To solve this we can treat these as two different investments and find their values at the end of the ten
years. The first investment has two parts.

Invests $25 a month for five years. Find the total after five years.
This total sits in that account for an additional five years.

The second investment only lasts for those last five years. Sarah deposits $100 a month into the account.

Part a. Solution:

We have an annuity problem. Find the value of that annuity after five years, or FV.

$\begin{array}{l} pymt = 25 \\ r = 0.055 \\ n = 12 \\ t = 5 \\ FV = \frac{25 ({(1 + \frac{0.055}{12})}^{5 \times 12} - 1)}{(\frac{0.055}{12})} = $ 1722.02 \end{array}$

After the first five years there is $1722.02 in the account. We will treat this as a lump sum deposit for an
additional five years earning 5.5% compounded monthly. Remember during those last five years she changed her
contribution to $100 and we will deal with that in part two.

$\begin{array}{l} P = $ 1722.02 \\ r = 0.055 \\ n = 12 \\ t = 5 \\ FV = 1722.02 {(1 + \frac{0.055}{12})}^{5 \times 12} = $ 2265.67 \end{array}$

So the first part yields a total of $2265.67 for ten years.

Part b. Solution:

We need to find the value of $100 a month for the last five years. Once we find the value we add the
solutions to a. with b.

$\begin{array}{l} pymt = 100 \\ r = 0.055 \\ n = 12 \\ t = 5 \\ FV = \frac{100 ({(1 + \frac{0.055}{12})}^{5 \times 12} - 1)}{(\frac{0.055}{12})} = $ 6888.08 \end{array}$

Adding the first and second part together gives
$$ 2, 265.67 + $ 6, 888.08 = $ 9, 153.75$
.

Example 2

Idina has been saving $250 a month into a retirement account for the last 25 years that earns an annual rate
of 6.3%. She would like to know how much she can withdraw each month from this account in order to deplete the
account after 20 years.

Solution

In this problem we need to first determine how much she has saved up over the first 25 years. This is an
annuity problem to solve. After we have that value we are trying to determine the monthly withdrawals (or
payouts) for that total to last 20 years. This is a payout annuity problem.

Solve for the value of the monthly payments of $250 at 6.3% for 25 years.

$\begin{array}{l} pymt = 250 \\ r = 0.063 \\ n = 12 \\ t = 25 \\ FV = \frac{250 ({(1 + \frac{0.063}{12})}^{25 \times 12} - 1)}{(\frac{0.063}{12})} = $ 181470.48 \end{array}$

Now we know that Idina is starting with $181,470.48 in the account and wants to make withdraws for 20 years.
To determine the withdrawal amount we will use the payout annuity formula.

$\begin{array}{l} P = $ 181470.48 \\ r = 0.063 \\ n = 12 \\ t = 20 \\ 181470.48 = \frac{pymt (1 - {(1 + \frac{0.063}{12})}^{- 20 \times 12})}{(\frac{0.063}{12})} \\ Solve for pymt \\ pymt = $ 1331.71 \end{array}$

Idina can withdraw $1331.71 over 20 years.

Example 3

Mike plans to make contributions to his retirement account for 20 years. After the last contribution,
he will start withdrawing $3,000 a month for 15 years. Assuming Mike’s account earns 8% compounded
monthly, how large must his monthly contributions be during the first 20 years, in order to accomplish his
goal?

Solution

In this situation we want to know how much Mike should be depositing each month for 20 years to reach his
goal of withdrawing $3,000 a month for 15 years. This deposit is the unknown for the annuity payments (the
time he is saving money up for retirement). In order to determine that deposit amount we first need to find
the amount of money needed at the start of the withdrawals, so we will work backwards by first finding out how
much we need in an account to fund the payouts.

The payout annuity account has an annual rate of 8% compounded monthly and he will be making $3,000
withdrawals each month. How much did the account need to begin with? This is the first part we will answer.

$P = \frac{3000 [1 - {(1 + \frac{.08}{12})}^{- 15 * 12}]}{\frac{.08}{12}} = 313921.78$

Mike needs $313,921.78 in the account at the start of retirement. This means his goal for savings over the
first 20 years is $313,921.78 in the annuity. Using this information we can now find the monthly deposit that
must be made into the account.

$\begin{array}{l} 313921.78 = \frac{pymt [{(1 + \frac{.08}{12})}^{12 * 20} - 1]}{\frac{.08}{12}} \\ pymt = 532.96 \end{array}$

Mike has to save $531.96 each month for the next 20 years to reach his retirement goal of being able to
withdraw $3,000 a month for 15 years.

Exercises

Some of these problems will be rather challenging while others will be straight forward review type problems.

