Question

A person borrows \[Rs{\text{ }}5000\] for two years at \[4\% \] per annum simple interest. He immediately lends to another person at \[6\dfrac{1}{4}\% \] per annum for \[2\] years. Find his gain in the transaction per year.
\[\left( 1 \right)\] \[Rs{\text{ }}226.50\]
\[\left( 2 \right)\] \[Rs{\text{ }}172.50\]
\[\left( 3 \right)\] \[Rs{\text{ }}112.50\]
\[\left( 4 \right)\] \[Rs{\text{ }}95.50\]

Answer 1

Hint: We have to find the gain in the transaction per year of the money a person borrowed and then lends to another person. We solve this question using the concept of finding the gain percentage and the formula for finding the simple interest. We will first calculate the amount after one year which the person had borrowed and then we would calculate the amount after one year of the money he had lent. We will then find the difference between the two amounts for the gain of the amount in the transaction per year.

Complete step-by-step solution:
Given :
\[Principal{\text{ }}amount = Rs{\text{ }}5000\]
\[Rate{\text{ }}of{\text{ }}borrowing = 4\% {\text{ }}per{\text{ }}annum\]
\[Rate{\text{ }}on{\text{ }}lending = 6\dfrac{1}{4}\% {\text{ }}per{\text{ }}annum\]
\[Time = 2{\text{ }}years\]
We also know that the formula for simple interest is stated as :
\[S.I. = \dfrac{{P \times R \times T}}{{100}}\]
Where \[P\] is the principal amount , \[R\] is the rate of interest per annum and \[T\] is the time in years.
Now using the formula , we get the simple interest of the amount borrowed as :
\[S.I{._B} = \dfrac{{P \times {R_B} \times T}}{{100}}\]
Where \[S.I{._B}\] is the simple interest of the amount borrowed , \[{R_B}\] is the rate of interest of the borrowed amount per annum and \[T\] is the time in years.
Putting the values , we get the simple interest of amount borrowed as :
\[S.I{._B} = \dfrac{{5000 \times 4 \times 2}}{{100}}\]
\[S.I{._B} = Rs400\]
Similarly , we will calculate the simple interest of amount lent as :
\[S.I{._L} = \dfrac{{P \times {R_L} \times T}}{{100}}\]
Where \[S.I{._L}\] is the simple interest of the amount lent , \[{R_L}\] is the rate of interest of the lent amount per annum and \[T\] is the time in years.
Putting the values , we get the simple interest of amount borrowed as :
\[S.I{._L} = \dfrac{{5000 \times 6\dfrac{1}{4} \times 2}}{{100}}\]
On simplifying , we get
\[S.I{._L} = \dfrac{{5000 \times 6.25 \times 2}}{{100}}\]
\[S.I{._L} = Rs625\]
We also know that the gain of transaction is given as :
\[gain = S.I{._L} - S.I{._B}\]
Substituting the values , we get
\[gain = 625 - 400\]
\[gain = 225\]
[This is the gain for the time period of \[2\] years]
The gain per transaction is given as :
\[gain{\text{ }}per{\text{ }}transaction = \dfrac{{gain}}{2}\]
Substituting the values , we get
\[gain{\text{ }}per{\text{ }}transaction = \dfrac{{225}}{2}\]
\[gain{\text{ }}per{\text{ }}transaction = Rs112.5\]
Hence , the gain in transaction per year is \[Rs112.5\]
Thus , the correct option is \[\left( 3 \right)\].

Note: We wrote the value of rate of interest of the amount lent \[6\dfrac{1}{4}\] as \[6.25\]. As using the property of writing the percentage of fixed fraction into its real number form.
The mixed fraction can be written as :
\[6\dfrac{1}{4} = \dfrac{{6 \times 4 + 1}}{4}\]
\[6\dfrac{1}{4} = \dfrac{{25}}{4}\]
And on solving , we get
\[6\dfrac{1}{4} = 6.25\]
We would have solved the question with a short trick as : directly finding the simple interest by the difference between the rate of interest of the amount borrowed and amount lent as :
\[S.I. = \dfrac{{P \times \left( {{R_L} - {R_B}} \right) \times T}}{{100}}\]
Putting the values in the formula , we would have found the amount of gain.