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Course: LSAT > Unit 1
Lesson 5: Analytical Reasoning – Worked examples- Ordering setup | Overview | Rules and deductions
- Ordering setup | Given info–basic 1 | Worked example
- Ordering setup | Given info–basic 2 | Worked example
- Ordering setup | Given info–could be true | Worked example
- Ordering setup | Given info–cannot be true 1 | Worked example
- Ordering setup | Given info–cannot be true 2 | Worked example
- Ordering setup | Given info–must be true | Worked example
- Ordering setup | New info–could be true 1 | Worked example
- Ordering setup | New info–could be true 2 | Worked example
- Ordering setup | New info–could be true 3 | Worked example
- Ordering setup | Completely determines | Worked example
- Ordering setup | New info-must be true | Worked example
- Grouping setup | Overview | Rules and deductions
- Grouping setup | Given info–basic | Worked example
- Grouping setup | Given info–could be true | Worked example
- Grouping setup | Given info–must be false | Worked example
- Grouping setup | Given info–must be true 1 | Worked example
- Grouping setup | Given info–must be true 2 | Worked example
- Grouping setup | New info–could be true 1 | Worked example
- Grouping setup | New info–could be true 2 | Worked example
- Grouping setup | New info–must be true | Worked example
- Grouping setup | "Completely determines" | Worked example
- Mixed setup | Overview | Rules and deductions
- Mixed setup | Given info–basic | Worked example
- Mixed setup | Given info–could be true 1 | Worked example
- Mixed setup | Given info–could be true 2 | Worked example
- Mixed setup | Given info–must be true | Worked example
- Mixed setup | Given info–cannot be true | Worked example
- Mixed setup | New info–could be true | Worked example
- Mixed setup | New info–must be true 1 | Worked example
- Mixed setup | New info–must be true 2 | Worked example
- Mixed setup | Rule substitution | Worked example
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Grouping setup | Given info–must be false | Worked example
Watch a demonstration of one way to approach a "Given info: CANNOT be true" question on a grouping setup on the analytical reasoning section of the LSAT. Created by Annie Hollister.
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I believe A could also be correct based on the setup of the question. In particular, rule 1 states that no two costumes can have the same color combination. If two costumes have indigo in them, then those same two costumes also have yellow, which would violate rule 1.(0 votes)
Video transcript
- [Narrator] Which one of
the follow must be false? With this type of question
we should be able to deduce the right answer from the
information we already have, and any inferences we made in the setup. So, first let's think
about what we already know. Rules two and three together
tell us that indigo and yellow are a pair and always appear together. Rule four says that red
can't be matched with indigo or green, and because
indigo and yellow are paired that also tells us that red
and yellow aren't together. That means that red can
only be matched with white and orange, and since each
costume has three colors in it that means that one costume
must be red, white, and orange. It also means that because
no two costumes have the same three colors, there's only one
costume that has red in it. So, that gives us a lot of information to think about what must be false, but looking at the
choices E says exactly two of the costumes have both
green and orange in them. We worked out that one of
the costumes is red, white, and orange which definitely
doesn't have green in it, and we also know that another one is indigo and yellow and some third color. Well that costume can't
have green and orange in it because the third color
is either green or orange, and the other two are indigo and yellow. That means that only one costume
could possibly have green and orange in it, and
that is that third costume that doesn't have red or the
indigo yellow pair in it. So, the answer is E.
Exactly two of the costumes have both green and orange in them. This must be false given
the rules of the setup. In the real exam you should just move on, but we might as well check
that A-D could be true given the rules of the setup. For instance A, exactly
two of the costumes have indigo in them. Well, that could be true if this third costume has indigo in it, which would also mean that
third costume has yellow in it, which means that one of these
costumes has green in it and the other has either orange or white. B says exactly two of the
costumes have white in them. Let's also find, for instance, if this costume is red, white and orange, this one is indigo, yellow, and white, and this costume is
indigo, yellow, and green. All three costumes have orange in them. Again, this is possible if the third color in this costume is orange, and this third costume is
orange, green, and white. Finally, exactly one of the costumes has both yellow and white in it. That can be true if this costume is indigo, yellow, and white, and this costume is say
indigo, yellow, and green. So all of these are possible, it's only E, exactly two of the costumes have both green and orange in
them that must be false.