NIMCET 2008 — Reasoning PYQ

To find the minimum number of moves, we need to track the state of both containers: the $9\text{ L}$ container and the $4\text{ L}$ container. We aim to reach a state where one container holds exactly $6\text{ L}$.

Initial State: $(0,0)$

Move #	Action	9 L Container	4 L Container
1	Fill the $9\text{ L}$ container	$9\text{ L}$	$0\text{ L}$
2	Pour from $9\text{ L}$ to $4\text{ L}$	$5\text{ L}$	$4\text{ L}$
3	Empty the $4\text{ L}$ container	$5\text{ L}$	$0\text{ L}$
4	Pour from $9\text{ L}$ to $4\text{ L}$	$1\text{ L}$	$4\text{ L}$
5	Empty the $4\text{ L}$ container	$1\text{ L}$	$0\text{ L}$
6	Pour from $9\text{ L}$ to $4\text{ L}$	$0\text{ L}$	$1\text{ L}$
7	Fill the $9\text{ L}$ container	$9\text{ L}$	$1\text{ L}$
8	Pour from $9\text{ L}$ to $4\text{ L}$ (filling it)	$6\text{ L}$	$4\text{ L}$

Explanation of the final move:

The $4\text{ L}$ container already had $1\text{ L}$. To fill it, we need exactly $3\text{ L}$ more. We pour that $3\text{ L}$ from the $9\text{ L}$ container.

Remaining in $9\text{ L}$ container: $9-3=6\text{ L}$.

The task is completed in exactly 8 moves.

Final Answer:

The smallest number of moves needed is 8.

The correct option is (a).