LSAT Logic Games: Grouping and Selection Games 2 — Questions and Answers
Question 1: A committee of exactly 3 must be chosen from 6 candidates: F, G, H, J, K, L. If F is selected, then G must also be selected. If G is selected, then H cannot be selected. Which group is an acceptable committee?
- F, G, H
- F, H, J
- F, G, J (Correct answer)
- G, H, K
Correct answer: F, G, J
F and G can serve together as long as H is excluded; F, G, J satisfies both constraints.
Question 2: In a grouping game where exactly 3 items are placed in Group 1 and exactly 2 in Group 2, a contrapositive is most useful when:
- You know which group an item is in
- A conditional rule has a known false consequent (Correct answer)
- Two items must always be together
- The number of items in each group changes
Correct answer: A conditional rule has a known false consequent
The contrapositive fires when the consequent is false, letting you conclude the antecedent is also false.
Question 3: Seven volunteers—1, 2, 3, 4, 5, 6, 7—are split into Team A (4 members) and Team B (3 members). Volunteer 3 and Volunteer 5 cannot be on the same team. If Volunteer 3 is on Team A, which of the following MUST be true?
- Volunteer 5 is on Team A
- Volunteer 5 is on Team B (Correct answer)
- Team B has exactly 2 members
- Volunteer 3 and 5 are on the same team
Correct answer: Volunteer 5 is on Team B
Because 3 and 5 cannot share a team and 3 is on Team A, 5 must be on Team B.
Question 4: In a selection game, which deduction technique is most efficient when two items are linked by an 'if and only if' (biconditional) rule?
- Treat the biconditional as two separate one-way conditionals (Correct answer)
- Ignore one direction of the biconditional
- Use the contrapositive of only the forward direction
- Assume both items are always selected
Correct answer: Treat the biconditional as two separate one-way conditionals
A biconditional (A↔B) means A→B AND B→A; treating it as two conditionals captures all inferences.
Question 5: A selection game requires choosing 4 from {M, N, O, P, Q, R}. The rule states: 'If M is not chosen, then both P and Q must be chosen.' If P is not chosen, what can be concluded?
- M must be chosen (Correct answer)
- Q must be chosen
- N must be chosen
- R cannot be chosen
Correct answer: M must be chosen
Via contrapositive of the rule (not-P or not-Q → M), since P is not chosen, M must be chosen.
Question 6: In a grouping game, 'at least one of A or B must be in Group X' is best represented as:
- A → not B
- not A → B (Correct answer)
- A and B → Group X
- A → B
Correct answer: not A → B
'At least one of A or B' means if A is absent then B must be present: not-A → B.
Question 7: Eight employees are divided into exactly two groups of 4. Employee T and Employee V must be in the same group. Employee V and Employee W cannot be in the same group. If T is in Group 1, which must be true?
- W is in Group 1
- W is in Group 2 (Correct answer)
- V is in Group 2
- T and W are in the same group
Correct answer: W is in Group 2
T is in Group 1, so V (who must be with T) is in Group 1, and W (who cannot be with V) must be in Group 2.
LSAT Logic Games: Grouping and Selection Games 2
A committee of exactly 3 must be chosen from 6 candidates: F, G, H, J, K, L.
If F is selected, then G must also be selected.
If G is selected, then H cannot be selected.
Which group is an acceptable committee?