October 1993 LSAT
Section 3
Question 7
If corsage 1 contains at least one gardenia and at least one violet, and if corsage 3 contains three different types ...
Replies
Sylvie on July 3, 2020
For clarification: the problem set in question appears to be drawn from the October 1992 LSAT (section 3, questions 4 and 7). I accessed them via the analytics tab by filtering for group games. Just wanted to be clear!Sylvie on July 3, 2020
*** October 1993 ***Apologies for the typo.
shunhe on July 5, 2020
Hi @Sylvie,Thanks for the question! So you already eliminated (B) because it has two orchids, which is great. Now let’s take a look at why you can eliminate (C), (D), and (E). It’s actually the same thing that eliminates all three of those answer choices.
So we know that based on the hypothetical presented in this question stem, corsage 1 is going to have at least one G and one V. Well, recall from rule 1 that corsage 1 has exactly two types of flowers. That means that corsage 1 is going to have Gs and Vs only, no Os or Rs in Corsage 1. OK, well, also recall that the nine flowers used are going to include at least one flower from each of the four types—in other words, we need at least one G, V, R, and O in the final diagram. But we know that corsage 3 doesn’t have any Os. Well, if corsage 1 doesn’t have any Os, and corsage 3 doesn’t have any Os, then corsage 2 has to have an O (since an O has to appear somewhere). But (C), (D), and (E) are all answer choices that don’t have Os! Which means we can eliminate all three of them for that very reason. Then, we eliminate (B) because it has two orchids, and that leaves us with (A).
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Sylvie on July 5, 2020
Thanks @shunhe ! I clocked the min. 2 requirement for the first corsage but overlooked the fact that it was *exactly* 2. Appreciate you clearing that up!shunhe on July 6, 2020
Glad you got it now! And yup, getting every detail is super important.