This is an explanation of the second logic game from Section III of LSAT Preptest 62, the December 2010 LSAT.
An artisan is creating three stained glass windows (1, 2, 3). Five colors will be used: green, orange, purple, rose, and yellow (G, O, P, R, Y). The artisan must use each color at least once, and she must use at least two colors of glass in each window.
Game Setup
This game is rather unique, I can’t think of any others like it. You have to choose colors for stained glass windows, in three groups.
One very important point: the order of the windows doesn’t matter. There is no difference between window 1 and window 3. This lets you be flexible in your diagrams.
For instance, I drew the first rule directly on the diagram:
Rule one says there’s exactly one GP. You can draw them in any group, because group order doesn’t matter. I chose group one.
Note that I’ve also drawn “not Y” beside this group. That comes from rule three, we’ll get to that soon.
Next, there are exactly two roses. I recommend just memorizing this rule. There’s no good way to draw it, and you’ll go faster if you know it cold.
I did draw one rose by itself as a reminder. Since there are two roses, at least one of the non-GP groups has a rose. This is an optional addition:
You should combined rules three and four. Both rules mention orange. You can almost always combine rules that mention the same variable. They say:
- If you have yellow, you don’t have green or orange.
- If you’re missing orange, you have purple.
You should always draw the contrapositive of any conditional diagrams, especially if they combine multiple statements:
Note that yellow always leads to purple. So there’s always at least one YP in the game. It’s also important to note that yellow can’t go with green or orange. That’s why I drew “not Y” beside GP on the main diagram.
The relationship between orange and purple (rule 4) is a bit special. Most students find this type of rule confusing, so the LSAT tests it mercilessly. If you can wrap your head around this kind of rule, you’ll do much better on in-out games.
So, the fourth rule says that if we don’t have purple, we do have orange. The contrapositive is that if we don’t have orange, we do have purple.
This means we always have at least one of purple or orange. In every window. This is crucial: remember this as a separate deduction.
Now, the part that people find confusing: orange and purple can both be in the same group. There’s no rule that says if you have orange, then you don’t have purple.
So if you’re missing one of O and P, you need the other. But you can have both O and P. Another way to think about it is that you need “purple or orange” and “or” is inclusive on the LSAT. “Inclusive” means that “or” includes the possibility of both.
Technically this is already covered in the conditional diagrams I drew earlier, but it’s something that most people miss if they just look at the diagram. The games test kind of rule so much that you need to be aware of it as a separate deduction. Think of it as “at least one and maybe both”.
Another way to think about this game is that there’s a minimum number of variables that you must place:
- GP
- Two R’s (in separate groups)
- At least one O (every variable need to be used)
- YP
You’ve got to fit those into the three groups. And YP can’t go with GP, or with O. This is hard to do. Whenever a question has you place some of these variables, you should ask yourself what’s left and where they can’t go.
It’s worth thinking about numbers. For instance, green, orange and yellow cannot be in all three groups. That’s because yellow can’t go with green and orange. Two questions test this deduction (11 and 12).
Note: There are a couple other rules. Every window needs at least two colors, and every color needs to be used at least once. You should just memorize these. I drew two spaces for every window, which helps me remember that. If those spaces were “only two”, I would have drawn a vertical line indicating those spaces were closed off.
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Sashauna Bent says
I’m slightly confused, the scenario says that AT LEAST two colours each, which to me reads that it doesn’t preclude more than 2 different colours being used as a possibility (esp. since each colour is used AT LEAST once). This gives me multiple answers to question 9. Can someone kindly explain?
MemberOrion says
It’s correct that the scenario says “at least two colours”, and your understanding is accurate. However, question 9 in particular states that two of the windows are made with exactly two colours of glass each, leaving only one window to potentially have more than two colours.
Iri says
I’m confused about the contrapositive for If Y, then not G nor O. Wouldn’t “nor” change to “and” in the contrapositive? So the If G AND O, then not Y? Thanks!
Iri says
Oh wait neither nor does imply both G and O cannot be present, so the contrapositive of that would be that if one is present, not Y. Think I answered my own question!?
TutorLucas (LSAT Hacks) says
Hah, yes that’s correct–“neither nor” means “not both”.
Y –> ~G + ~O
So,
G or O –> ~Y
Iri says
Thanks! :)