This is a question smaple from Atlus LSAT Prep.
A crayon manufacturer is designing a box with ten colors of crayons—M, N, O, P, R, S, T, V, W, and Y. The crayons will be arranged in two rows—front and back—and five columns, labeled 1-5 from left to right.
The following conditions apply:
P is in the same row as O, and there are exactly three crayons between them.
S is in the same column as W.
T is directly to the left of P.
If M is in the front row, W and Y are in the back row.
If W is in the front row, O is in the back row.
R is in the third column.
5. If V and Y are in the front row, which one of the following could be true?
(A) S is in the front row.
(B) O is in the front row.
(C) T is in the front row.
(D) M is in the front row.
(E) R is in the back row.
6. Which one of the following conditions, if true, would determine the complete order for at least one of the rows?
(A) T is in the fourth column in the back row.
(B) V is in the fourth column of the front row.
(C) S is in the second column in the front row.
(D) W is in the second column in the front row.
(E) P is in the fifth column of the front row.
Correct answers are 5. A
6. D
Can somebody explain to me why these are the correct choices? Thanks.
Logic Game Help? Forum
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FuManChusco
- Posts: 1217
- Joined: Tue Jan 26, 2010 8:56 pm
Re: Logic Game Help?
Rules:
O_ _ TP. O and P must have 3 crayons between them and T must be to the left of P.
S
W
Mf --> Wb Yb
Wf --> Ob
R3
The rules require that O must be in the 1st column, S and W must be in the 2nd column, R must be in the 3rd column, T must be in the 4th column, and P must be in the 5th column.
5.) This answer choice sets up V and Y in the front. This immediately eliminates answer choice (D) because if M is in the front then Y must be in the back, and the question states that it is not. If V and Y are in the front then the O_ _TP block must be in the back. If they weren't then there would be no place for the SW column block. So you know that OTPM and either S/W (because of the column block) is in the back row. This eliminates answers (B) (C) and (E) leaving only answer choice (A).
The set up looks like this.
_ S/W R _ _
O S/W M T P
V, Y, and N can be placed anywhere in the 3 remaining spots no that it matters.
6.) This one is a little trickier. Answer choice (A) reveals the spots of O and P because of the O_ _TP block, but does not force anything else. (B) reveals that V is in place and that OTP must be in the back row, but nothing else. (C) places S in the 2nd Column front row and therefore W is in the 2nd column back row. Nothing else can be inferred though. (E) is similar to (A). It forces O and T into the 1st and 4th spots in the front row respectively but nothing else is revealed. (D) places W in the 2nd column of the front row, which puts S in the 2nd column of the back row due to the SW block. It also places O in the back row due to the 5th rule, which subsequently places T and P as well. Lastly W being in the front row means, M must be in the back row because if M was in the front then W would have to be in the back which it is not.
So it looks like this.
_ W R _ _
O S M T P
V, Y and N can be placed anywhere but it doesn't really matter because the question is solved.
Problem 6 is moderately time consuming if you have trouble doing hypotheticals quickly in your head. Drawing it out is not the end of the world if you have time though. Hope this helps. I could've missed an inference that would have saved time in problem 6, but the main reason for the correct answers is there.
O_ _ TP. O and P must have 3 crayons between them and T must be to the left of P.
S
W
Mf --> Wb Yb
Wf --> Ob
R3
The rules require that O must be in the 1st column, S and W must be in the 2nd column, R must be in the 3rd column, T must be in the 4th column, and P must be in the 5th column.
5.) This answer choice sets up V and Y in the front. This immediately eliminates answer choice (D) because if M is in the front then Y must be in the back, and the question states that it is not. If V and Y are in the front then the O_ _TP block must be in the back. If they weren't then there would be no place for the SW column block. So you know that OTPM and either S/W (because of the column block) is in the back row. This eliminates answers (B) (C) and (E) leaving only answer choice (A).
The set up looks like this.
_ S/W R _ _
O S/W M T P
V, Y, and N can be placed anywhere in the 3 remaining spots no that it matters.
6.) This one is a little trickier. Answer choice (A) reveals the spots of O and P because of the O_ _TP block, but does not force anything else. (B) reveals that V is in place and that OTP must be in the back row, but nothing else. (C) places S in the 2nd Column front row and therefore W is in the 2nd column back row. Nothing else can be inferred though. (E) is similar to (A). It forces O and T into the 1st and 4th spots in the front row respectively but nothing else is revealed. (D) places W in the 2nd column of the front row, which puts S in the 2nd column of the back row due to the SW block. It also places O in the back row due to the 5th rule, which subsequently places T and P as well. Lastly W being in the front row means, M must be in the back row because if M was in the front then W would have to be in the back which it is not.
So it looks like this.
_ W R _ _
O S M T P
V, Y and N can be placed anywhere but it doesn't really matter because the question is solved.
Problem 6 is moderately time consuming if you have trouble doing hypotheticals quickly in your head. Drawing it out is not the end of the world if you have time though. Hope this helps. I could've missed an inference that would have saved time in problem 6, but the main reason for the correct answers is there.
