A dealer offers sells half of the eggs that he has and another half an egg to Anurag. Then he sells half of the balance eggs and another half an egg to Deepak. Then he sells half of the balance eggs and another half an egg to Sivani. In the end he is left with just 7 eggs and he claims that he never broke an egg. How many eggs did he start with?

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: $$\boxed{{\displaystyle 63\text{eggs}}}$$