by ohthatpatrick Tue Mar 05, 2019 9:34 am
We start off knowing this about each group:
Green: J/O
Red: L, O/J
If we framed the toss-up of 2 green, 3 red vs. 3 green, 2 red, we'd get a little more.
Green: J/O, __
Red: L, O/J, __
Green: J/O, K, M
Red: L, O/J
I wouldn't do Q2 until I had already done Q4 and Q5, because "If" questions give us possible scenarios to consider in order to eliminate answers on "must be true".
On must be true, we can eliminate any answer for which we have a counterexample. When we read each answer choice, we ask ourselves whether contradicting it is possible.
A) Could J be on green?
B) Could L be a non-facilitator.
C) Could O be on red?
D) Could both J and M be facilitators?
E) Could either J or K (or both) be facilitators?
Consider this one possible scenario
Green: J(f), K, M
Red: O(f), L
This allows us to eliminate (A), (B), (C), and (E).
Normally, we wouldn't be able to create one magic-bullet scenario that is a counterexample to all the wrong answers, but here we were.
(D) is the answer because it's impossible for us to make J and M both be facilitators, and that's essentially because there are only two facilitators and we were told that O must be one of the those two facilitators.
Hope this helps.