Each technician will be assigned to at least one computer. This means that we must use all five variables each time but some technicians may have to work on 2+ computers.
Each computer will be examined by exactly two technicians. This constrains the statement above. We must use all five variables each time but, as there are only six spots available, only one variable will repeat.
F: _ _ G: _ _ H: _ _
Now let's go through the conditions.
Rule 1 - If S examines F, X examines G
SF --> XG X not G --> S not F
Rule 2 - If W examines F, X examines H
WF --> XH X not H --> W not F
Rule 3 - No computer is examined by both S and W
S --> Not W W --> Not S
Rule 4 - Exactly one computer is examined by both R and T
There are several possibilities we can diagram from these rules.
Diagram 1: Suppose S examines F
X must examine G. R and T must examine H because it is the only computer with two available spots open. W must examine G because they cannot work with S.
F: S _ G: X W H: R T
Diagram 2: Suppose W examines F
X must examine H. R and T must examine G because it is the only computer with two available spots open. S must examine H because they cannot work with W.
F: W _ G: R T H: X S
Diagram 3: Suppose R and T examine F
S and W must examine different computers. It does not matter whether they examine G/H or H/G.
F: R T G: S/W _ H: W/S _
Now let's go through our answer choices. We are looking for the answer choice which cannot be true.
We can immediately eliminate answers (C) and (E) as our above diagrams show that these pairings are possible. S and X examine computer H in diagram 2. W and X examine computer G in diagram 1.
We can also eliminate answers (B) and (D). In diagrams 1 and 2, S and W are examining computer F alone. They need a partner and all five variables have already been assigned to a computer. The only rule constraining this is that S and W cannot work together. However, it is completely possible that T examines computer F with S or W in diagrams 1 and 2.
Therefore, based on our process of elimination, the answer should be (A). Based on our diagrams, we can see that this cannot be true. X is already assigned a computer in diagrams 1 and 2 and both diagrams only have one remaining opening, neither of which is working with R.
Diagram 3 does not allow for this pairing either as X can only work with S and/or W.
Hope this is helpful! Please let us know if you have any further questions.
