June 2007 LSAT Section 1 Question 21

June 2007 LSAT Section 1 Question 21

1 Reply

Ravi on June 28, 2019

@bobby,Great question. It doesn't declare that each of the 5 materials must

be recycled, so it's not necessary that this is the case.

Here are the rules of the game:

1) Max 3 materials at 1 center

2) W - >N

3) 2 - >1

4) Ex. 1 P

5) P - >/G (G - >/P)

For this question, we know that each center has 3 materials

1 2 3

_ _ _

_ _ _

_ _ _

This means 1 and 2 have the same materials. Since there's only 1 P, it

must go in 3

(A) is out because we know that 2 and 1 are recycling the same things

because of rule 3. Rule 3 also allows us to automatically eliminate

(C) and (E) since we know 2 and 1 are recycling the same things.

Now we're down to (B) and (D).

(B) says only 3 recycles N. Could this be true?

The problem with this is that if only 3 recycles N, that means that

neither 1 nor 2 can recycle W. We get this from the contrapositive of

rule 2 (/N - >/W).

So if P, N, and W can't go in 1 or 2, what're we left with? Just T,

but we have 3 total spots and with T and G, only 2 are occupied. Thus,

this can't be true.

Now we're left with (D).

If only 3 recycles T, then we have T and P in 3.

We could put W and N in 1 and 2 with G and then another N in 3, and

this would work.

1 2 3

G G P

N N T

W W N

Does this make sense? Let us know if you have any more questions!