Calculate Sum Lent at 5% P.A.

This problem involves calculating a specific portion of a total sum based on simple interest earned over time. We need to find the amount of money initially lent out at a 5% per annum interest rate.

Formulate Simple Interest Equations

Let the total sum be ₹9200.

Let the sum lent at 5% p.a. be $x$ (in ₹).

Let the sum lent at 8% p.a. be $y$ (in ₹).

We know that the total sum is the sum of the two parts:

$x + y = 9200 \quad (1)$

The time period is $T = 3$ years.

The simple interest (SI) formula is $SI = \frac{P \times R \times T}{100}$.

Interest earned from the sum $x$ at 5% p.a. is:

$SI_1 = \frac{x \times 5 \times 3}{100} = \frac{15x}{100} = 0.15x$

Interest earned from the sum $y$ at 8% p.a. is:

$SI_2 = \frac{y \times 8 \times 3}{100} = \frac{24y}{100} = 0.24y$

The total interest received is ₹1812. So:

$SI_1 + SI_2 = 1812$

$0.15x + 0.24y = 1812 \quad (2)$

Solve for the Principal Amount

We have a system of two linear equations:

1. $x + y = 9200$
2. $0.15x + 0.24y = 1812$

From equation (1), we can express $y$ in terms of $x$: $y = 9200 - x$

Substitute this expression for $y$ into equation (2):

$0.15x + 0.24(9200 - x) = 1812$

Distribute $0.24$:

$0.15x + (0.24 \times 9200) - 0.24x = 1812$

Calculate $0.24 \times 9200$:

$0.15x + 2208 - 0.24x = 1812$

Combine the terms involving $x$:

$(0.15 - 0.24)x + 2208 = 1812$

$-0.09x + 2208 = 1812$

Isolate the term with $x$:

$-0.09x = 1812 - 2208$

$-0.09x = -396$

Solve for $x$:

$x = \frac{-396}{-0.09}$

$x = \frac{396}{0.09}$

To simplify, multiply the numerator and denominator by 100:

$x = \frac{396 \times 100}{0.09 \times 100} = \frac{39600}{9}$

$x = 4400$

Thus, the sum lent at 5% p.a. was ₹4400.