#17 - Local, Could Be True, Except

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=6469)

The correct answer choice is (A)

This is the only question of the game to define the number of friends who can appear in a photograph; in this case, three. With only three spaces in the photograph, every variable becomes important, especially those that bring along other variables.

Because this is a Could Be True Except question, the correct answer choice Cannot Be True, and is one that provides a pair of friends that together result in four or more friends appearing in the photograph. Thus, you should immediately scan the answers for a variable such as U, who by itself brings S, W, and Y. Unfortunately, U does not appear in the answers (but it is worth taking the time to see if the question could be solved that easily).

Answer choice (A): This is the correct answer choice. When S appears in a photograph, then W and Y must also appear in that photograph. Since the answer choice also establishes that Z is in this photograph, that totals four friends (S, W, Y, and Z), and thus this answer cannot be true and is correct.

Answer choice (B): This answer choice is incorrect. T and Y could appear in a photograph of exactly three friends, as proven by the following hypothetical: T-Y-Z.

Answer choice (C): This answer choice is incorrect. W and S could appear in a photograph of exactly three friends, as proven by the following hypothetical: W-S-Y.

Answer choice (D): This answer choice is incorrect. Y and Z could appear in a photograph of exactly three friends, as proven by the following hypothetical: T-Y-Z.

Answer choice (E): This answer choice is incorrect. Z and R could appear in a photograph of exactly three friends, as proven by the following hypothetical: R-T-Z.

[nadiaguo]

Thanks for all the great help thus far...

I was wondering why 17 wasn't C. I understand A works too, but C also can't be true, because if W and S appear together, that means U would have to appear too, making it three people, but it would also mean that R WOULDN'T appear.

Rule 3 states that R appears in every photograph Y does not appear in, and the contrapositive of that would be that Y appears in every photograph R doesn't appear in, correct?

Not Y ---> R, Not R ---> Y

So if R doesn't appear with WSU, doesn't that force Y to also appear in the photograph, therefore making it impossible for there to be exactly three people in the photograph?

I must be getting something wrong here...

[Jon Denning]

Thanks for the question. Let's just go through the rules and the answer choices individually and examine why four answers are possible, and one would violate the rules (I'll give you a sample group of 3 for each of the incorrect answers):

Rule 1: S --> W
Rule 2: U --> S (these rules give U --> S --> W)
Rule 3: Not Y --> R; Not R --> Y (cannot both be absent, can both be included)
Rule 4: W --> Not T and Not R (note that Not R --> Y, so W --> Y and can be added to the USW chain above)


A. If S is in then W is in (1st rule), which means Y is in (3rd and 4th rules). So that's your three, which means you can't also have Z. So (A) is impossible and therefore correct.

B. T in means W is out (contra of rule 4), which means S and U are also out (contra rules 1 and 2). Y in tells us nothing. So the third here could be Z or R (note Y and R can be in together, they just can't both be out).

C. If S is in then W must be in, which also means Y is in (see discussion in (A) above). That could be your group of three. To think that includes U would be to make a mistaken reversal from S backwards to U. Don't do that. R could appear with the group SWY, but it wouldn't here since we're told only three people are in. Still, (C) is perfectly acceptable.

D. Y and Z could be included with a few other people: R, W, or T

E. Z and R could be included with a few other people: Y or T


I hope that helps!

[nadiaguo]

Thanks that helped a lot, I definitely made a mistaken reversal.