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 interests of the third year and the first year is INR $1, 105$ /- find the sum invested.

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Solution

According to the problem,
Principal ( Sum of invest) $= P$
Rate of interest $= r$
Time $= n$
The difference between third year and the first year intrest
implies that,
$\begin{matrix} P [(1 + 10 / 100)^{3} - 1] - P [(1 + 10 / 100)^{1} - 1] = 1105 P [(11 / 10)^{3} - 1] - P [(11 / 10)^{1} - 1] = 1105 P (1331 / 1000 - 1) - P (11 / 10 - 1) = 1105 P (331 / 1000) - P (1 / 10) = 1105 \end{matrix}$
Now take P as common and subtract the remaining part by taking LCM
So,
$\begin{matrix} P = 1105 * 1000 / 231 P = 4783.55 \end{matrix}$

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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 interests of the third year and the first year is INR 1,105/- find the sum invested.

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 interests of the third year and the first year is INR $1, 105$ /- find the sum invested.