You can afford to pay $300 per month for a car. What is the maximum price you can pay for a car if there is
an annual interest rate of 7% compounded monthly and you want to repay the loan in 4 years?
Answer

$\begin{array}{l} \end{array}$
This is a loan problem. Use the loan formula to find P₀.
$P_{0} = \frac{300 (1 - {(1 + \frac{0.07}{12})}^{- 4 \times 12})}{(\frac{0.07}{12})} = $ 12528.06$
A business needs $300,000 in 3 years. What lump sum should be put aside now in an account that pays an
annual rate of 5% compounding continuously so that three years from now the company will have $300,000?
Mr. Selznik bought his house in 1980. He had his loan financed for 30 years at an annual interest rate of
6.2% compounded monthly resulting in a monthly payment of $1200.
1. How much interest was paid over the life of the loan?
2. How much did the house cost if he had paid a down payment of 20%?
3. Solve for P₀.
  $P_{0} = \frac{1200 (1 - {(1 + \frac{0.062}{12})}^{- 30 \times 12})}{(\frac{0.062}{12})} = $ 195928.29$
  
  Next we find the total interest paid by adding up all payments for 30 years and subtracting the initial
  loan amount.
  
  $($ 1200 \times 12 \times 30) - $ 195928.29 = $ 236071.71$
  Total interest paid was $236071.71
4. We need to find the original price paid for the house. We know that after a down payment of 20% that
  the total was what was loaned ($195,928.29). Let us call x the original price, then (.20)x
  would be the down payment. This gives us that
  $x - (.20) x = $ 195928.29$
  or solving for x we find
  $x = $ 244910.36$
  .

Manuel borrows $800 for 6 months at an annual 18% simple interest. How much does he owe at the end of 6
months if has made no payments on this debt?
A car is sold for $2500 cash down and $425 per month for the next 5 years. Find the cash value of the car
when it was sold if the money is worth an annual rate of 6.3% compounded monthly.
Suppose you invest $50 a month for 5 years into an account at an annual rate of 8% compounded
monthly. fter 5 years, you leave the money, without making additional deposits, in the account for
another 25 years. How much will you have in the end?

Answer

First find the value after five years.

$\begin{array}{l} P_{5} = \frac{50 ({(1 + 0.08 / 12)}^{5 (12)} - 1)}{0.08 / 12} \\ P_{5} = $ 3673.84 \end{array}$

Next 25 years value is

$3673.84 {(1 + .08 / 12)}^{25 (12)} = $ 26, 966.65$

Mike plans to make contributions to his retirement account for 15 years. After the last contribution,
he will start withdrawing $10,000 a quarter for 10 years. Assuming Mike’s account at an annual rate of
8% compounded quarterly, how large must his quarterly contributions be during the first 15 years, in order
to accomplish his goal?

Answer

Working backwards,

P_{0} = \frac{10000 (1 - {(1 + 0.08 / 4)}^{- 10 (4)})}{0.08 / 4} = $ 273, 554.79

needed at retirement. To end up with that amount of money

273, 554.70 = \frac{d ({(1 + 0.08 / 4)}^{15 (4)} - 1)}{0.08 / 4}

. He’ll need to contribute d = $2398.52 a quarter.

Kendra wants to be able to make withdrawals of $60,000 a year for 30 years after retiring in 35
years. How much will she have to save each year up until retirement if her account at an annual rate
of 7% compounded annually?

Answer

This problem is a combination of a payout annuity during the retirement time period and an annuity while
saving for retirement. Since we know she wants $60,000 per year for 30 years we have to figure out what the
starting balance is in the payout annuity so that she can make those yearly withdrawals.
Payout Annuity (starting value):

$\begin{array}{l} P_{0} = \frac{d [1 - {(1 + \frac{r}{k})}^{- N k}]}{\frac{r}{k}} \\ P_{0} = \frac{60, 000 [1 - {(1 + \frac{.07}{1})}^{- 30 * 1}]}{\frac{.07}{1}} \\ P_{0} = $ 744, 542.27 \end{array}$

Kendra needs to have $744,542 in the account she will be withdrawing from by the time she retires in 35
years. She can make yearly payments into the account of what value to make this happen? This is the second
part we need to solve. Since we are not making a single lump sum payment and instead making yearly payments
this is an annuity problem:

$\begin{array}{l} P_{N} = \frac{d [{(1 + \frac{r}{k})}^{k * N} - 1]}{\frac{r}{k}} \\ 744, 542.27 = \frac{d [{(1 + \frac{.07}{1})}^{1 * 35} - 1]}{\frac{.07}{1}} \\ d = \frac{744, 542.27}{\frac{[{(1 + .07)}^{1 * 35} - 1]}{.07}} \\ d = $ 5, 385 \end{array}$

So in order to retire on yearly payments of $60,000 30 years from now she only needs to deposit roughly
$5,385 per year for the next 35 years.

