Question

In Sivakasi, each boy’s quota of matchsticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?

• A. 200
• B. 150
• C. 125
• D. 175

Hint:

Use trial and error with options when you get rational equations involving box and item relationships.

Answer 1

The Correct Option is B

Let the total number of sticks be $x \leq 200$, and the initial number of sticks per box be $y$. So: \[ \text{Number of boxes} = \frac{x}{y} \] If he reduces sticks per box by 25, he fills 3 more boxes: \[ \frac{x}{y - 25} = \frac{x}{y} + 3 \] Now test options: Try $x = 150$ Let’s test integer $y$ satisfying: \[ \frac{150}{y - 25} = \frac{150}{y} + 3 \] Multiply both sides: \[ \frac{150}{y - 25} - \frac{150}{y} = 3 \Rightarrow 150 \left(\frac{1}{y - 25} - \frac{1}{y} \right) = 3 \] \[ \Rightarrow \frac{150(y - (y - 25))}{y(y - 25)} = 3 \Rightarrow \frac{150(25)}{y(y - 25)} = 3 \Rightarrow \frac{3750}{y(y - 25)} = 3 \] \[ \Rightarrow y(y - 25) = 1250 \] Try $y = 50$: $50 \cdot 25 = 1250$ So $y = 50$, and $x = y \cdot \text{boxes} = 50 \cdot 3 = 150$ Valid! \[ \boxed{150} \]