Problem: Calculate Percent Above Cost for Listed Price

Given:

• Cash discount = 20%
• Profit = 20%
• Offers 16 articles for the price of 12 (i.e. sells 16 for price of 12)

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Step 1: Let the Cost Price per article = \( C \)

Step 2: Dealer makes 20% profit, so S.P. per article (normal) = \( 1.2 C \)

Step 3: But dealer gives 16 articles for price of 12, so effectively:

\[ \text{Effective S.P. per article} = \frac{12 \times \text{Listed Price per article} \times (1 - 0.20)}{16} \] Here, listed price per article = \( L \), and after 20% discount, price per article = \( 0.8 L \) So, \[ \text{Effective S.P. per article} = \frac{12 \times 0.8 L}{16} = 0.6 L \]

Step 4: Since effective S.P. per article = \(1.2 C\) (profit price), we have:

\[ 0.6 L = 1.2 C \implies L = \frac{1.2 C}{0.6} = 2 C \]

Step 5: The listed price \( L \) is 2 times the cost price \( C \), so:

\[ \text{Percent above cost} = \left( \frac{L - C}{C} \right) \times 100 = (2 - 1) \times 100 = 100\% \]

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Answer: The listed price was 100% above the actual cost price.

Solution:

Let the initial number of eggs = $x$.

First sale (to Anurag):
Sells half of the eggs plus half an egg:
$$\text{Eggs sold}=\frac{x}{2}+\frac{1}{2}$$ Remaining eggs: $$x-(\frac{x}{2}+\frac{1}{2})=\frac{x}{2}-\frac{1}{2}$$
Second sale (to Deepak):
Sells half of remaining eggs plus half an egg:
Remaining eggs: $$(\frac{x}{2}-\frac{1}{2})-(\frac{1}{2}(\frac{x}{2}-\frac{1}{2})+\frac{1}{2})=\frac{x}{4}-\frac{5}{4}$$
Third sale (to Shivani):
Sells half of remaining eggs plus half an egg:
Remaining eggs: $$(\frac{x}{4}-\frac{5}{4})-(\frac{1}{2}(\frac{x}{4}-\frac{5}{4})+\frac{1}{2})=\frac{x-9}{8}$$
Given that after all sales, 7 eggs are left: $$\frac{x-9}{8}=7$$ Multiplying both sides by 8: $$x-9=56$$ $$x=65$$
Verification:
Since dealer never broke an egg, and he must sell whole number of eggs each time:

Checking for $x=65$:
- After first sale: Remaining eggs = 32
- After second sale: Remaining eggs = 16
- After third sale: Remaining eggs = 7 (Not matching without breaking eggs). ❌

Checking for $x=63$:
- After first sale: Sells $\frac{63}{2}+\frac{1}{2}=32$ eggs, Remaining = 31 eggs ✅
- After second sale: Sells $\frac{31}{2}+\frac{1}{2}=16$ eggs, Remaining = 15 eggs ✅
- After third sale: Sells $\frac{15}{2}+\frac{1}{2}=8$ eggs, Remaining = 7 eggs ✅

Thus, the dealer originally had: $$\boxed{{\displaystyle 63\text{eggs}}}$$