#5 - Local, Suspension

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=1626)

The correct answer choice is (E)

The question stem asks you to suspend the third rule of the game and replace it with a condition that results in the same effect. Thus, you need an answer choice that effectively places H and P in separate groups.

Answer choice (A) addresses half of the rule requirement in that H and P cannot both be appointed to the appellate court, but, under this answer, H and P could both be appointed to the trial court, and thus this answer choice fails as a substitute for the third rule.

Answer choice (B) indicates that if H is missing from the trial court, then P must be appointed to the trial court. However, this rule does not preclude H and P from both being appointed to the trial court, and thus this answer does not result in the same restrictions as the third rule.

Answer choice (C) adds in J, but this restriction is not strong enough as H and P could still be appointed to the same court as long as J is appointed to the other court.

Answer choice (D) creates a conditional rule involving blocks:

PT63_Game_#1_#5_diagram 1.png

Because L and K are already assigned to the appellate court and the trial court, respectively, this rule is an equal substitute when H is assigned to the appellate court. However, if H is assigned to the trial court, P could also be assigned to the trial court, and thus this answer does not result in the same restrictions as the third rule.

In answer choice (E), three variables from the group of H, K, L, and P cannot be assigned to the same court, and thus, exactly two variables from the group are assigned to each court. Because L and K are already assigned to different courts from the first and second rules, the effective result is that one of H and P is assigned to the appellate court, and one of H and P is assigned to the trial court. Thus, this answer choice has the same operational result as the third rule, and is therefore the correct answer.

[elyssa1]

Hi - could you go through answer choice (E) in a little more detail? I first chose (B) and then (D) when I revisited the problem. These two answer choices left me with the same conditional: HA PT and the contrapositive PA HT. I understand my mistake on these (still leaves question if they can both be in T together).

I'm trying to understand (E). I understand that HKL or PKL would not work because of L and K already being in separate courts, but don't understand why LPH couldn't be in the appellate court together or KPH in trial court. Thank you!

[James Finch]

Hi Elyssa,

Sure! As you note, the problem with answer choices (B) and (D) is that both leave open the possibility that H and P are both appointed to the trial court; both answer choices would have the effect of requiring that at least one of P or H be appointed to the trial court. (E), however, interacts with the rules that place K and L in different groups to then force P and H to always be in different groups as well, as we would only have one slot in each group for P or H. For example, placing H in the appellate court would mean that we are already at the 2 of H-P-K-L limit in each group, forcing P to go with K in the trial court.

Hope this clears things up!