Finding the Cost Price of a Watch using Profit and Loss Percentages

This problem involves calculating the original cost price (CP) of a watch based on two different selling scenarios: one with a profit and another with a loss, where the difference in selling price is given.

Understanding the Concepts: Cost Price, Selling Price, Profit, and Loss

Let's first define the key terms:

• Cost Price (CP): The price at which an article is purchased. This is what we need to find.
• Selling Price (SP): The price at which an article is sold.
• Profit: Occurs when SP > CP. Profit = SP - CP.
• Loss: Occurs when CP > SP. Loss = CP - SP.
• Profit Percentage: $(\text{Profit} / \text{CP}) \times 100\%$.
• Loss Percentage: $(\text{Loss} / \text{CP}) \times 100\%$.

The selling price can be calculated from the cost price and profit/loss percentage using the following formulas:

• If there is a profit of $P\%$, then $SP = CP \times (1 + P/100)$.
• If there is a loss of $L\%$, then $SP = CP \times (1 - L/100)$.

Setting Up the Problem

Let the Cost Price (CP) of the watch be $x$ rupees.

Scenario 1: 25% Profit

The watch is sold at a profit of 25%. Using the formula for SP with profit:

$SP_{1} = x \times (1 + 25/100)$

$SP_{1} = x \times (1 + 0.25)$

$SP_{1} = 1.25x$

Scenario 2: 15% Loss

If the watch had been sold for Rs. 120 less than $SP_1$, there would have been a loss of 15%. Let the new selling price be $SP_2$.

$SP_2 = SP_1 - 120$

We also know that $SP_2$ corresponds to a 15% loss on the Cost Price $x$. Using the formula for SP with loss:

$SP_{2} = x \times (1 - 15/100)$

$SP_{2} = x \times (1 - 0.15)$

$SP_{2} = 0.85x$

Solving for the Cost Price (CP)

We now have two expressions for $SP_2$: $SP_1 - 120$ and $0.85x$. We also know $SP_1 = 1.25x$. Let's substitute $SP_1$ into the first expression for $SP_2$:

$SP_2 = 1.25x - 120$

Now, we can equate the two expressions for $SP_2$:

$1.25x - 120 = 0.85x$

To solve for $x$, let's rearrange the equation. Subtract $0.85x$ from both sides:

$1.25x - 0.85x - 120 = 0$

$0.40x - 120 = 0$

Add 120 to both sides:

$0.40x = 120$

To isolate $x$, divide both sides by 0.40:

$x = \frac{120}{0.40}$

To simplify the division, we can multiply the numerator and denominator by 100 to remove the decimal:

$x = \frac{120 \times 100}{0.40 \times 100}$

$x = \frac{12000}{40}$

$x = \frac{1200}{4}$

$x = 300$

So, the Cost Price (CP) of the watch is Rs. 300.

Verification

Let's verify the result:

• CP = Rs. 300
• Initial Profit = 25% of 300 = $0.25 \times 300 = 75$
• $SP_1 = CP + \text{Profit} = 300 + 75 = 375$
• $SP_2 = SP_1 - 120 = 375 - 120 = 255$
• Now check the loss percentage for $SP_2$: Loss = CP - $SP_2 = 300 - 255 = 45$
• Loss Percentage = $(\text{Loss} / \text{CP}) \times 100\% = (45 / 300) \times 100\% = (45/3) \% = 15\%$

The calculated loss percentage matches the problem statement (15% loss), so our calculated Cost Price of Rs. 300 is correct.

Step-by-Step Summary

1. Assume Cost Price = $x$.
2. Calculate Selling Price 1 ($SP_1$) with 25% profit: $SP_1 = 1.25x$.
3. Calculate Selling Price 2 ($SP_2$) with 15% loss: $SP_2 = 0.85x$.
4. Relate the two selling prices using the given difference: $SP_1 - SP_2 = 120$.
5. Substitute the expressions for $SP_1$ and $SP_2$ into the equation: $1.25x - 0.85x = 120$.
6. Solve the linear equation for $x$: $0.40x = 120 \implies x = 300$.
7. The Cost Price is Rs. 300.

Revision Table: Key Profit and Loss Formulas

Additional Information: Solving Profit and Loss Problems

Profit and loss problems often involve relating cost price, selling price, percentage profit or loss, and sometimes additional costs like overheads. A common approach is to set up equations based on the given information and solve for the unknown variable, which is usually the cost price or selling price.

In this specific problem, we used the percentage change relative to the cost price to express the selling prices. The difference between the two selling prices was given directly, providing a crucial link to form an equation to find the cost price.

Understanding the relationship between the percentage change (profit or loss) and the multiplier applied to the cost price to get the selling price ($1 + P/100$ or $1 - L/100$) is fundamental to solving such problems efficiently.

Practice with different types of profit and loss questions, including those with successive profits/losses, discounts, and marked prices, will help build proficiency.