Question

A sum of Rs. $10$ is lent to be returned in $11$ monthly installments of Rs. $1$ each, interest being simple. The rate of interest is:
(A) $9\dfrac{1}{{11}}\% $
(B) $10\% $
(C) $11\% $
(D) $21\dfrac{9}{{11}}\% $

Answer 1

Hint: Equate the total amount to be paid back (if paid at the end of $11$ months) and total effective amount (Rs.11 + Simple interest on Rs.1 for 10 months+ Simple interest on Rs.1 for 9 months+ Simple interest on Rs.1 for 8 months+……………….+ Simple interest on Rs.1 for 1 months) to find out the rate of rate of interest.

Complete step-by-step answer:
Given, amount borrowed= Rs. $10$
 Let the rate of interest be $R\% $ per annum.
Total amount to be paid back (if paid at the end of $11$ months)= Rs. $10$+Simple interest on Rs. $10$ for $11$ months = $10 + \dfrac{{P \times R \times T}}{{100}}$
Where\[P = Principal = Rs.10\],
\[\;R = Rate{\text{ }}of{\text{ }}Interest\]
\[T = Time = 11{\text{ }}months\]
Total amount to be paid back =$10 + \dfrac{{10 \times R \times \dfrac{{11}}{{12}}}}{{100}}$
$ \Rightarrow $ Total amount to be paid back =$10 + \dfrac{{11R}}{{120}}$
\[Total{\text{ }}effective{\text{ }}amount = \] (Rs.1+ Simple interest on Rs.1 for 10 months)+ (Rs.1+ Simple interest on Rs.1 for 9 months)+ (Rs.1+ Simple interest on Rs.1 for 8 months)+……………….+ (Rs.1+ Simple interest on Rs.1 for 1 months)+ Rs.1
\[ \Rightarrow \left( {1 + \dfrac{{1 \times R \times \dfrac{{10}}{{12}}}}{{100}}} \right) + \left( {1 + \dfrac{{1 \times R \times \dfrac{9}{{12}}}}{{100}}} \right) + \left( {1 + \dfrac{{1 \times R \times \dfrac{8}{{12}}}}{{100}}} \right) + ................. + \left( {1 + \dfrac{{1 \times R \times \dfrac{1}{{12}}}}{{100}}} \right) + 1\]
\[ \Rightarrow \left( {1 + \dfrac{{10R}}{{1200}}} \right) + \left( {1 + \dfrac{{9R}}{{1200}}} \right) + \left( {1 + \dfrac{{8R}}{{1200}}} \right) + ................. + \left( {1 + \dfrac{R}{{1200}}} \right) + 1\]
\[ \Rightarrow 11 + \dfrac{R}{{1200}}\left( {10 + 9 + 8 + ........ + 1} \right)\]
\[ \Rightarrow 11 + \dfrac{{55R}}{{1200}}\]
Now we have,
\[Total{\text{ }}effective{\text{ }}amount = \] \[Total{\text{ }}amount{\text{ }}to{\text{ }}be{\text{ }}paid{\text{ }}back{\text{ }}at{\text{ }}the{\text{ }}end{\text{ }}of\;11months\]
$ \Rightarrow 11 + \dfrac{{55R}}{{1200}} = 10 + \dfrac{{11R}}{{120}}$
$ \Rightarrow 11 + \dfrac{{11R}}{{240}} = 10 + \dfrac{{11R}}{{120}}$
$ \Rightarrow 11 - 10 = \dfrac{{11R}}{{120}} - \dfrac{{11R}}{{240}}$
$ \Rightarrow 1 = \dfrac{{22R - 11R}}{{240}}$
$ \Rightarrow \dfrac{{11R}}{{240}} = 1$
$ \Rightarrow R = \dfrac{{240}}{{11}}$
$ \Rightarrow R = 21\dfrac{9}{{11}}\% $
Hence, the rate of interest is $21\dfrac{9}{{11}}\% $.

Thus, option (D) is the correct answer.

Note: The formula for calculating the simple interest is given by, $S.I. = \dfrac{{P \times R \times T}}{{100}}$; where \[P = Principal,{\text{ }}R = Rate{\text{ }}of{\text{ }}interest,{\text{ }}T = Time.\] Also, the sum of \[10 + 9 + 8 + ........... + 1\] is calculated by using the formula $\dfrac{{n\left( {n + 1} \right)}}{2} = \dfrac{{10 \times 11}}{2} = 55$.