LSAT Logic Games : Solving grouping games

Study concepts, example questions & explanations for LSAT Logic Games

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Example Questions

Example Question #231 : Solving Grouping Games

There are exactly sixth different ingredients—Apple, Banana, Cucumber, Dragonfruit, Eggplant, and Fig—that can be used to make a smoothie. Each ingredient can only be used once. A smoothie is made with the follow conditions:  

           At least four ingredients are used.

           If Apples are used, then Dragonfruit is also used.

           If Eggplant is used, then Cucumber is not used.

           Fig must be used.

If Cucumber is used, how many possible acceptable combinations of smoothies are there?

Possible Answers:

6

2

5

4

3

Correct answer:

3

Explanation:

If Cucumber is used, Eggplant cannot be used. Thus the possible combinations are:

Cucumber, Apples, Dragonfruit, Fig

Cucumber, Banana, Dragonfruit, Fig

Cucumber, Apples, Bananas, Dragonfruit, Fig

Example Question #232 : Solving Grouping Games

Each of seven volunteers will be assigned to a region--East or West--to do volunteer work.  These volunteers are Haley, Jones, Klein, Lacey, Mullen, Ojai, and Peters.  Three volunteers will be assigned to the East and four will be assigned to the West, in accordance with the following conditions:

Lacey must be assigned to the East.

Klein must be assigned to the West.

Haley and Peters cannot be assigned to together.

Which one of the following CANNOT be true about the assignments?

Possible Answers:

Ojai and Peters are both assigned to the West.

Haley and Jones are both assigned to the West.

Mullen and Ojai are both assigned to the East.

Ojai and Peters are both assigned to the East.

Haley and Mullen are both assigned to the East.

Correct answer:

Mullen and Ojai are both assigned to the East.

Explanation:

This game emphasizes the importance of making deductions about the number of slots remaining after taking into account the given rules.  In this game, the East has only one open slot.  With that insight, we can quickly identify the impossibility of having Mullen and Ojai both in the East, since the East is occupied already by Lacey and Haley/Peters (one or the other).

Example Question #233 : Solving Grouping Games

Each of seven volunteers will be assigned to a region--East or West--to do volunteer work.  These volunteers are Haley, Jones, Klein, Lacey, Mullen, Ojai, and Peters.  Three volunteers will be assigned to the East and four will be assigned to the West, in accordance with the following conditions:

Lacey must be assigned to the East.

Klein must be assigned to the West.

Haley and Peters cannot be assigned to together.

Which one of the following CANNOT be true?

Possible Answers:

Mullen and Peters are both assigned to the West.

Mullen and Peters are both assigned to the East.

Jones and Mullen are both assigned to the East.

Jones and Mullen are both assigned to the West.

Mullen and Ojai are both assigned to the West.

Correct answer:

Jones and Mullen are both assigned to the East.

Explanation:

This question, like the other "CANNOT be true" question, tests whether the deduction was made regarding the single remaining slot in the East category.  Because only one slot remains open, Jones and Mullen cannot both be assigned to the East.  Two slots are already taken by Lacey and Haley/Peters (one or the other).

Example Question #234 : Solving Grouping Games

Each of seven volunteers will be assigned to a region--East or West--to do volunteer work.  These volunteers are Haley, Jones, Klein, Lacey, Mullen, Ojai, and Peters.  Three volunteers will be assigned to the East and four will be assigned to the West, in accordance with the following conditions:

Lacey must be assigned to the East.

Klein must be assigned to the West.

Haley and Peters cannot be assigned to together.

If Jones is assigned to the East, which one of the following must be true?

Possible Answers:

Ojai is assigned to the East.

Ojai is assigned to the West.

Haley is assigned to the East.

Peters is assigned to the East.

Peters is assigned to the West.

Correct answer:

Ojai is assigned to the West.

Explanation:

If Jones is in the East, then all three slots are taken:  Lacey, Haley/Peters, and Jones.  Therefore, Ojai cannot be assigned anywhere other than the West.

Example Question #235 : Solving Grouping Games

Each of seven volunteers will be assigned to a region--East or West--to do volunteer work.  These volunteers are Haley, Jones, Klein, Lacey, Mullen, Ojai, and Peters.  Three volunteers will be assigned to the East and four will be assigned to the West, in accordance with the following conditions:

Lacey must be assigned to the East.

Klein must be assigned to the West.

Haley and Peters cannot be assigned to together.

If Haley is assigned to the West and Jones is assigned to the East, which one of the following must be false?

Possible Answers:

Klein and Peters volunteer in different regions.

Jones and Ojai volunteer in the same region as Lacey.

Jones volunteers in a different region than does Klein.

Jones and Peters volunteer in the same region.

Jones volunteers in the same region as Lacey.

Correct answer:

Jones and Ojai volunteer in the same region as Lacey.

Explanation:

With Jones assigned to the East, all of the East slots are filled.  So no one other than Lacey, Peters, and Jones can be in the East category.  Thus, Jones and Ojai cannot be together in the East with Lacey.

Example Question #745 : Lsat Logic Games

Seven college football players—Fogel, Glenn, Hoskins, Innis, Wells, Xia, and Yates—have been drafted by a professional football team. One draftee is a kicker, three are offensive players, and three are defensive players. The following conditions apply in regards to which players fall within these three domains of the game:

  • Hoskins and Yates play the same position
  • Glenn does not play within the same domain as Fogel
  • If Xia plays defense, then Wells plays offense
  • Fogel plays offense

If Xia is drafted to play offense, then which one of the following could be true?

Possible Answers:

Glenn and Wells are both drafted to play offense.

Hoskins and Wells are both drafted to play offense.

Fogel and Yates are both drafted to play offense.

