Question

A man invested an amount at 12% per simple interest and
another amount at 10% per annum simple interest. He recieved an annual interest of 2600. But if he had interchanged the amounts
invested he would have recieved 140 less. What amounts did he invest
at the different rate?​

Answer 1

Let the amount bein ested at 12% be x and that invested at 10% be y. Then,

Total annual interest :

$\sf \dashrightarrow \bigg( \dfrac{x \times 12 \times 1}{100} + \dfrac{y \times 10 \times 1}{100} \bigg)$

$\sf \dashrightarrow \dfrac{12x + 10y}{100} = \dfrac{6x + 5y}{50}$

So,

$\sf \dashrightarrow \dfrac{6x + 5y}{50} = 2600$

$\sf \dashrightarrow 6x + 5y = 130000 \: - - - (i)$

Again, the amount invested at 12% is y and that invested at 10% is x.

Total annual interest at the new rates :

$\sf \dashrightarrow \bigg( \dfrac{y \times 12 \times 1}{100} + \dfrac{x \times 10 \times 1}{100} \bigg)$

$\sf \dashrightarrow \dfrac{12y + 10x}{100} = \dfrac{6y + 5x}{50}$

But, interest received at the new rates is,

$\sf \dashrightarrow (2600 - 140) = Rs \: \: 2460$

So,

$\sf \dashrightarrow \dfrac{6y +5x}{50} = 2460$

$\sf \dashrightarrow 5x + 6y = 123000 \: - - - (ii)$

Adding (i) and (ii), we get

$\sf \dashrightarrow 11x + 11y = 253000$

$\sf \dashrightarrow 11(x + y) = 253000$

$\sf \dashrightarrow x + y = 23000 \: - - - (iii)$

Subtracting (ii) from (i), we get

$\sf \dashrightarrow x - y = 7000 \: - - - (iv)$

Adding (iii) and (iv), we get

$\sf \dashrightarrow 2x = 30000$

$\sf \dashrightarrow x = 15000$

Putting x = 15000 in (i), we get

$\sf \dashrightarrow 15000 + y = 23000$

$\sf \dashrightarrow y = 23000 - 15000 = 8000$

$\sf \therefore x = 15000 \: \: and \: \: y = 8000$

Hence, the amount at 12% is ₹15000 and that at 10% is ₹8000.