Question

A sum of money amounts to $Rs.66836.70$ in $2$ years at $3\%$ p.a. compound interest. The sum is,

Answer 1

Hint: This is a problem based on compound interest. We are given the rate of interest, time period and the amount that is interest and principal together. We have to find the principal sum. That is the amount that was invested in this compound interest. So we will use the formula of compound interest and amount calculation to solve this problem.

Formula used:
\[A = p{\left[ {1 + \dfrac{R}{{100}}} \right]^t}\]
Where,
$A$ is the amount
$P$ is the principal sum
$R$ is the rate of interest per annum
$t$ is the time period in years

Complete step by step solution:
Given that, a sum of money amounts to Rs.66836.70 in 2 years at 3% p.a.
That is the amount after adding the interest is given . we have to find the principal. So we will use the formula above.
\[A = p{\left[ {1 + \dfrac{R}{{100}}} \right]^t}\]
Now substitute the given data,
\[66836.70 = p{\left[ {1 + \dfrac{3}{{100}}} \right]^2}\]
Now taking the LCM in the bracket,
\[66836.70 = p{\left[ {\dfrac{{103}}{{100}}} \right]^2}\]
\[\Rightarrow 66836.70 = p\left[ {\dfrac{{{{103}^2}}}{{{{100}^2}}}} \right]\]
On cross multiplying we get,
\[p = \dfrac{{66836.70 \times {{100}^2}}}{{{{103}^2}}}\]
Taking the squares,
\[p = \dfrac{{66836.70 \times {{100}^2}}}{{10609}}\]
On dividing by 10609 we get and also taking the square of 100,
\[p = 6.3 \times 10000\]
On multiplying the decimal will be removed,
\[p = 63000\]
This is the principal sum invested Rs.63000.

Note:
Note that they have given all the required data simply using the formula is enough. But the time period should be in years if it is not then do make it because the rate is per cent per annum.
If asked for the interest then we will subtract the principal from the amount sum.