Which one of the following is an acceptable selection of representatives for the committee?

Skylar January 12, 2020

@Vanessa-Ratcliff,

Question #13 asks "Which one of the following lists three representatives who could be selected together for the committee?" It is important to understand that the list given in the answer choices is not exhaustive- instead we must deduct the possibilities for the missing two representatives to determine if the given list would be viable. This is why (C) is correct even though it does not list Q for F. You also ask how (C) could be correct when there is no K for G, but G is not a variable listed in (C).

Let's walk through the answer choices to better understand this concept.

(A) If we are given F, G, J as variables that are in, what other variables can we deduce must also be in? Because of the rule F->Q, we know that Q must be in. Because of the rule G->K, we know that K must be in. Because of the rule J->M, we know that M must be in. This gives us a selection of: F, G, J, Q, K, M. However, this scenario cannot be correct because it requires six variables to be selected, and there are only five spots. Also, this scenario breaks the first rule that at least two tenants and two homeowners are selected, as Q is the only homeowner.

(B) If we are given F, G, M as variables that are in, what other variables can we deduce must also be in? Because of the rule F->Q, we know that Q must be in. Because of the rule G->K, we know that K must be in. Because of the rule M->J, we know that J must be in. This gives us a selection of: F, G, M, Q, K, J. Again, this scenario cannot be correct because it requires six variables to be selected despite there only being five spots.

(C) If we are given F, J, M as variables that are in, what other variables can we deduce must also be in? Because of the rule F->Q, we know that Q must be in. There are no other variables that we can deduce must be selected. In order to satisfy the rule of having at least two tenants and two homeowners, we know that the fifth variable selected will be a homeowner, and because of the rule that M and P cannot both be selected, we know that it will be either R or S. This gives us a selection of: F, J, M, Q, (R/S). This scenario is valid and aligns with our given rules, so (C) is correct.

(D) If we are given G, J, K as variables that are in, what other variables can we deduce must also be in? Because of the rule J->M, we know that M must be in. This gives us the four variables G, J, K, M. These variables are all tenants, and since there is only one spot left to fill, it is impossible to have two homeowners. This breaks our first rule and is invalid.

(E) If we are given G, J, M as variables that are in, what other variables can we deduce must also be in? Because of the rule G->K, we know that K must be in. This presents us with the same four variables as (D) did: G, J, M, K. Again, these four variables are all tenants, which makes it impossible for us to have five variables and two homeowners, so (E) is incorrect.

Does this help? Please let us know if you have any other questions and best of luck with your studies!