Comprehension
Each of nine persons, P, Q, R, S, T, U, V, W and X, lives in a different flat in an apartment building, which has six floors (excluding the ground floor, which is used only for parking) and three flats on each floor. The three flats on each floor are in a row and no two adjacent flats on a floor are occupied. At least one person lives on each floor.
Further the following information is known:
P and Q live on the same floor.
R and S live on different floors.
T lives in the middle flat on the fourth floor.
U lives on the sixth floor and V lives on the first floor.
W lives on the floor which is immediately above the floor on which X lives.
Question: 1
If W and U do not live on the same floor, then which of the following cannot be true?
If W and U do not live on the same floor, then which of the following cannot be true?
Show Hint
Always cross-reference with fixed placements before trying flexible ones. Fixed data like “U on 6th” helps eliminate.
Updated On: Jul 28, 2025
- V lives on the third floor
- Q lives on the third floor
- R lives on the second floor
- P lives on the second floor
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The Correct Option is A
Solution and Explanation
Given:
- P and Q live on the same floor
- R and S on different floors
- T lives on the middle floor → Floor 3
- U = Floor 6
- V = Floor 1
- W lives immediately above X
So: - Floors = 1 to 6
- T = Floor 3
- V = Floor 1
- U = Floor 6
- W = floor above X → W ≠ Floor 1
- W and U ≠ same floor → W ≠ Floor 6
Let’s check each option assuming W and U are not on the same floor:
(A) V on Floor 3 → Conflict!
- T is already on Floor 3
- V is given on Floor 1
So (A) is directly invalid regardless of W–U placement
Final Answer: \( \boxed{\text{A}} \)
- R and S on different floors
- T lives on the middle floor → Floor 3
- U = Floor 6
- V = Floor 1
- W lives immediately above X
So: - Floors = 1 to 6
- T = Floor 3
- V = Floor 1
- U = Floor 6
- W = floor above X → W ≠ Floor 1
- W and U ≠ same floor → W ≠ Floor 6
Let’s check each option assuming W and U are not on the same floor:
(A) V on Floor 3 → Conflict!
- T is already on Floor 3
- V is given on Floor 1
So (A) is directly invalid regardless of W–U placement
Final Answer: \( \boxed{\text{A}} \)
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Question: 2
If S and R are living on the first and sixth floor respectively, which of the following must be true?
If S and R are living on the first and sixth floor respectively, which of the following must be true?
Show Hint
Don’t overlook direct facts from instruction list. Even if conditions change, fixed clues like "T lives on middle floor" remain valid.
Updated On: Jul 28, 2025
- T is living on the same floor as X
- R is living on the second floor
- T is living on the third floor
- W is living alone on his floor
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The Correct Option is C
Solution and Explanation
From given:
- T lives on the middle floor → Floor 3 (this is already stated in fact iii)
Hence, regardless of where S or R live:
\[ T = \boxed{\text{Floor 3}} \text{ always} \Rightarrow \text{(c) is definitely true} \]
Final Answer: \( \boxed{\text{C}} \)
- T lives on the middle floor → Floor 3 (this is already stated in fact iii)
Hence, regardless of where S or R live:
\[ T = \boxed{\text{Floor 3}} \text{ always} \Rightarrow \text{(c) is definitely true} \]
Final Answer: \( \boxed{\text{C}} \)
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Question: 3
If Q lives on the third floor, then how many combinations of persons could live on the second floor?
If Q lives on the third floor, then how many combinations of persons could live on the second floor?
Show Hint
Always apply special pair constraints (like "W above X") while counting. Eliminate combinations that violate such relations.
Updated On: Jul 28, 2025
- 8
- 6
- 5
- 7
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The Correct Option is B
Solution and Explanation
Total persons = 9: P, Q, R, S, T, U, V, W, X
Given: - Q = Floor 3
- P and Q on same floor → P = 3
- T = Floor 3
So floor 3 = P, Q, T → Already full (3 people per floor)
V = Floor 1
U = Floor 6
W lives immediately above X → W ≠ Floor 1
So we need to find how many combinations of remaining persons can be placed on floor 2
Remaining: R, S, W, X
Try combinations of these 4 people taken 1 to 3 at a time on floor 2
Total possibilities (excluding invalids):
- \( \binom{4}{1} = 4 \)
- \( \binom{4}{2} = 6 \)
- \( \binom{4}{3} = 4 \)
But not all valid: W must be above X → if W and X both on floor 2 → invalid
So remove combinations where W & X both on floor 2
From all valid combinations, 6 are valid considering constraints Final Answer: \( \boxed{6} \)
Given: - Q = Floor 3
- P and Q on same floor → P = 3
- T = Floor 3
So floor 3 = P, Q, T → Already full (3 people per floor)
V = Floor 1
U = Floor 6
W lives immediately above X → W ≠ Floor 1
So we need to find how many combinations of remaining persons can be placed on floor 2
Remaining: R, S, W, X
Try combinations of these 4 people taken 1 to 3 at a time on floor 2
Total possibilities (excluding invalids):
- \( \binom{4}{1} = 4 \)
- \( \binom{4}{2} = 6 \)
- \( \binom{4}{3} = 4 \)
But not all valid: W must be above X → if W and X both on floor 2 → invalid
So remove combinations where W & X both on floor 2
From all valid combinations, 6 are valid considering constraints Final Answer: \( \boxed{6} \)
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