Note: I’m leaving this explanation as I wrote it, but commentor Adam Pan has a simpler explanation in a comment below.
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This question is a bit hard. You need to do some up front thinking. But then the most important thing is drawing. There’s actually very little room for thought on logic games, apart from deciding what to start drawing.
The question says no windows have both rose and orange. And there are two rose windows. So that fills all three groups.
That’s interesting. Let’s look at the base diagram again. GP is there because of rule 1:
We have to place two Rs and an O, across all three groups. This means there are only two scenarios: GP is either with one of the Rs, or with an O.
Let’s try putting one of the Rs with GP. We’ll place O in two (remember, window order doesn’t matter – O could equally go in 3).
Why start with GP and R? It doesn’t really matter. The main thing is to make a working scenario. If you make a working scenario, you can do two things:
- You can eliminate answers that don’t have to be true in the scenario.
- You can see if there’s another scenario. Then look for what must be true in both scenarios.
Now, if orange is in 2, rose is in windows 1 and 3.
Next, we need to place YP. YP can’t go with green or orange, so it must go with R in group 3:
Finally, every window needs two colors, so one of P/G needs to go with O in group 2:
This scenario eliminates answers A-D. None of those answers need to be true, because they’re not true or not necessarily true in this scenario. And E does have to be true in this scenario. So E is CORRECT.
Technically, we’ve just proven that A-D don’t have to be true. We can’t be 100% certain that E has to be true. Maybe there’s another possible scenario where it isn’t true. Maybe we’ve made a mistake.
This is unlikely, but if you want to be extra sure, you can draw a second scenario to check whether yellow, purple and rose must go together in that scenario. The other scenario will place O with GP, and R in the other two groups. Here’s the other one I drew:
This scenario still has YPR, so E is definitely true.
I recommend practicing making scenarios constantly. The more you do it, the faster you’ll get, and the better you’ll become at seeing how variables interact.
For instance, the reason YPR must go together is that O and the two R’s must go in different groups.
YP must go in one group. Since they can’t go with O, then they must go with R. No one else can go with YPR, since G can’t go with Y.
If that didn’t make sense, I recommend trying to draw it, and make a scenario that obeys the rules and doesn’t use YPR (it won’t work). That’s really the only way to understand these restrictions.
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Olivia says
“The question says no windows have both rose and orange. And there are two oranges. So that fills all three groups.”
Not to be super annoying, but I think it should be “there are two *roses*” instead of “oranges,” just to avoid confusion!
TutorLucas (LSAT Hacks) says
Thanks for catching that! The page has been updated.
Adam Pan says
I think there’s a much simpler explanation for this question:
We know that there are two panes that must include
GP
and
YP
We must have 2 panes of R, at least one pane of O. The question adds the extra constraint that R and O cannot coexist. Therefore the panel without O must have R, to satisfy these constraints.
Therefore, one pane must be
YPR
Note that Y prevents G and O from being used, so YPR cannot contain other colors of glass.
FounderGraeme Blake says
Ah, you’re right. That is simpler, and correct. I’m going to link to your comment from the top of the explanation.
However, I’m going to leave the explanation as is as well. Hopefully it’s a demonstration of how to solve a question when you don’t make the deduction.