The incomes of A and B are in ratio ` 4 : 3`. The expenditures of A and B are in the ratio ` 5 :2` . If B saves Rs. 3000, then which of the following cannot be the sacvings of A ?

Options

A. Rs. 1500
B. Rs.2500
C. Rs.3500
D. Rs. 4500

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Define the incomes of A and B
Given that the incomes of A and B are in the ratio \(4:3\), we can express their incomes as:
- Income of A = \(4x\)
- Income of B = \(3x\)

### Step 2: Define the expenditures of A and B
The expenditures of A and B are in the ratio \(5:2\), so we can express their expenditures as:
- Expenditure of A = \(5y\)
- Expenditure of B = \(2y\)

### Step 3: Determine B's savings
We know that B saves Rs. 3000. Savings can be calculated as:
- Savings of B = Income of B - Expenditure of B
- Therefore, \(3000 = 3x - 2y\)

### Step 4: Rearranging the equation for B's savings
From the equation \(3x - 2y = 3000\), we can express \(3x\) in terms of \(y\):
\[ 3x = 3000 + 2y \]
Now, dividing the entire equation by 3 gives us:
\[ x = 1000 + \frac{2}{3}y \]

### Step 5: Calculate A's savings
Now, we need to find A's savings. The savings of A can be calculated as:
- Savings of A = Income of A - Expenditure of A
- Therefore, Savings of A = \(4x - 5y\)

### Step 6: Substitute the value of \(x\) into A's savings
Substituting \(x\) from Step 4 into the savings of A:
\[ Savings\ of\ A = 4(1000 + \frac{2}{3}y) - 5y \]
Expanding this gives:
\[ = 4000 + \frac{8}{3}y - 5y \]
Now, we need to combine the terms:
\[ = 4000 + \frac{8}{3}y - \frac{15}{3}y \]
\[ = 4000 - \frac{7}{3}y \]

### Step 7: Determine the constraints on A's savings
Since savings cannot exceed income, we know:
\[ Savings\ of\ A \leq Income\ of\ A \]
Thus:
\[ 4000 - \frac{7}{3}y \leq 4000 \]
This implies:
\[ -\frac{7}{3}y \leq 0 \]
Hence, \(y \geq 0\).

### Step 8: Analyze possible savings of A
Now, we can analyze the possible savings of A based on different values of \(y\). Since \(y\) can take any non-negative value, we need to find the maximum savings of A:
- As \(y\) increases, the savings of A decreases.
- The maximum savings of A occurs when \(y = 0\), which gives:
\[ Savings\ of\ A = 4000 \]

### Conclusion
Since A's savings cannot exceed Rs. 4000, any option greater than Rs. 4000 cannot be A's savings.