- Demonstrate an understanding of how to solve word problems
- Learn how to solve compound interest word problems

## How to Solve Compound Interest Word Problems

### Compound Interest Formula

$$A=P\left(1+ \frac{r}{n}\right)^{nt}$$ The main thing for these word problems is to make sure you understand what each letter in the formula represents:- A is the future value or the account balance at the end of the investment period
- P is the principal, or amount invested, or the present value of the investment
- r is the annual interest rate as a decimal
- n is the number of compounding periods per year
- t is the number of years

Example #1: Solve each word problem. Round your answer to the nearest hundredth.

Heather invests $7,717 in a savings account with a fixed annual interest rate of 8% compounded 2 times per year. What will the account balance be after 7 years?

Let's just fill out our variables and plug into our formula:

$$P=\$7{,}717$$ $$r=.08$$ $$n=2$$ $$t=7$$ $$A=P\left(1+ \frac{r}{n}\right)^{nt}$$ $$A=7{,}717\left(1 + \frac{.08}{2}\right)^{2 \cdot 7}$$ $$A=13{,}363.35$$ After 7 years, the account balance will be about $13,363.35.

Example #2: Solve each word problem. Round your answer to the nearest hundredth.

Clark invests a sum of money in a 401(k) retirement account with a fixed annual interest rate of 2% compounded 4 times per year. After 19 years, the balance reaches $3,129.25. What was the amount of the initial investment?

Let's just fill out our variables and plug into our formula: $$P=\hspace{.1em}?$$ $$r=.02$$ $$n=4$$ $$t=19$$ $$A=\$3{,}129.25$$ $$A=P\left(1+ \frac{r}{n}\right)^{nt}$$ $$3{,}129.25=P\left(1+ \frac{.02}{4}\right)^{4 \cdot 19}$$ $$3{,}129.25=P\left(1+ \frac{.02}{4}\right)^{76}$$ $$P=\frac{3{,}129.25}{\left(1+ \frac{.02}{4}\right)^{76}}$$ $$P=\$2{,}142$$ The inital investment was about $2,142.

Example #3: Solve each word problem. Round your answer to the nearest hundredth.

Megan invests $2,496 in a money fund with a fixed annual interest rate compounded 3 times per year. After 4 years, the balance reaches $3,422.93. What is the interest rate of the account?

Let's just fill out our variables and plug into our formula: $$P=\$2{,}496$$ $$r=\hspace{.1em}?$$ $$n=3$$ $$t=4$$ $$A=\$3{,}422.93$$ $$A=P\left(1+ \frac{r}{n}\right)^{nt}$$ $$3{,}422.93=2{,}496\left(1+ \frac{r}{3}\right)^{3 \cdot 4}$$ $$3{,}422.93=2{,}496\left(1+ \frac{r}{3}\right)^{12}$$ $$\frac{3{,}422.93}{2{,}496}=\left(1+ \frac{r}{3}\right)^{12}$$ $$1+ \frac{r}{3}=\sqrt[12]{\frac{3{,}422.93}{2{,}496}}$$ Note: We don't need the plus or minus here as the negative can be disregarded. $$\frac{r}{3}=\sqrt[12]{\frac{3{,}422.93}{2{,}496}}- 1$$ $$r=3\left(\sqrt[12]{\frac{3{,}422.93}{2{,}496}}- 1\right)$$ $$r=.08$$ The interest rate is about 8%.

