Let the total investment be $P_{total} = ₹20,300$.
The investment is split into two parts:
We know that $P_1 + P_2 = ₹20,300$.
The time period for both investments is $T = 6$ years.
The formula for Simple Interest (SI) is: $SI = \frac{P \times R \times T}{100}$.
Interest earned from the first investment ($SI_1$): $SI_1 = \frac{P_1 \times R_1 \times T}{100} = \frac{P_1 \times 8 \times 6}{100} = \frac{48 P_1}{100}$
Interest earned from the second investment ($SI_2$): $SI_2 = \frac{P_2 \times R_2 \times T}{100} = \frac{P_2 \times 6 \times 6}{100} = \frac{36 P_2}{100}$
The problem states that the interests earned are equal: $SI_1 = SI_2$.
Therefore, $\frac{48 P_1}{100} = \frac{36 P_2}{100}$.
Simplifying the equation: $48 P_1 = 36 P_2$.
Dividing both sides by 12: $4 P_1 = 3 P_2$.
This gives us a relationship between $P_1$ and $P_2$: $P_2 = \frac{4}{3} P_1$.
Now substitute the expression for $P_2$ into the total investment equation ($P_1 + P_2 = 20300$):
$P_1 + \frac{4}{3} P_1 = 20300$
Combine the terms involving $P_1$: $\frac{3 P_1 + 4 P_1}{3} = 20300$ $\frac{7 P_1}{3} = 20300$
Solve for $P_1$: $7 P_1 = 20300 \times 3$ $P_1 = \frac{20300 \times 3}{7}$
Calculate the value: $P_1 = 2900 \times 3$ $P_1 = 8700$
The sum invested at the rate of 8% per annum is $₹8,700$.
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