rupsk wrote:What sum of money put at C.I amounts in 2 years to $8820 and in 3 years to $9261?

Let us assume the sum of money is S and rate of compound interest r percent per year.

After 1 year, sum of money = S(1 + r/100)
After 2 years, sum of money = S(1 + r/100)^2 = 8820 ............. (1)
After 3 years, sum of money = S(1 + r/100)^3 = 9261 ............. (2)

Dividing cube of (1) by square of (2), (S^3)/(S^2) = (8820^3)/(9261^2)
Hence, S = (20*(21^2))^3/(21^3)^2 = 20^3 = 8000

The correct answer is C.

rupsk wrote:thanks but how you can you do quick multiplication at the time of exam as number are huge. Do you have any method?

I won't call it a 'method' as it basically depends upon experience, observation etc.

I observed that 9261 is an weird looking number but it's pretty easy to notice that 8820 is divisible by 20. So I factored 8820 as 20*441. Now from experience I know 441 is nothing but 21^2. Again I observed that the options do not contain any fractions. So 9261 must be divisible by 4 or 5 or 3 or 7. But from its appearance I'm sure it is not divisible by 4 or 5. So I checked with 3 and 7, and it turns out that 9261 = 21*441 = 21^3

Hope that helps.