A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of ₹1300. If he had interchanged the amounts, he would have received ₹40 more as yearly interest. How much did he invest at different rates?

Let the amount invested at 12% S.I. be ₹x and at 10% S.I. be ₹y.

According to first condition,

$\Rightarrow \dfrac{12}{100} \times x + \dfrac{10}{100} \times y = 1300 \\[1em] \Rightarrow \dfrac{12x + 10y}{100} = 1300 \\[1em] \Rightarrow 12x + 10y = 130000 \\[1em] \Rightarrow 6x + 5y = 65000 …..(i)$

Now, let the amount invested at 12% S.I. be ₹y and at 10% S.I. be ₹x.

According to second condition,

$\Rightarrow \dfrac{12}{100}y + \dfrac{10}{100}x = 1340 \\[1em] \Rightarrow \dfrac{12y + 10x}{100} = 1340 \\[1em] \Rightarrow 12y + 10x = 134000 \\[1em] \Rightarrow 6y + 5x = 67000 ……(ii)$

Multiplying (i) by 6 and (ii) by 5, we have

⇒ 36x + 30y = 390000 …….(iii)

⇒ 30y + 25x = 335000 …….(iv)

Subtracting eq. (iv) from (iii) we get,

⇒ (36x + 30y) - (30y + 25x) = 390000 - 335000

⇒ 36x - 25x + 30y - 30y = 55000

⇒ 11x = 55000

⇒ x = 5000.

On substituting the value of x in (i) we get,

⇒ 6(5000) + 5y = 65000

⇒ 30000 + 5y = 65000

⇒ 5y = 65000 - 30000

⇒ 5y = 35000

⇒ y = 7000.

Hence, the investment at 12% is ₹5000 and the investment at 10% is ₹7000.