Question 85

If compound interest received on a certain amount in the 2nd year is Rs 1200, what will be the compound interest (in Rs) for the 4th year on the same amount at 10% rate of interest?

Solution

Let the sum = P = Rs. $$100x$$

Rate of interest = 10%

Compound interest in the 2nd year = (Amount received at the end of 2 years) - (Amount received after 1st year)

=> $$P(1+\frac{r}{100})^2-P(1+\frac{r}{100})^1=1200$$

=> $$[100x(1+\frac{10}{100})^2]-[100x(1+\frac{10}{100})]=1200$$

=> $$[100x(\frac{11}{10})^2]-[100x(\frac{11}{10})]=1200$$

=> $$121x-110x=1200$$

=> $$x=\frac{1200}{11}$$

$$\therefore$$ Compound interest for the 4th year

= $$P(1+\frac{r}{100})^4-P(1+\frac{r}{100})^3$$

= $$[100x(1+\frac{10}{100})^4]-[100x(1+\frac{10}{100})^3]$$

= $$[100x(\frac{11}{10})^4]-[100x(\frac{11}{10})^3]$$

= $$(100x\times \frac{11^3}{10^3})(\frac{11}{10}-1)$$

= $$(100x\times \frac{11^3}{10^3})(\frac{1}{10})$$

= $$[(100\times \frac{1200}{11})(\frac{11^3}{10000})]$$

= $$12\times121=Rs.$$ $$1452$$

=> Ans - (A)


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