LSAT Logic Games : Linear Games
Study concepts, example questions & explanations for LSAT Logic Games
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Example Questions
Example Question #11 : Linear Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest or the shortest.
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest
Which of the following must be true about acceptable height orders?
Ben is always at the back of the line, and Theresa is always at the front.
Dan is always between Corrin and Will.
Dan is always between Theresa and Will.
Will is always between Ben and Jonathan.
Will is always between Theresa and Corrin.
Ben is always at the back of the line, and Theresa is always at the front.
Theresa will always be at the front of the line, and Ben at the back. Theresa must be in front of Dan, Corrin, and Jonathan, and Will cannot be 1st.
Additionally, because of Will's interchangeability, there is no certainty that Dan will be between certain people or that Will is always between certain people. These answers are possibilities, but not certainties.
Example Question #11 : Lsat Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest or the shortest.
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest
Which of the following is a complete list of people who cannot be third in line?
Ben, Theresa, Corrin
Ben, Jonathan, Theresa
Ben, Theresa
Jonathan, Theresa, Dan
Jonathan, Theresa, Will
Ben, Jonathan, Theresa
Theresa has to be first, because she is shorter than Dan, Corrin, and Jonathan. Ben must always be 6th. Jonathan has to be either 4th or 5th, because he is after the girls, who surround Dan - which takes up 3 spots.
Example Question #12 : Lsat Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest.
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest
If Will was the shortest in line, what must the order be?
Will, Theresa, Jonathan, Corrin, Dan, Ben
Will, Theresa, Dan, Corrin, Jonathan, Ben
Will, Corrin, Dan, Theresa, Jonathan, Ben
Theresa, Will, Dan, Corrin, Jonathan, Ben
Will, Theresa, Dan, Corrin, Ben, Jonathan
Will, Theresa, Dan, Corrin, Jonathan, Ben
If Will is locked into the first position, all other positions are set and cannot be changed, according to the rest of the rules.
Example Question #13 : Lsat Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest or the shortest.
- No girl is taller than Jonathan.
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest.
Who are the only people that can be 5th in line?
Jonathan, Will, Theresa
Jonathan, Corrin, Will
Jonathan, Corrin, Ben
Jonathan and Will
Will, Dan
Jonathan and Will
Ben has to be 6th always. Theresa has to always be first. Corrin must always be earlier in line than Jonathan, so she can not be 5th. Will is interchangeable between the 2nd-5th spots, so he and Jonathan are the only ones that can be 5th.
Example Question #14 : Lsat Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest or the shortest.
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest
Who must be first in line?
Corrin
Jonathan
Theresa
Will
Dan
Theresa
Theresa must precede everyone. She has to be shorter than Dan and Corrin and Jonathan. Will cannot be shortest. Ben has to be the tallest. Thus, only Theresa can be 1st.
Example Question #15 : Lsat Logic Games
A gym teacher wants his 6 students to line up in height order, from shortest to tallest. Corrin and Theresa are girls. Ben, Jonathan, Will, and Dan are the boys.
- Will is not the tallest or the shortest.
- No girl is taller than Jonathan
- Dan is shorter than Corrin, but taller than Theresa.
- Ben is the tallest
What positions can Corrin fill?
2nd, 3rd, 4th
3rd, 4th, 5th
3rd, 4th
1st, 2nd, 3rd, 4th
4th, 5th
3rd, 4th
Corrin must follow Theresa and Dan, thus she cannot be 1st or second. She must also come before Jonathan and Ben, so she cannot be 5th or 6th. Thus, Corrin can only be 3rd or 4th.
Example Question #11 : Linear Games
A consultant has agreed to see each of his nine clients-- L, M, N, O, P, Q, R, S, T-- once in the next six days, from Monday through Saturday. He arranges his schedule so that he can see at least one of his clients each day, while maintaining the following conditions:
O is always scheduled on a day before R and M.
P is not scheduled for Saturday.
If T is scheduled on a day after O, then S is scheduled on a day after N.
If T is scheduled on a day before O, then R is scheduled on a day before L.
The consultant always sees less clients on Friday and Saturday combined than he sees on any other two days of the week combined.
If S must be scheduled for Saturday, which of the following must be true?
Q must be scheduled on a day earlier than L.
If L and M are scheduled together, than Q and O are scheduled together.
The consultant sees O no later than Wednesday.
There is no other client who can be scheduled on fewer days than M can be scheduled
If R is scheduled on the same day as another client, then M is scheduled alone.
The consultant sees O no later than Wednesday.
The consultant sees O no later than Wednesday.
This must be true. If O is scheduled for Thursday, then there is not enough room to schedule both R and M. O must be scheduled before R and M, but no more than one client may be scheduled for each of Friday and Saturday.
If R is scheduled on the same day as another client, then M is scheduled alone.
