CAT Interest Practice

On the Common Admissions Test (CAT), you will find numerous problems about money. This article will help you to understand how interest works. We will explain simple and compound interest, doubling time, and related topics that you’ll need to succeed on the CAT. Then you’ll have your chance to test your knowledge on an interest practice problem set!

handful of money

Here, have some money!

(Interest problems form an important part of the CAT Quantitative section. For more about what to expect on the CAT, check out this article: Math Syllabus for CAT Exam.)

The Basics: What is Simple Interest?

Suppose you lend $100 to your friend, but you know that it will take her a few months to pay it back. To give her some incentive to pay back the money sooner, you tell her: “Every month that you have the money out, you have to add $5 to the payback amount.”

So, the value of the loan after one month is $105.

After two months, the value is $110; after three months, $115 — you get the picture.

The extra money ($5 per month that the loan is out) is called interest and gets added to the original amount (the principle) when the loan comes due. We call this simple interest because the interest grows at a constant rate over time. In our little example, there is a monthly simple interest rate of 5%.

Formula for Simple Interest

Suppose the principle is P (dollars, pounds, euros, rupees, yen, or whatever the currency is in the problem). Let t stand for the time that the money is invested or lent (typically in years).

Finally, let r be the rate at which the interest grows. Make sure that r is expressed as a decimal rather than a percent. And check the units! If t is in years, then r must be an annual rate; if t is in months, then r should be a monthly rate.

Then we can compute the interest (I) and the total amount of the investment (A) using the following formulas:

  • Interest: I = Prt
  • Amount: A = P + Prt = P(1 + rt)

Let’s work out a simple interest practice problem.

Example 1

Suppose you lend $100 to your friend who agrees to pay you back with simple interest accruing at a monthly rate of 5%. If your friends does not pay any money back until a year later, what is the amount due at the end of that year?


Set up the simple interest formula. This problem asks for total amount due. Convert the rate of 5% to a decimal value: 0.05. The time will be t = 12 months.

A = P(1 + rt) = $100(1 + 0.05 · 12) = $100(1.6) = $160.

Compounding Interest

Simple interest may be fine for loans between friends, but no business or bank would ever use this formula when lending money. The problem is that simple interest does not grow fast enough to keep up with inflation.

In other words, there would be no incentive to pay back the amount in a timely manner — just wait until the value of money decreases enough!

Currencies from around the world

How does money grow with time? That’s an interesting question!

Comparing Simple and Compound Interest

Suppose that you loan another friend $100, again at 5% monthly interest, but this time you compute the amount due in a different manner. Every month that goes by without payment on the loan, you find the interest due and add it back to the principle.

For example, in the first month, both simple and compound interest loans will be the same: $105. Then in the second month, the first loan would just be $105 + $5 = $110, as we saw above. However, for the second friend, you would consider the new loan amount to be $105 and charge the amount of interest based on that new principle:

A = $105(1 + 0.05) = $105(1.05) = $110.25

It’s not a huge difference, $110.25 versus $110, but with monthly compounding, the amount will grow much quick over a long time period. The amount “snowballs” as higher balances will generate more and more interest over time.

Let’s see what the effect is after a full year.

End of Month AmountSimple InterestCompounding Interest

The difference is almost $20 by the end of the year. While that may not seem like very much, the real power of compounding shows up in long term loans. After five years, that same $100 loan would amount to $400 with simple interest and $1,867.92 using monthly compounding!

Formula for Compound Interest

In contrast to the formula for simple interest, which only requires three parameters (principle, time, and rate), the formula for compound interest requires four. We also need to know how often compounding should take place. It could be monthly, quarterly, yearly, even daily!

Let’s define our parameters:

  • P — the principle
  • r — the annual rate, or APR (sometimes labeled p.a.)
  • t — the time, in years
  • n — the number of compounding periods per year

Then the following two formulas compout the amount (A) of the loan after t years and the total interest (I) on the loan.

compound interest formula

Ready for some compound interest practice?

Example 2

Find the amount due in four years for a loan of $1,750 at 8% APR, compounded monthly. Then compute the total interest.


