June 2019 LSAT
Section 4
Question 20

# If Pang works on only one day, which one of the following must be true?

Reply

Irina on February 12, 2020

@Alexander-Blankers,The question asks us if P works on only one day, which of the following must be true. Since this game involves 5 volunteers and 9 slots, 3 on each day - we know that 4 volunteers must work for 2 of the days and 1 volunteer for 1, resulting in the following scenario:

2-2-2-2-1

This question tells us that P is the one that only works on one day:

2-2-2-2-P

Since P cannot work on Sat, we are left with two scenarios for this question:

(1) P - Th

(2) P - Fr

Let's consider both of them. I usually like to start with the more restrictive scenario and since we already know that N works on Fr, let's start with that:

1. P - F

Th: L M /N/Q

F: N P Q

S: L M /N /Q

We still have two L & M pairs and the only days they could go are Th & S since there is only one spot left on Fr. We still have two Q shifts and 1 N shift left - since N must go to either Th or S, Q cannot go on both of these days, meaning one of the Q shifts must fall on Friday, the remaining one could go on either day.

Let's briefly look at the answer choices and see if we can eliminate any of them. We can eliminate (A), (C), and (E). If I was running short on time here, I would pick (B) and move on because we know that (D) - P works on Friday is only one of our scenarios. Let's consider a case where P is on Th next, and if we can come up with a scenario that complies with all the rules when P is on Th, we can definitively say that the answer is (B).

2. P - Th

Th: P L M

F: N L M

S: Q N ????

Again, we have two L &M pairs - they cannot go on Th & F because it would leave no room for two Q shifts, thus we can conclude that L & M must definitely be on Sat in both scenarios and (B) is the correct answer.