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Mathematics

Find the difference between compound interest and simple interest on ₹ 12000 and in 1121\dfrac{1}{2} years at 10% compounded half-yearly.

Compound Interest

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Answer

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = 1121\dfrac{1}{2}

S.I.=P×R×T100=12000×10×32100=₹1800.S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{12000 \times 10 \times \dfrac{3}{2}}{100} \\[1em] = ₹ 1800.

Calculating C.I. :

When interest is compounded half-yearly :

A=P(1+r2×100)n×2=12000×(1+10200)32×2=12000×(1+120)3=12000×(2120)3=12000×92618000=277832=₹13891.50A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 12000 \times \Big(1 + \dfrac{10}{200}\Big)^{\dfrac{3}{2} \times 2} \\[1em] = 12000 \times \Big(1 + \dfrac{1}{20}\Big)^3 \\[1em] = 12000 \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] = 12000 \times \dfrac{9261}{8000} \\[1em] = \dfrac{27783}{2} \\[1em] = ₹ 13891.50

C.I. = A - P = ₹ 13891.50 - ₹ 12000 = ₹ 1891.50.

Difference between C.I. and S.I. = ₹ 1891.50 - ₹ 1800 = ₹ 91.50

Hence, required difference = ₹ 91.50

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