Question:

A dealer offers a cash discount of $20%$ and still makes a profit of $20%$, when he further allows $16$ articles to a dozen to a particularly sticky bargainer. How much percent above the cost price were his wares listed?

Show Hint

In multi-discount problems, always apply discounts and special offers in sequence to the marked price, and ensure the final selling price matches the profit condition.
Updated On: Aug 5, 2025
  • $100%$
  • $80%$
  • $75%$
  • $66\frac{2}{3}%$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Let the cost price of each article be $Rs.\ 1$.
Step 1: Selling price with profit
He makes a profit of $20%$, so his selling price per article (without discounts) should be: \[ SP = CP \times 1.20 = 1.20 \]
Step 2: Marked price with cash discount
A $20%$ cash discount means the selling price is $80%$ of the marked price. Let the marked price be $M$. Then: \[ 0.80M = 1.20 \quad \Rightarrow \quad M = \frac{1.20}{0.80} = 1.50 \] So the marked price is $Rs.\ 1.50$ per article.

Step 3: Effect of 16 for a dozen
If he allows $16$ articles for the price of $12$ articles at the marked price, then the effective selling price per article is: \[ \text{Price for 16 articles} = 12 \times 1.50 = 18.00 \] So price per article = $18 / 16 = Rs.\ 1.125$.

Step 4: Compare with cost price
Cost price per article = $1$. So profit = $1.125 - 1 = Rs.\ 0.125$.
Profit percentage = $(0.125 / 1) \times 100 = 12.5%$. But this is not matching the given profit of $20%$ — so our assumption needs to adjust: the given $20%$ profit is after both discounts.
Let $CP$ be $Rs.\ 100$. Then selling price after all discounts is $120$. Also: After cash discount of $20%$, price becomes $0.80M$. After giving $16$ for price of $12$, price per article becomes $(0.80M \times 12) / 16 = 0.60M$.
Since $0.60M = 120$, $M = 200$. This means marked price was $200$ when cost price was $100$.
Thus percentage above cost = $(200 - 100)/100 \times 100 = 100%$. So option (A) is correct.
Was this answer helpful?
0
0