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Mathematics

A person invests ₹10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹11200. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of second year.

Compound Interest

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Answer

(i) Let the rate of interest be R.

Given, amount at the end of first year = ₹11200.

Interest = Amount - Principal = ₹11200 - ₹10000 = ₹1200.

Interest = P×R×T100\dfrac{P \times R \times T}{100}

⇒1200=10000×R×1100⇒1200=100R⇒R=1200100⇒R=12%.\Rightarrow 1200 = \dfrac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100R \\[1em] \Rightarrow R = \dfrac{1200}{100} \\[1em] \Rightarrow R = 12\%.

Hence, the rate of interest is 12% per annum.

(ii) Amount after first year = ₹11200.

Interest for second year = 11200×12×1100=134400100\dfrac{11200 \times 12 \times 1}{100} = \dfrac{134400}{100} = ₹1344.

Amount at the end of second year = ₹11200 + ₹1344 = ₹12544.

Hence, the amount at the end of second year = ₹12544.

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