GMAT Math : Calculating compound interest

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Example Questions

Example Question #21 : Interest Problems

Casey deposits $\$1000$ in his savings account that pays $10\%$ interest compunded yearly. Two years later, he deposits $\$100$ more into the same saving account. How much money is in Casey's account three years after he started his account?

Possible Answers:

$\$1400$

$\$1441$

$\$1500$

$\$1464.10$

$\$1844.10$

Correct answer:

$\$1441$

Explanation:

We will make use of the formula A = P(1+i)^t where is the accumulated amount, is the starting amount, is the rate of interest, and is the time in year the money is invested.

At the beginning, Casey starts with $1000, at an interest rate of 10% (or .1) and saves his money for 2 years. So after 2 years, he has

1000(1+.1)^2 = 1000(1.1)^2 = 1000(1.21) = 1210 dollars in his account.

After he sees the $1210, he deposits $100 more, and then waits one more year.

Now becomes 1310. And after this 3rd year, Casey has

1310(1.1)^1 = 1441 in his account.

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Example Question #22 : Interest Problems

If you invest $\$10,000$ today into a fund which earns a $10\%$ annually compounded interest, what amount of money will you have in the fund years from now?

Possible Answers:

$\$14,461$

$\$12,100$

$\$13,310$

$\$14,520$

$\$13,000$

Correct answer:

$\$13,310$

Explanation:

The accumulated amount at the end of 3 years will be $10,000 x (1.1)^3 .

It is easier to find the correct answer by using the following approach:

Calculate the amount accumulated at the end of each year. (Note that the interest is compounded, so use the amount accumulated at the end of the previous year to calculate the interest for the next year.)

At the end of year 1

$Amount\:accumulated= 10000 x (1+0.1) = 10000+1000=11,000$

At the end of year 2

$Amount\:accumulated= 11000 x (1+0.1) = 11000+1100=12,100$

At the end of year 3

$Amount\:accumulated= 12100 x (1+0.1) = 12100+1210=13,310$

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Example Question #31 : Interest Problems

$\$45,000$ is invested at $6\%$ compounded annual interest rate, how much will the investment yield after one year?

Possible Answers:

$\$4,254$

$\$2,700$

$\$3,250$

$\$5,400$

$\$1,350$

Correct answer:

$\$2,700$

Explanation:

This problem simply ask for the amount of interest received one year after having invested this money. We could, for this problem, either use the simple interest formula or the compounded interest formula, which is $P\left(1+\frac{r}{c}\right)^{n\cdot c}-P$ , where is the principal, the rate, the number of compounding periods and the number of years for which we invest. We substract because we only want the amount of interest, not the total value at the end of periods. Note that is positive for increases and negative for decreases.

Applying this formula we get,

$45,000\left(1+\frac{0.06}{1}\right)^{1\cdot 1}-45,000$ , giving us 2700 .

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Example Question #11 : Calculating Compound Interest

$\$45,000$ is deposited in an account paying $6\%$ compounded annual interest rate, how much will there be on the account after two years?

Possible Answers:

$\$51,789$

$\$50,100$

$\$47,700$

$\$50,562$

$\$50,132$

Correct answer:

$\$50,562$

Explanation:

We apply the compound interest rate formula

A=P(1+r)^t

where P=principal, r=rate, and t=time.

Pluggin in our values we get

$45,000(1+0.06)^{2}$ , or 50,562 .

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Example Question #31 : Interest Problems

We invest $\$100,000$ for two years in an investment paying interest rates. At the end of the two years we end up with $\$132,250$ . What is ?

Possible Answers:

$14\%$

$13\%$

$11\%$

$12\%$

$15\%$

Correct answer:

$15\%$

Explanation:

The way we should treat this problem is as an equation and plug in what we know in the compounded interest formula, as follows: $100,000(1+r)^{2}=132,250$ . By manipulating the terms we get: $(1+r)^{2}= 1.3225$ . Now, since it would be too complicated to solve this quadratic equation, we should just try with the values in the answers. For example let's try with the square of 1.12 , which turns out to be 1.2544 , therefore it is too small and we should look for a larger rate. Let's try until we find the right answer. Remember when you test answers to find the right answer, make sure you go slow so you don't have to test twice in case you would make an error.

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Example Question #14 : Calculating Compound Interest

Ten years ago today, Geri's grandmother deposited some money into a college fund that yielded interest at a rate of 3.6% compounded monthly. There is now $6,400 in the account. Assuming that no money has been deposited or withdrawn, which of the following expressions must be evaluated in order to determine the amount of money originally deposited?

Possible Answers:

$6,400 e ^{0.36}$

$6,400 - 6,400 \cdot 0.036 \cdot 10$

$\frac{6,400}{\left $1 .003 \right $^{12 0}}$

$6,400 \cdot 1 .003 ^{12 0}$

$\frac{6,400 }{e ^{0.36}}$

Correct answer:

$\frac{6,400}{\left $1 .003 \right $^{12 0}}$

Explanation:

The formula for compound interest is

$A = P \left $1+ \frac{r}{n} \right $^{nt}$ ,

where is the current, or accrued, value of the investment, is the initial amount invested, or principal, is the annual rate expressed as a decimal, is the number of periods per year, and is the number of years.

