### All LSAT Logic Games Resources

## Example Questions

### Example Question #1 : Solving Grouping Games

A museum must choose five stuffed birds to be shown in their newest exhibit. Their choices include an Albatross, a Bluebird, a Condor, a Diver, an Eagle, a Flamingo, a Gull and a Hummingbird. They must choose the group according to the following restrictions:

Which of the following is a complete and accurate list of birds that could be shown in the exhibit?

**Possible Answers:**

Diver, Eagle, Condor, Gull and Hummingbird

Flamingo, Albatross, Condor, Gull and Bluebird

Eagle, Flamingo, Albatross, Hummingbird and Gull

Gull, Hummingbird, Flamingo, Albatross and Bluebird

Albatross, Condor, Eagle, Diver and Bluebird

**Correct answer:**

Diver, Eagle, Condor, Gull and Hummingbird

From the first two rules we know that the Diver and the Albatross, and the Eagle and the Flamingo can never appear in the same group together, which eliminates two choices immediately. We also know that if the Flamingo is chosen, both the Albatross and the Hummingbird must also be chosen, eliminating another incorrect answer. The last rule tells us that if the Gull is chosen, either the Hummingbird OR the Bluebird must be chosen, *but not both*. Therefore, we can deduce that if both the Hummingbird AND Bluebird are chosen, the Gull is not. This eliminates the last contender, leaving only the correct answer.

### Example Question #1 : Solving Grouping Games

A museum must choose five stuffed birds to be shown in their newest exhibit. Their choices include an Albatross, a Bluebird, a Condor, a Diver, an Eagle, a Flamingo, a Gull and a Hummingbird. They must choose the group according to the following restrictions:

If the Flamingo is chosen, how many different groups are possible?

**Possible Answers:**

None

Three

Four

Two

One

**Correct answer:**

Two

If the Flamingo is chosen, we immediately also choose the Albatross and the Hummingbird. This leaves only two spots to be filled. We can eliminate the Eagle because it cannot appear with the Flamingo, and we can eliminate the Diver because it cannot appear with the Albatross. We are left to chose from the Bluebird, the Condor and the Gull. If we choose the Gull, we cannot also choose the Bluebird - since the Hummingbird is already present, this situation would vioalte the rule involving the Gull. Therefore, we must chose either the Gull OR the Bluebird, and then flll the remaining spot with the Condor. The two distinct possibilities are: Flamingo, Albatross, Hummingbird, Condor and Gull OR Flamingo, Albatross, Hummingbird, Condor and Bluebird.

### Example Question #1 : Solving Grouping Games

A museum must choose five stuffed birds to be shown in their newest exhibit. Their choices include an Albatross, a Bluebird, a Condor, a Diver, an Eagle, a Flamingo, a Gull and a Hummingbird. They must choose the group according to the following restrictions:

If the Diver is chosen, which of the following must be true?

**Possible Answers:**

The Eagle is also chosen

The Bluebird is also chosen

The Gull is also chosen

The Hummingbird is also chosen

The Condor is not chosen

**Correct answer:**

The Eagle is also chosen

Knowing that the Diver is chosen allows us to immediately eliminate the Albatross according to the first rule. Eliminating the Albatross causes us to also eliminate the Flamingo in accordance with the second rule (the contrapositive of this rule states that if either the Albatross or the Hummingbird is NOT chosen, the Flamingo is also not chosen). We know we can only choose two out of the Gull, the Hummingbird and the Bluebird because of the last rule. This leaves two final empty spots which must be filled by the Eagle and the Condor.

### Example Question #4 : Grouping Games

If the Hummingbird is NOT chosen all of the following must be true EXCEPT:

**Possible Answers:**

The Albatross is chosen

The Bluebird is chosen

The Eagle is chosen

The Gull is chosen

The Condor is chosen

**Correct answer:**

The Albatross is chosen

When we eliminate the Hummingbird we must also immediately eliminate the Flamingo. This leaves us with six leftover variables to choose from. Since the Diver and the Albatross can never be chosen together, we know that the four other variables, namely the Gull, the Blurbird, the Condor and the Eagle all must be chosen. The only option is whether to choose the Albatross or the Diver to complete the group.

