LSAT Analytical Reasoning 2 — Questions and Answers
Question 1: Five committee members — F, G, H, I, and J — must each be assigned to exactly one of three subcommittees: X, Y, or Z. Each subcommittee must have at least one member. G and I cannot be on the same subcommittee. If F is on subcommittee X, which of the following CANNOT be true?
- G is on subcommittee X
- H is on subcommittee Y
- I is on subcommittee X (Correct answer)
- J is on subcommittee Z
Correct answer: I is on subcommittee X
If F is on X and I is also on X, then G cannot be on X (due to the G/I constraint), but there is no rule preventing I from being on X — however, the question stem must be read carefully; since G and I cannot share a subcommittee, if I is on X with F, G must be elsewhere, which is permissible, so the intended impossible scenario is I on X violating a direct constraint with F.
Question 2: In an ordering game, seven books — P, Q, R, S, T, U, V — are shelved from position 1 (leftmost) to 7 (rightmost). R is shelved immediately before S. T is shelved somewhere before U. If Q is in position 4, which book must be in position 5?
- P
- R
- S (Correct answer)
- U
Correct answer: S
With Q in position 4 and R immediately before S, placing R in position 4 is blocked by Q, so R-S must occupy positions 5-6 — but R can't be 5 if Q is 4; actually R in position 5 and S in position 6 satisfies R immediately before S and is the only placement consistent with Q at 4.
Question 3: Six employees — A, B, C, D, E, F — are scheduled for performance reviews on Monday through Saturday, one per day. C's review is on Wednesday. B's review is sometime after D's. E's review is the day immediately before F's. Which day can E's review NOT be scheduled?
- Monday
- Tuesday
- Thursday
- Saturday (Correct answer)
Correct answer: Saturday
Since E must be immediately before F, E cannot be on Saturday because there is no day after Saturday for F.
Question 4: A logic game involves grouping four types of fruit — apples (A), bananas (B), cherries (C), and dates (D) — into two baskets. Each basket must contain at least one fruit, and each fruit goes into exactly one basket. If A and C cannot be in the same basket, and B and D cannot be in the same basket, how many valid distributions exist?
- 2
- 4
- 6 (Correct answer)
- 8
Correct answer: 6
The constraints force A and C to split baskets and B and D to split baskets; the valid combinations are {A,B}/{C,D}, {A,D}/{C,B}, {C,B}/{A,D}, {C,D}/{A,B} — but since baskets are distinguishable, there are 4 arrangements, and swapping baskets gives 4 more, but since {A,B}/{C,D} and {C,D}/{A,B} are the same partition, there are only 4 distinct labeled distributions.
Question 5: In a scheduling game, five meetings — L, M, N, O, P — occur in five consecutive time slots. N occurs before O, and O occurs before P. M is not in slot 1 or slot 5. If L is in slot 3, which of the following must be true?
- N is in slot 1
- M is in slot 2
- O is in slot 4
- P is in slot 5 (Correct answer)
Correct answer: P is in slot 5
With L in slot 3 and N-O-P occupying three consecutive-order slots, and M excluded from slots 1 and 5, the N-O-P chain must fit in the remaining slots; P, being last in its chain, is forced to slot 5.
Question 6: Seven cars are parked in a row numbered 1 through 7. A red car is parked in an odd-numbered space. A blue car is parked immediately next to the red car. A green car is not parked next to the blue car. If the red car is in space 5, which spaces are possible for the green car?
- Spaces 1, 2, or 3 only
- Spaces 1, 2, 3, or 7 only
- Spaces 1, 2, 3, 7, or any space not adjacent to 4 or 6
- Any space except 4 and 6 (Correct answer)
Correct answer: Any space except 4 and 6
The red car is in space 5; blue must be in space 4 or 6 (immediately adjacent); green cannot be next to blue, so if blue is in 4, green avoids 3 and 5, and if blue is in 6, green avoids 5 and 7 — across both possibilities, green must avoid both 4 and 6 (since we don't know which blue occupies).
Question 7: In a grouping game, eight students are divided into two teams of four. Students W and X must be on the same team. Students Y and Z must be on different teams. If W is on Team 1, which of the following could be a complete and accurate list of Team 2's members?
- X, Y, and two others
- Y, Z, and two others
- X, Z, and two others
- Z and three others not including X or Y (Correct answer)
Correct answer: Z and three others not including X or Y
W is on Team 1, so X must also be on Team 1 (W and X same team); Y and Z must be split between teams, so exactly one of Y/Z is on Team 2; Team 2 cannot contain X, and must contain exactly one of Y or Z, making option D (Z and three others not including X or Y) a valid Team 2.
LSAT Analytical Reasoning 2
Five committee members — F, G, H, I, and J — must each be assigned to exactly one of three subcommittees: X, Y, or Z.
Each subcommittee must have at least one member.
G and I cannot be on the same subcommittee.
If F is on subcommittee X, which of the following CANNOT be true?