In this scenario,

A = 6,400 , r=0.036, n = 12, t = 10 ,

so the equation becomes

$6,400 = P \left $1+ \frac{0.036}{12} \right $^{12 \cdot 10}$

$6,400 = P \left $1+0.003 \right $^{12 0}$

$6,400 = P \left $1 .003 \right $^{12 0}$

$P = \frac{6,400}{\left $1 .003 \right $^{12 0}}$

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Example Question #15 : Calculating Compound Interest

Veronica's aunt invested $4,000 in some corporate bonds for her niece the day Veronica was born; the bonds paid 4% annual interest compounded continuously. No money was deposited or withdrawn over the next eighteen years.

Which of the following expressions is equal to the amount of money in the account on Veronica's eighteenth birthday?

Possible Answers:

$\frac{4,000 }{e^{0.72}}$

$\frac{4,000 }{e^{1.72}}$

$4,000 e^{0.72}$

4,000 (1 .72)

$4,000 e^{1.72}$

Correct answer:

$4,000 e^{0.72}$

Explanation:

The formula for continuously compounded interest is

$A = P e^{rt}$

where is the current, or accrued, value of the investment, is the initial amount invested, or principal, is the annual rate expressed as a decimal, and is the number of years.

In this scenario,

P = 4,000 , r=0.04 , t = 18 ,

$A = 4,000 e^{0.04 \cdot 18} = 4,000 e^{0.72}$

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Example Question #16 : Calculating Compound Interest

Carl's uncle invested money in some corporate bonds for his nephew the day Carl was born; the bonds paid 4% annual interest compounded continuously. No money was deposited or withdrawn over the next fifteen years. The current value of the bonds is $5,000.

Which of the following expressions is equal to the amount of money Carl's uncle invested initially?

Possible Answers:

$\frac{5,000 }{e^{0.6 }}$

$\frac{5,000 }{e^{1.6 }}$

$5,000 e^{0.6 }$

$5,000 \cdot 1 .04 ^{15 }$

$\frac{5,000}{ 1 .04 ^{15 }}$

Correct answer:

$\frac{5,000 }{e^{0.6 }}$

Explanation:

The formula for continuously compounded interest is

$A = P e^{rt}$

where is the current, or accrued, value of the investment, is the initial amount invested, or principal, is the annual rate expressed as a decimal, and is the number of years.

In this scenario,

A = 5,000 , r = 0.04, t = 15

The equation becomes

$5,000 = P e^{0.04 \cdot 15 }$

$5,000 = P e^{0.6 }$

$P = \frac{5,000 }{e^{0.6 }}$

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Example Question #17 : Calculating Compound Interest

An amount of money is invested in corporate bonds that pay 6% interest compounded monthly. Which of the following expressions would give the amount of time it would take for the money to double?

Possible Answers:

$\frac{ \ln 2}{0.06}$

$\frac{ \ln 2}{\ln 1.06}$

$\frac{ \ln 2}{1.06}$

$\frac{12 \ln 2}{\ln 1.005}$

$\frac{ \ln 2}{12 \ln 1.005}$

Correct answer:

$\frac{ \ln 2}{12 \ln 1.005}$

Explanation:

The formula for compound interest is

$A = P \left $1+ \frac{r}{n} \right $^{nt}$ ,

In this scenario,

A=2P , r = 0.06, n = 12 , and is unknown.

The equation becomes

$2P = P \left $1+ \frac{0.06}{12} \right $^{12 t}$

$\frac{2P }{P}= \frac{P \left $1+ \frac{0.06}{12} \right $^{12 t}}{P}$

$2 = \left $1+ \frac{0.06}{12} \right $^{12 t}$

$2 = \left $1+ 0.005 \right $^{12 t}$

$2 = \left $1 .005 \right $^{12 t}$

$\ln 2 = \ln \left [ \left $1 .005 \right $^{12 t} \right ]$

$\ln 2 =12 t \ln 1.005$

$\frac{ \ln 2}{12 \ln 1.005} =\frac{12 t \ln 1.005}{12 \ln 1.005}$

$t = \frac{ \ln 2}{12 \ln 1.005}$

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Example Question #18 : Calculating Compound Interest

Five years ago today, Jimmy's grandfather deposited $5,000 into a college fund that yielded interest at an annual rate of 4.8% compounded monthly.

Assuming that no money has been deposited or withdrawn, which of the following expressions would have to be evaluated in order to calculate the amount of money in the account now?

Possible Answers:

$\frac{ 5,000 }{1.048^{5}}$

$5,000\cdot e^{0.24}$

$\frac{ 5,000 }{1.004 ^{60}}$

$5,000\cdot 1.004^{60}$

$5,000 \cdot 1.048 ^{5}$

Correct answer:

$5,000\cdot 1.004^{60}$

Explanation:

The formula for compound interest is

$A = P \left $1+ \frac{r}{n} \right $^{nt}$ ,

In this scenario,

P = 5,000, r = 0.048, n = 12, t = 5

Therefore,

$A = 5,000 \left $1+ \frac{0.048}{12} \right $^{12 \cdot 5}$

$= 5,000 \left $1+ 0.004 \right $^{60}$

$=5,000\cdot 1.004^{60}$

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GMAT Math : Calculating compound interest

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #21 : Interest Problems

Example Question #22 : Interest Problems

Example Question #31 : Interest Problems

Example Question #11 : Calculating Compound Interest

Example Question #31 : Interest Problems

Example Question #14 : Calculating Compound Interest

Example Question #15 : Calculating Compound Interest

Example Question #16 : Calculating Compound Interest

Example Question #17 : Calculating Compound Interest

Example Question #18 : Calculating Compound Interest

All GMAT Math Resources

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