Question

If the lion is in enclosure 1 and the tiger is in enclosure 3, and the lion is to be moved to enclosure 7, the tiger could be in which of the following enclosures when all of the transfers have been completed?

• A. 1
• B. 3
• C. 4
• D. 5
• E. 6
• A. 2
• B. 3
• C. 4
• D. 6
• J. 7
• A. The lion is transferred to enclosure 3 at some time during the move.
• B. The tiger is transferred to enclosure 5 twice.
• C. One of the two animals is transferred to enclosure 3 twice.
• D. Three transfers to enclosure 5 are made.
• O. At least one transfer is made to either enclosure 2 or enclosure 4.
• A. Four
• B. Five
• C. Six
• D. Seven
• T. Eight
• A. The lion is transferred to enclosure 2 in the first transfer.
• B. The lion is transferred to enclosure 3 in the second transfer.
• C. The lion is transferred to enclosure 4 in the second transfer.
• D. The tiger is transferred to enclosure 5 in the first transfer.
• Y. The tiger is transferred to enclosure 3 in the second transfer.
• A. lion to enclosure 2
• B. lion to enclosure 5
• C. tiger to enclosure 4
• D. tiger to enclosure 5
• ^. tiger to enclosure 7
• A. Exactly five enclosures are used in the move
• B. One animal is transferred exactly twice as many times as the other animal.
• C. All of the transfers of the lion are completed before any transfer of the tiger occurs.
• D. At one point one of the animals is transferred to either enclosure 2 or enclosure 4.
• c. At one point neither the lion nor the tiger is in enclosure 3, enclosure 5, or enclosure 6

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to move the Lion (L) from 1 to 7 using the minimum number of total transfers. The Tiger (T) starts at 3, which is on the shortest path for L. This means T must be moved out of the way. We need to find a possible final location for T after L reaches its destination.
Step 2: Detailed Explanation:
\begin{enumerate} \item Shortest Path for L: The shortest path for L from 1 to 7 is 1 \(\rightarrow\) 3 \(\rightarrow\) 5 \(\rightarrow\) 7. This takes 3 transfers. \item The Blockage: T is in enclosure 3, so L cannot move from 1 to 3. The path is blocked. \item Minimal Solution: To clear the path with the minimum number of moves, T must first move out of enclosure 3. T has adjacent enclosures \{1, 2, 4, 5\}. T cannot move to 1 (L is there). T can move to 2, 4, or 5. Let's try moving T to a "parking" spot that doesn't interfere further. \item Scenario 1: \begin{itemize} \item Transfer 1: T moves 3 \(\rightarrow\) 4. (State: L=1, T=4). The path for L is now clear. \item Transfer 2: L moves 1 \(\rightarrow\) 3. (State: L=3, T=4). \item Transfer 3: L moves 3 \(\rightarrow\) 5. (State: L=5, T=4). \item Transfer 4: L moves 5 \(\rightarrow\) 7. (State: L=7, T=4). \end{itemize} \item The goal is achieved. L is in 7. The total number of transfers is 4. T's final position is 4. This is a possible outcome. Any path requiring more than 4 transfers would not be minimal, as we have found a 4-transfer solution. \item For example, if T moved 3 \(\rightarrow\) 5, it would block L's path again later, requiring T to move a second time and resulting in more than 4 total transfers. \end{enumerate} Step 3: Final Answer:
We have found a minimal 4-transfer sequence where the lion reaches enclosure 7 and the tiger ends up in enclosure 4. Therefore, it is possible for the tiger to be in enclosure 4.

