#16 - Local, Could Be True, Except

[Dave Killoran]

This game is also discussed in our Podcast, at the 36:54 mark: LSAT Podcast Episode 37: The November 2019 LSAT Logic Games Section

Complete Question Explanation

(The complete setup for this game can be found here: lsat/viewtopic.php?t=31747)

The correct answer choice is (B).

W can examine H only under Template #1, modified to show W in H:


Template #1: RT block assigned to F

This is the most open of the three templates. when S and W are assigned to G and H, there is no rule governing what occurs, and thus X could be assigned to either G or H:

• _R_ _X/_ _/X_
  
  
  _T_ _S_ _W_
  
   F   G       H

The uncertainty now revolves around the open space in G and H, and thus in this Could Except question answers containing G and H are less likely to be correct. Standing out is answer choice (B), which is the only answer to address F. As it turns out, since F has already been assigned R and T, S cannot examine F, and (B) cannot occur.


Answer choice (A): This can occur, and X would examine G.

Answer choice (B): This is the correct answer.

Answer choice (C): This can occur, and X would examine H.

Answer choice (D): This can occur, and X would examine G.

Answer choice (E): This can occur, and the empty space on H would then be any of several different variables.

[flexbubbleboi]

I'm confused about why answer C is acceptable. If W examines H, and T examines G, that would put S on F. Doesn't that mean X would have to be examining G, according to the first rule? That would put RTX all on G, which would violate the two technicians per computer rule, so it seems like C can't work either:

If T examines G, we have:

G: RT (because we need one RT according to the last rule)

The stimulus stipulates W examines H, which would force S onto F:

F : S
G: RT
H: W

But, according to Rule 1, if S examines F, X examines G, which leaves us with:

F: S
G: RTX
H: W

Which doesn't work! What am I missing here?

[Adam Tyson]

You may be forgetting that one technician has to be in two different groups, flexbubbleboi, and that tech could be T. In this scenario, there is no way that RT is at G, because that would force S to F, which forces X to G - too many things in group G. But we COULD have ST at G and RT at F (RT must go to F), and then put X at H with W:

F: RT
G: ST
H: WX

The rule about RT examining one computer together does not mean that neither of them could examine a second computer with someone else! This makes answer C acceptable, and therefore incorrect.

[flexbubbleboi]

that makes perfect sense -- thank you!

[genuischan]

Hi,

I noticed that scenario #1 is used here, with the diagram below
F: RT
G: S
H: W

However, I was wondering if we look at rule 1 and rule 2, will they be contradictory?
Rule 1 states that if S examines F, X will examine G; which implied that if X does not examine G, S will not examine F as well.
While Rule 2 states that if W examines F, X will examine H; which implied that if X does not examine H, W will not examine F as well.
Referring back to the diagram and question 16, if we apply rule 1, and X does not examine G, this implies that X will examine H. Doesn't this contradict with rule 2, as it states that if X doesn't examine H, W will not examine F.
In this case, X is examining H, but, W is not examining F.

May I please get your advice on this? Thank you so much!

[Beth Hayden]

Hi Geniuschan,

I'm not sure I fully understand your concern here---rules 1 and 2 don't come into play in this question because the sufficient conditions will never be satisfied. Since RT fills the F block, S will never examine F and W will never examine F. So whether X does or does not examine G/H really doesn't matter. X could examine both G and H, but if it picks one and we have to invoke the contrapositive, that's fine because the sufficient conditions will never be true (neither S nor W will ever examine F in this problem).

S(F) --> X (G)
X(G) --> S(F)

W(F) --> X(H)
X(H) --> W(F))

I hope that helps, but please let me know if I've misunderstood!