#22 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8976)

The correct answer choice is (E)

This is the only Local question of the game. Given the information in the question stem, according to the second rule L must be advertised during week 4:

Since only G, K, and M are available to fill the second advertising slot in week 4, it can be inferred that either G, K, or M must be advertised with L. It follows that either answer choice (A), (D), or (E) is correct. Since G must be advertised along with J or O, G must be advertised in week 2 or 3, and thus G cannot be advertised with L. Consequently, answer choice (A) can be eliminated. Since L is the doubled product and K must then be advertised in week 1 or 2 only, it follows that K cannot be advertised with L, and answer choice (D) can be eliminated. Answer choice (E) is thereby proven correct by process of elimination.

Note how much easier this question is to solve with the knowledge that J must be advertised during week 2.

[dimi.wassef@yahoo.com]

Hello, why can't J be in Week 3?

[Adam Tyson]

That's a Global inference, Dimi, and it's true for a combination of reasons.

First, there's the rule that G has to be with either J or else O. Then there's the rule that O is in week 3. When we put those together, putting J in week 3 with O causes a problem for G, because then it couldn't be with either one of them.

But, you may ask, what about the fact that one variable will be advertised twice? Couldn't J or O be used again and be paired with G?

The answer is found in the second rule, which tells us that the product advertised twice is not advertised in week 3. Thus, anything we put in week 3, like O, can never be used again. Now putting J in week 3 makes it impossible for G to ever be paired with O or with J, so we cannot comply with that rule. Since putting J in week 3 forces us to break at least one rule, it cannot happen!

A related inference is that J cannot be the variable that is used twice, because if we used it twice we would also have to place H immediately before it both times. That can't happen because only one variable gets used twice! That leads to yet another inference about J, that it can never be in week 4, either. If we put J in week 4, H would go in week 3. Then, since G needs to go with O or else J, we have to put G in week 4. But neither G nor J could be used again in this case, and we end up breaking the rule that the variable that goes twice is in week 4.

Ultimately, all of this should lead us to the inference that J must always go in week 2, which means H must always be in week 1. If we get those back during the original setup, we'll be in great shape for the rest of the game!