Tom bought his house in 1985, and financed the loan for 30 years at an annual interest rate of 7.8%
compounded monthly. His monthly payment was $1260. In 1995, Tom decided to pay off the loan. Find the
balance of the loan he still owes.

Answer

This problem is a combination of a payout annuity during the retirement time period and an annuity while

We can treat this as a loan problem where the monthly payments are $1260 with 20 years for payments (since
he has already paid for 10 years on his 30 year loan). Find P₀.

$\begin{array}{l} d = 1260 \\ r = 0.078 \\ k = 12 \\ N = 20 \\ P_{0} = \frac{1260 (1 - {(1 + \frac{0.078}{12})}^{- 20 \times 12})}{(\frac{0.078}{12})} = $ 152905.93 \end{array}$

A corporation estimates it will need $1,300,000 in 8 years to replace its existing machinery. How much
should it deposit each quarter at an annual rate of 4.4% compounded quarterly to meet this obligation?
An OB/GYN physician in private practice knows her malpractice insurance will be $85,000 and is due in one
year. How much should she deposit into an account each month if that account has an annual rate of 3.7%
compounded monthly and she wishes to pay it in full knowing that the insurance company drops the price by 2%
if paid in full?

Answer

We can treat this an an annuity problem where we are looking to solve for the payments, d, made into the
account. Since she wants to pay the account in full we know the future value after 1 year will be
$P_{1} = $ 85, 000 - (.02) $ 85, 000 = $ 83, 300$
. Solve for d in the annuity formula.

$\begin{array}{l} $ 83, 300 = \frac{d ({(1 + \frac{0.037}{12})}^{1 \times 12} - 1)}{(\frac{0.037}{12})} \\ d = $ 6824.73 \end{array}$

You are told that City Bank pays an annual interest rate of 6% compounded annually, while Western Bank pays
an annual rate of 5.8% compounded continuously. Which one is a better deal if the money sits in the account
for more than one year?

Answer

City Bank. Examine the total amount of money in each account for various initial deposits.

Mrs. Brown needs $5,000 in three years. If the annual interest rate is 4.9% compounding weekly, how much
should she deposit today in that account to have $5000 three years from now?

Answer

This is a fixed lump sum deposit compounding interest problem. We are looking for that initial deposit
knowing that the future amount will be $5000 in three years in an account with an annual rate of 4.9%
compounded weekly. Use the compounding interest formula to solve for P₀.

$\begin{array}{l} 5000 = P_{0} {(1 + \frac{0.049}{52})}^{3 \times 52} \\ P_{0} = $ 4317.77 \end{array}$

Mrs. Brown will need to deposit $4317.77.

Two friends decide to save money for retirement. Tom puts away $100 a month for thirty years into an
account with an annual rate of 5% compounded monthly. His friend Jerry decides to wait ten years before
investing. How much must Jerry deposit each month to catch up to Tom when they both retire in thirty years
(assume Jerry will be depositing into a similar account).

Answer

Both Tom and Jerry are putting a fixed payment away each month. This means we have an annuity problem for
each of them. What we need to find is the monthly deposit Jerry makes after waiting ten years to start so
that he has the same ending balance as Tom during that time period. To solve this we can first figure out
how much Tom will save up over the 30 years making those $100 monthly deposits.
Tom’s Total:

$\begin{array}{l} P_{N} = \frac{d [{(1 + \frac{r}{k})}^{k * N} - 1]}{\frac{r}{k}} \\ P_{30} = \frac{100 [{(1 + \frac{.05}{12})}^{12 * 30} - 1]}{\frac{.05}{12}} \\ P_{0} = $ 83225.86 \end{array}$

We know Jerry has to end up with that same total of $83,225.86 after only 20 years of a fixed deposit. This
is still an annuity problem, but we need to find that fixed deposit that gives us this total after 20 years.

$\begin{array}{l} P_{N} = \frac{d [{(1 + \frac{r}{k})}^{k * N} - 1]}{\frac{r}{k}} \\ 83225.86 = \frac{d [{(1 + \frac{.05}{12})}^{12 * 20} - 1]}{\frac{.05}{12}} \\ d = \frac{83225.86}{\frac{[{(1 + \frac{.05}{12})}^{12 * 20} - 1]}{\frac{.05}{12}}} \\ d = $ 202.48 \end{array}$

By waiting those ten years we see Jerry has to make a deposit almost twice as big as Tom’s to reach that
same goal at the end of the initial 30 years.

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

6.6 Applications

5.7

Which Equation to Use

Learning Objectives

Which equation to use?

Try it Now

Answer

Example 1

Example 2

Example 3

Exercises

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

Answer

License

Share This Book