A shopkeeper sold 12 cameras at a profit of 20% and 8 cameras at a profit of 10%. If he had sold all the 20 cameras at a profit of 15%, then his profit would have been reduced by Rs.36. What is the cost price of each camera?

Options

A. 100
B. 150
C. 180
D. 220

Answer

To find the cost price of each camera, we can set up the problem step by step as follows:

### Step 1: Define the variables
Let the cost price of each camera be \( CP \).

### Step 2: Calculate the selling price for the cameras sold at different profits
1. **For 12 cameras sold at 20% profit:**
- Selling Price (SP) = Cost Price + Profit
- Profit = 20% of CP = \( 0.20 \times CP \)
- Therefore, SP for 12 cameras = \( 12 \times (CP + 0.20 \times CP) = 12 \times 1.20 \times CP = 14.4 \times CP \)

2. **For 8 cameras sold at 10% profit:**
- Profit = 10% of CP = \( 0.10 \times CP \)
- Therefore, SP for 8 cameras = \( 8 \times (CP + 0.10 \times CP) = 8 \times 1.10 \times CP = 8.8 \times CP \)

### Step 3: Calculate the total selling price from both transactions
- Total SP from both transactions = \( 14.4 \times CP + 8.8 \times CP = 23.2 \times CP \)

### Step 4: Calculate the selling price if all cameras were sold at 15% profit
- Total SP for 20 cameras at 15% profit:
- Profit = 15% of CP = \( 0.15 \times CP \)
- Therefore, SP for 20 cameras = \( 20 \times (CP + 0.15 \times CP) = 20 \times 1.15 \times CP = 23 \times CP \)

### Step 5: Set up the equation based on the problem statement
According to the problem, the profit would have been reduced by Rs. 36 if all cameras were sold at 15% profit:
- The difference in profit = Total SP from selling at 20% profit - Total SP from selling at 15% profit
- This gives us the equation:
\[ 23.2 \times CP - 23 \times CP = 36 \]

### Step 6: Simplify the equation
- Simplifying the left side:
\[ (23.2 - 23) \times CP = 36 \]
- This simplifies to:
\[ 0.2 \times CP = 36 \]

### Step 7: Solve for CP
- Dividing both sides by 0.2:
\[ CP = \frac{36}{0.2} = 180 \]

### Conclusion
The cost price of each camera is Rs. 180.

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