A man invested certain amount of money in two schemes A and B which offer interest at the rate of 8% per annum and 9% per annum respectivley. He received ₹ 1,860 as annual interest. However, if he had interchanged the amount of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme ?

Let ₹ x will be invested in first scheme and ₹ y will be invested in second scheme.

Interest of rate of 2 schemes = 8% and 9%

Sum of interest = ₹ 1,860

$\text{Interest} = \dfrac{P \times R \times T}{100}$

$\Rightarrow \dfrac{x \times 8 \times 1}{100} + \dfrac{y \times 9 \times 1}{100} = 1,860\\[1em] \Rightarrow \dfrac{8x}{100} + \dfrac{9y}{100} = 1,860\\[1em] \Rightarrow 0.08x + 0.09y = 1,860 …………………..(1)$

When the amount of investments are interchanged,

$\Rightarrow \dfrac{x \times 9 \times 1}{100} + \dfrac{y \times 8 \times 1}{100} = 1,880\\[1em] \Rightarrow \dfrac{9x}{100} + \dfrac{8y}{100} = 1,880\\[1em] \Rightarrow 0.09x + 0.08y = 1,880 …………………..(2)$

Multiplying 9 in equation (1) and 8 in equation (2), we get

⇒ (0.08x + 0.09y = 1,860) x 9

⇒ 0.72x + 0.81y = 16,740 ……………….(3)

And, (0.09x + 0.08y = 1,880) x 8

⇒ 0.72x + 0.64y = 15,040 ……………….(4)

Subtracting equation (3) and (2), we get:

$\begin{matrix} & 0.72x & + & 0.81y & = & 16,740 \ & 0.72x & + & 0.64y & = & 15,040 \ & - & &- & & - \ \hline & & & 0.17y & = & 1,700 \ \Rightarrow & & & y & = & \dfrac{1700}{0.17} \ \end{matrix}$

⇒ y = 10,000

Putting the value of y in equation (1), we get

⇒ 0.08x + 0.09 $\times$ 10,000 = 1,860

⇒ 0.08x + 900 = 1,860

⇒ 0.08x = 1,860 - 900

⇒ 0.08x = 960

⇒ x = $\dfrac{960}{0.08}$

⇒ x = 12,000

Hence, ₹ 12,000 and ₹ 10,000 are invested in each scheme.