6.6: Applications

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.

1. Loan Problems
   • Key Words: Loan, amortize, finance (a car), mortgage.
   • Characteristics: Involve regular payments to repay borrowed money.
   • Example: Calculating monthly payments on a home mortgage, find the amount of a loan a person can afford based on a payment amount.
2. Compound Interest Problems
   • Key Words: Compounding (monthly, weekly, daily), compounding exponentially.
   • Characteristics: One time deposit where money grows in an account over time due to interest being reinvested.
   • Example: Determining the future value of a one-time investment in a savings account with compounding interest.
3. Simple Interest Problems
   • Key Words: Simple interest, bond, simple interest add-on loan.
   • Characteristics: Interest is not reinvested; often used for short-term investments.
   • Example: Calculating the interest on a short term simple interest loan.
4. Annuity Problems
   • Basic Annuity
     • Key Words: Repeated deposits, ordinary annuity.
     • Characteristics: Regular payments into an account, starting with zero balance.
     • Example: Saving for retirement through monthly contributions.
   • Payout Annuity
     • Key Words: Repeated withdrawals, payout annuity.
     • Characteristics: Regular withdrawals from an account, ending with zero balance.
     • Example: Retirement income planning.

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. A good first step is rewriting down the given information to keep track of important details (like what the interest rate is, the amount of time, principal or payment amount), etc…

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

Exercises

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

1. 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 (click to Show/Hide)

   This is a loan problem. Use the loan formula to find $P$.

   $$\begin{array}{l}P=\frac{300\left(1-{\left(1+\frac{0.07}{12}\right)}^{\left(-12\cdot 4\right)}\right)}{\left(\frac{0.07}{12}\right)}\\ P=\$12,528.06\end{array}$$

2. 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?
3. 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%?
   Answer (click to Show/Hide)

   Start with determining the initial loan amount $P$. After $P$ is found you can answer parts $a$ and $b$.

   1. Solve for $P$.

      $$\begin{array}{l}P=\frac{1200\left(1-{\left(1+\frac{0.062}{12}\right)}^{\left(-12\cdot 30\right)}\right)}{\left(\frac{0.062}{12}\right)}\\ P=\$195,928.29\end{array}$$

      Next we find the total interest paid by adding up all payments for 30 years and subtracting the initial loan amount.

      $\left(\$1200\times 12\times 30\right)-\$195928.29=\$236071.71$

      Total interest paid was $236071.71

   2. 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 $\left(0.20\right)x$ would be the down payment. This gives us that MathJax Original Source $x-\left(.20\right)x=\$195928.29$ or solving for $x$ we find $x=\$244910.36$.
4. 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?
5. 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.
6. Suppose you invest $50 a month for 5 years into an account at an annual rate of 8% compounded monthly. After 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 (click to Show/Hide)

   Find the value after five years of $50 a month deposits.

   $$\begin{array}{l}P=\frac{pymt\left(1-{\left(1+\frac{r}{n}\right)}^{\left(-n\cdot t\right)}\right)}{\left(\frac{r}{n}\right)}\\ P=\frac{50\left({\left(1+\frac{0.08}{12}\right)}^{\left(12\cdot 5\right)}-1\right)}{\left(\frac{0.08}{12}\right)}\\ P=\$3673.84\end{array}$$

   The additional 25 years with that money set aside in the account is:

   $$\begin{array}{l}FV=P{\left(1+\frac{r}{n}\right)}^{\left(n\cdot t\right)}\\ FV=3673.84{\left(1+\frac{0.08}{12}\right)}^{\left(12\cdot 25\right)}\\ FV=\$26,966.65\end{array}$$

7. 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 (click to Show/Hide)

   Working backwards we can find the amount needed in the account at the time for retirement:

   $$\begin{array}{l}P=\frac{10000\left(1-{\left(1+\frac{0.08}{4}\right)}^{\left(-4\cdot 10\right)}\right)}{\left(\frac{0.08}{4}\right)}\\ P=\$273,554.79\end{array}$$

   Next find the amount needed to be contributed each quarter to reach the goal of $273,554.70 at the time of retirement:

   $$\begin{array}{l}pymt=\frac{273,554.79\left(\frac{0.08}{4}\right)}{\left({\left(1+\frac{0.08}{4}\right)}^{\left(15\cdot 4\right)}-1\right)}\\ pymt\approx \$2398.52\end{array}$$

   Needs to contribute $2398.52 per quarter.

