Correct Option

Let the monthly income of Peter be 4x and Paul be 3x, based on their income ratio of 4:3.

Let the monthly expenses of Peter be 3y and Paul be 2y, based on their expense ratio of 3:2.

Given that each saves Rs 6,000 at the end of the month, we can form the following equations:

For Peter: Income - Expenses = Savings

4x - 3y = 6000 (Equation 1)

For Paul: Income - Expenses = Savings

3x - 2y = 6000 (Equation 2)

To solve this system of linear equations, multiply Equation 1 by 3 and Equation 2 by 4 to eliminate x:

$(4 x - 3 y = 6000) \times 3 ⟹ 12 x - 9 y = 18000$

$(3 x - 2 y = 6000) \times 4 ⟹ 12 x - 8 y = 24000$

Subtract the first modified equation from the second modified equation:

$(12 x - 8 y) - (12 x - 9 y) = 24000 - 18000$

$12 x - 8 y - 12 x + 9 y = 6000$

$y = 6000$

Substitute the value of y into Equation 1:

$4 x - 3 (6000) = 6000$

$4 x - 18000 = 6000$

$4 x = 6000 + 18000$

$4 x = 24000$

$x = \frac{24000}{4}$

$x = 6000$

Now, calculate their monthly incomes:

Peter's monthly income = 4x = 4 × 6000 = Rs 24,000
Paul's monthly income = 3x = 3 × 6000 = Rs 18,000

Thus, their monthly incomes are Rs 24,000 and Rs 18,000 respectively.

Incorrect Options

Options 2, 3, and 4 are incorrect because the income values provided in these options do not satisfy the given conditions of the problem, specifically the income and expense ratios, and the constant savings of Rs 6,000 for both individuals, when subjected to the same system of linear equations derived from the problem statement.