Question

A shopkecper 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? (a) 100 (b) 150 (c) 180 (d) 220

Short answer

Answer: Rs. 180

Step-by-step solution

Calculate the actual profit

Let's first find out the actual profit the shopkeeper made by selling 12 cameras at 20% profit and 8 cameras at a 10% profit. Let the cost price of each camera be x. The total cost price of 20 cameras would be 20x. Profit on 12 cameras = (20% of x) * 12 Profit on 8 cameras = (10% of x) * 8 Now, add the profit on both sales: Total actual profit = (20% of x) * 12 + (10% of x) * 8

Calculate the hypothetical profit

Now, we need to find the profit the shopkeeper would have made if he had sold all the 20 cameras at a profit of 15%. To do this, we will multiply the 15% profit by the total cost price: Hypothetical profit = (15% of x) * 20

Find the difference in profits and set up the equation

According to the problem, the actual profit was Rs. 36 more than the hypothetical profit. Therefore, we can set up the following equation: Total actual profit - Hypothetical profit = 36 Substituting the expressions we found in Steps 1 and 2: ((20 \% of x) * 12 + (10 \% of x) * 8) - (15 \% of x) * 20 = 36

Solve the equation

Now, we'll solve the equation for the cost price of each camera, x: (0.20x * 12 + 0.10x * 8) - (0.15x * 20) = 36 2.4x + 0.8x - 3x = 36 0.2x = 36 Divide both sides by 0.2: x = 180 So, the cost price of each camera is Rs. 180, and the correct answer is (c) 180.

Key concepts

Cost Price Calculation

When dealing with problems related to profit and loss, calculating the cost price is essential. Cost price (CP) is the original amount paid for an item before any profit or loss occurs. In this exercise, we are given the task to find the cost price of one camera when sold under different profit scenarios.

First, identify the total cost price when the shopkeeper buys multiple items. In this problem, if the cost price of each camera is assumed to be \( x \), then the total cost price for 20 cameras is simply \( 20x \). This simple multiplication step forms the basis for computing profits made from these cameras.

Once you have the total cost price, you can move on to calculating individual profits or losses which are percentage-based calculations over this base cost price.

Percentage Profit

Percentage profit is a measure that indicates how much profit or loss is obtained as a percentage of the cost price. It allows the comparison of profitability across different transactions. To calculate percentage profit, use the formula:

• Percentage Profit = \( \frac{\text{Profit}}{\text{Cost Price}} \times 100\% \)

In the original exercise, profit percentages for different batches of cameras are provided. For instance, a 20% profit means that the profit made is 20% of the cost price of those cameras.

For each camera sold at different rates, this percentage helps calculate the actual profit made. For instance, the profit from selling 12 cameras at a 20% profit rate is \( (0.20 \times x) \times 12 \). Calculating this for both sets of camera sales gives the actual profit derived from all cameras sold.

Equations in Profit and Loss

To solve profit and loss problems effectively, equations are essential, especially when comparing scenarios. You set up equations to encapsulate all given conditions and solve for the unknown variable, which in this case was the cost price \( x \) of each camera.

From our problem, there are two profit scenarios. First, actual profits from selling cameras at 20% and 10%, and second, the hypothetical scenario of selling all at 15%. By representing each profit scenario as an equation, you establish the basis to find the cost price.

• Equation for actual profit: \( (0.20x \times 12 + 0.10x \times 8) \)
• Equation for hypothetical profit: \( 0.15x \times 20 \)

The difference given (Rs. 36) allows us to construct a single linear equation based on these expressions:
\[ (0.20x \times 12 + 0.10x \times 8) - (0.15x \times 20) = 36 \]
By solving this equation, we derive the cost price \( x \) as Rs. 180, reinforcing the value of algebraic methods in profit and loss scenarios.