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Mathematics

The compound interest on ₹1000 at 10% p.a. compounded annually for 2 years is

  1. ₹190

  2. ₹200

  3. ₹210

  4. ₹1210

Compound Interest

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Answer

C.I. = P[(1+r100)n−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]

Putting values in formula we get,

C.I.=1000×[(1+10100)2−1]=1000×[(110100)2−1]=1000×[(1110)2−1]=1000×[121100−1]=1000×21100=₹210.C.I. = 1000 \times \Big[\Big(1 + \dfrac{10}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{110}{100}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\Big(\dfrac{11}{10}\Big)^2 - 1\Big] \\[1em] = 1000 \times \Big[\dfrac{121}{100} - 1\Big] \\[1em] = 1000 \times \dfrac{21}{100} \\[1em] = ₹210.

Hence, Option 3 is the correct option.

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