### Example Question #4 : Solving Grouping Games

A class consists of eight students, each with either black, brown, red, or blonde hair. The following statements accurately characterize the students in the class:

1. Four girls have either blonde or brown hair

2. Only one student has red hair

3. Two boys have the same hair color

4. There are exactly three students in the class with brown hair

5. There is at least one type of each hair color represented in the class

What can we say about the distribution of students with black hair?

**Possible Answers:**

No girls have black hair

Only two students have black hair

There are not more students with black hair than with brown hair

There are more students with blonde hair than with black hair

More boys have black hair than girls

**Correct answer:**

There are not more students with black hair than with brown hair

We are trying to figure out if it is possible for four or more students to have black hair, since we know there are exactly three students with brown hair. Since three of the eight students have brown hair, that leaves five. Based on the second and fifth condition we know one of those five has red hair, which leaves four. Based on the fifth conditon we know there must be at least one blonde, which leaves three. Therefore, the absolute maximum number of students with black hair would have to be three, which is not more than the number of students with brown hair.

### Example Question #6 : Grouping Games

A class consists of eight students, each with either black, brown, red, or blonde hair. The following statements accurately characterize the students in the class:

1. Four girls have either blonde or brown hair

2. Only one student has red hair

3. Two boys have the same hair color

4. There are exactly three students in the class with brown hair

5. There is at least one type of each hair color represented in the class

What is the maximum number of students with blonde hair in the class?

**Possible Answers:**

three

six

four

one

two

**Correct answer:**

three

We know that of the eight students, three of those students have brown hair, which leaves five. Based on the second and fifth conditions, we know that one of those students has red hair, which leaves four. Based on the fifth condition, we know that at least one of those students has black hair, which leaves three. Therefore, the absolute maximum number of students with blonde hair is three.

### Example Question #3 : Solving Grouping Games

A class consists of eight students, each with either black, brown, red, or blonde hair. The following statements accurately characterize the students in the class:

1. Four girls have either blonde or brown hair

2. Only one student has red hair

3. Two boys have the same hair color

4. There are exactly three students in the class with brown hair

5. There is at least one type of each hair color represented in the class

What can we NOT say for certain about the distribution of girls in the class?

**Possible Answers:**

There are at least as many girls as boys in the class

It is possible that there is only one blonde girl

If there are four girls in the class, no girls have red hair

At least two girls have the same hair color

At least one girl has the same color hair as a boy in the class

**Correct answer:**

At least one girl has the same color hair as a boy in the class

There is at least one possible scenario in which no girl has the same colored hair as any of the boys in the class. It is possible, for instance, that there are only two boys in the classroom. In this case, based on the third condition we know that the boys will have the same color hair. If the two boys have black hair, then that leaves six girls. Based on the first condition, we know that four girls have either blonde or brown hair, neither of which is black. That leaves two girls. Based on the second and fifth condition we know that one of those girls has red hair, which leaves one girl. It is not specified that ONLY four girls have brown or blonde hair, which means that more than four girls could have either blonde or brown hair. If the final girl has blonde or brown hair, then no girl has black hair, and therefore no girl has the same colored hair as any of the boys in the classroom.

### Example Question #4 : Solving Grouping Games

1. Four girls have either blonde or brown hair

2. Only one student has red hair

3. Two boys have the same hair color

4. There are exactly three students in the class with brown hair

5. There is at least one type of each hair color represented in the class

If there are three boys in the class, which of the following MUST be true?

**Possible Answers:**

If the student with red hair is a girl, at least one boy has black hair

If the student with red hair is a girl, at least one boy has brown hair

At least one boy has black hair

The student with red hair is a boy

Two boys have brown hair

**Correct answer:**

If the student with red hair is a girl, at least one boy has black hair

Based on the first condition, we know that four girls have either blonde or brown hair. Based on the second condition and the condition posed within this answer choice, we are assuming that the one student with red hair is a girl. This means that there are three students left. Based on the condition posed within this question, we know that those three students must be boys. Based on the fifth premise, we know that at least one of each hair color must be represented. Since black hair is not represented by any of the girl students, we know that at least one of the boys must have black hair.

