#9 - Local, Must Be True

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=11881)

The correct answer choice is (D)

The question stem creates a rotating OS block. You are then tasked with finding the two businesses that must be next to this block. Because of the wording of the question (“must”), there is only one pair of businesses that can be next to this block, and thus any single hypothetical will reveal the answer to this question.

The easiest hypothetical is one that uses P as an endpoint and places O next to P, thus satisfying the third rule. Templates #1 and #2 act in identical ways (just in reverse order), so the choice of which template to use is immaterial. We will use Template #1:

• Template #1:
  
  In this template, when O is in space 2, then S must be in space 3. R1 must then be in space 4 in order to comply with the second rule:

PT66_J12_Game_#2_#9_diagram 1.png

• However, this leaves T and V (in either order) in spaces 5 and 6, a violation of the fourth rule. Thus, we can infer that P can never be one of the businesses that borders the SO block. Why? We’ve shown that in Template #1 it cannot occur, but because Template #2 is the same setup in reverse, it also cannot occur in Template #2 (try it—the results are identical to the above). Thus, any answer choice that contains P cannot be correct, and answer choices (A) and (B) can be eliminated.

Because neither POS nor SOP will work for this question, from the third rule we can infer that O cannot be next to P, and therefore V must be next to P. Consequently, let’s try a different hypothetical, one where V is in the space next to P. Again, we can use either template since they are functionally identical; we will use Template #1 again:

Template #1:

• In this template, V is now in space 2. At first it appears that R1 could be in space 3 or 4, but if it is placed in space 4 then the SO block will occupy spaces 5 and 6, leaving T in space 3, a violation of the fourth rule. Thus, let’s place R1 in space 3 instead:

PT66_J12_Game_#2_#9_diagram 2.png

• Now, the SO block (in whatever order) must occupy either spaces 4-5 or spaces 5-6. The remaining space—4 or 6—must be occupied by T. These scenarios produce no violations, and show that the SO block will be bordered by T and an R. Thus, due to the nature of the question stem, answer choice (D) is correct.

[Basia W]

Hello,

I was at a bit of a loss on how to a attack this question- this may be due to a lack of inferences on my part in my initial diagram. But because of the flexibility of P being next to either V/O I wasn't sure how S factored in.

Thank you for your help!

Best,

Basia

[Emily Haney-Caron]

Hi Basia,

This is definitely a tricky one. For this question, you have fewer possibilities that you have to try out than you think, because it doesn't really matter whether P is in spot 1 and R is in spot 7, or the other way around; it's essentially reversible (because it doesn't change what is next to the OS block regardless of which direction you arrange it). So, just pick one for this question. Let's put P first and R last. We know either O or V needs to be next to P, so let's try putting O there:
P O S __ __ __ R
We know that the Rs need to be at least 2 spaces apart, so we need to put R right next to S. However, that means that T and V would be next to each other, which violates a rule. That must mean that V will be next to P, rather than O:
P V __ __ __ __ R
T can't go next to P, so that leaves the OS block or R. If we put the OS block next to V, we don't have enough room left to make the 2 Rs 2 spaces from each other. That must mean we have to put R next to V:
P V R __ __ __ R
Then, no matter where we put the OS block and where we put T, there will be T on one side of the block and R on the other.

Hope that helps!

[Basia W]

Yes it does! I am somewhat relieved that this was more of a trial and error based question rather than me lacking some fundamental inference. Thank you for going through it with me!

Best,

Basia