by timmydoeslsat Wed Apr 18, 2012 11:35 pm
I agree that if you did not make the two inferences of:
X in a facility, that facility must do Z the next year
Z in a facility, that facility must do W the next year
...This game would take a little longer. However, those inferences just almost jump off the page at us! If V and W cant go in a facility the year after X was done...who is left to use? There are only 4 models. We also know that models cant repeat in the same facility in consecutive years.
And you can make the further inference that you know, in the first two years, that x cannot be the variable that repeats due to that producing multiple Z's in the following year.
In this local question, we are asked who could be assembled in factories F and G in the second year provided that in the first year F has V and G has W.
F G
V W
So we know that any answer choice that lists V on the left hand side or W on the right hand side would be incorrect due to elements repeating in consecutive years. Also any answer choice that contains 2 Z's, as we know only 1 Z can go per year.
This eliminates A, D, and E. I imagine most people are down to B and C to consider. And the only difference between these 2 answer choices is the factory of G. I know that X has not been placed yet. I know that when I do place X, and I will have to eventually because each model is done at least once per year, I know that a X will follow. So I know that there is no way I could waste a Z with it following the W in this case. If I had Z follow W in factory G for year 2, I would have to necessarily place 2 Zs in year 2 because once I place X in year 1, Z will follow in year 2 in that factory.
Answer B wins by process of elimination.