The game involves 6 sets of train tickets - 3 sets each month - J and F for lines 1, 2, 3. Each of the sets of ticket is one of the following colors - G P R Y. We can diagram the initial setup as follows:
J._. _. _ F _. _. _ 1. 2. 3
The rules tell us that:
(1) exactly one set of J tickets is red. (2) no Feb tickets are purple (3) J tickets for line 2 are purple (4) For line 3 either J or F tickets are green but not both
J. _. P __. 1R F _. _. __. ~P 1 2. 3 1G
The rules also tell us that for each line tickets are different colors, and for each month tickets are different colors.
The question asks us if none of the tickets are purple, which of the following could be true? Let's start by removing rule (3) from our diagram:
J. _. __ __. 1R. ~P F _. _. __. ~P 1 2. 3 1G
Now, none of the tickets are purple and we only have three color options to assign to the six sets of tickets - G Y R. We can infer that in each month at least one set of tickets must be G, one set Y, and one set R but the tickets for any single line must also differ in color in J and F, so if let's say F3 is G, J3 must be Y or R. With this in mind, let's look at the answer choices:
(A) None of the J tickets are green.
Incorrect. We only have three possible colors to assign to three lines, thus we must use one of each color - G Y R.
(B) None of the F tickets are green.
Incorrect. Same as for J, we must use one of each color to comply with the rule saying that for each line, tickets for each month must be in different colors.
(C) None of line 2 tickets are green.
Let's consider this scenario. We already know that one set of tickets for line 3 is G and since none are G for line 2, 1 must be G for line 1 in a month that is different from G tickets for line 3. Let's say J3 is G, then F1 must be G and vice versa. Then we just need to make sure that line 2 tickets are Y and R in either order, and the remaining J1 and F3 tickets are Y or R as long as the color is different from the color for line 2.
J. Y/R R/Y /G. F /G. Y/R R/Y 1 2. 3
There are total of four possible combinations that could be true if none of the tickets on line 2 are G:
(1)
J. R Y G. F G. R Y 1 2. 3
(2)
J. Y. R G. F G. Y R 1 2. 3
(3)
J. G Y. R F Y R G 1 2. 3
(4)
J. G R Y F R Y G 1 2. 3
Of course on a real test, one scenario is sufficient to determine that this could be true without violating any of the rules and conclude that (C) is the correct answer choice.
(D) No line 1 or 2 tickets are yellow.
Impossible. Since we must have at least 1 Y ticket in J and 1 Y ticket in F, if no line 1 and 2 tickets are yellow, both J and F line 3 tickets will have to be yellow in violation of the rule telling us that for each line J tickets are different color than F ones.
(E) No line 2 or 3 tickets are red.
Impossible. For the same reason as above, line 1 tickets for J and F would have to be red in this scenario in violation of the rules.
Let me know if this makes sense and if you have any further questions.
megmcdermottOctober 7, 2019
This was extremely helpful! I had trouble diagramming this game so seeing it written out helped me a lot! Thank you!