Answer 2

Step 1: Understanding the Concept:
The question asks which destination for the Tiger (T) would result in a process that takes exactly two total transfers. We start with T=5 and L=3. We need to test the options as destinations.
Step 2: Detailed Explanation:
Let's analyze the number of transfers required to move T from 5 to each of the target enclosures. \begin{itemize} \item Destinations 6 and 7: T can move directly from 5 to 6 or from 5 to 7. The path is not blocked by L. This move takes exactly one transfer. So, options (D) and (E) are incorrect. \item Destination 3: The path for T is 5 \(\rightarrow\) 3. This is a single step. However, L starts in enclosure 3, blocking the destination. For T to move to 3, L must first move out of 3. \begin{itemize} \item Transfer 1: L moves out of 3. L can move to 1, 2, or 4. Let's say L moves 3 \(\rightarrow\) 1. (State: T=5, L=1). \item Transfer 2: Now enclosure 3 is empty. T moves 5 \(\rightarrow\) 3. (State: T=3, L=1). \end{itemize} The task of moving T to 3 is completed in exactly two transfers. This matches the question. So, option (B) is likely correct. \item Destinations 2 and 4: The shortest paths for T from 5 are 5 \(\rightarrow\) 3 \(\rightarrow\) 2 and 5 \(\rightarrow\) 3 \(\rightarrow\) 4. These are two-step paths for the tiger. But the intermediate step, enclosure 3, is blocked by L. \begin{itemize} \item Transfer 1: L must move out of 3. (e.g., L: 3 \(\rightarrow\) 1). \item Transfer 2: T can now move into 3. (T: 5 \(\rightarrow\) 3). \item Transfer 3: T moves out of 3 to its destination. (e.g., T: 3 \(\rightarrow\) 2). \end{itemize} This process takes a total of three transfers. So, options (A) and (C) are incorrect. \end{itemize} Step 3: Final Answer:
Only the task of moving the tiger to enclosure 3 requires a total of exactly two transfers (one for the lion to clear the space, and one for the tiger to move in).

Answer 3

Step 1: Understanding the Concept:
This question asks for a necessary action ("must be true") during a swap of the lion (L) and tiger (T) from enclosures 6 and 7. Since these are adjacent only to enclosure 5, they must "dance" around each other using a temporary "parking" spot. We must find the minimal sequence for this swap and identify an action that occurs in all minimal versions of the swap.
Step 2: Detailed Explanation:
The enclosures 6 and 7 are like two rooms off a single hallway (enclosure 5). To swap occupants, one animal must leave its room, go down the hall, and park somewhere else, letting the second animal walk down the hall and into the first animal's empty room. Then the first animal can return and enter the second's now-empty room. \begin{enumerate} \item Minimal Parking Spot: The only exit from the \{5, 6, 7\} area is through enclosure 3. The closest, and therefore minimal, parking spot is enclosure 3. \item Scenario 1: L moves first. \begin{itemize} \item 1. L moves out: 6 \(\rightarrow\) 5. \item 2. L parks: 5 \(\rightarrow\) 3. \item 3. T moves to hall: 7 \(\rightarrow\) 5. \item 4. T enters L's old room: 5 \(\rightarrow\) 6. (T is done). \item 5. L returns to hall: 3 \(\rightarrow\) 5. \item 6. L enters T's old room: 5 \(\rightarrow\) 7. (L is done). \end{itemize} This minimal sequence takes 6 transfers. \item Scenario 2: T moves first. \begin{itemize} \item 1. T moves out: 7 \(\rightarrow\) 5. \item 2. T parks: 5 \(\rightarrow\) 3. \item 3. L moves to hall: 6 \(\rightarrow\) 5. \item 4. L enters T's old room: 5 \(\rightarrow\) 7. (L is done). \item 5. T returns to hall: 3 \(\rightarrow\) 5. \item 6. T enters L's old room: 5 \(\rightarrow\) 6. (T is done). \end{itemize} This also takes 6 transfers. \end{enumerate} Now let's evaluate the options against BOTH minimal scenarios. A "must be true" statement must hold for all possibilities. \begin{itemize} \item (A) The lion is transferred to enclosure 3. This is true in Scenario 1, but false in Scenario 2 (where the tiger goes to 3). So, this is not a "must". \item (B) The tiger is transferred to enclosure 5 twice. In both scenarios, the tiger moves into 5 once and out of 5 once. Not twice. False. \item (C) One of the two animals is transferred to enclosure 3 twice. False. In each case, one animal moves into 3 once and out of 3 once. \item (D) Three transfers to enclosure 5 are made. This means three moves that END in enclosure 5. \begin{itemize} \item In Scenario 1: L(6\(\rightarrow\)5), T(7\(\rightarrow\)5), L(3\(\rightarrow\)5). Yes, three transfers are made TO 5. \item In Scenario 2: T(7\(\rightarrow\)5), L(6\(\rightarrow\)5), T(3\(\rightarrow\)5). Yes, three transfers are made TO 5. \end{itemize} Since this is true in both minimal scenarios, it must be true. \item (E) At least one transfer to 2 or 4. False. The parking maneuver only needs to use enclosure 3. Going further is not minimal. \end{itemize} Step 3: Final Answer:
Regardless of which animal moves first to park in enclosure 3, the sequence of moves to complete the swap requires exactly three transfers into enclosure 5.

