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What sum of money borrowed on ${{24}^{th}}$ May will amount to Rs. 10210.2 on ${{17}^{th}}$ October of the same year at $5%$ per annum simple interest?

Answer
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Hint: We first find out the total time span the principal was under the interest rate. We find that in a unit of days. Then we convert it in years. We use the formula of simple interest to make the linear equation after we assume the principal value. We get a linear equation. We solve the equation to find the solution of the problem.

Complete step-by-step answer:
The simple interest rate is $5%$ per annum.
Let’s assume the principal value was Rs. x. This amount was under interest rate from ${{24}^{th}}$ May to ${{17}^{th}}$ October of the same year.
So, we need to calculate the number of days there in that time period. We will calculate month wise.
Month of May has $\left( 31-24 \right)=7$ days.
The Months of June, July, August, September have 30, 31, 31, 30 days respectively.
Now, in October we have till ${{17}^{th}}$ which means 17 days.
We add up all these days to count the total number of days.
Total number of days will be $\left( 7+30+31+31+30+17 \right)=146$.
We convert the days into units of years. So, $146=\dfrac{146}{365}=\dfrac{2}{5}=0.4$ years.
So, effectively Rs. x was under interest rate is $5%$ per annum for 0.4 years.
Now we apply the formula of simple interest on the problem.
If P be the principal, t be the time span, r be the rate of interest then the total outcoming amount will be A through the formula $A=P\left( 1+\dfrac{nr}{100} \right)$.
Now we put the values for our problem where $A=10210.2,P=x,n=0.4,r=5$.
We need to find the value of x.
So, $10210.2=x\left( 1+\dfrac{0.4\times 5}{100} \right)$.
We solve the equation to find the value of x.
$\begin{align}
  & 10210.2=x\left( 1+\dfrac{0.4\times 5}{100} \right) \\
 & \Rightarrow 10210.2=x\left( 1+\dfrac{2}{100} \right) \\
 & \Rightarrow 10210.2=x\left( \dfrac{102}{100} \right) \\
\end{align}$
Now we cross multiply to find out the value.
$\begin{align}
  & 10210.2=x\left( \dfrac{102}{100} \right) \\
 & \Rightarrow x=\dfrac{100\times 10210.2}{102}=\dfrac{100\times 102102}{10\times 102} \\
 & \Rightarrow x=10\times 1001=10010 \\
\end{align}$
So, the principal amount was 10010.

Note: In this kind of problem we need to remember that the unit of interest rate and time span rate should be the same. In our problem interest rate was $5%$ per annum and the total time span was also in years. The formula would have changed for quarterly or half-yearly interest rate.