### How to Derive the Compound Interest Formula

Where does the given compound interest formula come from? Let's recall the formula for simple interest earned: $$I=Prt$$ Where I is the amount of simple interest earned, P is the principal or amount invested, r is the annual interest rate as a decimal, and t is the time in years.To get our compound interest formula, we need to rewrite the "t" using the number of compounding times per year. $$I=Pr\left(\frac{1}{n}\right)$$ $$I=P\left(\frac{r}{n}\right)$$ Let's stop for a moment and address this change with a quick example. Suppose you invest $1,000 in an account with a fixed annual interest rate of 6% compounded 12 times per year. How much interest was earned after 1 month? For the first month, we can just use the simple interest formula as we haven't yet received any interest. We just need to adjust the formula for the given time "t". Here t is given in years. Since each year is 12 months and we want 1 month, we use 1/12. $$I=1{,}000(.06)\left(\frac{1}{12}\right)=5$$ Now, let's return to our formula: $$I=P\left(\frac{r}{n}\right)$$ From here, we need to realize that after one compounding period, we would have an account balance (A). This account balance (A) is coming from the principal or amount invested plus the interest we earned. $$A=P + I$$ Plug in for I using the given formula: $$I=P\left(\frac{r}{n}\right)$$ $$A=P + P\left(\frac{r}{n}\right)$$ Rewrite using the multiplication symbol: $$A=P + P\cdot \frac{r}{n}$$ Going back to our example, you would have $1,005 in your account after one month or one compounding period. This comes from the principal ($1,000) plus the interest ($5). $$A=1{,}000 + 1{,}000 \cdot \frac{0.06}{12}$$ $$A=1{,}005$$ After the first interest is deposited in the account, we have to adjust our formula. Our principal is changing and we are receiving interest on the new principal, not the original one. How do we represent this change? Let's start with our account balance after our first compounding period. $$A=P + P\cdot \frac{r}{n}$$ Factor out the P from the right side: $$A=P\left(1 + \frac{r}{n}\right)$$ Apply the formula to the next compound period: $$A=P + I$$ $$A=P\left(1 + \frac{r}{n}\right) + P\left(1 + \frac{r}{n}\right) \cdot \frac{r}{n}$$ Let's factor out the P(1 + r/n) from the right side: $$A=P\left(1 + \frac{r}{n}\right)\left(1 + \frac{r}{n}\right)$$ $$A=P\left(1 + \frac{r}{n}\right)^2$$ Going back to our example, after two months, we would have: $$A=1{,}005 + 1{,}005(0.06)\left(\frac{1}{12}\right)$$ $$A=1{,}005\left(1 + \frac{0.06}{12}\right)$$ $$A ≈ 1{,}010.02$$ The $1,005 is the account balance after the first month. After the second month, we obtain our account balance by adding the previous account balance ($1,005) and the interest (which is now earned on the new balance of $1,005).

If we match this up with our formula from above, we would obtain the same amount: $$A=1{,}000\left(1 + \frac{0.06}{12}\right)^2$$ $$A ≈ 1{,}010.02$$ That extra $0.02 is due to compound interest. With simple interest, you would only get exactly $5 each month. In the second month, you actually got about $5.02.

If we continue with this pattern, after the third month, we would have: $$A=P\left(1 + \frac{r}{n}\right)^3$$ So after n compounding periods in 1 year, the account balance (A) is given by: $$A=P\left(1 + \frac{r}{n}\right)^n$$ And because t years will contain n • t compounding periods: $$A=P\left(1 + \frac{r}{n}\right)^{nt}$$

#### Skills Check:

Example #1

Solve each word problem. Round your answer to the nearest hundredth.

Malorie invests $8,441 in a savings account with a fixed annual interest rate of 6% compounded 12 times per year. What will the account balance be after 5 years?

Please choose the best answer.

Example #2

Solve each word problem. Round your answer to the nearest hundredth.

Jennifer invests a sum of money in a retirement account with a fixed annual interest rate of 2.74% compounded 12 times per year. After 14 years, the balance reaches $9,278.25. What was the amount of the initial investment?

Please choose the best answer.

Example #3

Solve each word problem. Round your answer to the nearest hundredth.

Edgar invests $1,332 in a retirement account with a fixed annual interest rate compounded 6 times per year. After 19 years, the balance reaches $7,271.65. What is the interest rate of the account?

Please choose the best answer.

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