A sum of money when invested for 7.2 years gives ₹5,768 as the amount on maturity. If the simple rate of interest was 2% per annum more than what it was, the amount on maturity would have been ₹6,272. The original sum invested was:

Correct option is A

Given:

Time (T) = 7.2 years

Amount 1 (A1) = ₹5,768

Amount 2 (A2) = ₹6,272

Rate difference = 2% per annum

Formula Used:

Simple Interest (SI) =$$\frac{ (P × R × T) }{ 100}$$​​

Amount (A) = Principal (P) + Simple Interest (SI)

Solution:

Let P be the principal and R be the original rate of interest.

Equation 1: 5768 = P + $$\frac{(P × R × 7.2) }{ 100}$$​​

Equation 2: 6272 = P +$$\frac{ (P × (R + 2) × 7.2) }{ 100}$$​​

A2 - A1 = 6272 - 5768 = 504

This difference of 504 is the difference of the simple interest earned because of the 2% increase in rate.

The difference in interest is caused by the 2% increase in rate. Thus.

504 =$$\frac{P \times 2 \times 7.2 }{ 100}$$​​

50400 = P $$\times$$​ 14.4

P =$$\frac{50400 }{ 14.4}$$​​

P = 3500

Therefore, the original sum invested was ₹3,500.

Alternate Method: 

$$7.2\times 2 = 14.4\% \rightarrow 504 \\ \ \\100\% = \frac{504}{14.4}\times 100$$   

= 3500