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 Jon Denning
PowerScore Staff
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  • Joined: Apr 11, 2011
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#37267
The final question, #16 for this third game is another Local question, again asking what Must Be True. Let's apply the local condition given to what we know from our main diagram and see just what we can prove.

We're told here that R and V are both ahead of T, so let's imagine once again what that will mean for our most severely restricted variable, S.

S can either be in 3 or 4, so consider each.

If S is in 3 and we try to put R and V ahead of T we can attempt a few different things. We could put both R and V ahead of S, in 1-2...ahh but that breaks the RV Not Block rule. We could try to put R, V, and T all after S 3, with R and V in 4-5 ahead of T in 6...but again that puts R and V together and breaks a rule. Lastly, we could split R and V, with one ahead of S 3 and the other after (with T), somewhere in 4-6. We've avoided putting them side by side, but in doing so we've caused a different problem: now there's no place for the WZ block! If R or V takes a space in 1-2, and R or V, along with T, take two spaces in 4-6, then there's a single spot open in 1-2, and another single spot open in 4-6...but no two consecutive spaces to accommodate the WZ block. So this option won't work either!

What's at the heart of those three failed attempts? S in 3!

What does that mean for S? S much be in 4! And that is answer choice (A).

That's as far as I'd go for this question on the actual test, but for the purposes of really illustrating what is at work here let me show you the only two different arrangements where R and V are both ahead of T (and you'll see in each case that S is 4):

..... W Z R/V S V/R T

..... R/V W Z S V/R T

The other four answer choices for #16, like we saw with question 15, don't have to be true. They are either possibilities—V in 5, W in 1, Z in 3—or they cannot be true—T in 5 (T in 5 needs a single thing after it, which would have to be either R or V breaking the condition given in this question).

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