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June 2017 LSAT
Which one of the following, if true, completely determines the assignment of managers to cities?
on July 4, 2019
What is the best way to set this game up?
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on July 5, 2019
We know that a company has four product managers: Fan, Gleeson, Haley, and Ibanez
F, G, H, I
The company is sending these four product managers to visit three cities: Manila, Sydney, and Tokyo
M, S, T
Each manager will visit at least one of the cities, and each city will be visited by exactly two of the managers. We know that the arrangement of cities does not change but the assignment of managers to cities does. Therefore we can set up the game as such:
M: _ _
S: _ _
T: _ _
Now let's go through the conditions.
1) Ibanez must visit exactly two of the cities.
I must visit a combination of either MS, ST, or MT
2) Fan and Haley cannot visit the same city as each other.
F - > not H
H - > not F
3) If Manila is visited by Gleeson, then Tokyo must be visited by Haley.
MG - > TH
Not TH - > Not MG
We also know that if Haley visits a city then Fan cannot visit that city. So, if Manila is visited by Gleeson, then Tokyo must be visited by Haley and, therefore, Fan cannot visit Tokyo.
MG - > TH - > not TF
4) Gleeson cannot visit Sydney.
Therefore, the only possibilities are MG or TG
Now we can use this set-up and conditions to address the questions. Let's go through the first question together so that you can see how this set-up is employed.
Question 1 asks: "which one of the following, if true, completely determines the assignment of managers to cities?"
Three out of the five answer choices state that a manager visits exactly two of the cities. The remaining two answer choices place two managers in one city.
Let's start with the answer choices which place two managers in one city (D and E). These are essentially the same because the only thing that we know about F and H is that they cannot visit the same city as each other. H is also limited if G visits M; however, both these answer choices place G in Tokyo so we can ignore this condition.
As F and H are limited by the same condition, we can swap either of them out and get the same result. Therefore, both D and E must be incorrect. If F and G visiting Tokyo completely determine the assignment of managers to cities, then H and G must also do so because F and H are essentially the same variable. Both D and E cannot be correct, therefore both answer choices are incorrect.
Now let's address the remaining three answer choices. Again, we can see that F and H are the variables in two of the three choices (A and C). As F and H are essentially the same variable, we can also eliminate answer choices A and C.
This leaves B as the correct answer. Let's double check using the set-up outlined above.
We know that Gleeson cannot visit Sydney. So, G must visit M and T.
We also know that if G visits M, then H must visit T.
M: G _
S: _ _
T: G H
We know that Ibanez must visit exactly two of the cities. So, I must visit M and S as they are the only two cities remaining with an empty slot.
M: G I
S: I _
S: G H
We also know that each manager must visit at least one of the cities. Fan has not yet been assigned to a city, so F must visit S.
We cannot move five of the six variables which forces F to visit S locking in all six variables. Therefore, the assignment of managers has been completely determine if G visits exactly two of the cities making B the correct answer.
Hope this is helpful! Please let us know if you have any further questions.