Question

A certain sum of money is invested at the rate of 10% per annum compound interest compounded annually.if the difference between the interests of third year and first year is 1105,find the sum invested.

Answer 1

To Calculate the interest:

P[(1+r/100)^n-1], Where P = Principle(Sum invested), r = rate of interest and n= Time period.

In the given question, r = 10%

the difference between the interests of third year and first year is 1105.

P[(1+10/100)^3-1] - P[(1+10/100)^1-1] = 1105

P[(11/10)^3-1] - P[(11/10)^1-1] = 1105

P(1331/1000-1) - P(11/10-1) = 1105

P(331/1000) - P(1/10) = 1105

Now take P as common and subtract the remaining part by taking LCM

P(231/1000) = 1105

P = 1105*1000/231

P = 4783.55

So, Rs 4783.55 is the sum invested.

Answer 2

In the attachments I have answered this problem.

I have applied the compound interest formula to calculate interest.

See the attachment for detailed solution.