Solution:

Let the initial number of eggs =

x

.

First sale (to Anurag):
Sells half of the eggs plus half an egg:

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:

63 eggs

Question Practice

Discussion

Join the Conversation