Question

Find the sum invested at 10% compounded annually in which the interest for the third year exceeds the interest of the first year by rs 252.

Answer 1

Hey mate!

Here's your answer!!

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.

$p(1 + \frac{10}{100} ) {}^{3} - p(1 + \frac{10}{100}) {}^{2}$

$p = ( \frac{11}{10} ) {}^{3} - (\frac{11}{10} ){}^{2}$

$p ( \frac{11}{10} ){}^{2} ( \frac{11}{10} - 1)$

$= p \times \frac{121}{100} \times \frac{1}{10}$
= 121/1000p

C. I for first year = p × 10 × 1 / 100 = p/10

According to question, 121/1000p - p/10 = 252

$( \frac{121 - 100}{1000}) p = 252$
$\frac{21}{100} p = 252$
P = ₹12000

Answer 2

heya......


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/



tysm....#gozmit