How to solve problems on Simple and Compound Interest: CAT Quantitative Aptitude

The concept of Simple and Compound Interest is usually introduced in 8th standard. Concept being simple doesn`t get much attention while aspirants prepare for CAT but this concept gets applied indirectly in many questions. And for a holistic preparation, it becomes imperative to know all the concepts.

Let us go through some of the application problems where the concept of Simple and Compound Interest has been used.

Example:

The compound interest on a certain sum at a certain rate of interest for the 2nd year is Rs. 2,200 and for the 3rd year is Rs. 2,420. Find the principal and rate of interest.

Solution:

You should note that the problem gives the compound interest for the second and the third years but NOT for two and three years.

We know that the difference between the compound interest for (k + 1)^th year and that for the kth year is equal to the interest for one year on the compound interest for the kth year.
 
Hence the difference between the two figures i.e., 2420 and 2200 — which  is 220 will be equal to interest for one year on 2200.
2200 × r/100 = 220  r = 10%

The compound interest for the 2^nd year will be equal to the difference between the amount at end of 2nd year and the amount at the end of 1st year i.e.,
P(1 + r)² – P(1 + r) = 2200
  P(1 + r) (1 + r -1) = 2200
  P r(1 + r) = 2200  P 0.1 (1.1) = 2200
  P(0.11) = 2200
  P = 20,000
The principal and the rate of interest is Rs. 20,000 and 10% p.a. respectively.

Example:

Rs.25,000 is invested for 3 years at 12% compound interest p.a. What is the interest in the third year?

[1] Rs.3,763
[2] Rs.3,110
[3] Rs.3,000
[4] Rs.3,450

Solution:

To solve this problem, we should progress to find the interest in the following way:

Interest for 3 years = P   - P = 25000 [(1.12)³ – 1] = Rs.10123

Interest for 2 years = P - P = 25000[(1.12)2 – 1] = Rs.6360

Interest for third year = 10123 - 6360 = 3763. Hence, [1] is the correct option.