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PrepTest 79, Game 2, Question 10


If K is assigned to area 2, then which one of the following could be true? So we're gonna go ahead and drop K into area 2. Of course, we already have M in area 3, since that's one of our major rules. So if we put K in area 2, we know that L cannot go in area 1. Why is that the case? Well, because of this rule here.

So L must be paired with M or L must be paired with K. Cannot be paired with both, but it needs to be paired with one of them. If M is in 3, K is in 2, L is going to need to go in 2 or 3. Meaning, of course, that it cannot go in 1. So the correct answer is E, P is assigned to area 3. So remember, you were just trying to find a valid arrangement here in this type of question.

Four of your answer choices are not going to give that to you, answer choice E does. So we can just drop P in 3, bring O along with it, have K and L together to meet our requirement here and then J can hang out by itself here in space 1. We're not breaking any rules. Everything looks good. This is a possible arrangement.

Let's look at some of the other answer choices here, because they were a bit tricky in terms of figuring out whether or not they could be true. So answer choice A, J is assigned to area 2. So if we brought J down here with K, we're gonna need to bring O along with it, since J and K are together. Now, our rule tells us that if O is in 2, then J and K must be together, and if O is not in 2, then J or K cannot be together.

So it sort of covers all the bases there. So we're either gonna have O in 2 and JK, or O not in 2 and not JK. So because we have JK, O's gotta go in 2. Then we need to fill our requirement here with L and M or L and K. There's no room here in area 2, so L's gonna have to go down here in 3. So now we're trying to place P.

We cannot put it in 3, so then we're left with P. So while it may be tempting just to drop it here in 3, remember that each area needs to have at least one variable assigned to it. If we put P in 3, then 1 is going to be empty. But the problem is, is that we cannot put P in 1. So this is not going to be a valid arrangement.

There's no way that we can possibly distribute all the variables without breaking at least one rule. Answer choice B, you run into a similar problem. So if Jefferson is assigned to area 3, we've already placed K in 2 based on the question task. That means that J and K are not together, which means that O is not in 2.

So O is gonna need to go here in 3. Now, to meet our LM or LK requirement, we're going to need the drop L into 2. We're running into the same problem we did with answer choice A. P doesn't have a home. So we could put it in 2, but that's gonna leave 1 empty and that's not valid. So there's not a valid arrangement here without breaking at least one of the rules.

Answer choice C reverts back to sort of the inference that we made in the very beginning, L cannot be assigned to that first area. And this is what happens when we try to do that, we're gonna end up with L, K, and M in separate areas, and we're not gonna meet that requirement that L be paired with M or K. And then answer choice D, O is assigned to area 2.

So if we assign O to area 2, then J or K's gonna come along with it. So that means we're gonna need to drop L down here in area 3, and since P can't go here, we could try to drop it in area 3, but it's gonna create an empty area 1. So we're running into the same problem with P in almost all of these answer choices. So we're running sort of a spatial problem, is that by putting K in 2, it's restricting the movement of a lot of the other variables.

So we just wanna be mindful of how that plays out, and this is where it's helpful to have sort of our rules written out on our diagram, so that we're not just sort of in a panic placing P in 1, and thinking that we've solved our problem. So P can't go in 1, it ends up being a big problem for most of these answer choices.

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