#15 - Local, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=1007)

The correct answer choice is (D)

If Alicia takes S3 and G, the initial diagram appears as:

Because G is selected, from the fifth rule W cannot be selected, and answer choice (E) can be eliminated.

Because P requires S9 from the fourth rule, and Alicia already takes S3 (which from the condition in the game scenario eliminates her ability to take S9), P cannot be selected. Therefore, answer choice (C) can be eliminated.

The remaining two incorrect answers are more difficult to eliminate. Consider for a moment the courses that have been eliminated:


W, P, and S9 each cannot be selected.


Thus, three courses have already been eliminated, which in an 8-into-4 scenario means that four of the remaining five courses must be selected. Let us examine those remaining five courses more closely. They are:

S3, G, J, R, and M


S3 and G are already selected. The remaining two selections must then come from the group of J, R, and M (remember, both J and R can be selected; the first rule just addresses what occurs when one is not selected). However, from the second rule, J and M cannot be selected, and thus, by Hurdling the Uncertainty, we can infer that R must be selected. Thus, answer choice (D) must be true and is correct.

[SherryZ]

Hi there, thank you guys for your generous help!

Dec 2009 LSAT, Sec 1 LG, Game 3, Q15:

I don't know why D is the correct answer?

Base on the question stem, since G is in, then W CANNOT be in. In addition, P CANNOT be in because if P is in, then Alicia will take S at 9am, which violates the question stem. Now, there are 2 remain spots. I don't know how to decide which class must to be taken

Please help me out and if my logic above is wrong, please correct me.

Thank you so much! Have a nice day!

---Sherry

[KelseyWoods]

Hi Sherry!

So you're right that W and P are definitely out. That leaves us with two available spots and 3 options to fill them: M, R, J.

By combining the first two rules, we learn that if she takes M, she can't take J, (second rule) and if she can't take J, she must take R (contrapositive of the first rule).

So: M /J R
M R
contrapositive: /R /M

Since we now know that if we have M, we have to have R, that means that R has to be in. With W and P out, there's only one more "out" slot. If R is out, M would have to be out as well, but we don't have enough out slots for that. So R must be in. For the final "in" slot, it could be M or J, either one would work.

That gives us answer choice (D): Alicia must take Russian.

Hope that helps!

Best,
Kelsey

[Basia W]

Good afternoon,

I went through this question and am at a loss of why Russian is the topic she must take. If she takes statistics at 3, and G, we know that she cannot have W. But because NOT W is at the end of the P S9 chain we cannot infer any further.

Thank you for your help!

Best,

Basia

[Emily Haney-Caron]

We know she definitely takes: S3, G
We know she definitely cannot take: S9, H
We now need to use 2 of P, J, M, and R to fill her schedule. If she takes P, she has to take S9, and we know she can't - so now we know P is also out. That leaves J, M, and R. She cannot take both J and M together, so she HAS to take R.

[Jeanel]

Hi!

I just had a quick question about this. I was able to get this question right, but I was just wondering if someone could explain to me why B) was an incorrect answer? I am probably overthinking it but I just want to make sure. I was able to narrow down that R, J, and M were the pieces that had to fit into the group, but aren't R and M two pieces that equally MUST be chosen? Since if you choose M, you must choose R and if you choose R you must choose M for this specific situation? If J is chosen, the course schedule would not be filled.

Thank you so much!

[Clay Cooper]

Hi Jeanel,

Thanks for your question, and welcome to the forum! I hope you will find it useful.

It looks to me like you have misinterpreted a rule here. I'm not sure what you mean when you say that R, J, and M must be included; in fact, J and M can never both be included (due to the second rule: M ~J). While it is true that, If you choose M, you must choose R as well, the reverse is not true; that is, if you choose R, it is not the case that you must choose M.

Choosing R is not the sufficient condition in any of our rules. We can deduce from the local rule that we must leave out W (because G is chosen) and the early Statistics (because the late statistics is chosen); furthermore, we know that since the early statistics is left out, Psychology must be too. That leaves only one open space in the out-group, which we know from the second rule must be filled by either J or M. There is no reason to suppose, however, that it could not be M in this space; in other words, we could choose G, the late Statistics, J, and R, without breaking any rules.

Does that help?