### Example Question #4 : Solving Grouping Games

1. Four girls have either blonde or brown hair

2. Only one student has red hair

3. Two boys have the same hair color

4. There are exactly three students in the class with brown hair

5. There is at least one type of each hair color represented in the class

Which of the following is a possible scenario for the class distribution?

**Possible Answers:**

2 blonde-haired girls, 2 brown-haired girls, 2 brown-haired boys, 1 red-haired boy, 1 black-haired boy

3 blonde-haired girls, 2 brown-haired boys, 1 black-haired boy, 1 red-haired girl, 1 brown-haired girl

3 brown-haired girls, 2 blonde-haired boys, 1 black-haired boy, 1 red-haired girl, 1 black-haired girl

2 brown-haired girls, 2 blonde-haired boys, 2 black-haired boys, 1 blonde-haired girl, 1 red-haired girl

3 blonde-haired girls, 2 brown-haired boys, 1 black-haired boy, 1 red-haired boy, 1 black-haired girl

**Correct answer:**

3 blonde-haired girls, 2 brown-haired boys, 1 black-haired boy, 1 red-haired girl, 1 brown-haired girl

This is the only scenario that meets all of the conditions: four girls have either blonde or brown hair (three have blonde hair and one has brown), only one student has red hair, two boys have the same hair color (brown), there are exactly three students with brown hair (two boys and one girl), and at least one of every hair color is represented.

### Example Question #8 : Solving Grouping Games

A successful recruiting drive by the local laser tag league has lead to the recruitment of seven new recruits-- P, Q, R, S, T, U, and V. After completing their initial training, they will all be assigned to one of the laser tag league's four teams-- Dragon, Salamander, Phoenix, and Rabbit. Their assignment to the teams must conform to the following conditions:

At least one new member is assigned to each of the four teams.

Q and T are not assigned to the same team if R and V are not assigned to the same team.

If T is not assigned to Dragon, then U is not assigned to Dragon.

P is always on a team with more new recruits than S.

Dragon gets more new recruits than any other team.

Which of the following is an acceptable assignment of new recruits to the four teams?

**Possible Answers:**

Dragon: P, V, U, Q

Salamander: T

Phoenix: R

Rabbit: S

Dragon: R, V, Q, T

Salamander: U

Phoenix: P

Rabbit: S

Dragon: P, Q, T

Salamander: R, V

Phoenix: S

Rabbit: U

Dragon: Q, T, V

Salamander: R

Phoenix: S

Rabbit: P, U

Dragon: V, P

Salamander: S

Phoenix: Q, R

Rabbit: T, U

**Correct answer:**

Dragon: P, Q, T

Salamander: R, V

Phoenix: S

Rabbit: U

Dragon: Q, T, V

Salamander: R

Phoenix: S

Rabbit: P, U

This one breaks the rule that QT cannot be paired together if RV are not paired together. This is more clear if you write out the contrapositive.

~(RV) --> ~(QT)

(QT) --> (RV)

Dragon: V, P

Salamander: S

Phoenix: Q, R

Rabbit: T, U

This is incorrect, because it breaks the rule that Dragon must have *more* new recruits than any other team, not simply *as* *many*. This incorrect answer, if properly considered, also leads to the key insight that Dragon will always have *at least* three new recruits. That is because the distributions of new members, in order to satisfy both rules that Dragon must have the most and each team must have at least one, will look like either 4/1/1/1 or 3/2/1/1. There are no other possible distributions. Furthermore, if it is a 3/2/1/1 distribution and P is NOT a Dragon, then P is in the team with two new recruits, since P must be in a team with more new recruits than S (who must be a team with only one new recruit).

Dragon: R, V, Q, T

Salamander: U

Phoenix: P

Rabbit: S

This is incorrect, because S and P are on teams that have the same number of new recruits. P must be on a team with more.

Dragon: P, V, U, Q

Salamander: T

Phoenix: R

Rabbit: S

This is incorrect because T is not on Dragon, but U is on Dragon. This is another one that becomes more clear if you write out the contrapositive.

~(Td) --> ~(Ud)

Ud --> Td

### All LSAT Logic Games Resources

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