June 2011 LSAT Section 2 Question 18
If there are exactly two white balls, then which one of the following boxes could contain a green ball?
1 Reply

Irina on August 5 at 01:27AM
@ ali,The setup for this game is fairly straightforward even though there are few initial deductions:
We have six boxes with one ball each, either g, r, or w
___ ___ ___ ___ ___ ___
1 2 3 4 5 6
(1) There are more red balls than white
R >W
(2) There is a box containing a green ball that is lower than any box that contains a red ball
G(1) -> R
What this rule is telling us is that there is at least 1 green ball that is below all the red ones. There could be more green balls that are above red ones, hence we cannot rule out box #6 as not green, but we can conclude that box #1 is not red.
___ ___ ___ ___ ___ ___
1 2 3 4 5 6
~R
(3) There is a white ball in a box that is immediately below a box with a green ball.
WG
Now, the question asks us if we have two white balls, then which one could contain a green ball?
Let's think about this setup:
Per rule (1), if we have 2 W, 3R, and 1G ball. Per rules (2) and (3) we can conclude that this G ball must be below all the R balls, and directly above the W ball.
/R
/G /G /W /W /W
W /W /W R R R
___ ___ ___ ___ ___ ___
1 2 3 4 5 6
These restrictions demonstrate that the lowest possible box for the G ball could be #3 if all 3 R balls are consecutive and are in boxes #3, #4 and #5.
W W G R R R
___ ___ ___ ___ ___ ___
1 2 3 4 5 6
The only other alternative for the G ball is the box #2 in a scenario where boxes 3,4,5, and 6 are filled with some combination of 3 red and 1 white ball, the exact order of which is not restricted by any of the rules Box #1 must be the W ball per rule (3).
/R /W /W /W
W G /W /R /R / R
___ ___ ___ ___ ___ ___
1 2 3 4 5 6
Does this make sense?
Let me know if you have any further questions.