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

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Rs. 3250

Rs. 3500

Rs. 3750

Rs. 4250

answer is C.

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Detailed Solution

Given,
Difference in interest for two periods = Rs. 412.50
For 1st simple interest

(S . I_{1})

Principle amount = Rs. X (say)
Rate = 11%
Time period =

4 \frac{1}{2}

= 4.5 years
Using formula for simple interest

= \frac{P \times T \times R}{100}

Put the value

((S . I_{1})) = \frac{X \times 11 \times 4.5}{100}

For 2nd simple interest
Principle amount = Rs. X
Rate = 11 %
Time period =

3 \frac{1}{2}

= 3.5 years
Using formula for simple interest

(S . I_{2})

= \frac{P \times T \times R}{100}

Put the value

(S . I_{2}) = \frac{X \times 11 \times 3.5}{100}

Now calculate the both simple interest

Difference = S . I_{1} - S . I_{2}

Now, simplify

412.50 = \frac{X \times 11 \times 4.5}{100} - \frac{X \times 11 \times 3.5}{100}

99 X - 77 X = 412.50 \times 200

\Rightarrow 22 X = 82500

\Rightarrow X = \frac{82500}{22}

\Rightarrow X = 3750

Hence, the value of each sum of money = Rs. 3750.
Option 3 is correct.

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