Question

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received ₹1,350 as annual interest. Had he interchanged the amounts invested, he would have received ₹45 less as interest. What amounts did he invest at different rates?

Answer 1

Let the the amounts invested at 10% and 8% be ₹x and ₹y respectively.
Then as per the question
$$\begin{gathered} \frac{x\times 10\times 1}{100}+\frac{y\times 8\times 1}{100}=1350 \\ 10x+8y=135000.....\left(i\right) \end{gathered}$$
After the amounts interchanged but the rate being the same, we have
$$\begin{gathered} \frac{x\times 8\times 1}{100}+\frac{y\times 10\times 1}{100}=1350-45 \\ 8x+10y=130500.....\left(ii\right) \end{gathered}$$
Adding (i) and (ii) and dividing by 9, we get
$2x+2y=29500.....\left(iii\right)$
Subtracting (ii) from (i), we get
$2x-2y=4500.....\left(iv\right)$
Now, adding (iii) and (iv), we have
$$\begin{gathered} 4x=34000 \\ \Rightarrow x=\frac{34000}{4}=8500 \end{gathered}$$
Putting x = 8500 in (iii), we get
$$\begin{gathered} 2\times 8500+2y=29500 \\ \Rightarrow 2y=29500-17000=12500 \\ \Rightarrow y=\frac{12500}{2}=6250 \end{gathered}$$
Hence, the amounts invested are ₹8,500 at 10% and ₹6,250 at 8%.