What is the compound interest on a sum ₹10000 for: for 234\frac{3}{4}43 years at 12% p.a., the interest is compounded annually? (nearest to integer) A₹4049 B₹3637 C ₹3673 D ₹3649 Show answer & explanationCorrect answer: CExplanationThe correct answer is Option 3 i.e. ₹3673. A=P(1+R100)n(1+ab×R100)A = P \left(1 + \frac{R}{100}\right)^n \left(1 + \frac{a}{b} \times\frac{R}{100}\right)A=P(1+100R)n(1+ba×100R) Given : P=10000, R=12%, n=2, ab=34P = 10000,\ R = 12\%,\ n = 2,\ \frac{a}{b} = \frac{3}{4}P=10000, R=12%, n=2, ba=43 A =10000×(1.12)2×(1+34×12100)=10000×1.2544×1.09=₹13672.96= 10000 \times (1.12)^2 \times \left(1 + \frac{3}{4} \times \frac{12}{100} \right) = 10000 \times 1.2544 \times 1.09 = ₹13672.96=10000×(1.12)2×(1+43×10012)=10000×1.2544×1.09=₹13672.96 CI = 13672.96 - 10000 = ₹3673