September 2016 LSAT
Section 3
Question 13

# Which one of the following could be the complete assignment of assistants to Pricing?

Replies

cdichiar on September 19, 2019

Please help with game setupIrina on September 19, 2019

@Gianna-Rossi,Let's look a the setup. The game involves an economics department assigning 6 TAs - R S T V Y Z to 3 courses - L M P. Each TA is assigned to exactly one course an, and each course will have at least one assistant assigned to it. This rule tells us that every assistant must be assigned, at there must be at least one assistant per course, but there is no upper limit, so we could have a 1 -1 -4 scenario or 1 -2-3 or 2 -2 -2.

Let's look at the rules:

(1) M must have exactly 2 assistants assigned to it.

L:

M: 2

P:

This rule tells us that 1-1-4 scenario is out since at least M must have exactly 2 assistants, we are down to 1-2-3 or 2-2-2 scenarios.

(2) S & T must be assigned to the same course as each other.

ST

Let' s remember this rule for now.

(3) V & Y cannot be assigned to the same course as each other.

This rule tells us if V is assigned to a course, then Y cannot and vice versa.

V -> ~Y

Y -> ~V

(4) Y & Z must both be assigned to pricing if either one of them is.

This rule tells us that either both YZ must be assigned to P or none of them.

We can also infer that if either Y or Z are assigned to pricing, then neither S nor T could be assigned to pricing because we cannot have more than 3 TAs per class, and we can infer that V cannot be assigned per pricing per rule (3).

Y v Z (P) -> ~(S/T/V) (P)

Considering there a fair number of restrictions on who might be assigned to pricing, let's consider a couple of possible scenarios:

(1) Y &Z assigned to pricing.

1-2-3 scenario

L: V

M: S T

P: Y Z R

2-2-2 scenario

L: ST/ VR

M: ST/ VT

P: Y Z

(2) Y & Z are not assigned to pricing.

In a 1-2-3 scenario, S &T must be assigned to pricing because either Y or Z or both must be assigned to M since there is only one spot in L and neither of them can be assigned to P. Also, either V or R must be assigned to P in this scenario along with S & T.

L: 1 of Y/Z/V/R

M: 2 of Y/Z/V/R

P: S T + /V/R

One possible option could be:

L: Y

Z: Z V

P: STR

In a 2-2-2 scenario, ST must still be assigned as a group to either of the classes, YZ could be split across L & M as long as Y is not assigned to any class with V or assigned as a group to L or M. For example:

L: ST

M: YZ

P: VR

or

L: YR

M: ZV

P: ST

The point is not to list our every possible scenario but to see the impact of the most restrictive rule on the setup.

The question asks us which of the following could be the complete assignment of assistants to P?

We can see right away that all the options involve 3+ TAs, so we know that it is a 1 -2-3 scenario and we can right away eliminate (C) because we know that 1 -1 -4 is impossible. Let's look at our 1-2-3 scenarios above -

If Y Z are assigned to pricing, then R must be assigned with them.

If Y Z are not assigned to pricing, then ST must be assigned to pricing along with V or R.

Let's look at the answer choices:

(A) R Y Z

Correct. This is one of our scenarios above.

(B) S T Y

Incorrect. Z&Y are a package deal for this class per rule (4), so we cannot have only one of them assigned to P.

(C) S T Y Z

Incorrect. We have already eliminated (C) because it is impossible to have more than 3 assistants per class.

(D) T Y Z

Incorrect. ST must be assigned to the same class per rule (2).

(E) V Y Z

Incorrect. V & Y cannot be assigned to the same class per rule (3).

Let me know if this makes sense and if you have any further questions.