Which one of the following, if substituted for the condition that Molina must be assigned to exactly one committee, w...

Melissa on September 1, 2019

any guidance on how to arrive at the correct answer for this question

kindly show me how to arrive at the correct answer for this question

1 Reply

Irina on September 1, 2019

@Tony,

Let's quickly look at the setup:

5 volunteers - H J K M N are assigned to X Y Z.

Each committee will have 3 volunteers assigned to it with each volunteer holding one of the three positions: leader, secretary, or treasurer.

We thus have 5 people to fill in 9 slots.

X ___ ___ ___
Y ___ ___ ___
Z ___ ___ ___
L S T

The following conditions apply:

(1) N -> N (L)

(2) M = 1

(3) K must be assigned to Y but cannot be assigned to Z.

X ___ ___ ___
Y ___ ___ ___ K
Z ___ ___ ___ ~K
L S T

(4) J must be the secretary for Y but cannot be assigned to X or Z

X ___ ___ ___ ~J
Y ___ J ___ K
Z ___ ___ ___ ~K ~J
L S T

These rules tell us that out of 9 slots, 1 of them must be filled with J, one with M, and two with K per rule (3) - K must be assigned to Y but cannot be assigned to Z. We still have 5 slots to fill and only 2 volunteers left - N & H, meaning they must either be assigned to 3 or 2 committees each.

Note that we cannot have K assigned only to Y because we would not be able to fill the rest of the roles as N & H cannot both be assigned to all 3 committees due to J & K already being assigned to committee Y. Thus, we could only have either H or J on committee Y, but not both.

The number of potential positions each volunteer could occupy could thus be summarized as:

M =1
J =1
K = 2
N = 3 or 2
H = 3 or 2

The question asks us what rule would have the same effect as M=1?

The rule must lead to the same inference as in our initial setup that M could only take on one position. Let's look at the answer choices:

(A) H must be assigned to more committees than M

Incorrect. This rule only allows us to infer to M could be assigned to 2 committees at most, not 1.

(B) J must be assigned to more committees than M.

Incorrect. This rule would result in M being assigned to 0 committees because J is assigned to 1, which is impossible as everyone must be assigned to at least 1 committee to fill all the positions.

(C) K must be assigned to more committees than M.

Correct. We can see that K must be assigned to 2, and this rule would mean M has to be assigned to 1.

(D) M must be assigned to more committee than H.

Incorrect. This would lead to H being assigned to 0 committees, which is impossible.

(E) N must be assigned to more committees than M.

Incorrect. This rule only allows us to infer to M could be assigned to 2 committees at most, not 1.

Does this make sense?

Let me know if you have any further questions.