#15 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=362&t=3362)

The correct answer choice is (C)

If three children are assigned to boat 1, then only one adult is assigned to boat 1, and the other two adults are assigned to boat 2. If only adult is assigned to boat 1, then from the contrapositive of rule 2, that adult cannot be G (otherwise F would also have to be assigned to boat 1). Thus, G must be assigned to boat 2. The other two adults—F and H—then rotate in a dual-option:

G3-Q15-d1.png

As in question #14, the remaining unassigned people are V, W, and Y. Because of the limited number of remaining spaces in boat 2, V and W can never both be in boat 2, and from the third rule, when one of them is assigned to boat 1, the other is assigned to boat 2. Thus, V and W effectively form a rotating dual-option in boats 1 and 2, and the only remaining space for Y is in boat 1:

G3-Q15-d2.png

Answer choices (A) and (D) are both incorrect because they reference pairs of variables in dual-options. Answer choices (B) and (E) each contain Y, a person who is known to be in boat 1. Answer choice (C) is the correct answer as shown by the diagram above.

[ellenb]

Dear Powerscore,

In # 15 do we have two templates?
1 x/z Y V/w H 1 X/z V/W F Y
2 x/z V/w F G and 2 X/z V/W G H

one of them must be wrong ( not sure which one) because if both are right than we have two right answers.

Thanks!

[Adam Tyson]

Your templates look good to me, Ellen - the key is knowing that the one adult in boat 1 has to be either F or H (knocking out answer A - those two can't both be in boat 2, since one of them must be in boat 1). Y will have to be in boat 1, otherwise you couldn't come up with any three kids that will play nice together in that boat. That will eliminate answers B and E, because we only want pairs that can be in boat 2 together (not pairs that could be in the same boat as each other, as was asked in the previous question).

Answer D is knocked out by the templates - V and W must be split in this case, so that one of them can be in boat 1. That should leave you with just answer C, which is supported by your templates.

Let me guess - you were also liking answer E, because your templates showed that W and Y could be in the same boat as each other? They can, but only if it's in boat 1, and the question asked about boat 2.

Perfect job on the templates for this question! Just have to be sure you read the question carefully.

Adam M. Tyson
PowerScore SAT and LSAT Instructor

[Kp13]

Hi, I have a question about the logic used to answer question 15 for this game (the game is about grouping adults and children to two boats).

What is the thought process used to derive the correct answer for this question? I think I am confused by the conditional rules in this game and how they should be applied to find the right answer.

Thank you!

[Dave Killoran]

Hi Kp,

Thanks for the question. This question relies on you Hurdling the Uncertainty to get to the final answer. But first, you mention confusion over the conditional rules. Let's talk about those because they obviously have a big impact on the game.

Let's start with the second rule, which is initially diagrammed as:


F2 G2


Of course, because the game is a two-value system, when we take the contrapositive, we can convert the negatives into positives by changing the boat number (if G is not in boat 2, then G must be in boat 1, etc ):


G1 F1


Note that this rule (and its contrapositive) does not imply that F and G are always in the same boat. For example, G could be in boat 2 and F could be in boat 1. Thus, this rule does not create a block (and this is where many people fall into a trap).

Ok, please take a look at that and let me know if that looks any different from how you had interpreted it. And, if you want, would you like to apply that reasoning approach to the third rule, which is also conditional? If you post your analysis, I'll let you know if you are on the money, and then we can take a look at how these rules help solve question #15.

Thanks!

[Kp13]

Thanks Dave for the quick reply.

I diagrammed the conditional rules the following way:

F2 --> G2

Because the 1st rule states that each boat is assigned at least one adult, I made an inference that if both F and G are in in boat 2, H must then be placed in boat 1:

F2 --> G2 --> H1

and arrived at a contra-positive:

H2 -->G1--->F1

For the 3rd rule:

V1 -->W2
W1 -->V2

And finally:

X <--/-->Z (these can never be together).

Question 15:
If exactly 3 children are assigned to boat 1, which one of the following is a pair of people who could both be assigned to boat 2?

Boat 1:
X/Z, V/W, Y, H
Boat 2:
X/Z, V/W, F, G

OR

Boat 1:
X/Z, V/W, Y, F
Boat 2:
X/Z, V/W, H, G.

The reason why I put "H" in boat 1 is because, I can't put it in boat 2. My inference above shows that if H2 -->G1-->F1 and I can't have 2 adults in the first boat, as it leaves no room for the three kids. So, I am putting only H in boat 1 and F and G in boat 2. I can also put F in boat 1, and H and G in boat 2.

So, I guess, I can see the logic now... the correct answer is that H and W could both be assigned to boat 2 under the question requirements.

Can you give me any advise on how I can move through these kinds of questions faster in the future? Currently, I am very slow and really take the time to double and triple check to make sure I am placing the right variables based on the conditional rules of the game...I guess more drills and logic game practice is what's needed here.

[Dave Killoran]

Hi Kp,

Thanks for the reply. Yes, it looks like you figured out the logic to #15. Part of preparing for the LSAT is working through problems like this, because the next time you see something similar, you will intuitively have a better handle on it.

I will say that practice is definitely going to make you faster and better at these questions

Please let me know if that helps at all. Thanks!

[Kp13]

Thanks Dave,

It is clearer now. I just need to keep on practicing these.

[tae.chung5]

Would you please help me understanding how you can put H & G in boat 2 together?

As somebody mentioned above, F2 G2 H1. By contrapositive, H2 G1 F1.

So, the moment H goes in boat 2, it triggers that relationship and both G and F should go in boat 1. Am I missing something here?

Please help me. Thanks!

[Adam Tyson]

Thanks for asking, tae.chung5! The diagram above was strictly about the inference that can be drawn when F is in Boat 2. The rules tell us that when F is in 2, G must be in 2, but they also say that at least one adult (FGH) must be in each boat. So, if F is in 2, G is in 2 AND H is in 1. But that's not the same as saying that WHENEVER G is in 2, H must be in 1. That was only true because of F being in Boat 2. A better diagram there might have been:

F2 G2 + H1

G2 wasn't sufficient for H1; F2 was sufficient for both G2 and H1.

Try another scenario, with F in Boat 1 instead. At that point, G is free to go in either boat, and so is H, as long as at least one of them is in Boat 2. Putting BOTH of them in boat 2 won't violate any rule! We would have adults in both boats, and that rule about F and G simply no longer applies.

I hope that clears it up for you! Keep at it!