Question

If U is displayed in gallery 2, which of the following must be true?

• A. P is displayed in gallery 1.
• B. R is displayed in gallery 2.
• C. S is displayed in gallery 1.
• D. T is displayed in gallery 2.
• E. W is displayed in gallery 1.
• A. P and Q
• B. P and T
• C. Q and T
• D. T and W
• J. U and W
• A. Q and R
• B. Q and T
• C. Q and U
• D. R and U
• O. S and T
• A. R
• B. S
• C. T
• D. U
• T. W
• A. P is displayed in gallery 1.
• B. Q is displayed in gallery 1.
• C. R and U are displayed in the same gallery as each other.
• D. P and Q are not displayed in the same gallery as each other.
• Y. Q and R are not displayed in the same gallery as each other.
• A. P and S
• B. Q and R
• C. Q and W
• D. R and U
• ^. T and W
• A. P is displayed in gallery 1.
• B. R is displayed in gallery 2.
• C. Q and S are displayed in gallery 2.
• D. P is displayed in the same gallery as W.
• c. R is displayed in the same gallery as U.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We are given a new condition and asked to find a necessary consequence.

Step 2: Key Formula or Approach:
Apply the new condition and see what other rules are immediately triggered.

Step 3: Detailed Explanation:
1. The condition is that U is in gallery 2.
2. Rule 1 states that U and W cannot be in the same gallery.
3. Therefore, if U is in gallery 2, W must be in gallery 1.
4. This is a direct and necessary consequence of the initial condition and Rule 1. Let's check the other options to ensure they are not also "must be true."
  - We can construct two valid scenarios with U in gallery 2:
    - Scenario A: G1 = {W, R, P, Q}, G2 = {U, S, T}. Here, S and T are in G2.
    - Scenario B: G1 = {W, S, T, P}, G2 = {U, R, Q}. Here, S and T are in G1.
  - Since the placement of P, R, S, and T varies between these valid scenarios, none of the options (A), (B), (C), or (D) can be a "must be true" statement. Only (E) holds in both cases.

Step 4: Final Answer:
The rule that U and W must be in different galleries directly forces W into gallery 1 when U is placed in gallery 2.

Answer 2

Step 1: Understanding the Concept:
Given S is in gallery 2, we need to find a possible pair to complete the gallery.

Step 2: Key Formula or Approach:
Apply the condition S \(\in\) G2 and follow the chain of deductions. Then, test the options to see which pair can validly fill the remaining two spots in gallery 2.

Step 3: Detailed Explanation:
1. If S is in gallery 2, then R must be in gallery 1 (Rule 2).
2. If R is in gallery 1, then T must be in gallery 2 (Rule 2).
3. So, we know gallery 2 must contain S and T. The gallery is G2 = {S, T, ?}.
4. The third statue must be one of the remaining four: {P, Q, U, W}. The remaining three will go to gallery 1 with R. So, G1 = {R, ...}.
5. We must also satisfy the rule that U and W are in different galleries.
  - Let's test putting U in G2: G2 = {S, T, U}. Then G1 = {R, P, Q, W}. This is valid.
  - Let's test putting W in G2: G2 = {S, T, W}. Then G1 = {R, P, Q, U}. This is valid.
  - Let's test putting P in G2: G2 = {S, T, P}. Then G1 = {R, Q, U, W}. This puts U and W together, violating Rule 1. So P cannot be the third statue in G2.
  - Let's test putting Q in G2: G2 = {S, T, Q}. Then G1 = {R, P, U, W}. This also puts U and W together, violating Rule 1. So Q cannot be the third statue in G2.
6. Therefore, gallery 2 must be either {S, T, U} or {S, T, W}.
7. The question asks for the "other two statues," which are T and the third statue. So the pair can be {T, U} or {T, W}.
8. Option (D), T and W, is one of these valid pairs.

Step 4: Final Answer:
Deductions show that gallery 2 must contain S, T, and either U or W. Therefore, the pair T and W is a possible combination for the other two statues.

Answer 3

Step 1: Understanding the Concept:
This is an "EXCEPT" question, which means we are looking for the one pair that CANNOT be in gallery 1 under the given conditions. Four of the pairs will be possible.

