Question 141

The compound interest on a certain sum invested for 2 years at 10% per annum is ₹1,522.50, the interest being compounded yearly. The sum is:

Solution

Let the amount =P

So, $$A=P(1+\dfrac{R}{100})^n$$

$$A=P(1+\dfrac{10}{100})^2$$

$$A=P\dfrac{11\times 11}{100}------(i)$$

Given that $$A-P=1522.5-----(ii)$$

From the equation (i) and (ii)

$$\Rightarrow P\dfrac{121}{100}-P=1522.5$$

$$\Rightarrow \dfrac{121P-100P}{100}=1522.5$$

$$\Rightarrow 21P=152250$$

$$\Rightarrow P=\dfrac{152250}{21}=Rs. 7250$$

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