#8 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8467)

The correct answer choice is (A)

After completing the first two questions with relative ease, you should arrive at this question feeling confident. Since this is a Local question, you should as always make a mini-diagram next to the question. The “if” statement in the question, in combination with the second rule, produces the following setup:

pt2_o91_g2_q8.png

Several of the answer choices can be eliminated by using the rule that states that each floor has either one or two apartments. Since both J and P live on the fourth floor, no other residents can live on the fourth floor, and answer choices (B) and (D) can be eliminated. Since K lives on the fifth floor, the M and N block cannot live on the fifth floor, and answer choice (C) can be eliminated. Answer choice (E) can be eliminated since P must live on the fourth floor. Accordingly, answer choice (A) is correct.

[aessy77]

Hi,

I set two diagrams for this question. (1) J4 -> K5 -> P4 and (2) P4X -> K5X -> J4X.

According to the first diagram, the answer should be (A), but according to the second diagram:

5 M N
4 K Q
3 P J
2 L
1 O

So the answer (A), (B), (C), and (E) all make sense.

Can you please help me solving this problem?

Thank you.

[Jonathan Evans]

Hi, aessy,

Let's look at the two ways that this could work. The deciding factor will be the location of Q.

Start with the core setup for this local question:

• 5 K
  4 J _
  3 _
  2 _
  1 _

We know that J must be on a floor with exactly one other and P is immediately beneath K. Therefore we know:

• 5 K
  4 J P
  3 _
  2 _
  1 _

Now the next most restricted item is Q because Q cannot be on the 1st or 2nd floors. Floor 4 is finished with J and P, so Q must be on either 5 or 3. Let's look at both:

• Scenario α: Q on 5
  5 K Q
  4 J P
  3 _
  2 _
  1 _
  
  Scenario ß: Q on 3
  5 K
  4 J P
  3 Q
  2 _
  1 _

Since we have a 2 – 2 – 2 – 1 – 1 distribution, let's consider the implications of this distribution on both scenarios:

• Scenario α: Q on 5
  5 K Q
  4 J P
  3 _ → (one or two slots)
  2 _ → (exactly one slot, per the rule)
  1 _ → (one or two slots)
  
  Either 3 or 1 must have M ↔ N in it. This leaves us with 0 and L. L is by itself. It could be in 3, 2, or 1 by itself. O is in the remaining floor by itself.
  
  Thus, we have to Hurdle the Uncertainty™ for floors 1, 2, and 3. The items on these floors will be M & N, L, or O with the restriction that M & N cannot be on the 2nd floor.
  
  Scenario ß: Q on 3
  5 K O
  4 J P
  3 Q
  2 L
  1 M N
  
  In this scenario, M & N still must be together. Since the 2nd floor has exactly one apartment, M & N must be on the remaining empty floor, the 1st floor. L needs to be by itself, so the only option for L is the 2nd floor. O Q so O must be with K on the 5th floor.

Now we can note that only answer choice (A) could be true.

Good question. I hope this helps!