find the sum invested at 10% compounded annually on which the interest for the third year exceeds the interest of the first year by 252
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Hi
Interest for the first year
=PR/100
= (P*10)/100
= P/10
Interest for the third year
= P{(1+10/100)^3 - 1}
=P (1331/1000 - 1)
= P (331/1000)
By the problem,
331P/1000 - P/10=252
=>231 P = 252000
=>P = Rs. 1,090.90
Thus the the sum invested is Rs.1090.90
Hope this helps. :)
Interest for the first year
=PR/100
= (P*10)/100
= P/10
Interest for the third year
= P{(1+10/100)^3 - 1}
=P (1331/1000 - 1)
= P (331/1000)
By the problem,
331P/1000 - P/10=252
=>231 P = 252000
=>P = Rs. 1,090.90
Thus the the sum invested is Rs.1090.90
Hope this helps. :)
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