GMAT Math : Interest Problems

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Example Questions

Example Question #1 : Calculating Compound Interest

Scott wants to invest $1000 for 1 year. At Bank A, his investment will collect 3% interest compounded daily while at Bank B, his investment will collect 3.50% interest compounded monthly. Which bank offers a better return? How much more will he receive by choosing that bank over the other?

Possible Answers:

$Bank\ A;\ \$1005.12$

$Bank\ A;\ \$5.12$

$Bank\ B;\ \$5.12$

$Bank\ B;\ \$1005.12$

Correct answer:

$Bank\ B;\ \$5.12$

Explanation:

Calculate the total amount from each bank using the following formula:

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

Bank A:

$A=1000\left(1+\frac{0.03}{365}\right)^{(1)(365)}=1030.45$

Bank B:

$A=1000\left(1+\frac{.035}{12}\right)^{(1)(12)}=1035.57$

1035.57-1030.45=5.12

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Example Question #1 : Calculating Compound Interest

Bryan invests $8,000 in both a savings account that pays 3% simple interest annually and a certificate of deposit that pays 8% simple interest anually. After the first year, Bryan has earned a total of $365.00 from these investments. How much did Bryan invest in the certificate of deposit?

Possible Answers:

$\$4,500$

$\$1,500$

$\$3,500$

$\$5,500$

$\$2,500$

Correct answer:

$\$2,500$

Explanation:

Let be the amount Bryan invested in the certificate of deposit. Then he deposited $\$ 8,000-C$ in a savings account. 8% of the amount in the certificate of deposit is 0.08C , and 3% of the amount in the savings account is $0.03 \left ( 8,000-C \right )$ ; add these interest amounts to get $365.00. Therefore, we can set up and solve the equation:

$0.08C + 0.03 \left ( 8,000-C \right ) = 365$

$0.08C + 0.03 \cdot 8,000 -0.03 \cdot C = 365$

0.05C + 240 = 365

0.05C + 240-240 = 365 -240

0.05C = 125

$0.05C \div 0.05 = 125 \div 0.05$

C = 2,500

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Example Question #1 : Calculating Compound Interest

Barry invests $9000 in corporate bonds at 8% annual interest, compounded quarterly. At the end of the year, how much interest has his investment earned?

Possible Answers:

$\$734.40$

$\$741.89$

$\$800.00$

$\$747.00$

$\$720.00$

Correct answer:

$\$741.89$

Explanation:

Use the compound interest formula

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

substituting P = 9,000 (principal, or amount invested), r = 0.08 (decimal equivalent of the 8% interest rate), n = 4 (four quarters per year), t = 1 (one year).

$A =9,000 \left ( 1+ \frac{0.08}{4}\right )^{4 \cdot 1}$

$=9,000 \left ( 1.02 \right )^{4 }$

$\approx 9,741.89$

Subtract 9,000 from this figure - the interest earned is $741.89

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Example Question #1 : Calculating Compound Interest

Tom deposits his $10,000 inheritance in a savings account with a 4% annual interest rate, compounded quarterly. He leaves it there untouched for six months, after which he withdraws $5,000. He leaves the remainder untouched for another six months.

How much interest has Tom earned on the inheritance after one year?

Possible Answers:

$\$306.08$

$\$305.54$

$\$306.58$

$\$300.00$

$\$304.00$

Correct answer:

$\$305.54$

Explanation:

Since in each case the interest is compounded quarterly, the annual interest rate of 4% is divided by 4 to get 1%, the effective quarterly interest rate.

The $10,000 remains in the savings account six months, or two quarters, so 1% is added twice - equivalently, the $10,000 is multiplied by 1.01 twice:

$\$ 10,000.00 \times 1.01 = \$ 10,100.00$

$\$ 10,100.00 \times 1.01 = \$ 10,201.00$

$5,000 is withdrawn from the savings account, leaving $\$ 10,201.00 - \$ 5,000.00 = \$ 5,201.00$

This money is untouched for six months, or two quarters, so again, we multiply by 1.01 twice:

$\$ 5,201.00 \times 1.01 = \$ 5,253.01$

$\$ 5,253.01 \times 1.01 =\$ 5,305.54$

Subtract $5,000 to get the interest:

$\$ 5,305.54 - \$5,000.00 = \$305.54$

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Example Question #1 : Calculating Compound Interest

On January 1, Gary borrows $10,000 to purchase an automobile at 12% annual interest, compounded quarterly beginning on April 1. He agrees to pay $800 per month on the last day of the month, beginning on January 31, over twelve months; his thirteenth payment, on the following January 31, will be the unpaid balance. How much will that thirteenth payment be?

