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Mathematics

A man invests ₹ 7,000 for three years, at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 7,980.

Calculate :

(i) the rate of interest per annum.

(ii) the interest accrued in the second year.

(iii) the amount at the end of the third year.

Compound Interest

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Answer

(i) Let R be the rate of interest.

For the first year :

P = ₹ 7,000, R = R %, T = 1 year, A = ₹ 7,980

Amount at the end of first year = P + I

⇒ 7,980 = ₹ 7,000 + I

⇒ I = 980

Interest for first year=P×R×T100980=7,000×R×1100R=980×1007000\text{Interest for first year} = \dfrac{P \times R \times T}{100}\\[1em] 980 = \dfrac{7,000 \times R \times 1}{100}\\[1em] R = \dfrac{980 \times 100}{7000}

Hence, the rate of interest p.a. = 14 %.

(ii) For the second year :

P = ₹ 7,980, R = 14 %, T = 1 year

Interest for second year=P×R×T100=7,980×14×1100=1,11,720100=1,117.2\text{Interest for second year} = \dfrac{P \times R \times T}{100}\\[1em] = \dfrac{7,980 \times 14 \times 1}{100}\\[1em] = \dfrac{1,11,720}{100}\\[1em] = 1,117.2

Hence, the interest accrued in the second year = ₹ 1,117.2.

(iii) Principal for third year = P + I

= ₹ 7,980 + ₹ 1,117.2

= ₹ 9,097.2

P = ₹ 9,097.2, R = 14 %, T = 1 year

Interest for third year=P×R×T100=9,097.2×14×1100=1,27,360.8100=1,273.60\text{Interest for third year} = \dfrac{P \times R \times T}{100}\\[1em]= \dfrac{9,097.2 \times 14 \times 1}{100}\\[1em] = \dfrac{1,27,360.8}{100}\\[1em] = 1,273.60

Amount for the third year = P + I

= ₹ 9,097.2 + ₹ 1,273.60

= ₹ 10,370.80

Hence, the amount at the end of the third year = ₹ 10,370.80.

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