A shopkeeper marks up the price of an article *P* by 20% and then offers a discount of *d*%. As a result, he incurs a loss of *(d - 11)*%. If the cost price of another article *Q* is Rs. 200 and the shopkeeper gains 1.5 *d*% on it, what is the selling price of article *Q*?

Question

Accepted Answer

**Solution:**

1. Let the cost price of article *P* be Rs. 100x.

2. The marked price after a 20% markup is:
   $$	ext{Marked Price} = 100x 	imes \frac{120}{100} = 120x$$

3. After giving a discount of *d*%, the selling price becomes:
   $$120x 	imes \frac{100 - d}{100}$$

4. According to the problem, the shopkeeper incurs a loss of *(d - 11)*%, so the selling price is also:
   $$100x 	imes \frac{100 - (d - 11)}{100} = 100x 	imes \frac{111 - d}{100}$$

5. Equate the two expressions for selling price:
   $$120x 	imes \frac{100 - d}{100} = 100x 	imes \frac{111 - d}{100}$$

6. Simplify by canceling 100x:
   $$120 	imes (100 - d) = 100 	imes (111 - d)$$

7. Expanding both sides:
   $$12000 - 120d = 11100 - 100d$$

8. Rearranging terms:
   $$12000 - 11100 = 120d - 100d$$
   $$900 = 20d$$

9. Solving for *d*:
   $$d = \frac{900}{20} = 45\%$$

10. Now, for article *Q*, the cost price is Rs. 200 and profit is 1.5 *d*% = 1.5 × 45 = 67.5%.

11. Selling price of *Q* is:
    $$200 	imes \left(1 + \frac{67.5}{100}ight) = 200 	imes 1.675 = Rs. 335$$

**Answer:** Rs. 335

Answer

Answer

Answer

Answer