Question

Two equal sums of money were invested, one at 4% and the other at $4\dfrac{1}{2}\%$ . At the end of 7 years, the simple interest received from the latter exceeded than that received from the former by Rs.31.50. What was each sum?
(a) Rs.900
(b) Rs.950
(c) Rs.850
(d) Rs.800

Answer 1

Hint:Here, first we have to take each sum of money invested as x. Therefore, the principal is x. The time period given is 7 years .Then, apply the formula for the simple interest, $SI=\dfrac{P\times R\times T}{100}$. First calculate $S{{I}_{1}}$ for $4\%$ rate of interest and $S{{I}_{2}}$ for $4\dfrac{1}{2}\%$ rate of interest. It is also given that $S{{I}_{2}}-S{{I}_{1}}=31.50$. From this equation we will get the value of x.

Complete step-by-step answer:
Here, we are given that two equal sums invested one at $4\%$ and the other at $4\dfrac{1}{2}\%$. At the end of 7 years, the simple interest received from the latter exceeded than that received from the former by Rs.31.50.
Now, we have to find each sum of money invested.
So, now, we can take each sum of money invested be x.
Here, the simple interest is calculated. We, know that the simple interest is given by the formula:
$SI=\dfrac{P\times R\times T}{100}$, where P is the principal amount, R is the rate of interest and T is time period.
Now, here one of the sum is invested at the rate of $4\%$ per annum. So, by the end of 7 years the interest will be:
$\Rightarrow S{{I}_{1}}=\dfrac{x\times 4\times 7}{100}$
$\Rightarrow S{{I}_{1}}=\dfrac{28x}{100}$ ……. (1)
Now, the other sum of money is invested at the rate of $4\dfrac{1}{2}\%$, which,
$\begin{aligned} & \Rightarrow 4\dfrac{1}{2}=\dfrac{4\times 2+1}{2} \\\ & \Rightarrow 4\dfrac{1}{2}=\dfrac{8+1}{2} \\\ & \Rightarrow 4\dfrac{1}{2}=\dfrac{9}{2} \\\ & \Rightarrow 4\dfrac{1}{2}=4.5 \\\ \end{aligned}$
Therefore, the rate of interest is $4.5\%$.
Hence, the simple interest calculated at the end of 7 years,
$\Rightarrow S{{I}_{2}}=\dfrac{x\times 4.5\times 7}{100}$
$\Rightarrow S{{I}_{2}}=\dfrac{31.5x}{100}$ ……… (2)
We also have that the simple interest received from the latter exceeded than that received from the former by Rs.31.50.
Therefore, we will get:
$S{{I}_{2}}-S{{I}_{1}}=31.50$
Now, from equation (1) and equation (2) we will obtain:
$\dfrac{31.5x}{100}-\dfrac{28x}{100}=31.50$
Next, by taking LCM:
$\begin{aligned} & \Rightarrow \dfrac{31.5x-28x}{100}=31.50 \\\ & \Rightarrow \dfrac{3.5x}{100}=31.50 \\\ \end{aligned}$
Now, by cross multiplying,
$\begin{aligned} & \Rightarrow x=\dfrac{31.5\times 100}{3.5} \\\ & \Rightarrow x=\dfrac{31.5\times 100}{3.5} \\\ \end{aligned}$
In the next step, by cancellation,
$\begin{aligned} & \Rightarrow x=9\times 100 \\\ & \Rightarrow x=900 \\\ \end{aligned}$
Therefore, we can say that each sum of money invested is Rs.900.
Hence, the correct answer for this question is option (a).

Note: Here, the sum invested is the principal amount. It is given that the principal amount is the same. Therefore, you can take the same principal amount x for both the simple interests. Then, you have to find the value of x from the given data.