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    The total amount payable at maturity on a certain sum invested at simple interest at 8% p.a. for 3123 \frac12321​​ years exceeds the simple inter
    Question

    The total amount payable at maturity on a certain sum invested at simple interest at 8% p.a. for 3123 \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:

    A.

    7,800

    B.

    8,000

    C.

    7,500

    D.

    8,200

    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 = P×R×T100\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;

    SI1=P×8100×3.5=P×0.28\text{SI}_1 = P \times \frac{8}{100} \times 3.5 = P \times 0.28

    A1=P+SI1=P+P×0.28=P(1+0.28)=P×1.28A_1 = P + \text{SI}_1 = P + P \times 0.28 = P(1 + 0.28) = P \times 1.28 

    For second condition; 

    SI2=P×10.5100×12=P×1.26\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:

    A1SI2=156A_1 - \text{SI}_2 = 156

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

    P×(1.281.26)=156P \times (1.28 - 1.26) = 156​​

    P×0.02=156P \times 0.02 = 156​​

    P=1560.02=7800P = \frac{156}{0.02} = 7800 

    Thus, Principal be ₹7800

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