A shopkeeper sells his goods at 13.33% loss on selling price.find his loss Percentage on cost price

To solve this problem, let the **selling price (SP)** be denoted as \( x \).

### Given:
- The loss on the **selling price** is 13.33%, meaning the shopkeeper sells his goods at a 13.33% loss on the selling price.
  
  This implies the **loss** is:
  \[
  \text{Loss} = 13.33\% \times x = 0.1333 \times x
  \]

### Now, we can calculate the **cost price (CP)**.
The cost price is the selling price plus the loss:
\[
\text{Cost Price} = \text{Selling Price} + \text{Loss} = x + 0.1333x = 1.1333x
\]

### To find the loss percentage on the cost price:
The **loss percentage** is given by:
\[
\text{Loss Percentage} = \left( \frac{\text{Loss}}{\text{Cost Price}} \right) \times 100
\]
Substituting the values:
\[
\text{Loss Percentage} = \left( \frac{0.1333x}{1.1333x} \right) \times 100
\]
The \( x \) cancels out, so:
\[
\text{Loss Percentage} = \left( \frac{0.1333}{1.1333} \right) \times 100 \approx 11.76\%
\]

### Final Answer:
The loss percentage on the cost price is approximately **11.76%**.

Would you like further details or have any questions?

Here are 5 related questions:
1. How do we calculate profit percentage on cost price?
2. If the shopkeeper sold at a 15% loss on cost price, what would the selling price be?
3. What is the formula for calculating loss percentage when both CP and SP are known?
4. How can we convert profit or loss percentages between cost price and selling price?
5. How does changing the selling price affect the loss or profit percentage?

**Tip:** Always remember that loss percentages calculated on selling price and cost price are different, so it's important to know which one you're dealing with.

A shopkeeper sells his goods at 13.33% loss on selling price. Find his loss percentage on cost price.

Learn how to calculate the loss percentage on the cost price when a shopkeeper sells goods at a 13.33% loss on the selling price. This problem covers key percentage concepts and profit and loss calculations, making it suitable for students in Grades 8-10.

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