LSAT Logic Games: Grouping and Selection Games 5

Question 1

Nine items are distributed into 3 groups of 3. Item X and Item Y must be in the same group. Item Y and Item Z must be in different groups. If X is in Group 1, which must be true?

Accepted Answer

Y is in Group 2 or 3

Answer

X in Group 1 forces Y to Group 1; Y and Z cannot share a group, so Z must be in Group 2 or 3—meaning Y is in Group 1 (not 2 or 3), but Z is not in Group 1, confirming Z is in Group 2 or 3.

Question 2

A selection game picks 4 from {A, B, C, D, E, F, G}. Rule: 'If A is selected, then B is selected. If B is selected, then C is selected.' If C is NOT selected, what follows?

Accepted Answer

B and A are both not selected

Answer

Contrapositive chain: not C → not B (from B→C), and not B → not A (from A→B), so both A and B must be excluded.

Question 3

In a grouping game, 'the number of items in Group 1 is greater than the number in Group 2' restricts distributions to which kind?

Accepted Answer

Distributions where Group 1 has more items than Group 2

Answer

The constraint directly states Group 1 must have more items, eliminating equal or Group-2-majority distributions.

Question 4

Seven candidates—P, Q, R, S, T, U, V—are selected or rejected for a panel. Exactly 4 are selected. Rule: P and Q are either both selected or both rejected. Rule: R is selected if and only if S is rejected. If P is rejected, which is a valid selection?

Accepted Answer

Q, S, T, U

Answer

P rejected → Q rejected (both-or-neither rule); R selected ↔ S rejected—so Q, S, T, U works with R excluded (S in means R out).

Question 5

Which of the following best describes an 'in/out' grouping game on the LSAT?

Accepted Answer

Items are assigned to one of two categories: selected or not selected

Answer

An in/out game divides elements into exactly two buckets: those chosen (in) and those not chosen (out).

LSAT - Law School Admission Test Practice Test

LSAT - Law School Admission Test Practice Test

LSAT Logic Games: Grouping and Selection Games 5