PrepTest 58 (September 2009 LSAT), Game 4, gave a lot of test-takers trouble.

It's a Grouping: In-Out / Selection game, but it's not like most In-Out logic games. In-Out games often involve both of the following types of rules:

1. A ---> NOT B

2. NOT C ---> D

However, PrepTest 58, Game 4, only involves the first kind of rule, giving you a ton of "double-not arrows" to work with (read the post linked above if you don't know what those are). These make the game seem deceptively easy - it's actually a bit more complicated.

I consider this game to be the hardest out of the 4 in that section. In this blog post, I'll explain the game from start to finish.

Those of you who haven't seen this game before, please don't look at the rest of this post until the week before your exam. You'll want to save PrepTest 58 until then.

The rest of you, read on.

Also see: Explanations for Recent LSAT Logic Games

Due to LSAC's copyright rules, I'm forced to leave out certain details of the game. You'll need a copy of the exam in order to follow along.

First off, the rules:

H ---> NOT S and NOT M

If H is chosen (IN), then neither S nor M is chosen (so they must be out). This means if either S or M is chosen, H must be not be chosen. However, I wouldn't take the time to write out the contrapostive. I would immediately jump to diagramming it with a double-not arrow, as I have below.

The remaining two rules should be treated the same way:

M ---> NOT P and NOT T

W ---> NOT P and NOT S

The three rules can be diagrammed as follows:

This means:

H conflicts with both S and M.

M conflicts with both P and T.

W conflicts with both P and S.

This means, among other things, that we must always lack at least one of H and S, one of M and T, and one of W and P. In other words, we must always lack at least 3 variables out of the 7 total.

Since the initial paragraph of the game also says that we must always have at least 3 variables in, this, when combined with the info that we must always lack at least 3 variables, means we will either have "4 in, 3 out" or "3 in, 4 out". ("4 in, 3 out" is demonstrated by the first diagram below.)

Of course, there are other pairs of variables that conflict with each other (pairs from which we must always lack at least one). However, I'm choosing these particular pairs because none of these conflicts overlap.

What I mean is - none of these conflicts mention the same variables - none of the variables are involved multiple times in the conflicts that I've chosen.

By choosing these particular conflicts out of all the conflicts mentioned in the rules, I can create the following 2 main templates / options:

Both diagrams indicate that we must always lack at least one of H and S, one of M and T, and one of W and P. The "4 in, 3 out" diagram is nearly complete because the conflicts I've chosen fill all the "out" slots, so everything else must be in, including L. The "3 in, 4 out" diagram is a bit more ambiguous because we still have 1 "out" slot remaining, and we don't know which variable will fill it.

It's important to remember that these diagrams don't address all of the rules - only the ones I handpicked. We still have to deal with a few conflicts, which are:

Let's keep these in mind as we go through the game.

Question 18

Typical List/Acceptability question

We're always going to lack at least one of H/S. Choice A violates this.

We're always going to lack at least one of M/T. Choice E violates this.

We're always going to lack at least one of P/W. Choice D violates this.

Looking at the conflicts not included on the diagram, we're always going to lack at least one of M/H. Choice B violates this.

By elimination, Choice C is our answer.

Question 19

Knowing that we must have at least 3 in and at least 3 out, the max we can have in is 4. We already know this from the inferences we made before starting the questions.

Question 20

If both P and W are out, our diagram becomes:

Choice A: We can't lack both H and L because L is definitely in. Eliminated.

Choice B: We CAN lack both H and M because there's no reason we can't have both S and T. This would create:

IN: STL

OUT: HMPW

This doesn't violate any of the remaining conflicts not included in the main diagram, so it's our answer.

I'll go through the rest anyway.

Choice C: We can't lack both H and S because there's only space in the out column to lack one of the two.

Choice D: We can't lack both L and M because L is definitely in.

Choice E: We can't lack both S and T because this would create:

IN: HML

OUT: STPW

H and M can't both be in because this violates one of the remaining conflicts not included in the main diagram.

Question 21

This is a local question asking what must be true if M is in. Remember that there are two major possibilities: having 3 variables in and 4 variables out, or having 4 variables in and 3 variables out.

In both possibilities, if M is in, H, P, and T will automatically be out. H and P will be out because M conflicts with both in the "remaining rules" not included on the main diagram. T will be out because we must always lack at least one of M/T according to the main diagram.

This fully determines the "4 in, 3 out diagram", creating:

This diagram is invalid because due to the "remaining rules," S and W can't both be in.

