https://file.aiquickdraw.com/mathbot/file/172914951219904bavi0w.jpg

To solve this problem, let's analyze the information given step by step.

### Given:
1. **Scenario 1:**
   - The shopkeeper marks up an item by **x%** and then gives a **y%** discount.
   - Result: He earns a profit of **20%**.

2. **Scenario 2:**
   - The shopkeeper marks up an item by **y%** and gives a **x%** discount.
   - Result: He incurs a **loss of 40%**.

We need to find **x + y**.

### Approach:

1. **Let the cost price be \( C \).**
   
   **Scenario 1:**
   - **Marked Price (MP1):** After marking up by **x%**, the new price becomes:
     \[
     MP1 = C \left(1 + \frac{x}{100}\right)
     \]
   - **Selling Price (SP1):** After giving a discount of **y%**, the price becomes:
     \[
     SP1 = MP1 \left(1 - \frac{y}{100}\right) = C \left(1 + \frac{x}{100}\right) \left(1 - \frac{y}{100}\right)
     \]
   - Since there is a **20% profit**, we can write:
     \[
     SP1 = 1.2C
     \]
   - Therefore:
     \[
     C \left(1 + \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) = 1.2C
     \]
     \[
     \left(1 + \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) = 1.2 \quad \text{(Equation 1)}
     \]

2. **Scenario 2:**
   - **Marked Price (MP2):** After marking up by **y%**, the new price becomes:
     \[
     MP2 = C \left(1 + \frac{y}{100}\right)
     \]
   - **Selling Price (SP2):** After giving a discount of **x%**, the price becomes:
     \[
     SP2 = MP2 \left(1 - \frac{x}{100}\right) = C \left(1 + \frac{y}{100}\right) \left(1 - \frac{x}{100}\right)
     \]
   - Since there is a **40% loss**, we can write:
     \[
     SP2 = 0.6C
     \]
   - Therefore:
     \[
     C \left(1 + \frac{y}{100}\right) \left(1 - \frac{x}{100}\right) = 0.6C
     \]
     \[
     \left(1 + \frac{y}{100}\right) \left(1 - \frac{x}{100}\right) = 0.6 \quad \text{(Equation 2)}
     \]

### Solving Equations:
From Equations 1 and 2:
\[
\left(1 + \frac{x}{100}\right) \left(1 - \frac{y}{100}\right) = 1.2
\]
\[
\left(1 + \frac{y}{100}\right) \left(1 - \frac{x}{100}\right) = 0.6
\]

Let’s simplify these:
- Expanding Equation 1:
  \[
  1 + \frac{x}{100} - \frac{y}{100} - \frac{xy}{10000} = 1.2
  \]
  \[
  \frac{x - y - xy/100}{100} = 0.2 \quad \text{(Equation 3)}
  \]

- Expanding Equation 2:
  \[
  1 + \frac{y}{100} - \frac{x}{100} - \frac{xy}{10000} = 0.6
  \]
  \[
  \frac{y - x - xy/100}{100} = -0.4 \quad \text{(Equation 4)}
  \]

Solving Equations 3 and 4 simultaneously gives **x = 50%** and **y = 100%**.

### Conclusion:
\[
x + y = 50 + 100 = 150
\]

The final answer is:
\[
\boxed{150}
\]

Would you like further explanation or have any questions?

### Related Questions:
1. If the profit percentage is changed, how will it affect the values of **x** and **y**?
2. What happens if the discount percentages are swapped in another way?
3. Can you solve similar problems with different profit or loss percentages?
4. How does a change in the cost price affect the overall calculation of markup and discounts?
5. What would be the result if both **x%** and **y%** were the same in both scenarios?

### Tip:
Always convert percentages into fractions or decimals for easier handling in equations.

A shopkeeper marks up an item by x% and then provides a discount of y% on the marked up price to earn a profit of 20%. If he marks up by y% and provides a discount of x%, he suffers a loss of 40%. What is the value of x + y?

Learn how to solve a profit and loss problem involving markup and discount percentages. This problem helps you understand key mathematical concepts like percentage markup, profit margins, and discounts. Includes detailed equations and step-by-step calculations to find the solution.

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