Possible Answers:

$\$1,122.25$

$\$1,020.79$

$\$448.00$

$\$913.16$

$\$661.07$

Correct answer:

$\$913.16$

Explanation:

12% annual interest compounded quarterly is, effectively, 3% interest per quarter.

Over the course of one quarter, Gary pays off $\$800 \times 3 = \$2,400$ , and the remainder of the loan accruses 3% interest. This happens four times, so we will subtract $2,400 and subsequently multiply by 1.03 (adding 3% interest) four times.

First quarter:

$\$10,000 - \$2,400 = \$7,600$

$\$7,600 \times 1.03 = \$7,828$

Second quarter:

$\$7,828 - \$2,400 = \$5,428$

$\$ 5,428 \times 1.03 = \$5,590.84$

Third quarter:

$\$5,590.84 - \$2,400 =\$ 3,190.84$

$\$ 3,190.84 \times 1.03 = \$3,286.57$

Fourth quarter:

$\$3,286.57 - \$2,400 =\$ 886.57$

$\$ 886.57 \times 1.03 = \$913.16$

The thriteenth payment, with which Gary will pay off the loan, will be $913.16.

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Example Question #1 : Calculating Compound Interest

Jessica deposits $5,000 in a savings account at 6% interest. The interest is compounded monthly. How much will she have in her savings account after 5 years?

Possible Answers:

$\$6744$

None of the other answers are correct.

$\$1500$

$\$6500$

$\$2744$

Correct answer:

$\$6744$

Explanation:

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

where is the principal, is the number of times per year interest is compounded, is the time in years, and is the interest rate.

$A=5000(1+\frac{0.06}{12})^{(12)(5)}=6744$

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Example Question #2 : Calculating Compound Interest

A real estate company is considering whether to accept a loan offer in order to develop property. The principal amount of the loan is $400,000, and the annual interest rate is 7% compounded semi-anually. If the company accepts the loan, what will be the balance after 4 years?

Possible Answers:

491,720

414,000

615,130

14,000

526,720

Correct answer:

526,720

Explanation:

Recall the formula for compound interest:

$Amount=P(1+r/n)^{nt}$ , where n is the number of periods per year, r is the annual interest rate, and t is the number of years.

Plug in the values given in the question:

$400,000(1+0.07/2)^{2\cdot 4}=400,000(1.035)^{8}=$

400,000(1.3168)=526,720

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Example Question #2 : Calculating Compound Interest

Nick found a once-in-a-lifetime opportunity to buy a rare arcade game being sold at a garage sale for $5730. However, Nick can't afford that right now, and decides to take out a loan for $1000. Nick didn't really read the fine print on the loan, and later figures out that the loan has a 30% annualy compounded interest rate! (A very dangerous rate). How much does Nick owe on the loan 2 years from the time he takes out the loan? (Assume he's lazy and doesn't pay anything back over those 2 years.)

Possible Answers:

$\$3000$

$\$1690$

$\$1600$

$\$5070$

$\$152.10$

Correct answer:

$\$1690$

Explanation:

For compound interest, the amount Nick owes is

P(1+i)^t

where is the principal, or starting amount of the loan ($1000), is the interest rate per year (30% = .3). and is the time that has passed since Nick took out the loan. (2)

We have

1000(1+.3)^2 =1000(1.3)^2 = 1000(1.3)(1.3)=1000(1.69)=1690

Hence our answer is $1690.

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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 #21 : 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:

$\$13,310$

$\$14,520$

$\$13,000$

$\$12,100$

$\$14,461$

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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GMAT Math : Interest Problems

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Calculating Compound Interest

Example Question #1 : Calculating Compound Interest

Example Question #1 : Calculating Compound Interest

Example Question #1 : Calculating Compound Interest

Example Question #1 : Calculating Compound Interest

Example Question #1 : Calculating Compound Interest

Example Question #2 : Calculating Compound Interest

Example Question #2 : Calculating Compound Interest

Example Question #21 : Interest Problems

Example Question #21 : Interest Problems

All GMAT Math Resources

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