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प्रश्न
Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs. 252.
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उत्तर
Let the sum of money be Rs.100.
Rate of interest = 10% p.a.
Interest at the end of 1st year = 10% of Rs. 100 = Rs. 10
Amount at the end of 1st year = Rs. 100 + Rs. 10 = Rs. 110
Interest at the end of 2nd year = 10% of Rs. 110 = Rs. 11
Amount at the end of 2nd year = Rs. 110 + Rs. 11 = Rs.121
Interest at the end of 3rd year =10% of Rs. 121= Rs. 12.10
Difference between interest of 3rd year and 1st year
= Rs. 12.10 - Rs. 10 = Rs. 2.10
When difference is Rs. 2.10, principal is Rs. 100
When difference is Rs. 252, principal = `[100 xx 252]/2.10` = Rs.12,000.
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