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June 1995 LSAT
If Gene is not assigned to a team, then Jill must be assigned to a team with
on September 30 at 12:07AM
Why is A correct? Why is C incorrect?
on September 30 at 05:38PM
Let's first briefly look at the setup for this game. The game requires us to assign five experienced - F G J K M and four inexperienced plumbers
- R S T V - to four work teams, exactly two plumbers each. Since we have 8 spots total, and 9 plumbers, one of them is going to be unassigned.
E___ ___ ___ ___
I ___ ___ ___ ___
1 2 3 4
The following rules apply:
(1) Each plumber is assigned to at most one team.
This rule tells us that we must pick 8 out of 9 plumbers, no more no less.
(2) At least one plumber on each team must be experienced.
This rule leaves open the possibility that both plumbers on a team are experienced, and presents two possible scenarios:
E E E E
E I I I
E E E E
I I I I
In the first scenario, we have to leave out one inexperienced plumber, and in the second scenario, we have to leave out one experienced plumber.
(3)Neither M nor R nor V can be assigned to a team with F.
This rule tells us that if F is assigned to a team, M/R/V cannot be assigned to the same team, and vice versa.
F-> ~ M/R/V
(4) If T is assigned to a team, either G or K must be assigned to that team. We can infer that if neither of them is assigned to a team, then T is not assigned to that team.
T-> G v K
~G & ~K -> ~ T
(5) J cannot be assigned to a team with R.
J -> ~R
The question asks us if G is not assigned to a team, then J must be assigned to a team with.
Since G is an experienced plumber and is left out, we know that each team must have one experienced and one inexperienced plumber (scenario 2 above).
Let's consider the assignments in this scenario. First, we know that F cannot be assigned with any experienced plumbers and cannot be assigned with R or V per rule (3) or with T per rule (4) as T can only be assigned with G or K. This means that the only inexperienced plumber left to pair up with F is S, thus F and S must be on the same team.
F __ __ __
S __ __ __
Next, we know that per rule (4) T must be assigned with G or K, but G is unassigned in this hypothetical, hence T must be assigned with K.
F K __ __
S T __ __
Now we have to experienced plumbers left to assign - J and M, and two inexperienced - R and V. Per rule (5) J cannot be assigned with R, so she must be assigned with V. and the only remaining pair of plumbers that must be paired up are M & R.
F K J M
S T V R
We can thus conclude that (A) is the correct answer, and (C) is impossible because each team must have one experienced and one inexperienced plumber if an experienced plumber is left out.
Let me know if this makes sense and if you have any further questions.
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