### All LSAT Logic Games Resources

## Example Questions

### Example Question #51 : Solving Three Variable Logic Games

At an art exhibit six sculptures, identified by the first letter of the artist's last name, will be displayed sequentially, from front to back. The sculptures are L, M, O, S, V, and Z. The front-to-back display must accord with the following restrictions:

M must be displayed closer to the front than S.

S must be displayed closer to the front than both L and V.

V and Z each must be displayed closer to the front than O.

Which one of the following cannot be true?

**Possible Answers:**

V is displayed closer to the front than L.

Z is displayed closer to the front than M.

L is displayed closer to the front than V.

O is displayed closer to the front than S.

S is displayed closer to the front than Z.

**Correct answer:**

O is displayed closer to the front than S.

Our flow chart, constructed by blending the rules together, reveals that O must always be somewhere behind V, and V must always be somewhere behind S. thus, O can never be ahead of S.

### Example Question #51 : Solving Three Variable Logic Games

At an art exhibit six sculptures, identified by the first letter of the artist's last name, will be displayed sequentially, from front to back. The sculptures are L, M, O, S, V, and Z. The front-to-back display must accord with the following restrictions:

M must be displayed closer to the front than S.

S must be displayed closer to the front than both L and V.

V and Z each must be displayed closer to the front than O.

Exactly how many different spots could be the spot in which Z is displayed?

**Possible Answers:**

five

three

two

six

four

**Correct answer:**

five

In the flow chart generated by blending the rules, Z is only restricted by the rule that it must precede O. Thus, Z could be first, second, third, fourth, or fifth. Thus, the answer is five.

### Example Question #52 : Solving Three Variable Logic Games

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?

**Possible Answers:**

Chicago is visited third.

Albany is visited sixth.

Fairbanks is visited first.

Glendale is visited fifth.

Denver is visited fourth.

**Correct answer:**

Fairbanks is visited first.

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.

### Example Question #51 : Solving Three Variable Logic Games

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.

Given the conditions, what are all of the possible positions in which Albany can be visited?

**Possible Answers:**

Second, Fifth, Sixth

First, Second, Fifth, Seventh

Second, Third, Fourth, Sixth

First, Second, Fourth, Sixth

Second, Fourth, Sixth

**Correct answer:**

Second, Fourth, Sixth

The game can be mapped out into these four possible worlds:

1. _________ _________ Glendale _________ Chicago _________ _________

2. _________ _________ Chicago _________ Glendale _________ _________

3. Glendale _________ Chicago _________ _________ _________ Fairbanks

4. Fairbanks _________ _________ _________ Chicago _________ Glendale

In worlds 1 and 2, Boston and El Paso can be either first/second or sixth/seventh, and Fairbanks is ether first or last whichever is still available. This however; makes the first and last positions filled in all possible worlds.

Additionally, either Boston or El Paso must be placed fifth in world 3, and third in world 4, making the third and fifth position filled in all the possible worlds.

This leaves only the second, fourth, and sixth position available for Albany.

### Example Question #51 : Three Variable

A school has seven sports teams---baseball, football, golf, hockey, karate, swimming, and track. The new school gymnasium has three floors. Each floor can allow for a maximum of four sports teams to do indoor practicing or working out. No sports team can practice or work out on more than one floor. Assignment of the sports teams to the appropriate floor of the gym is subject to the following conditions:

Karate and track must conduct their practices or workouts on the same floor.

Football must conduct practices or workouts on the floor immediately above golf.

Hockey must conduct practices or workouts on its own floor without any other sports team.

Which one of the following CANNOT be the assignment of sports teams to floor of the gymnasium?

**Possible Answers:**

Karate, swimming, track, and hockey.

Swimming, track, karate, and golf

Baseball, track, golf, and karate.

Football, track, and karate

Swimming, track, karate, and football.

**Correct answer:**

Karate, swimming, track, and hockey.

This question punishes over-thinking. The correct answer simply tests the rule that hockey must occupy its own floor. It must be false that hockey is with three other sports team on a particular floor.

