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- Mon Oct 02, 2017 4:28 pm
#40239
Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8562)
The correct answer choice is (B)
If L is in bouquet 1, so is R (fourth rule). Bouquet 3, therefore, can have neither L nor R in compliance with the first rule:
Next, recall that bouquets 2 and 3 must have exactly two kinds of flowers in common, which also means that each of these bouquets must have at least two kinds of flowers in it. In this question, the options for bouquet 3 are quite limited: that bouquet must have at least one flower other than S, and the only remaining flowers are T and P. Thanks to the contrapositive of the last rule, we know that if P is not selected, T cannot be selected:
Without P in bouquet 3, however, we would be left with only one flower in that bouquet (S). This would violate the second rule. Therefore, bouquet 3 must contain P:
This prephrase immediately proves that answer choice (B) is correct.
As a side note, just because bouquets 2 and 3 must have exactly two flowers in common does not mean that these flowers will be P and S. Bouquet 3 can also contain T, and as a result the exact composition of flowers in bouquet 2 is unknown. Nevertheless, observant test takers will notice that bouquet 2 must contain P, because without P (and therefore without T), that bouquet cannot have two flowers in common with bouquet 3. This deduction was immaterial to the solution of Question #20.
Answer choice (A) is incorrect, because there is no reason to expect that L must be in bouquet 2.
Answer choice (B) is the correct answer choice, as it agrees with our prephrase above.
Answer choice (C) is incorrect, because R need not be in any bouquet other than bouquet 1.
Answer choice (D) is incorrect, because T need not be in any of the bouquets. It is entirely possible that the two flowers shared by bouquets 2 and 3 are P and S.
Answer choice (E) is incorrect, because T could be in bouquet 3, but it does not have to be. To conclude otherwise would be a Mistaken Reversal of the last rule.
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8562)
The correct answer choice is (B)
If L is in bouquet 1, so is R (fourth rule). Bouquet 3, therefore, can have neither L nor R in compliance with the first rule:
Next, recall that bouquets 2 and 3 must have exactly two kinds of flowers in common, which also means that each of these bouquets must have at least two kinds of flowers in it. In this question, the options for bouquet 3 are quite limited: that bouquet must have at least one flower other than S, and the only remaining flowers are T and P. Thanks to the contrapositive of the last rule, we know that if P is not selected, T cannot be selected:
Without P in bouquet 3, however, we would be left with only one flower in that bouquet (S). This would violate the second rule. Therefore, bouquet 3 must contain P:
This prephrase immediately proves that answer choice (B) is correct.
As a side note, just because bouquets 2 and 3 must have exactly two flowers in common does not mean that these flowers will be P and S. Bouquet 3 can also contain T, and as a result the exact composition of flowers in bouquet 2 is unknown. Nevertheless, observant test takers will notice that bouquet 2 must contain P, because without P (and therefore without T), that bouquet cannot have two flowers in common with bouquet 3. This deduction was immaterial to the solution of Question #20.
Answer choice (A) is incorrect, because there is no reason to expect that L must be in bouquet 2.
Answer choice (B) is the correct answer choice, as it agrees with our prephrase above.
Answer choice (C) is incorrect, because R need not be in any bouquet other than bouquet 1.
Answer choice (D) is incorrect, because T need not be in any of the bouquets. It is entirely possible that the two flowers shared by bouquets 2 and 3 are P and S.
Answer choice (E) is incorrect, because T could be in bouquet 3, but it does not have to be. To conclude otherwise would be a Mistaken Reversal of the last rule.
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