#18 - Local, Must 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 (A)

G3-Q18-d1.png

The remaining unassigned people are F, G, V, and W, which is an interesting set of four variables remaining because those four are all addressed in the second and third rules. As you might imagine, those rules play a role in determining what occurs.

From the second rule, if F is assigned to boat 2, then G is assigned to boat 2. But, under the condition in this question, that would fill boat 2, forcing V and W into boat 1, a violation of the third rule. Thus, F cannot be in boat 2 and must be in boat 1. Thus, answer choice (A) is correct.

[ellenb]

Dear Powerscore,

I still have a question for this game for Quest 18.

Why does F has to be in boat one? Is it because if we have F1 none of the rules get activated? I was just curious.

Thanks

Ellen

[Adam Tyson]

There are a few ways to approach this question, and we can look at more than one, but one thing they both have in common is that we can look back to our previous work for help. That, by the way, is a great strategy in a lot of games - use what you already know to figure things out.

For question 18, take a look back at what happened in question 14. In that question, F was assigned to boat 2, and what happened? Just that, by itself, ended up forcing H and Y both into boat 1. Boat 2 had F, G, X/Z and V/W, while boat 1 had H, Y, Z/X and W/V. Think of it conditionally - if F is in 2, H and Y are in 1. The contrapositive, then, is if H and Y are not both in 1 (and if they are not in the same boat, that sufficient condition is met) then F is not in boat 2, meaning F must be in boat 1, right?

Another approach to q18 is to try templates. Put H in 1 and Y in 2 and see how it works out. Then flip them and try again. In both cases you should find that F is forced into boat 1.

Using prior solutions is fantastic - it allows you to get more bang for your buck and save time to boot. I often find myself on the last question or two looking back to see what I have already done, just in case something there solves it for me without doing anything more.

Adam M. Tyson
PowerScore SAT and LSAT Instructor

[Kp13]

Hi, I have a question about the logic used to answer question 18 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 questions. 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 #18.

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 18
If H is assigned to a different boat than Y, which one of the following must be assigned to boat 1?

H in B1/Y in B2

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

OR

Y in B1/H in B2 (#1)

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

The correct answer is that F must be assigned to boat 1.

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 #18. 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.

Question #18 is designed to be hard, and designed to require a higher level of abstract thinking. So, the fact that you struggled with this one isn't a surprise (everyone did). The takeaway form this question for me is the nature of the variables they put into play, and which variables that leaves yet to be placed. With X and Z, and H and Y split between the two boats, the remaining unassigned people are F, G, V, and W, which is an interesting set of four variables remaining because those four are all addressed in the second and third rules. As you might imagine, those rules then play a role in determining what occurs. And that's the key here--not just who has been placed, but who is yet to be placed, and what connections exists between those variables. It doesn't make this question easy, but it does make it easier.

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.

[bella243]

Based on KP's most recent post: in this template, Y and H are not necessarily fixed to boat 1 and 2, respectively, right? Meaning, H could be boat 1, and Y could be boat 2, right?
Y in B1/H in B2 (#1)

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

[Adam Tyson]

That's correct, and your diagram here works fine, bella243! While we are at it, that old post has one error in it that I can see, and that is that H being in Boat 2 is not sufficient to prove that G is in Boat 1, because G in 2 does not prove that H is in 1. It's the combination of F and G BOTH being in 2 that forces H to 1. But there is a viable solution with H and G in Boat 2 and F in Boat 1:

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

As I mentioned earlier in this thread, I would start my attack on this question with a quick review of any local diagrams I had already done (and that's a bunch of diagrams, because every question other than the list question was a local one). If I had done templates I would have reviewed those as well. And then, if I still had more than one contender, I would just test one and see what I got.