This leaves us with the "3 in, 4 out" diagram, which is a little ambiguous. We will again lose H, P, and T for the same reasons as above. However, there's still one more "out" slot to take care of. It's addressed by the fact that, in our "remaining rules", S and W conflict with each other. At least one of the two must be out, and since there's only one more "out" slot, whichever one of S and W isn't out must be in.

The diagram becomes:

In this diagram, L must be in, so L is our answer.

Question 22

The most efficient method here is to use previous work (previous valid scenarios).

In Q21's "3 in, 4 out" diagram, we could have lacked both H and S, so Choice A is eliminated.

In Q18, we lacked both L and W, so Choice C is eliminated.

In Q20's diagram, we could have lacked both M and P, so Choice D is eliminated.

In Q21's "3 in, 4 out" diagram, we could have lacked both T and W, so Choice E is eliminated.

By elimination, Choice B is our answer.

Question 23

We already know that M conflicts with H, P, and T. This means that the only courses out of the 7 it could ever be with are L, S, and W. Choice B says this, so it's our answer.

Choice B is just another way of saying something we already know, so if B is true instead of M conflicting with P and T, it'll have the same effect.

Photo created at www.addletters.com/bart-simpson-generator.htm

It's a Grouping: In-Out / Selection game, but it's not like most In-Out logic games. In-Out games often involve both of the following types of rules:

1. A ---> NOT B

2. NOT C ---> D

However, PrepTest 58, Game 4, only involves the first kind of rule, giving you a ton of "double-not arrows" to work with (read the post linked above if you don't know what those are). These make the game seem deceptively easy - it's actually a bit more complicated.

I consider this game to be the hardest out of the 4 in that section. In this blog post, I'll explain the game from start to finish.

Those of you who haven't seen this game before, please don't look at the rest of this post until the week before your exam. You'll want to save PrepTest 58 until then.

The rest of you, read on.

Also see: Explanations for Recent LSAT Logic Games

Due to LSAC's copyright rules, I'm forced to leave out certain details of the game. You'll need a copy of the exam in order to follow along.

First off, the rules:

H ---> NOT S and NOT M

If H is chosen (IN), then neither S nor M is chosen (so they must be out). This means if either S or M is chosen, H must be not be chosen. However, I wouldn't take the time to write out the contrapostive. I would immediately jump to diagramming it with a double-not arrow, as I have below.

The remaining two rules should be treated the same way:

M ---> NOT P and NOT T

W ---> NOT P and NOT S

The three rules can be diagrammed as follows:

This means:

H conflicts with both S and M.

M conflicts with both P and T.

W conflicts with both P and S.

This means, among other things, that we must always lack at least one of H and S, one of M and T, and one of W and P. In other words, we must always lack at least 3 variables out of the 7 total.

Since the initial paragraph of the game also says that we must always have at least 3 variables in, this, when combined with the info that we must always lack at least 3 variables, means we will either have "4 in, 3 out" or "3 in, 4 out". ("4 in, 3 out" is demonstrated by the first diagram below.)

Of course, there are other pairs of variables that conflict with each other (pairs from which we must always lack at least one). However, I'm choosing these particular pairs because none of these conflicts overlap.

What I mean is - none of these conflicts mention the same variables - none of the variables are involved multiple times in the conflicts that I've chosen.

By choosing these particular conflicts out of all the conflicts mentioned in the rules, I can create the following 2 main templates / options:

Both diagrams indicate that we must always lack at least one of H and S, one of M and T, and one of W and P. The "4 in, 3 out" diagram is nearly complete because the conflicts I've chosen fill all the "out" slots, so everything else must be in, including L. The "3 in, 4 out" diagram is a bit more ambiguous because we still have 1 "out" slot remaining, and we don't know which variable will fill it.

It's important to remember that these diagrams don't address all of the rules - only the ones I handpicked. We still have to deal with a few conflicts, which are:

Let's keep these in mind as we go through the game.

Question 18

Typical List/Acceptability question

We're always going to lack at least one of H/S. Choice A violates this.

We're always going to lack at least one of M/T. Choice E violates this.

We're always going to lack at least one of P/W. Choice D violates this.

Looking at the conflicts not included on the diagram, we're always going to lack at least one of M/H. Choice B violates this.

By elimination, Choice C is our answer.

Question 19

Knowing that we must have at least 3 in and at least 3 out, the max we can have in is 4. We already know this from the inferences we made before starting the questions.

Question 20

If both P and W are out, our diagram becomes:

Choice A: We can't lack both H and L because L is definitely in. Eliminated.

