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Two equal sums of money were lent at simple interest at $11\% $p.a. for $3\dfrac{1}{2}$ years and $4\dfrac{1}{2}$ years respectively. If the difference in interest for two periods was $Rs.412.50$, find the sum.

  A)Rs.3250
  B)Rs.3500
  C)Rs.3750
  D)Rs.4250

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Last updated date: 27th May 2024
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Answer
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Hint: Use simple interest formula to find simple interest -$1$ and simple interest-$2$ with given respective years of time period.
                    Simple interest =$\dfrac{{P \times T \times R}}{{100}}$

Complete step-by-step answer:
Here we need to find simple interests for the given years of time period.
According to data given let us find individual simple interest for given years of time period.
Let the each sum of money be $Rs.X$
Required data for simple interest-$1$
Principle =$Rs.X$
Rate = $11\% $p.a.
Time =$4\dfrac{1}{2}$ years = $4.5$years
Simple interest = $\dfrac{{P \times T \times R}}{{100}}$
Then,
Simple interest - $1$ ($S.{I_1}$) = $\dfrac{{X \times 11 \times 4.5}}{{100}}$
Required data for simple interest -$2$
Principle =$Rs.X$
Rate = $11\% $p.a.
Time = $3\dfrac{1}{2}$ years = $3.5$ years
Simple interest = $\dfrac{{P \times T \times R}}{{100}}$
Then,
Simple interest - $2$($S.{I_2}$) = $\dfrac{{X \times 11 \times 3.5}}{{100}}$
Given, difference in interest for two periods = $Rs.412.50$

Difference = $S.{I_1}$ - $S.{I_2}$
 $412.50$ = $\dfrac{{X \times 11 \times 4.5}}{{100}}$ - $\dfrac{{X \times 11 \times 3.5}}{{100}}$
On simplification, we get
$
   \Rightarrow 99X - 77X = 412.50 \times 200 \\
   \Rightarrow 22X = 82500 \\
   \Rightarrow X = \dfrac{{82500}}{{22}} \\
   \Rightarrow X = 3750 \\
 $
As we consider that the each sum of money as = $Rs.X$
Hence the value of each sum of money is $Rs.3750$
Option C is the correct answer.

NOTE: Here we have to find the simple interest individually for given periods. As we know that simple interest for a given time period will be different. And after finding simple interest-1 and simple interest-2, we have to find the difference between them.
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