Question

Mr. Jose buys some eggs. After bringing the eggs home, he finds two to be rotten and throws them away. Of the remaining eggs, he puts five-ninth in his fridge, and brings the rest to his mother’s house. She cooks two eggs and puts the rest in her fridge. If her fridge cannot hold more than five eggs, what is the maximum possible number of eggs bought by Mr. Jose?

• A. 9
• B. 17
• C. 11
• D. 20
• E. 29

Answer 1

The Correct Option is C

Step 1: Let the total eggs bought = \(N\) 

After discarding 2 rotten eggs: \[ \text{Good eggs} = N - 2 \]

Step 2: Eggs in Mr. Jose’s fridge

He keeps \(\tfrac{5}{9}\) of the remaining eggs: \[ \text{Jose’s fridge} = \frac{5}{9}(N-2) \]

Step 3: Eggs taken to his mother’s house

The rest are: \[ \text{Mother’s house} = (N-2) - \frac{5}{9}(N-2) = \frac{4}{9}(N-2) \]

Step 4: After cooking 2 eggs

His mother keeps: \[ \frac{4}{9}(N-2) - 2 \] eggs in her fridge.

Step 5: Constraint on mother’s fridge

Mother’s fridge can hold at most 5 eggs: \[ \frac{4}{9}(N-2) - 2 \leq 5 \] \[ \frac{4}{9}(N-2) \leq 7 \] \[ N - 2 \leq \frac{63}{4} = 15.75 \] \[ N \leq 17.75 \] So maximum integer \(N = 17\).

Step 6: Check divisibility condition

Since Jose divided the eggs in a \(5:4\) ratio (\(\tfrac{5}{9}\) and \(\tfrac{4}{9}\)), \((N-2)\) must be divisible by 9.

Step 7: Possible values of \((N-2)\)

The multiples of 9 ≤ 15.75 are: 9, 18 (but 18 is too big). So, \((N-2) = 9 \Rightarrow N = 11\).

Step 8: Verification

• Total bought = 11, rotten = 2 → good eggs = 9
• Jose keeps \(\tfrac{5}{9} \times 9 = 5\) eggs
• Mother receives \(\tfrac{4}{9} \times 9 = 4\) eggs
• She cooks 2, puts 2 in her fridge (≤ 5 ✅)

Final Answer:

\[ \boxed{11} \]