The total amount payable at maturity on a certain sum invested at simple interest at 8% p.a. for $$3 \frac12$$​ years exceeds the simple interest payable on the same sum at 10.5% p.a.for 12 years by ₹156. The sum (in ₹)is:

Correct option is A

Given:

Simple interest rate 1 =  8% p.a. for 3.5 years

Simple interest rate 2 = 0.5% p.a. for 12 years

The difference between total amount at maturity for first case and interest in second case is ₹156.

Formula Used:

SI = $$\frac{P \times R \times T}{100}$$​

Total amount (A) at maturity:
A = P + SI

Solution: 

Let the Principal be P , then

For condition 1;

​$$\text{SI}_1 = P \times \frac{8}{100} \times 3.5 = P \times 0.28$$​

$$A_1 = P + \text{SI}_1 = P + P \times 0.28 = P(1 + 0.28) = P \times 1.28$$ 

For second condition; 

​$$\text{SI}_2 = P \times \frac{10.5}{100} \times 12 = P \times 1.26$$​​

Now,  the total amount at maturity for the first case exceeds the simple interest in the second case by ₹156:

$$A_1 - \text{SI}_2 = 156$$​

​$$P \times 1.28 - P \times 1.26 = 156$$​​

​$$P \times (1.28 - 1.26) = 156$$​​

​$$P \times 0.02 = 156$$​​

​$$P = \frac{156}{0.02} = 7800$$ 

Thus, Principal be ₹7800