find the sum invested at 10% compounded annually on which the interest for the third year exceeds the interest of the first year by rupees 252
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Let the sum be P.
Rate = 10%
C.I. for third year = Amount at the end of 3rd year – Amount at the end of 2nd year
= p [1+ 10/ 100]3 - p [ 1+10/100]2
=p { (11/10)3 - (11/10)2}
= p (11/10)2 (11/10- 1)
=p x 121/100 x 1/10
= 121/1000 P
C.I.for first year =
Px10x1 / 100 = p / 100
( S. I. = PxRxT/100)
According to the question
121/1000 P - P/100=252
(121-100 / 1000) P = 252
21/100 P = 252
Ans is, P= 12000/-
please tell me the third question answer please
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