If Tran is not trained in month 2, which one of the following could be true?
aseikhon11on July 18, 2020
How is C the right answer?
If T cant be second, then Q is not first.
If 2 is fourth then the only option left is Q T for month 5 & 7 because Rule #1 says no ones in month 6. But Rule #2 says Q has to be the month prior to T which would be 5 &6 or 6 & 7? Confused. Please help
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To test (C), we start out with: __ __ __ S __ -- __ 1 2 3 4 5 6 7 ~T
Rule #3 says that we must have a QT block in consecutive spaces. The only open adjacent spaces are 1-3, and we know that T cannot go in spot 2. Therefore, we place Q in spot 2 and T in spot 3. This gives us: __ Q T S __ -- __ 1 2 3 4 5 6 7
Rule #4 tells us that S is the third agent trained. In our current diagram, this is true. To preserve this, we should block out spot 1 as a month in which no agent is trained. We now have: -- Q T S __ -- __ 1 2 3 4 5 6 7
Rule #2 tells us that U must precede R. We only have two open spots left, so we should place U in the earliest one and R in the latest. This gives us a final diagram of: -- Q T S U -- R 1 2 3 4 5 6 7
Does that make sense? Please let us know if you have any other questions!
Ecochranon September 29, 2020
@Skylar according to how you diagram, that puts S in the fifth month, not the fourth, as the correct answer stipulates. Also, how do you have R beyond the 7th month?
Jesseniaon September 15, 2021
I am confused as to why B is eliminated. The explanation states that B is eliminated because only R can be placed in month 5 in scenario A. Would anyone be able to provide a more in depth explanation on why B is eliminated? Thanks
Emil-Kunkinon September 22, 2022
Hi, there are only two places that T can go, because it must be in a block with Q, and there are only two open blocks of two, 12 and 45.
In this question, we know that T cannot go in 2, so the QT block must go in 4 and 5. So, R cannot also go in 5.
