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Question
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.
Solution
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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