Answer 4

Step 1: Understanding the Concept:
We need to find the minimum number of transfers to move L from 3 to 5, and T from 4 to 7. We must identify path conflicts and find the most efficient sequence of moves.
Step 2: Detailed Explanation:
\begin{enumerate} \item Analyze Paths and Conflicts: \begin{itemize} \item Lion's Path (L): 3 \(\rightarrow\) 5. A single step. \item Tiger's Path (T): 4 \(\rightarrow\) 3 \(\rightarrow\) 5 \(\rightarrow\) 7. A three-step path. \item Conflicts: L starts at 3, which is on T's path. L's destination is 5, which is also on T's path. \end{itemize} \item Strategy 1: L moves to its destination first. \begin{itemize} \item 1. L moves 3 \(\rightarrow\) 5. (L is done). State: L=5, T=4. \item 2. T begins its path: 4 \(\rightarrow\) 3. State: L=5, T=3. \item 3. T needs to move 3 \(\rightarrow\) 5, but L is blocking the destination. L must move out of the way. \item 4. L parks: 5 \(\rightarrow\) 6. State: L=6, T=3. \item 5. T continues: 3 \(\rightarrow\) 5. State: L=6, T=5. \item 6. T continues: 5 \(\rightarrow\) 7. (T is done). State: L=6, T=7. \item 7. L must return to its destination: 6 \(\rightarrow\) 5. (L is done). State: L=5, T=7. \end{itemize} This sequence takes 7 transfers. This might not be minimal. \item Strategy 2: L parks to let T pass first. \begin{itemize} \item 1. L moves from 3 to a parking spot to clear T's path. Let's choose 2. L moves 3 \(\rightarrow\) 2. State: L=2, T=4. \item 2. T moves 4 \(\rightarrow\) 3. State: L=2, T=3. \item 3. T moves 3 \(\rightarrow\) 5. State: L=2, T=5. \item 4. T moves 5 \(\rightarrow\) 7. (T is done). State: L=2, T=7. \item 5. Now L moves from its parking spot (2) to its destination (5). Path: 2 \(\rightarrow\) 3 \(\rightarrow\) 5. \item 6. L moves 2 \(\rightarrow\) 3. State: L=3, T=7. \item 7. L moves 3 \(\rightarrow\) 5. (L is done). State: L=5, T=7. \end{itemize} This sequence also takes 6 transfers. Let's re-count: L(3-2), T(4-3), T(3-5), T(5-7), L(2-3), L(3-5). That's 6 transfers. \end{enumerate} Comparing the two strategies, 6 transfers is fewer than 7. It is the minimal number required to resolve the conflicts.
Step 3: Final Answer:
The most efficient method requires the lion to first move out of the way, allowing the tiger to complete its journey, and then for the lion to proceed to its destination. This process takes a minimum of six transfers.

