Question

A man invested an amount at 12% per annum simple interest and another amount at 10% per annum simple interest. He received an annual interest of Rs 1145. But, if he had interchanged the amounts invested, he would have received Rs 90 less. What amounts did he invest at the different rates?

Answer 1

Let the amounts invested at 12% per annum and 10% per annum be Rs. x and Rs. y​, respectively.
Then, we have:
Simple interest on Rs. x at 12% p.a. for 1 year = Rs. $\left(\frac{x\times 12\times 1}{100}\right)$ = Rs. $\left(\frac{3x}{25}\right)$
Simple interest on Rs. y at 10% p.a. for 1 year = Rs. $\left(\frac{y\times 10\times 1}{100}\right)$ = Rs. $\left(\frac{y}{10}\right)$
Given:
Total simple interest = Rs. 1145
Rs. $\left(\frac{3x}{25}\right)$ + Rs. $\left(\frac{y}{10}\right)$ = Rs. 1145
⇒ $\left(\frac{6x+5y}{50}\right)=1145$
⇒ (6x + 5y) = 57250 ....(i)

Again, we have:
​Simple interest on Rs. x at 10% p.a. for 1 year = Rs. $\left(\frac{x\times 10\times 1}{100}\right)$ = Rs. $\left(\frac{x}{10}\right)$
Simple interest on Rs. y at 12% p.a. for 1 year = Rs. $\left(\frac{y\times 12\times 1}{100}\right)$ = Rs. $\left(\frac{3y}{25}\right)$
Given:
Total simple interest = Rs. (1145 − 90) = Rs. 1055
Rs. $\left(\frac{x}{10}\right)$ + Rs. $\left(\frac{3y}{25}\right)$ = Rs. 1055
⇒ $\left(\frac{5x+6y}{50}\right)=1055$
⇒ (5x + 6y) = 52750 ....(ii)
On multiplying (i) by 6 and (ii) by 5, we get:
36x + 30y = 343500 ....(iii)
25x + 30y = 263750 ....(iv)
On subtracting (iv) from (iii), we get:
11x = (343500 − 263750) = 79750
⇒ x = 7250
On substituting x = 7250 in (i), we get:
6 × 7250 + 5y = 57250
⇒ 43500 + 5y = 57250
⇒ 5y = (57250 − 43500) = 13750
⇒ y = 2750
∴ Amount invested at 12% = Rs. 7250
And, amount invested at 10% = Rs. 2750