8. 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 (click to Show/Hide)

   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):

   $$\begin{array}{l}P=\frac{60000\left(1-{\left(1+\frac{0.07}{1}\right)}^{\left(-1\cdot 30\right)}\right)}{\left(\frac{0.07}{1}\right)}\\ P=\$744,542.27\end{array}$$

   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:

   $$\begin{array}{l}pymt=\frac{744,542.27\left(\frac{0.07}{1}\right)}{\left({\left(1+\frac{0.07}{1}\right)}^{\left(1\cdot 35\right)}-1\right)}\\ pymt\approx \$5,385\end{array}$$

   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.

9. 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 (click to Show/Hide)

   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 P.

   $$\begin{array}{l}pymt=\$1260\\ r=0.078\\ n=12\\ t=20\\ P=\frac{1260\left(1-{\left(1+\frac{0.078}{12}\right)}^{\left(-12\cdot 20\right)}\right)}{\left(\frac{0.078}{12}\right)}\\ P=\$152,905.93\end{array}$$

10. 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?
11. 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 (click to Show/Hide)

    We can treat this an an annuity problem where we are looking to solve for the payments made into the account. Since she wants to pay the account in full we know the future value after 1 year will be:

    $$\begin{array}{l}FV=85,000-\left(.02\right)85,000\\ FV=\$83,300\end{array}$$

    The future value of the annuity is known, so solve for the payments needed to reach that goal of $83,300.

    $$\begin{array}{l}pymt=\frac{83,300.00\left(\frac{0.037}{12}\right)}{\left({\left(1+\frac{0.037}{12}\right)}^{\left(12\cdot 1\right)}-1\right)}\\ pymt\approx \$6,824.73\end{array}$$

12. 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 (click to Show/Hide)

    City Bank. Examine the total amount of money in each account for various initial deposits and lengths of time is one approach to compare how the two accounts differ.

    Another approach looks at the formulas for continuous compounding and discrete compounding where we deposit the same amount of money over the same period of time (this means to treat $P$ and $t$ in both formulas to be the same values)

    City Bank (Annual Compounding)	Wester Bank (Continuous Compounding)
    $$\begin{array}{l}FV=P{\left(1+\frac{r}{n}\right)}^{\left(n\cdot t\right)}\\ FV=P{\left(1+\frac{0.06}{1}\right)}^{\left(1\cdot t\right)}\\ FV=P{\left(1.06\right)}^t\end{array}$$	$$\begin{array}{l}FV=P{e}^{\left(r\cdot t\right)}\\ FV=P{e}^{\left(0.58\cdot t\right)}\\ FV=P{\left(e^{.058}\right)}^t\\ FV\approx P{\left(1.0597\right)}^t\end{array}$$

    Now in both formulas we have the same structure, but one base is higher than the other. That higher base will always give a greater value in the account at any given time since you end up multiplying by a larger number. This tells us that the 6% compounding annually by City Bank is a better deal.

13. 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 (click to Show/Hide)

    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 $P$

    $$\begin{array}{l}5000=P{\left(1+\frac{0.049}{52}\right)}^{\left(3\cdot 52\right)}\\ P=\$4317.77\end{array}$$

    Mrx. Brown will need to deposit $4,317.77

14. 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 (click to Show/Hide)

    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:

    $$\begin{array}{l}FV=\frac{100\left({\left(1+\frac{0.05}{12}\right)}^{\left(12\cdot 30\right)}-1\right)}{\left(\frac{0.05}{12}\right)}\\ FV=\$83225.86\end{array}$$

    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.

    $$\begin{array}{l}pymt=\frac{83225.86\left(\frac{0.05}{12}\right)}{\left({\left(1+\frac{0.05}{12}\right)}^{\left(12\cdot 20\right)}-1\right)}\\ pymt=\$202.48\end{array}$$

    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.

Attributions

This page contains modified content from David Lippman, “Math In Society, 2nd Edition.” Licensed under CC BY-SA 4.0.
This page contains content by Robert Foth, Math Faculty, Pima Community College, 2021. Licensed under CC BY 4.0.