# XAT 2015 Quant & DI Q12: Compound Interest

This aptitude question appeared as a part of the quantitative reasoning and data interpretation section of the XAT 2015. Question 12 of a total of 33 questions that appeared in this section in XAT 2015. This one is a medium level difficulty question and tests the concept of computing the principal invested in compound interest.The explanation walks you through an interesting way to assigning variables and solving them effectively.

## Question

In the beginning of the year 2004, a person invests some amount in a bank. In the beginning of 2007, the accumulated interest is Rs.10,000 and in the beginning of 2010, the accumulated interest becomes Rs.25,000. The interest rate is compounded annually and the annual interest rate is fixed. The principal amount is _____:

1. Rs.16,000
2. Rs.18,000
3. Rs.20,000
4. Rs.25,000
5. None of the above

Correct Answer      Choice C. The principal invested is Rs. 20,000.

#### Given Data

Interest from the start of 2004 to the start of 2007 (for 3 years) = Rs.10,000
Interest from the start of 2004 to the start of 2010 (for 6 years) = Rs.25,000

#### Variables Assigned

Let the principal = x
And annual interest rate = 100r%

The formula to compute the Amount when a sum is invested in compound interest is given as follows:
A = $P $left[1 + \frac{r}{100} \right]^n \\$ Where A is the amount, P is the principal, r is the rate of interest per annum in % and n is the number of years We have assigned x as the principal and 100r% as the rate of interest. ∴ The amount for 3 years A3 = $x \left[1 + \frac{100r}{100} \right]^3 \\$ = x$1 + r)3 = x + 10000 (Amount = Principal + Interest)

And amount for 6 years A3 = x(1 + r)6 = x + 25000

Further let us take, (1 + r)3 = a
Then, x(1 + r)3 = xa = x + 10000 ..... (1)
And x(1 + r)6 = xa2 = x + 25000 ..... (2)

#### Solve the two equations to find x

xa = x + 10000 ..... (1)
xa2 = x + 25000 ..... (2)

Bringing x terms to one side in both the equations, we can rewrite the two equations as
x(a – 1) = 10000
x(a2 – 1) = 25000

Equation (2) can therefore be expressed as x(a2 – 1) = x(a - 1) (a + 1) = 25000

We know from equation (1) x(a – 1) = 10000
Substitute x(a – 1) = 10000 in equation (2). ∴ x(a - 1) (a + 1) = 10000 (a + 1) = 25000

Or, (a + 1) = $$frac{25000}{10000}\\$ = 2.5 If a + 1 = 2.5, a – 1 =$a + 1 – 2) = 2.5 – 2 = 0.5

From equation (1) we know x(a – 1) = 10000
We have computed (a – 1) = 0.5

So, x = $\frac{10000}{0.5}\\$ = 20000

The correct answer is Choice C.

## Video Explanation

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