Question

If W and U do not live on the same floor, then which of the following cannot be true?

• A. V lives on the third floor
• B. Q lives on the third floor
• C. R lives on the second floor
• D. P lives on the second floor
• A. T is living on the same floor as X
• B. R is living on the second floor
• C. T is living on the third floor
• D. W is living alone on his floor
• A. 8
• B. 6
• C. 5
• D. 7

Answer 1

The Correct Option is A

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: \( \boxed{\text{A}} \)

Answer 2

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: \( \boxed{\text{C}} \)

Answer 3

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: \( \boxed{6} \)