First, plug P = 1750, r = 0.08, t = 4, and n = 12 into the compounding interest formula to find the amount A. Note, we use n = 12 because there are 12 months in a year.

It’s very important to use as much precision as possible during the intermediate steps. Do not round anything until the final step.

Example 2 worked out

So we have a total amount of $2,407.42. To compute the interest, simply take the principle amount away.

I = $2,407.42 – $1,750 = $657.42

Simple and Compound Interest Practice Problems

Here’s a set of interest practice questions modeled from actual CAT problems. Scroll down to check your answers.

  1. What amount will be available after six years if $5200 is invested in an account bearing 4.5% interest compounded quarterly?

    A. $6,801.55   B. $6,942.12    C. $7,209.07   D. $7,551.19

  2. Suppose that a particular investment grows your money to 121% of its original value at the end of two years, using annual compounding. How many years will it take to double your money if invested into an account with the same APR but earning simple interest?

    A. 7   B. 9    C. 10  D. 11

  3. Sarah invests Rs.8000 for 10 years at 7% p.a. compound interest, where compounding occurs semi-annually. She decides to withdraw half of her accrued interest every six months for personal expenses, leaving the remaining amount invested. How much will her investment be worth at the end of the 10-year period?

    A. Rs.10,981.37   B. Rs.11,318.23    C. Rs.11,527.99  D. Rs.11,778.62

  4. Mike bought a new house and secured a mortgage from a bank for $250,000 to pay for it. Due to unforeseen circumstances, Mike could not begin paying the mortgage back until one year later. In the mean time, the mortgage accrued interest, compounded monthly. After the initial year, the mortgage balance was $259,667.22. What was the annual rate that this bank charged?

    A. 3.4%   B. 3.5%    C. 3.7%  D. 3.8%

  5. Two sisters, Lisa (age 9) and Reesa (age 11), have the same birth day, which happens to be today. Their family sets up two investment accounts today for the girls totaling $12,000. If both accounts earn 5% interest compounded annually, how much money will each girl have if the money was split so that they would have the same amount at the age of 18?

    A. $8,769.01   B. $8,854.19    C. $8,892.32  D. $8,909.17

Solutions to CAT Interest Practice Problems

  1. A. $6,801.55.

    Be sure to use the compounding interest formula. Here, P = 5200, r = 0.045, t = 6, and n = 4 (quarterly means 4 times per year).
    interest problem 1 - solution

  2. C. 10.

    This problem has two main parts. First we have to determine the rate of the original investment. Use the compounding formula with n = 1 (annual compounding).

    Interest Problem 2 - solution, part A

    Next, use r = 0.1 to set up an equation involving the simple interest formula. Solve for the unknown time t. Again, the principle amount (P) cancels from the equation, so we never needed to know exactly how much money was invested in the first place.

    P(1 + 0.1t) = 2P

    1 + 0.1t = 2

    t = 10

  3. B. Rs.11,318.23

    Semi-annual compounding implies that n = 2 in the formula. However, each compounding period (6 months), half of the accrued interest is diverted out of the investment. So the effective rate of return would be: r = 7%/2 = 3.5% = 0.035. Now plug everything into the compounding interest formula.

    Interest Problem 3 - solution

  4. D. 3.8%

    Generally, when there is an unknown rate in a compounding interest problem, you’ll have to use logarithms to solve the equation. If you need a refresher, check out: CAT Logarithm Practice.

    Set up the compounding interest formula with all of the parameters given in the problem and solve for r.

    Interest Problem 4 - solution

    Therefore, the rate is 3.8%.

  5. B. $8,854.19

    This problem requires some algebra! Let x be the amount that is initially invested in Lisa’s account. Then Reesa’s account would start with 12000-x. Furthermore, the time of investment will be different for each girl: Lisa’s account will be for 18-9 = 9 years, and Reesa’s will be for 18-11 = 7 years.

    First solve for x to find out the initial investment.

    Interest Problem 5 - solution, part A

    Finally, use x = 5707.49 as the principle amount in Lisa’s account, for t = 9 years.

    Interest problem 5 - solution, Part B

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