Question

Q for math lovers...


here is your question.

a certain sum of money is invested at the rate of 10% per annum compound interest the interest compounded annually.if the difference between the interest of 3rd year and 1st year is rupees 1105 find the sum invested

please ans .....

Answer 1

For Annual compound intrest,Amount A at the end of n years is given by 
  A = P (1+r)^{n}P(1+r)n where, r is rate of intrest expressed as  percentage 

Rate of intrest on a amount C =c*r where r is rate of intrest expressed as  percentage
Let , the invested amount be P
rate of intrest r=10/100=0.1 

Intrest of first year = P * 10/100 = 0.1P
Amount at the end of second year = A2=P (1+0.1)^{2}P(1+0.1)2 
Intrest of third year = A2*0.1

Given , difference between intrest of third year and first year is rs.1105

So, 0.1*P*(1.1)^{2}(1.1)2   - 0.1P = 1105

0.1P(1.1^{2}-11.12−1 ) = 1105

0.1P(0.21) = 1105

P = Rs.52619

Answer 2

Given, R = 10%, Time n = 1 years.

(i) 1st year:

Interest = (x * 10 * 1)/100

              = 10x/100

              = 0.1 x.

Amount = x + 0.1 x

              = 1.1 x.

(ii) 2nd year:

Interest = (1.1x + 10 * 1)/100

             = 0.11 x.

Amount = 1.1x + 0.11x

             = 1.21 x.

(iii) 3rd year:

Interest = (1.21x + 10 * 1)/100

             = 0.121 x

Amount = 1.21 x + 0.121 x

              = 1.331 x.

Now,

Given Difference between the interest of 3rd year and 1st year is 1105.

⇒ 0.121x - 0.1x = 1105.

⇒ x = 52619.047.

Therefore, the sum is 52619.04 (or) ~52619.

Hope this helps!