Three-Variable
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Annie is packing her things to move. She has six objects left to pack: a globe, a microscope, a telescope, a fossil, a Bunsen burner, and an hourglass. She has three boxes left and wants to put exactly two objects in each box. Her packing choices are restricted to the following limits:
The microscope and the telescope cannot be in the same box.
The hourglass must be placed in the box directly after the globe.
If the fossil is in the second box, the Bunsen burner must be in the first box.
The first box cannot contain the fossil.
If the fossil is in the second box, which of the following could be true?
The microscope is in the first box.
The telescope is in the second box.
The globe is in the first box.
The hourglass is in the second box.
The microscope is in the second box.
Explanation
Since the fossil is in the second box, that means that the Bunsen burner must be in the first box.
Let's now look at our answer choices. If the globe is in the first box with the Bunsen burner, that means that the hourglass must be in the second box with the fossil, and thus the microscope and telescope would be in the third box together. But we know that the microscope and telescope cannot be in the same box, so we see that two of our answer choices are not possible.
Now let's put the telescope in the second box with the fossil. That means that the globe is either in the first box with the Bunsen burner or the third box. But since the hourglass must be placed in the box directly after the globe, we know that the globe cannot be placed in the third box; so it must be placed in the first. But that is also impossible, because the hourglass would then have to go in the third box, meaning that it is not placed in the box directly after the globe. The same problem arises when we place the microscope in the second box. The hourglass does not directly follow the globe.
Let's place the microscope in the first box. Then we can place the globe in the second box with the fossil, and the hourglass in the third box in the telescope. This works. Thus, it is the correct answer.
A quarterback has time left in the game to call exactly seven plays—L, M, N, O, P, S, and T—one at a time and in any order according to the following conditions:
- Play P is called either first or last.
- The quarterback calls L at some point before calling N.
- The quarterback calls M at some point before calling T.
- The quarterback calls exactly one play between calling play L and play O.
- The quarterback calls exactly one play between calling play M and play P.
If N is called fourth, all of the following must be false EXCEPT .
L is called first
L is called second
M is called third
O is called fifth
S is called first
Explanation
L cannot be called second, because then O would have no place to go, placeholder 4 already being filled by N. M cannot be placed third because then P would come first and L would come in second which, as discussed above, does not work. O cannot be placed fifth because then M must be placed third, P first and thus L second which, again, does not work. If S is first, P comes last, M fifth, T sixth, and L and O must be placed next to each other in the second and third positions which again, violates our conditions. The only answer that violates no conditions is L in the first position.
Last week, Sara, Tony, Ulrich, Victor, and Wynne each saw La Traviata once at the opera house. The group divided up five tickets—one for the Thursday performance, one for the Friday performance, two for the Saturday performance, and one for the Sunday matinee—according to the following conditions:
- Neither Sara nor Victor attend Saturday performances
- If Wynne sees the opera before Victor, Wynne sees the first performance.
- Ulrich does not see a performance until Tony has also seen a performance.
Which of the following is a complete and accurate listing of the people who can attend the Thursday show?
Sara, Tony, Victor
Sara, Tony, Ulrich, Victor, Wynne
Sara, Tony, Victor, Wynne
Sara, Tony, Ulrich, Wynne
Sara, Tony, Wynne
Explanation
Ulrich cannot see the first performance by the third rule in the problem. Neither can Wynne: if Wynne sees the Thursday show, there are only two non-Saturday tickets, which must be used by Sara and Victor. This leaves Ulrich and Tony to see the show at the same time, which violates this third rule again.
Annie is packing her things to move. She has six objects left to pack: a globe, a microscope, a telescope, a fossil, a Bunsen burner, and an hourglass. She has three boxes left and wants to put exactly two objects in each box. Her packing choices are restricted to the following limits:
The microscope and the telescope cannot be in the same box.
The hourglass must be placed in the box directly after the globe.
If the fossil is in the second box, the Bunsen burner must be in the first box.
The first box cannot contain the fossil.
If the fossil is in the second box, which of the following could be true?
The microscope is in the first box.
The telescope is in the second box.
The globe is in the first box.
The hourglass is in the second box.
The microscope is in the second box.
Explanation
Since the fossil is in the second box, that means that the Bunsen burner must be in the first box.
Let's now look at our answer choices. If the globe is in the first box with the Bunsen burner, that means that the hourglass must be in the second box with the fossil, and thus the microscope and telescope would be in the third box together. But we know that the microscope and telescope cannot be in the same box, so we see that two of our answer choices are not possible.
Now let's put the telescope in the second box with the fossil. That means that the globe is either in the first box with the Bunsen burner or the third box. But since the hourglass must be placed in the box directly after the globe, we know that the globe cannot be placed in the third box; so it must be placed in the first. But that is also impossible, because the hourglass would then have to go in the third box, meaning that it is not placed in the box directly after the globe. The same problem arises when we place the microscope in the second box. The hourglass does not directly follow the globe.
Let's place the microscope in the first box. Then we can place the globe in the second box with the fossil, and the hourglass in the third box in the telescope. This works. Thus, it is the correct answer.
Seven retired professional football players---identified as A, B, C, D, E, F, and G to preserve their anonymity from the press---received votes to the Hall of Fame. Because only four can actually be inducted in this particular year, they must be ranked in terms of votes from lowest to highest. The ranking accords with the following specifications:
B and C received less votes than A.
B received more votes than E.
F and G received less votes than C.
D and F received less votes than E.
F did not receive the least amount of votes.
