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 rupees 252

Answer 1

Let intial amount is P
Given,
rate of interest , r = 10%
C.I for the 3rd year = 252 + C.I for the 1st year

Now, C.I for the 3rd years = amount after 3 years - amount after 2 years
= P(1 + r/100)³ - P(1 + r/100)²
= P(1 + 10/100)³ - P(1 + 10/100)²
= P(11/10)³ - P(11/10)²
= P(1331/1000 - 121/100)
= 121P/1000

And C.I for 1st year = S.I of 1st year = P × r × 1/100
= P × 10/100 = P/10

Now, 121P/1000 = 252 + P/10
⇒ 121P/1000 - P/10 = 252
⇒(121 - 100)P/1000 = 252
⇒ 21P = 252 × 1000
⇒ P = 12000

Hence initial amount = 12000 Rs

Answer 2

Answer:

Let the sum be P.

R%= 10%

T= 3years

For the first year:

C.I. = P*10*1/100 = P/10

Amount = P/10 + P = P +10P/10 =11P/10

For the second year:

C.I.= 11P*10*1/100*10 = 11P/100

Amount = 11P/100 +11P/10 = 11P +110P/100 =121P/100

For the third year:

C.I.= 121P*10*1/100*100 = 121P/1000

Given:

Interest for 1st year + Rs.252 = Interest for 3rd year

or, P/10 +Rs.252 = 121P/1000

or, P +Rs.2520/10 =121P/1000

or, P + Rs.2520 = 121P/100

or,100(P +Rs.2520) = 121P

or, 100P + Rs.252000 =121P

or, Rs.252000 = 121P - 100P

or, Rs.252000 =21 P

or, Rs.252000/21 = P

or, Rs. 12000 =P

The sum =P =Rs.12000