A farmer sold a cow and a calf for Rs 760 and got a profit of 10% on the cow and 25% on the calf. If he sells the cow and the calf for Rs 767.50 and gets a profit of 25% on the cow and 10% on the calf, find the individual cost price of the cow and the calf.

Answer

To solve the problem, we need to find the cost prices of the cow and the calf based on the given selling prices and profit percentages. Let's denote:

- Cost price of the cow = \( C_c \)
- Cost price of the calf = \( C_k \)

### Step 1: Set up the equations based on the first sale
From the problem, we know:
- The farmer sold the cow and calf for Rs. 760.
- He made a profit of 10% on the cow and 25% on the calf.

Using the profit formula:
- Selling Price (SP) = Cost Price (CP) + Profit
- Profit = CP × Profit Percentage

For the cow:
- Selling Price of the cow = \( C_c + 0.10 C_c = 1.10 C_c \)

For the calf:
- Selling Price of the calf = \( C_k + 0.25 C_k = 1.25 C_k \)

Thus, we can write the equation for the total selling price:
\[ 1.10 C_c + 1.25 C_k = 760 \quad \text{(1)} \]

### Step 2: Set up the equations based on the second sale
In the second scenario:
- The farmer sold the cow and calf for Rs. 767.50.
- He made a profit of 25% on the cow and 10% on the calf.

For the cow:
- Selling Price of the cow = \( C_c + 0.25 C_c = 1.25 C_c \)

For the calf:
- Selling Price of the calf = \( C_k + 0.10 C_k = 1.10 C_k \)

Thus, we can write the equation for the total selling price:
\[ 1.25 C_c + 1.10 C_k = 767.50 \quad \text{(2)} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \( 1.10 C_c + 1.25 C_k = 760 \) (Equation 1)
2. \( 1.25 C_c + 1.10 C_k = 767.50 \) (Equation 2)

We can solve these equations simultaneously.

**From Equation (1)**:
\[ 1.10 C_c + 1.25 C_k = 760 \]
Rearranging gives:
\[ C_k = \frac{760 - 1.10 C_c}{1.25} \quad \text{(3)} \]

**Substituting (3) into Equation (2)**:
\[ 1.25 C_c + 1.10 \left(\frac{760 - 1.10 C_c}{1.25}\right) = 767.50 \]

Multiplying through by 1.25 to eliminate the fraction:
\[ 1.25^2 C_c + 1.10(760 - 1.10 C_c) = 767.50 \times 1.25 \]
Calculating \( 767.50 \times 1.25 = 959.375 \):
\[ 1.5625 C_c + 836 - 1.21 C_c = 959.375 \]
Combining like terms:
\[ 0.3525 C_c + 836 = 959.375 \]
Subtracting 836 from both sides:
\[ 0.3525 C_c = 123.375 \]
Dividing by 0.3525:
\[ C_c = \frac{123.375}{0.3525} \approx 350 \]

### Step 4: Find the cost price of the calf
Substituting \( C_c = 350 \) back into Equation (3):
\[ C_k = \frac{760 - 1.10 \times 350}{1.25} \]
Calculating:
\[ C_k = \frac{760 - 385}{1.25} = \frac{375}{1.25} = 300 \]

### Final Answer
- Cost price of the cow \( C_c = Rs. 350 \)
- Cost price of the calf \( C_k = Rs. 300 \)