October 2013 LSAT
Section 3
Question 7

October 2013 LSAT
Section 3
Question 7

Reply

Ravi on June 3, 2019

@msaber,Happy to help. Let's look at (E) and (A).

In looking at our original rules, we have

V - Z - X and W - X

U must be in 4/5/6

U having to be in 4/5/6 means that V must be in 1/2/3 because V has to

come before Z and X and if V were in 4, then we wouldn't have enough

room for Z, X, and U.

V must be in 1/2/3

Y must be in 1/2/3

Y having to be in 1/2/3 means that X has to go in 5/6 because V, Z,

and W come before it and if X were in 4, then we wouldn't have enough

room for V, Z, W, and Y.

X 5/6

The question tells us we are subbing a new rule for the rule that W

and Z are before X.

(E) says X performs in either slot 5 or 6.

The problem with (E) is that if we substituted this rule in, we could

have W and X swapping places in 5 and 6, whereas in the old world W

had to come before X.

With (E), we could have a game board that showed

Y V Z U X W

This board wouldn't be possible with our original rules, so (E) is out.

(A) says only U can perform in a later slot than X.

This means V, Z, W, and Y have to go before X and U is the only piece

that can go after.

This restores the W before X from our original rules. It also restores

Z going before X. V was already before X, so that stays the same. And

Y before X? Yes, that was true in the original rules because we know

that Y had to go in 1/2/3 and X was in 5/6, so (A) is a perfect rule

substitution. In our original rules, U, having the ability to go in

4/5/6 was the only piece that could go after X in 5/6.

Hope this helps. Let us know if you have any more questions!

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