Step 2: Key Formula or Approach:
Start with the initial conditions and create the possible valid scenarios. Then, check which of the listed pairs is not found together in gallery 1 in any valid scenario.

Step 3: Detailed Explanation:
1. We are given P \(\in\) G1 and W \(\in\) G2.
2. From W \(\in\) G2, Rule 1 implies U \(\in\) G1.
3. So far: G1 = {P, U, ?, ?}, G2 = {W, ?, ?}.
4. Now we must place the {R, S, T} block.
  - Scenario A: R \(\in\) G1. Then S and T must be in G2. G2 = {W, S, T}, which is full. G1's remaining spot must be filled by Q. So, G1 = {P, U, R, Q}. This is a valid world.
  - Scenario B: R \(\in\) G2. Then S and T must be in G1. G1 = {P, U, S, T}, which is full. G2's remaining spot must be filled by Q. So, G2 = {W, R, Q}. This is a valid world.
5. Now we check the options to see which pair is impossible for G1. Our possible compositions for G1 are {P, U, R, Q} or {P, U, S, T}.
  - (A) Q and R: This pair is in G1 in Scenario A. Possible.
  - (B) Q and T: In Scenario A, G1 has Q but not T. In Scenario B, G1 has T but not Q. In no scenario are Q and T in gallery 1 together. This is the answer.
  - (C) Q and U: This pair is in G1 in Scenario A. Possible.
  - (D) R and U: This pair is in G1 in Scenario A. Possible.
  - (E) S and T: This pair is in G1 in Scenario B. Possible.

Step 4: Final Answer:
Under the given conditions, Q and T are always in separate galleries. Therefore, they cannot be displayed together in gallery 1.

Answer 4

Step 1: Understanding the Concept:
Given P and Q are in gallery 1, we need to find another statue that is forced to be in gallery 1 as a result.

Step 2: Key Formula or Approach:
Place P and Q in gallery 1 and then test the two possible placements for R (in G1 or G2). One of these placements should lead to a contradiction, proving the other is necessary.

Step 3: Detailed Explanation:
1. We are given P and Q are in gallery 1. G1 = {P, Q, ?, ?}.
2. Let's test the possibility that R is in gallery 2.
  - If R \(\in\) G2, then S and T must be in G1 (Rule 2).
  - This would make G1 = {P, Q, S, T}, which is now full.
  - The remaining statues (R, U, W) must go into gallery 2. So G2 = {R, U, W}.
  - However, this places U and W in the same gallery, which violates Rule 1.
  - This leads to a contradiction, so our initial assumption (R is in gallery 2) must be false.
3. Therefore, R must be in gallery 1.
4. The question asks what statue must also be in gallery 1. We have just proved it must be R.

Step 4: Final Answer:
Placing P and Q in gallery 1 and R in gallery 2 leads to a violation of the rule separating U and W. Therefore, R must be in gallery 1.

Answer 5

Step 1: Understanding the Concept:
Given S is in gallery 1, we must find a statement about the arrangement that is always true.

Step 2: Key Formula or Approach:
Apply the condition S \(\in\) G1 and follow the chain of deductions to define the structure of the galleries.

Step 3: Detailed Explanation:
1. If S is in gallery 1, R must be in gallery 2 (Rule 2).
2. If R is in gallery 2, T must be in gallery 1 (Rule 2).
3. So far: G1 = {S, T, ?, ?}, G2 = {R, ?, ?}.
4. The remaining four statues are {P, Q, U, W}. They must fill the remaining four slots (two in each gallery).
5. We know U and W must be in different galleries (Rule 1). So, one goes to G1 and one goes to G2.
6. This leaves P and Q. There is one spot left in G1 and one spot left in G2. P and Q must fill these two spots.
7. Therefore, P and Q must be in different galleries from each other.
8. Option (D) states that P and Q are not displayed in the same gallery. This matches our deduction and must be true.

Step 4: Final Answer:
The condition that S is in gallery 1 forces R into gallery 2 and T into gallery 1. After placing U and W in separate galleries, there is one spot left in each gallery, which must be filled by P and Q, forcing them to be apart.