This is a trick for those who saw that O is restricted, but who make the mistake that O somehow must be scheduled for Wednesday instead of just on or before Wednesday. This is true if O is on Wednesday thanks to the restrictions on seeing clients on Friday and Saturday.
Q must be scheduled on a day earlier than L.
There are a number of valid schedules where this is not true. For example:
Mon: T
Tues: O, P
Wed: N, R
Thur: L, M
Fri: Q
Sat: S
If L and M are scheduled together, than Q and O are scheduled together.
The previous diagram shows a schedule where this is not true.
There is no other client who can be scheduled on fewer days than M can be scheduled
R and M are restricted in that they cannot be placed on Monday, since O must come before both of them. This gives them each 4 possible days to be scheduled. O must be scheduled no later than Thursday, which gives it 3 possible days to be scheduled. No other client is restricted whatsoever, except by the fact that S already occupies Saturday. This can be easiliy seen by the fact that N will always come before S, which makes any configuration where O comes before T valid. This leaves virtually every spot open for every letter, since there are no further restrictions that come after O > T. The only restriction is that T cannot be on Monday if O > T, but you can easily make a T > O diagram that is legal as well to make up for that shortcoming.
Example Question #11 : Determining Sequence In Linear Games
A consultant has agreed to see each of his nine clients-- L, M, N, O, P, Q, R, S, T-- once in the next six days, from Monday through Saturday. He arranges his schedule so that he can see at least one of his clients each day, while maintaining the following conditions:
O is always scheduled on a day before R and M.
P is not scheduled for Saturday.
If T is scheduled on a day after O, then S is scheduled on a day after N.
If T is scheduled on a day before O, then R is scheduled on a day before L.
The consultant always sees less clients on Friday and Saturday combined than he sees on any other two days of the week combined.
If there are exactly four days between the days that the consultant schedules P and T (not necessarily in that order), then what is the maximum number of days that can be between S and R?
2
4
1
0
3
2
The easiest way to approach this question is to simply diagram what is being asked. Four days between P and T places them at Monday and Saturday. P cannot be placed on Saturday, so that means that the diagram so far is P_ _ _ _T.
If T is the last client to be seen, then O obviously comes before it, which means that N comes before S for this question. Lastly, O always comes before R and M.
That means that we have an NS block and and ORM block. Neither S nor R can appear earlier than Tuesday. Neither S nor R can appear later than Thursday (only one client can be on Saturday, and that's T). The number of days between Tuesday and Friday is 2.
Here is an example diagram:
Mon: O, P
Tues: R, M
Wed: L, Q
Thur: N
Fri: S
Sat: T
Example Question #12 : Lsat Logic Games
A consultant has agreed to see each of his nine clients-- L, M, N, O, P, Q, R, S, T-- once in the next six days, from Monday through Saturday. He arranges his schedule so that he can see at least one of his clients each day, while maintaining the following conditions:
O is always scheduled on a day before R and M.
P is not scheduled for Saturday.
If T is scheduled on a day after O, then S is scheduled on a day after N.
If T is scheduled on a day before O, then R is scheduled on a day before L.
The consultant always sees less clients on Friday and Saturday combined than he sees on any other two days of the week combined.
If O is scheduled for Thursday, then for how many other clients do you know the precise day on which they are scheduled?
4
1
2
0
3
1
If O is on Thursday, then Friday and Saturday must be some combination of R and M.
_ _ _ O R/M R/M
T cannot go after O, because Friday and Saturday can only have one client a piece. T cannot go before O either, because that would require L to come after R. There is no space to do that.
If T cannot go before OR after O, then it must be on the same day. There is no other option.
If T and O are on the same day, then the two conditional rules (If O before T and If T before O) do not apply. Thus, you could place any of L, N, P, Q, S in any combination or order in the first three days. R and M are interchangeable on Friday and Saturday. Thus, only the position of T is fixed by scheduling O on Thursday.
Example Question #16 : Lsat Logic Games
A Halloween parade consists of six floats. The floats are of a goblin, a witch, a pumpkin, a skeleton, a mummy, and a piece of candy. The floats will look best when arranged in a certain order according to the following rules:
1. The goblin and the pumpkin must be consecutively ordered.
2. The piece of candy cannot be first.
3. The skeleton and mummy cannot be consecutively ordered.
4. If the pumpkin is second, the witch must not be fourth.
Which of the following is a possible order for the floats?
goblin, pumpkin, skeleton, witch, mummy, candy
pumpkin, candy, skeleton, witch, goblin, mummy
goblin, pumpkin, skeleton, mummy, witch, candy
pumpkin, goblin, skeleton, witch, candy, mummy
candy, pumpkin, goblin, skeleton, witch, mummy
pumpkin, goblin, skeleton, witch, candy, mummy
This arrangement satisfies all of the conditions: the goblin and the pumpkin are consecutively ordered, the piece of candy isn't first, the skeleton and mummy are not consecutively ordered, and since the pumpkin isn't second the witch can be fourth.
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