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.