Answer 6

Step 1: Understanding the Concept:
Given T \(\in\) G2, we need to find a pair that must be in separate galleries. This is a "must be false" question for them being together.

Step 2: Key Formula or Approach:
Establish the possible scenarios when T is in G2. Then, check each pair in the options to see if they are ever together in any scenario. The pair that is never together is the answer.

Step 3: Detailed Explanation:
1. If T is in gallery 2, R must be in gallery 1 (Rule 2).
2. If R is in gallery 1, S must be in gallery 2 (Rule 2).
3. So far: G1 = {R, ?, ?, ?}, G2 = {T, S, ?}.
4. The remaining four statues {P, Q, U, W} must fill the remaining four slots (three in G1, one in G2).
5. We also know U and W must be in different galleries.
  - Scenario A: The last statue in G2 is U. G2 = {T, S, U}. Then G1 = {R, P, Q, W}. This is valid.
  - Scenario B: The last statue in G2 is W. G2 = {T, S, W}. Then G1 = {R, P, Q, U}. This is valid.
  - The last statue in G2 cannot be P or Q, because that would force U and W into G1 together.
6. So we have two possible worlds:
  - World A: G1 = {R, P, Q, U}, G2 = {S, T, W}
  - World B: G1 = {R, P, Q, W}, G2 = {S, T, U}
7. Now check the pairs: which pair is never together?
  - (A) P and S: In World A, P is in G1, S is in G2. In World B, P is in G1, S is in G2. They are always in different galleries. This is the answer.
  - (B) Q and R: They are together in G1 in both worlds.
  - (C) Q and W: They are together in G1 in World B.
  - (D) R and U: They are together in G1 in World A.
  - (E) T and W: They are together in G2 in World A.

Step 4: Final Answer:
In all valid scenarios where T is in gallery 2, P ends up in gallery 1 and S ends up in gallery 2. Thus, they can never be together.

Answer 7

Step 1: Understanding the Concept:
Given that Q and S are grouped together, we must find a necessary consequence.

Step 2: Key Formula or Approach:
Test the two main possibilities: {Q, S} are in gallery 1, or {Q, S} are in gallery 2. One case should lead to a contradiction.

Step 3: Detailed Explanation:
1. Case 1: Assume Q and S are in gallery 2.
  - If S \(\in\) G2, then R must be in G1 (Rule 2).
  - If R \(\in\) G1, then T must be in G2 (Rule 2).
  - This would make G2 = {Q, S, T}, which is full.
  - The remaining statues {R, P, U, W} must be in G1. So G1 = {R, P, U, W}.
  - However, this places U and W in the same gallery, violating Rule 1.
  - Therefore, Case 1 is impossible. Q and S cannot be in gallery 2.
2. Deduction: Since they cannot be in gallery 2, Q and S must be in gallery 1.
3. Now let's proceed with Q and S in gallery 1.
  - If S \(\in\) G1, then R must be in gallery 2 (Rule 2).
  - This is a necessary consequence of the initial condition. Let's check if it's an answer choice.
  - Option (B) is "R is displayed in gallery 2." This matches our deduction perfectly.
4. To be complete, let's determine the rest of the possible arrangements:
  - Q \(\in\) G1, S \(\in\) G1 → R \(\in\) G2.
  - R \(\in\) G2 → T \(\in\) G1.
  - So far: G1 = {Q, S, T, ?}, G2 = {R, ?, ?}.
  - Remaining: {P, U, W}. One goes to G1, two to G2.
  - Since U and W must be separate, one goes to G1 and one to G2. The last spot in G2 must be P.
  - Scenario A: G1 = {Q, S, T, U}, G2 = {R, P, W}.
  - Scenario B: G1 = {Q, S, T, W}, G2 = {R, P, U}.
  - Checking the options against these two worlds confirms that only (B) is always true.

Step 4: Final Answer:
The assumption that Q and S are in gallery 2 leads to a contradiction. Therefore, Q and S must be in gallery 1, which in turn forces R to be in gallery 2.