GMAT Interest Practice Questions

Question 1 of 3 free Easy

Jolene entered an 18-month investment contract that guarantees to pay 2 percent interest at the end of 6 months, another 3 percent interest at the end of 12 months, and 4 percent interest at the end of the 18 month contract. If each interest payment is reinvested in the contract, and Jolene invested $10,000 initially, what will be the total amount of interest paid during the 18-month contract?

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Correct answer: E

In this question, interest is accruing on previous interest because interest payments are being reinvested into the contract. We cannot use a standard compound interest formula here because we don't have one fixed rate to compound.

Step 1: Calculate the interest earned after the first 6 months.

$2\% \times \$10{,}000 = \$200$

Balance after 6 months: $\$10{,}000 + \$200 = \$10{,}200$

Step 2: Calculate the interest earned after the next 6 months (months 7-12).

$3\% \times \$10{,}200 = \$306$

Balance after 12 months: $\$10{,}200 + \$306 = \$10{,}506$

Step 3: Calculate the interest earned after the final 6 months (months 13-18).

$4\% \times \$10{,}506 = \$420.24$

Balance after 18 months: $\$10{,}506 + \$420.24 = \$10{,}926.24$

Step 4: Calculate the total interest paid.

Total interest = Final balance - Initial investment = $\$10{,}926.24 - \$10{,}000 = \$926.24$

Alternatively: Total interest = $\$200 + \$306 + \$420.24 = \$926.24$

The correct answer is E.

Question 2 of 3 free Easy

The population of locusts in a certain swarm doubles every two hours. If 4 hours ago there were 1,000 locusts in the swarm, in approximately how many hours will the swarm population exceed 250,000 locusts?

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Correct answer: D

The population of locusts doubles every 2 hours. We are told that 4 hours ago, there were 1,000 locusts.

First, determine the current population:
4 hours ago: $1{,}000$
2 hours ago: $1{,}000 \times 2 = 2{,}000$
Now: $2{,}000 \times 2 = 4{,}000$

Next, track the population forward from now until it exceeds 250,000:
In 2 hours: $4{,}000 \times 2 = 8{,}000$
In 4 hours: $8{,}000 \times 2 = 16{,}000$
In 6 hours: $16{,}000 \times 2 = 32{,}000$
In 8 hours: $32{,}000 \times 2 = 64{,}000$
In 10 hours: $64{,}000 \times 2 = 128{,}000$
In 12 hours: $128{,}000 \times 2 = 256{,}000$

At 12 hours from now, the population reaches 256,000, which exceeds 250,000.

Alternatively, using the geometric progression formula $a_n = a_1 \times r^{n-1}$, where $a_1$ is the first term and $r$ is the common ratio:
Here, $a_1 = 4{,}000$ (current population) and $r = 2$. We need to find the smallest $n$ such that $4{,}000 \times 2^{n-1} > 250{,}000$.

Dividing both sides by 4,000: $2^{n-1} > 62.5$
Since $2^5 = 32$ and $2^6 = 64$, we need $n - 1 = 6$, so $n = 7$.

This means 6 doubling cycles (each 2 hours) are needed: $6 \times 2 = 12$ hours.

The correct answer is D.

Question 3 of 3 free Medium

Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?

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Correct answer: A

If we decide to find a constant multiple by the hour, then we can say that the population was multiplied by a certain number three times from 1 p.m. to 4 p.m.: once from 1 to 2 p.m., again from 2 to 3 p.m., and finally from 3 to 4 p.m.

Let's call the constant multiple x.

$ 2000(x)(x)(x) = 250,000 $
$ 2000(x^3) = 250,000 $
$ x^3 = \frac{250,000}{2,000} = 125 $
$ x = 5 $

Therefore, the population gets five times bigger each hour.
At 3 p.m., there were $ 2000(5)(5) = 50,000 $ bacteria.

The correct answer is A.

Question 4 of 5 preview Medium

Donald plans to invest x dollars in a savings account that pays interest at an annual rate of $ 8\% $ compounded quarterly. Approximately what amount is the minimum that Donald will need to invest to earn over $ \$100 $ in interest within 6 months?

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Question 5 of 5 preview Hard

Louie takes out a three-month loan of $ \$1000 $. The lender charges him $ 10\% $ interest per month compounded monthly. The terms of the loan state that Louie must repay the loan in three equal monthly payments. To the nearest dollar, how much does Louie have to pay each month?

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