Question

A man distributed 43 chocolates to his children. How many of his children are more than five years old?

• A. The question can be answered with the help of statement I alone.
• B. The question can be answered with the help of statement II alone.
• C. Both statements I and II are needed to answer the question.
• D. The question cannot be answered even with the help of both the statements.
  \textbf{Statements:}
  I. A child older than five years gets 5 chocolates.
  II. A child 5 years or younger in age gets 6 chocolates.

Hint:

In DS problems with distributions, define variables for each group and use the total. Modular checks often confirm uniqueness of integer solutions.

Answer 1

The Correct Option is C

Step 1: Set variables and equation (using both statements).
Let $x =$ number of children older than $5$ years, $y =$ number of children $\le 5$ years.
Using I and II with total $43$: \quad $5x + 6y = 43$. Step 2: Show each statement alone is insufficient.
I alone: only tells older children get $5$ each—no info for younger $\Rightarrow$ \emph{cannot form a total}.
II alone: only tells younger get $6$ each—no info for older $\Rightarrow$ \emph{cannot form a total}.
Thus neither alone suffices. Step 3: Solve with both.
$5x + 6y = 43 \Rightarrow 6y \equiv 43 \pmod{5} \Rightarrow y \equiv 3 \pmod{5}$.
Try $y=3 \Rightarrow 5x = 43 - 18 = 25 \Rightarrow x=5$.
Next $y=8$ makes $5x=43-48$ impossible.
So the unique solution is $(x,y)=(5,3)$. \[ \boxed{\text{Number older than five} = x = 5} \]