### Example Question #52 : Solving Three Variable Logic Games

A school has seven sports teams---baseball, football, golf, hockey, karate, swimming, and track. The new school gymnasium has three floors. Each floor can allow for a maximum of four sports teams to do indoor practicing or working out. No sports team can practice or work out on more than one floor. Assignment of the sports teams to the appropriate floor of the gym is subject to the following conditions:

Karate and track must conduct their practices or workouts on the same floor.

Football must conduct practices or workouts on the floor immediately above golf.

Hockey must conduct practices or workouts on its own floor without any other sports team.

If karate conducts practice or workouts on the second floor along with exactly two other sports teams, then which one of the following must be true?

**Possible Answers:**

Baseball conducts practice or workouts on the same floor as swimming.

Karate conducts practice or workouts on the same floor as football.

Baseball conducts practice or workouts on the floor immediately above football.

Baseball conducts practice or workouts on the floor immediately below hockey.

karate conducts practice or workouts on the same floor as golf.

**Correct answer:**

Baseball conducts practice or workouts on the same floor as swimming.

The question stipulates that karate is on the second floor with exactly two other teams. Karate must be with track, according to the first rule. The second rule effectively places either football or golf with karate. That means the configuration must be one of two options: karate, track, and football; or karate, track, and golf. Baseball and swimming, therefore, must be together on another floor (just not the one with hockey).

### Example Question #57 : Solving Three Variable Logic Games

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

Which of the following is a complete and accurate list of how the toys could be arranged in the window?

**Possible Answers:**

seal, flamingo, dog, koala, cat

flamingo, seal, dog, koala, cat

seal, flamingo, cat, koala, dog

flamingo, seal, dog, koala, cat

seal, flamingo, koala, dog, cat

**Correct answer:**

seal, flamingo, dog, koala, cat

This is a simple "grab-a-rule" question. The condition that if the seal is first then the koala is fourth eliminates one choice. The condition that if the flamingo is first then the dog is fourth eliminates two choices. And the fact that the dog must come before the cat eliminates the final wrong choice.

### Example Question #51 : Solving Three Variable Logic Games

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?

**Possible Answers:**

The dog is third and the cat is fifth

The koala is pink and is fourth

The seal is blue and is first

The seal is pink and is first

The flamingo is pink and in the fourth spot

**Correct answer:**

The flamingo is pink and in the fourth spot

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).

### Example Question #52 : Solving Three Variable Logic Games

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 a blue cat is the second toy in the line, which of the following could be true?

**Possible Answers:**

There are three total blue toys

A blue koala is fourth

The seal and koala are both blue

The seal and koala are both pink

The flamingo and the dog are the only pink toys

**Correct answer:**

A blue koala is fourth

If a blue cat is second, then we know that a pink dog must be first (because the dog must come before the cat, and they must be different colors). Since there must be at least two blue toys, the second blue toy is either fourth or fifth. We are left with three toys (flamingo, seal, koala) the only rule concerning them that we need to worry about is that the flamingo is pink, and therefore cannot go in whatever spot is designated for the second blue toy.

If the seal and koala are both pink, it forces the flamingo to be blue, which does not work. If the seal and the koala are both blue that means there are three blue toys total, which does not work. Same thing is the cat and flamingo are the only pink toys.

### Example Question #52 : Solving Three Variable Logic Games

- 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 the flamingo is first in the line, all of the following could be true EXCEPT:

**Possible Answers:**

The cat is third and is blue

The koala is second and is pink

The dog is fourth and is blue

The koala is second and is blue

The seal is third and is blue

**Correct answer:**

The cat is third and is blue

If the flamingo is first we know that there are only two blue toys, since three blue toys 2-5 would force at least two of them to be adjacent. We also know that the dog must be in the fourth spot according to the second conditional. If the dog is fourth, then we know that the cat must be fifth, since it has to come later in the line than the dog. This leaves the koala and the seal to interchangeably fill the second and third spaces.

The two blue toys bust be spread out among the second through fifth spaces. Therefore the blue toys can be second and fifth, second and fourth, or third and fifth.

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