Question

A man distributed 43 chocolates to his children. How many of his children are more than five years old?
I. A child older than five years gets 5 chocolates.
II. A child 5 years or younger in age gets 6 chocolates.

• A. if the question can be answered with the help of statement I alone
• B. if the question can be answered with the help of statement II alone
• C. if both statement I and statement II are needed to answer the question
• D. if the statement cannot be answered even with the help of both the statements

Hint:

When two categories are involved, you often need distribution rules for both to form a solvable equation.

Answer 1

The Correct Option is C

Let the number of children older than five years be $x$ and the number of children aged 5 years or younger be $y$. From the problem, $5x + 6y = 43$.
From Statement I alone: We know older children get 5 chocolates, but without knowing chocolates for younger children, we cannot form a unique equation to solve $x$.
From Statement II alone: We know younger children get 6 chocolates, but without the distribution rule for older children, we again cannot solve uniquely.
Combining both statements: Older children get 5 chocolates, younger ones get 6 chocolates, and total chocolates are 43. The equation $5x + 6y = 43$ along with $x$ and $y$ being non-negative integers leads to a unique solution for $x$.
Thus, both statements are needed.