A farmer sold a calf and a cow for Rs 7,600 thereby making a profit of 25 % on the calf and 10% on the cow. By selling them for Rs 7,675, he would have realised a profit of 10% on the calf and 25% on the cow. Find the cost price of each.

Answer

To solve the problem step by step, we will denote the cost price of the calf as \( X \) and the cost price of the cow as \( Y \).

### Step 1: Set up the equations based on the given information.

1. **Selling Price of the Calf**:
The selling price of the calf when sold for Rs. 7,600 with a profit of 25%:
\[ SP_{\text{calf}} = X + 0.25X = 1.25X = \frac{5X}{4} \]

2. **Selling Price of the Cow**:
The selling price of the cow when sold for Rs. 7,600 with a profit of 10%:
\[ SP_{\text{cow}} = Y + 0.10Y = 1.10Y = \frac{11Y}{10} \]

3. **Total Selling Price**:
The total selling price of both animals is given as Rs. 7,600:
\[ \frac{5X}{4} + \frac{11Y}{10} = 7600 \]

### Step 2: Clear the fractions by multiplying through by a common denominator.

The common denominator for 4 and 10 is 20. Multiply the entire equation by 20:
\[ 20 \left(\frac{5X}{4}\right) + 20 \left(\frac{11Y}{10}\right) = 20 \times 7600 \]
This simplifies to:
\[ 25X + 22Y = 152000 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation based on the second selling price.

1. **Selling Price of the Calf**:
The selling price of the calf when sold for Rs. 7,675 with a profit of 10%:
\[ SP_{\text{calf}} = X + 0.10X = 1.10X = \frac{11X}{10} \]

2. **Selling Price of the Cow**:
The selling price of the cow when sold for Rs. 7,675 with a profit of 25%:
\[ SP_{\text{cow}} = Y + 0.25Y = 1.25Y = \frac{5Y}{4} \]

3. **Total Selling Price**:
The total selling price of both animals is given as Rs. 7,675:
\[ \frac{11X}{10} + \frac{5Y}{4} = 7675 \]

### Step 4: Clear the fractions by multiplying through by a common denominator.

The common denominator for 10 and 4 is 20. Multiply the entire equation by 20:
\[ 20 \left(\frac{11X}{10}\right) + 20 \left(\frac{5Y}{4}\right) = 20 \times 7675 \]
This simplifies to:
\[ 22X + 25Y = 153500 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations.

Now we have two equations:
1. \( 25X + 22Y = 152000 \) (Equation 1)
2. \( 22X + 25Y = 153500 \) (Equation 2)

We can solve these equations using the method of substitution or elimination. Here, we will use elimination.

1. Multiply Equation 1 by 25 and Equation 2 by 22 to align the coefficients of \( Y \):
\[ 625X + 550Y = 3800000 \quad \text{(Equation 1 multiplied by 25)} \]
\[ 484X + 550Y = 3377000 \quad \text{(Equation 2 multiplied by 22)} \]

2. Subtract the second equation from the first:
\[ (625X + 550Y) - (484X + 550Y) = 3800000 - 3377000 \]
\[ 141X = 422000 \]
\[ X = \frac{422000}{141} = 3000 \]

### Step 6: Substitute \( X \) back to find \( Y \).

Substituting \( X = 3000 \) back into Equation 1:
\[ 25(3000) + 22Y = 152000 \]
\[ 75000 + 22Y = 152000 \]
\[ 22Y = 152000 - 75000 \]
\[ 22Y = 77000 \]
\[ Y = \frac{77000}{22} = 3500 \]

### Final Solution:
- Cost Price of the Calf \( (X) = Rs. 3000 \)
- Cost Price of the Cow \( (Y) = Rs. 3500 \)