Two equal sums of money were invested, one at $4 %$ and the other at $4 \frac{1}{2}$ %. At the end of $7$ years, the simple interest received from the latter exceeded that received from the former by Rs. $31.50$ Each sum was

Rs.

1000

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Rs.

500

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Rs.

750

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900

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Solution

The correct option is D Rs $900$
Change in S.I. $=$ Rs. $31.50$
$R_{1} = 4, R_{2} = 4 \frac{1}{2}, T_{1} = 7, T_{2} = 7$
Let the principal be $P$ ,
Thus, change in SI $= \frac{P}{100} (R_{2} T 2 - R_{1} T_{1})$
$\Rightarrow 31.50 = \frac{P}{100} (4 \frac{1}{2} \times 7 - 4 \times 7)$
$\Rightarrow 3150 = 7 P (\frac{9}{2} - 4)$
$\Rightarrow 450 = P (\frac{9 - 8}{2})$
$\Rightarrow P = 450 \times 2$
$\Rightarrow P = 900$
Thus, the sum $=$ Rs. $900$

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Two equal sums of money were invested, one at 4% and the other at 412%. At the end of 7 years, the simple interest received from the latter exceeded that received from the former by Rs. 31.50 Each sum was

The correct option is D Rs 900Change in S.I. = Rs. 31.50R1=4,R2=412,T1=7,T2=7Let the principal be P,Thus, change in SI =P100(R2T2−R1T1)⇒31.50=P100(412×7−4×7)⇒3150=7P(92−4)⇒450=P(9−82)⇒P=450×2⇒P=900Thus, the sum = Rs. 900

Two equal sums of money were invested, one at $4 %$ and the other at $4 \frac{1}{2}$ %. At the end of $7$ years, the simple interest received from the latter exceeded that received from the former by Rs. $31.50$ Each sum was