Question

A sum of Rs 1000 is lent to be returned in 11 monthly installments of Rs 100 each, interest being simple. The rate of interest:
(a) $9\dfrac{1}{11}%$
(b) $10%$
(c) $11%$
(d) $21\dfrac{9}{11}%$

Answer 1

Hint: We know that the formula of simple interest is equal to $S.I.=\dfrac{P\times R\times T}{100}$ where “P” stands for the principal amount, “R” stands for the annual rate of interest and “T” stands for the time in years. Principal is equal to 1000 and time is given in months so convert into years first then substituting these values in the simple interest formula. Amount paid at the end of 11 months is equal to the addition of principal amount and the simple interest. Mark this as the eq. (1). Then the effective payment at the end of 11 months with the monthly installment of 100 each is calculated by adding the amount at the end of 10 months with the amount at the end of 9 months till the amount at the end of 1st month. Mark this as eq. (2). Now, equate eq. (1) and eq. (2) to get the rate of interest.

Complete step-by-step answer:
It is given that the principal amount is Rs 1000. And the amount at the end of 11 months is calculated by adding principal with the simple interest at the end of 11 months.
The formula for simple interest is given by:
$S.I.=\dfrac{P\times R\times T}{100}$
In the above formula, “P” stands for principal, “R” stands for the annual rate of interest and “T” stands for the time in years. Principal is given as Rs 1000; time is 11 months which on converting to years will be equal to $\dfrac{11}{12}$. Substituting these values in the above formula we get,
$S.I.=\dfrac{1000\times R\times }{100}\left( \dfrac{11}{12} \right)$
Amount that needs to pay at the end of 11 months is equal to:
$\begin{align}
  & 1000+\dfrac{1000\times R}{100}\left( \dfrac{11}{12} \right) \\
 & \Rightarrow 1000\left( 1+\dfrac{11R}{1200} \right).......eq.(1) \\
\end{align}$
Now, we have given that monthly installment is given as Rs 100 each so the effective payment at the end of 11 months is calculated by:
Amount to be paid at the end of 1 month is given as Rs 100.
Amount to be paid at the end of 2 months is given by the addition of 100 with the simple interest on Rs 100 for 1 month which is given as:
$\begin{align}
  & 100+\dfrac{100\times R\times \dfrac{1}{12}}{100} \\
 & =100\left( 1+\dfrac{R}{1200} \right) \\
\end{align}$
Amount to be paid at the end of 3 months is calculated in the same way as we have shown above for 2 months. The only difference is that here the time is for 2 months.
$\begin{align}
  & 100+\dfrac{100\times R\times \dfrac{2}{12}}{100} \\
 & =100\left( 1+\dfrac{2R}{1200} \right) \\
\end{align}$
Similarly, you can find the amount for 4 months, 5 months till 11 months. In the below, we are showing the amount paid at the end of 11 months which is:
$\begin{align}
  & 100+\dfrac{100\times R\times \dfrac{10}{12}}{100} \\
 & =100\left( 1+\dfrac{10R}{1200} \right) \\
\end{align}$
Now, adding all the months’ amount that we have shown above to get the effective payment at the end of 11 months we get,
$\begin{align}
  & 100+100\left( 1+\dfrac{R}{1200} \right)+100\left( 1+\dfrac{2R}{1200} \right)+........+100\left( 1+\dfrac{10R}{1200} \right) \\
 & =100\left( 11 \right)+100\left( \dfrac{\left( 1+2+3+......+10 \right)R}{1200} \right) \\
\end{align}$
The summation of 1 + 2 + 3 + 4…..10 is done by adding the numbers from 1 to 10 like adding 1 and 2 which is 3 when add this number to 4 repeat this process till the number 10. After adding these numbers we get the summation value as 55.
Substituting this summation value in the above equation we get,
$\begin{align}
  & 100\left( 11 \right)+100\left( \dfrac{\left( 55 \right)R}{1200} \right) \\
 & =100\left( 11+\dfrac{55R}{1200} \right).........eq.(2) \\
\end{align}$
From the above, the effective payment at the end of 11 months is equal to $100\left( 11+\dfrac{55R}{1200} \right)$.
Equating eq. (1) and eq. (2) we get,
$1000\left( 1+\dfrac{11R}{1200} \right)=100\left( 11+\dfrac{55R}{1200} \right)$
Dividing 100 on both the sides we get,
\[\begin{align}
  & 10\left( 1+\dfrac{11R}{1200} \right)=\left( 11+\dfrac{55R}{1200} \right) \\
 & \Rightarrow 10+\dfrac{110R}{1200}=11+\dfrac{55R}{1200} \\
\end{align}\]
Rearranging the above equation we get,
$\begin{align}
  & \dfrac{110R}{1200}-\dfrac{55R}{1200}=1 \\
 & \Rightarrow \dfrac{55R}{1200}=1 \\
\end{align}$
On cross multiplication of the above equation we get,
$55R=1200$
$\begin{align}
  & \Rightarrow R=\dfrac{1200}{55} \\
 & \Rightarrow R=\dfrac{240}{11}\% \\
\end{align}$
We can also write $\dfrac{240}{11}\%$ as $21\dfrac{9}{11}\%$. From the above solution, the annual rate of interest is equal to $21\dfrac{9}{11}\%$.
Hence, the correct option is (d).

Note: The most plausible mistake that could happen is the calculation mistake. In the above solution we have written the sum of first 10 consecutive numbers as 55. In the below, we have shown a trick to calculate the sum of first 10 consecutive numbers starting from 1.
We know that summation of first n consecutive terms i.e. $1+2+3+.......+n$ is equal to:
$\dfrac{n\left( n+1 \right)}{2}$
As we have to find the sum of first 10 consecutive terms so substituting the value of “n” as 10 in the above formula we get,
$\begin{align}
  & \dfrac{10\left( 11 \right)}{2} \\
 & =55 \\
\end{align}$