GMAT Math : Interest Problems

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Example Questions

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 #131 : Word 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:

$15\%$

$12\%$

$13\%$

$11\%$

$14\%$

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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Example Question #141 : Problem Solving Questions

Darin invested $4,000 in some corporate bonds that pay 6% annual interest compounded semiannually. what will be the value of the bonds after one year (nearest dollar)?

Possible Answers:

$\$4\textup{,}494$

$\$4\textup{,}244$

$\$4\textup{,}240$

$\$4\textup{,}120$

$\$4\textup{,}247$

Correct answer:

$\$4\textup{,}244$

Explanation:

The formula for compound interest is

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

In this scenario,

P = 4,000 , r=0.06, n = 2, t = 1 ,

so the equation to solve is

$A = 4,000 \left $1+ \frac{0.06}{2} \right $^{2 \cdot 1}$

$A = 4,000 \cdot 1.03 ^{2}$

$= 4,000 \cdot 1.0609$

=4,243.60

Round this to $4,244.

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Example Question #142 : Problem Solving Questions

Phillip invests $5,000 in a savings account at 5.64% per year interest, compounded monthly. If he does not withdraw or deposit any money, how much money will he have in the account at the end of five years?

Possible Answers:

$\$6,628.89$

$\$6,578.27$

$\$6,624.52$

$\$6,410.00$

None of the other answers are correct.

Correct answer:

$\$6,624.52$

Explanation:

Use the compound interest formula

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

where P=5,000 , r=0.0564 , t=5 , and n=12

$A = 5,000\cdot \left ( 1+ \frac{0.0564}{12} \right )^{12 \cdot 5}$

$A = 5,000\cdot 1.0047 ^{60}$

$A = 5,000\cdot 1.3249 \approx 6624.52$

Phillip will have $6,624.52 in his account.

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GMAT Math : Interest Problems

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #31 : Interest Problems

Example Question #11 : Calculating Compound Interest

Example Question #131 : Word 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

Example Question #141 : Problem Solving Questions

Example Question #142 : Problem Solving Questions

All GMAT Math Resources

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