#20 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=182&p=88548#p88548)

The correct answer choice is (E).

If M’s delivery is first and L’s delivery is third, the following initial diagram results:

G4-Q20-d1.png

This scenario can only occur under Template #1. In that template, F and H are delivered earlier than G K. Thus, with only four available spaces, F and H must be delivered second and fourth (not necessarily in that order), and G and K must be delivered fifth and sixth:

G4-Q20-d2.png

Regardless of the placement of F and H, the third rule will satisfied.

Accordingly, answer choice (E) must be true and is correct.

[paullmatt]

September 2007 LSAT, Section 2, Question 20

For the life of me I cannot figure out where I went wrong. I marked B but the correct answer is being shown as E.

It appears to me that both B & E Must Be True. Any assistance solving this would be much appreciated.

[Dave Killoran]

Hi Paul,

Thanks for the question. H's delivery could be second or fourth, and thus while (B) could be true, it does not have to be true. Here's what happens:

If M’s delivery is first and L’s delivery is third, the following initial diagram results:


M _ L _ _ _
1 2 3 4 5 6


This scenario can only occur under Template #1 of the game (where M > H). In that template, F and H are delivered earlier than G > K. Thus, with only four available spaces, F and H must be delivered second and fourth (not necessarily in that order), and G and K must be delivered fifth and sixth:


M F/H L H/F G K
1 2 3 4 5 6


Regardless of the placement of F and H, the third rule will satisfied.

Accordingly, answer choice (E) must be true and is correct.

[paullmatt]

Thank you for the clarification. My mistake was that I was taking the contrapositive of if F>M ---> L>H to mean that L could not come before H at all...