Hoskins and Wells are both drafted to play defense.

Fogel and Hoskins are both drafted to play defense.

Correct answer:

Hoskins and Wells are both drafted to play defense.

Explanation:

If Xia is drafted to play offense, then we have 2 offensive players locked in, Xia and Fogel.

We also know that Hoskins and Yates play the same position, and they cannot play kicker (only 1 spot by rule) and they cannot play offense (2 of the 3 spots are now taken) so they must play defense.

There is now one spot available in each of the different domains. Instead of going any further, we can plug in the answer choices to see which are eliminated by this devleoped scenario.

3 of the 5 answers are eliminated because Yates cannot play offense, Fogel cannot play defense, and Hoskins cannot play offense.

A fourth answer is eliminated because Glenn can't play in the same domain as Fogel, who plays offense.

Thus the correct answer is "Hoskins and Wells are both drafted to play defense." We don't need to try to prove this correct, as our devolped scenario makes it POSSIBLE, and the other four choices are IMPOSSIBLE.

Example Question #746 : Lsat Logic Games

A company sells gum in eight sticks per pack. Each pack contains at least one of the company's four gum flavors, which are apple, grape, watermelon, and lychee. Each individual stick is wrapped in either green, red, or yellow paper. For every pack of gum, the following conditions are met:

1. Apple and grape are only wrapped in green or red

2. Watermelon is never wrapped in green

3. If there are more than three reds in a pack, there must be at least two yellows

4. The number of sticks of watermelon is always less than the number of sticks of lyche

There is a manufacturing error and a pack contains all yellow-wrapped sticks. If the requirement that there must be one of each flavor is thus remeoved, but all the conditions otherwise remain the same - that is, conditions 1-4 still apply - which of the following would be false?

Possible Answers:

The pack would mostly contain lychee

The pack could contain a maximum of four watermelon

The pack would contain no apple or grape

The pack could contain a maximum of seven lychee

None of the above

Correct answer:

The pack could contain a maximum of four watermelon

Explanation:

If there are eight yellow wrappers, we know from the first premise that none of them can be apple or grape. This means the entire pack only has watermelon and lychee. According to the fourth premise we know that there must always be more lychee than watermelon. If there are four watermelon and there are eight sticks total, there would only be four lychee. Since there would not be more lychee than watermelon, we know that it is therefore impossible to have four watermelon.

Example Question #236 : Solving Grouping Games

Eight students-- Alyssa, Brian, Craig, Denzel, Erica, Felicia, Ginny, and Harmony-- decide to break into two groups to study for their upcoming exam. Alyssa and Ginny are the only freshmen; everyone else is a sophomore, junior, or senior. Each group has four students, and they must abide by the following requirements:

  • Craig cannot be in a group with Harmony.
  • Erica and Ginny must be in the same group.
  • There can be no more than one freshman in each group.
  • Denzel is in Group 2.

If Craig is in Group 2, which of the following is not possible?

Possible Answers:

Brian and Felicia are in the same group.

Felicia and Harmony are in the same group.

Alyssa and Denzel are in the same group.

Denzel and Harmony are in the same group.

Erica and Harmony are in the same group.

Correct answer:

Denzel and Harmony are in the same group.

Explanation:

Craig and Denzel are both in Group 2. Since Craig and Harmony cannot be in the same group, Harmony must be in Group 1. Therefore, Denzel and Harmony are not in the same group.

Example Question #237 : Solving Grouping Games

Eight students-- Alyssa, Brian, Craig, Denzel, Erica, Felicia, Ginny, and Harmony-- decide to break into two groups to study for their upcoming exam. Alyssa and Ginny are the only freshmen; everyone else is a sophomore, junior, or senior. Each group has four students, and they must abide by the following requirements:

  • Craig cannot be in a group with Harmony.
  • Erica and Ginny must be in the same group.
  • There can be no more than one freshman in each group.
  • Denzel is in Group 2.

Each of the following could constitute Group 2 except:

Possible Answers:

Craig, Denzel, Erica, Ginny

Alyssa, Craig, Denzel, Ginny

Alyssa, Denzel, Craig, Felicia

Alyssa, Denzel, Felicia, Harmony

Denzel, Erica, Ginny, Harmony

Correct answer:

Alyssa, Craig, Denzel, Ginny

Explanation:

Alyssa and Ginny are both freshmen, and one of the requirements states: "There can be no more than one freshman in each group." Therefore, Alyssa and Ginny cannot be in the same group.

Example Question #238 : Solving Grouping Games

Eight students-- Alyssa, Brian, Craig, Denzel, Erica, Felicia, Ginny, and Harmony-- decide to break into two groups to study for their upcoming exam. Alyssa and Ginny are the only freshmen; everyone else is a sophomore, junior, or senior. Each group has four students, and they must abide by the following requirements:

  • Craig cannot be in a group with Harmony.
  • Erica and Ginny must be in the same group.
  • There can be no more than one freshman in each group.
  • Denzel is in Group 2.

If Alyssa is in Group 1, which of the following must be true?

Possible Answers:

Brian is in Group 1.

Felicia is in Group 2.

Harmony is in Group 1.

Craig is in Group 1.

Erica is in Group 1.

Correct answer:

Brian is in Group 1.

Explanation:

If Alyssa is in Group 1, then Ginny must be in Group 2 because Alyssa and Ginny are both freshmen and, therefore, cannot be in the same group. Since Ginny is in Group 2, Erica must also be in Group 2. We already know Denzel is in Group 2, and Craig and Harmony cannot be in the same group, so Group 2 can only be made up of: Craig, Denzel, Erica, Ginny -- OR -- Denzel, Erica, Ginny, Harmony.

Since Brian is not in either possibility for Group 2, Brian must be in Group 1.

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