Choice B: We CAN lack both H and M because there's no reason we can't have both S and T. This would create:

IN: STL

OUT: HMPW

This doesn't violate any of the remaining conflicts not included in the main diagram, so it's our answer.

I'll go through the rest anyway.

Choice C: We can't lack both H and S because there's only space in the out column to lack one of the two.

Choice D: We can't lack both L and M because L is definitely in.

Choice E: We can't lack both S and T because this would create:

IN: HML

OUT: STPW

H and M can't both be in because this violates one of the remaining conflicts not included in the main diagram.

Question 21

This is a local question asking what must be true if M is in. Remember that there are two major possibilities: having 3 variables in and 4 variables out, or having 4 variables in and 3 variables out.

In both possibilities, if M is in, H, P, and T will automatically be out. H and P will be out because M conflicts with both in the "remaining rules" not included on the main diagram. T will be out because we must always lack at least one of M/T according to the main diagram.

This fully determines the "4 in, 3 out diagram", creating:

This diagram is invalid because due to the "remaining rules," S and W can't both be in.

This leaves us with the "3 in, 4 out" diagram, which is a little ambiguous. We will again lose H, P, and T for the same reasons as above. However, there's still one more "out" slot to take care of. It's addressed by the fact that, in our "remaining rules", S and W conflict with each other. At least one of the two must be out, and since there's only one more "out" slot, whichever one of S and W isn't out must be in.

The diagram becomes:

In this diagram, L must be in, so L is our answer.

Question 22

The most efficient method here is to use previous work (previous valid scenarios).

In Q21's "3 in, 4 out" diagram, we could have lacked both H and S, so Choice A is eliminated.

In Q18, we lacked both L and W, so Choice C is eliminated.

In Q20's diagram, we could have lacked both M and P, so Choice D is eliminated.

In Q21's "3 in, 4 out" diagram, we could have lacked both T and W, so Choice E is eliminated.

By elimination, Choice B is our answer.

Question 23

We already know that M conflicts with H, P, and T. This means that the only courses out of the 7 it could ever be with are L, S, and W. Choice B says this, so it's our answer.

Choice B is just another way of saying something we already know, so if B is true instead of M conflicting with P and T, it'll have the same effect.

Photo created at www.addletters.com/bart-simpson-generator.htm

Hi Steve,

ReplyDeleteThanks for all you do! I have a couple of questions if you have time to get back to this..

In #21, your 4-3 diagram incudes both S and W. Considering that S and W cannot both be in, can we use that diagram to draw a valid conclusion that can be used to answer the question?

And in regards to #23, answer B certainly says what we are looking for, but is the effect not the same as that given by answer A? And how do we differentiate between these two answers? I assume LSAC is looking for the BEST answer always, but they seem to have the "same effect," as the question asks for.

Thanks a ton for your hard work!

Also, what are some other games similar to this one we can practice with? I like your idea to set up the variables without multiple conflicts and then leaving two extra rules to remember at the bottom of your diagram. That certainly helps weed through all the details. Thanks again!

ReplyDeleteGlad you're enjoying the blog!

ReplyDeleteAs you noted for #21, the 4-3 diagram is invalid because S and W can't both be in. This means that we can only consider the 3-4 diagram for this question, where M, one of S/W, and L are all in. I've made this more explicit in the explanation above in case anyone was confused.

For #23, answer choice A does not have the same effect as answer choice B. If answer choice A *replaces* the rule that M conflicts with both P and T, we will now be able to have situations like:

IN: MPS

OUT: HLTW

This would not have been acceptable under the original rules, but it is acceptable when A replaces the M <--/--> P, T rule. Therefore, it does not have the same effect.

Hope this helps!

As for your final question, I can't recall any games quite like this one.

ReplyDeleteHi Steve,

ReplyDeleteI just found this blog and have been preparing for the June LSAT. I had a quick question on this game. How did y ou determine which variable combos to put in the "Out" Group - ie the w/p, m/t, h/s options. I cant say that I would have done the same, and looking at your explanation, doing so would have saved me a lot of time.

Thank you!

Hey briahn,

ReplyDeleteAs I said above:

I'm choosing these particular pairs because none of these conflicts overlap.

What I mean is - none of these conflicts mention the same variables - none of the variables are involved multiple times in the conflicts that I've chosen.

Hope this helps.

What do you mean none of variables are involved multiple times? Aren't both S and P mentioned in two not-both relationships?

Delete