Q17

[ShehryarB30]

Could you pls explain how to do this one?

[ohthatpatrick]

R S T V Y Z

L: __+
M: __ __
P: __+

(ST)
~(VY)
YZ both in P or both not in P

Sure thing.

The condition is that R is alone is some group. He could either be alone in L or in P (can't be alone in M, since M has to have two members. So I would start by framing those two worlds. Since we know that M has exactly two members, if R is alone in a group, then the remaining group has the other three people.

L: r
M: __ __
P: __ __ __

or

L: __ __ __
M: __ __
P: r

From there, we start asking ourselves about the ST chunk, splitting up the VY enemies, and whether YZ are both in P or both not in P.

The most definitive thing we can do, when we're down to two available groups and have an enemy rule, is to split up the enemies.

L: r
M: v/y __
P: y/v __ __

or

L: v/y __ __
M: y/v __
P: r

This helps to clarify where the ST chunk could fit.

L: r
M: v/y __
P: y/v, s, t

or

L: v/y, s, t
M: y/v __
P: r

In the first frame, there's no way for Y and Z to both be in P, so they have to both not be in P. Thus, that frame looks like

L: r
M: y, z
P: v, s, t

for the other one, we just know that Z is the final member of group M, but it doesn't matter whether we have put V or Y into groups L or M.

L: v/y, s, t
M: y/v, z
P: r

Since the condition of this problem led to two worlds, and they're asking what must be true (i.e. "what is always true"), we can look at our two worlds for a common denominator.

It looks like in both cases Z is in Markets. So that should be the answer. Scan for it and pick (D).

(A) frame 1 is a counterexample
(B) frame 1 is a counterexample
(C) in frame 2, Y could be assigned to Labor
(E) frame 2 is a counterexample

Hope this helps.