If B gets more votes than C, and F gets more votes than D, then which one of the following must be true?
B received the second most votes.
B received the third most votes.
C received the third most votes.
E received the third most votes
D received the least amount of votes.
Explanation
We know that B or C must be second, based upon all of our deductions:
A . . . B/C
A . . . C . . . F/G
B . . . E . . . D/F
By combining the latter two sequences, we can establish that B or C must take the second slot. Since the question posits that B received more votes than C, we can quickly arrive at the correct answer: B must be second.
Seven retired professional football players---identified as A, B, C, D, E, F, and G to preserve their anonymity from the press---received votes to the Hall of Fame. Because only four can actually be inducted in this particular year, they must be ranked in terms of votes from lowest to highest. The ranking accords with the following specifications:
B and C received less votes than A.
B received more votes than E.
F and G received less votes than C.
D and F received less votes than E.
F did not receive the least amount of votes.
If B gets more votes than C, and F gets more votes than D, then which one of the following must be true?
B received the second most votes.
B received the third most votes.
C received the third most votes.
E received the third most votes
D received the least amount of votes.
Explanation
We know that B or C must be second, based upon all of our deductions:
A . . . B/C
A . . . C . . . F/G
B . . . E . . . D/F
By combining the latter two sequences, we can establish that B or C must take the second slot. Since the question posits that B received more votes than C, we can quickly arrive at the correct answer: B must be second.
Last week, Sara, Tony, Ulrich, Victor, and Wynne each saw La Traviata once at the opera house. The group divided up five tickets—one for the Thursday performance, one for the Friday performance, two for the Saturday performance, and one for the Sunday matinee—according to the following conditions:
- Neither Sara nor Victor attend Saturday performances
- If Wynne sees the opera before Victor, Wynne sees the first performance.
- Ulrich does not see a performance until Tony has also seen a performance.
Which of the following is a complete and accurate listing of the people who can attend the Thursday show?
Sara, Tony, Victor
Sara, Tony, Ulrich, Victor, Wynne
Sara, Tony, Victor, Wynne
Sara, Tony, Ulrich, Wynne
Sara, Tony, Wynne
Explanation
Ulrich cannot see the first performance by the third rule in the problem. Neither can Wynne: if Wynne sees the Thursday show, there are only two non-Saturday tickets, which must be used by Sara and Victor. This leaves Ulrich and Tony to see the show at the same time, which violates this third rule again.
A quarterback has time left in the game to call exactly seven plays—L, M, N, O, P, S, and T—one at a time and in any order according to the following conditions:
- Play P is called either first or last.
- The quarterback calls L at some point before calling N.
- The quarterback calls M at some point before calling T.
- The quarterback calls exactly one play between calling play L and play O.
- The quarterback calls exactly one play between calling play M and play P.
If N is called fourth, all of the following must be false EXCEPT .
L is called first
L is called second
M is called third
O is called fifth
S is called first
Explanation
L cannot be called second, because then O would have no place to go, placeholder 4 already being filled by N. M cannot be placed third because then P would come first and L would come in second which, as discussed above, does not work. O cannot be placed fifth because then M must be placed third, P first and thus L second which, again, does not work. If S is first, P comes last, M fifth, T sixth, and L and O must be placed next to each other in the second and third positions which again, violates our conditions. The only answer that violates no conditions is L in the first position.
Exactly seven cities—Albany, Boston, Chicago, Denver, El, Paso, Fairbanks, and Glendale—are visited on a politician’s campaign trail. Each city is visited only once during the campaign. No cities are visited at the same time as each other. The following conditions apply to the campaign:
Chicago is either the third or the fifth city visited.
Boston is visited immediately before El Paso.
There is exactly one city visited between Chicago and Glendale.
Fairbanks is either the first or the last city visited.
If Boston is visited second, which of the following must be true?
Fairbanks is visited first.
Chicago is visited third.
Denver is visited fourth.
Albany is visited sixth.
Glendale is visited fifth.
Explanation
If Boston is second, El Paso must be third. Since El Paso is third, Chicago must fifth and Glendale must be last. Since Glendale is last, Fairbanks must be first.
The game is represented below and any city filled-in "must" be true.
Fairbanks Boston El Paso Chicago Glendale
Thus Fairbanks is visited first.
A children's toy store is creating a display of plush baby toys for the front window. There are five different toy animals: a flamingo, a koala, a seal, a cat, and a dog. Each toy is either blue or pink. The toys are to be arranged in a line, according to the following restrictions:
- There are at least two blue toys, and no blue toys can be adjacent to each other
- The flamingo must be pink
- If the seal is first, the koala is fourth
- If the flamingo is first, the dog is fourth
- The dog must come earlier in the line than the cat
- Either the cat or the dog is blue, but not both
If there are exactly three blue toys, which of the following could be true?
The flamingo is pink and in the fourth spot
The seal is blue and is first
The dog is third and the cat is fifth
The seal is pink and is first
The koala is pink and is fourth
Explanation
If there are three blue toys they must be spread out, in spots one, three and five, since no blue toys can be adjacent to each other. This leaves only two pink toys, one of which is the flamingo. Knowing this we can assume two different diagrams: one in which the flamingo is second, and one in which it is fourth. In this second diagram the flamingo would, as always, be pink, which is why this is the correct answer.
The seal cannot be first because either the flamingo is in the fourth spot where the koala should be, or if the flamingo is second and the koala can go fourth, that forces the dog and cat to both be blue. This is also the reason why the dog and cat cannot be third and fifth, respectively (they would both be blue).