Answer 5

Step 1: Understanding the Concept:
This is a "CANNOT be true" question, constrained by the rule of using the minimum number of transfers. We must find the minimal path for the given goal (L: 1\(\rightarrow\)3, T: 7\(\rightarrow\)1) and then determine which of the options describes an event that does not happen in any minimal sequence.
Step 2: Finding the Minimal Path:
\begin{enumerate} \item Analyze Paths and Conflicts: \begin{itemize} \item L's unconflicted path: 1 \(\rightarrow\) 3 (1 step). \item T's unconflicted path: 7 \(\rightarrow\) 5 \(\rightarrow\) 3 \(\rightarrow\) 1 (3 steps). \item Conflicts: L starts at T's destination (1). L's destination (3) is on T's path. \end{itemize} \item Determine Minimal Transfers: L must move out of 1 to let T in. T needs to pass through 3. A shuffle is required. Let's find the shortest sequence. \item Optimal Sequence: The most efficient way is for one animal to park while the other completes its journey. \begin{itemize} \item 1. L moves to a parking spot: 1 \(\rightarrow\) 2. (State: L=2, T=7). \item 2. T starts its journey: 7 \(\rightarrow\) 5. (State: L=2, T=5). \item 3. T continues: 5 \(\rightarrow\) 3. (State: L=2, T=3). \item 4. T continues: 3 \(\rightarrow\) 1. (T is done). (State: L=2, T=1). \item 5. L moves from parking to its destination: 2 \(\rightarrow\) 3. (L is done). (State: L=3, T=1). \end{itemize} This sequence takes 5 transfers. Any other sequence, like L moving to 3 first, takes 6 transfers and is not minimal. Other minimal 5-step sequences exist (e.g., L parking in 4, or T moving first), but they have a similar structure. \end{enumerate} Step 3: Evaluating the Options Against Minimal Paths:
The possible first transfers are L(1\(\rightarrow\)2), L(1\(\rightarrow\)4), or T(7\(\rightarrow\)5). The possible second transfers are T(7\(\rightarrow\)5) (if L moved first) or L(1\(\rightarrow\)2)/L(1\(\rightarrow\)4) (if T moved first). Let's check the options: \begin{itemize} \item (A) L to 2 in the first transfer. This is a possible first move in a 5-step solution. (CAN be true). \item (B) L to 3 in the second transfer. Let's see if this can happen. 1st move: T(7\(\rightarrow\)5). State is L=1, T=5. 2nd move: L(1\(\rightarrow\)3). State is L=3, T=5. This is a possible 2-move start. However, this path leads to a 6-step solution, which is not minimal. The question requires using the minimal number of transfers. Therefore, this move CANNOT happen in a minimal path. \item (C) L to 4 in the second transfer. Yes. 1st move: T(7\(\rightarrow\)5). 2nd move: L(1\(\rightarrow\)4). This is a possible start to a minimal 5-step path. (CAN be true). \item (D) T to 5 in the first transfer. This is a possible first move in a 5-step solution. (CAN be true). \item (E) T to 3 in the second transfer. For T to be transferred to 3, it must be in an adjacent enclosure (1, 2, 4, or 5). T starts at 7. In one transfer, T can only get to 5. Therefore, it is physically impossible for the tiger to be transferred to enclosure 3 in the second transfer. It needs at least two transfers (7\(\rightarrow\)5\(\rightarrow\)3) just to get there. \end{itemize} Step 4: Final Answer:
Both (B) and (E) describe events that do not occur in a minimal path. However, (E) is impossible under any circumstances (minimal or not), as the tiger cannot reach enclosure 3 from its starting point in a single move, which would be required for it to then be transferred to 3 in the second transfer. (B) is only impossible because of the minimality rule. Option (E) describes a geographical impossibility within the given timeframe. It is the strongest "CANNOT be true" statement.

Answer 6

Step 1: Understanding the Concept:
This is a logic puzzle involving the movement of two animals between numbered enclosures. The rules are that an animal can only be moved to an adjacent, empty enclosure, and two animals cannot occupy the same enclosure at the same time. We need to find a possible second move in a sequence that takes the animals from their starting to their final positions. The problem asks what the second transfer could be, implying we need to find at least one valid sequence of moves where the given option is the second step.
Step 2: Detailed Explanation:
Let's analyze the initial and final states:
\begin{itemize} \item Initial State: Lion (L) at enclosure 1, Tiger (T) at enclosure 3. \item Final State: Lion (L) at enclosure 6, Tiger (T) at enclosure 5. \end{itemize} The core of the problem is that the animals' paths must cross, so they need to be moved strategically to avoid blocking each other. We will test the viability of the options by seeing if they can be the second move in a logical sequence. Let's analyze option (D).
Can "tiger to enclosure 5" be the second transfer?
For this to be the second transfer, the tiger must move from its current location to enclosure 5. The tiger starts at enclosure 3. So, the transfer is T: 3 \(\rightarrow\) 5.
This requires a preceding first transfer. Let's consider a logical first move:
\begin{itemize} \item Move 1: To clear a path and begin the process, a reasonable first move is to move the Lion out of its starting position. Let's move the Lion from enclosure 1 to an adjacent enclosure, say 2.
State after Move 1: Lion is at 2, Tiger is at 3. (L@2, T@3).
\item Move 2: Now, can the Tiger be transferred to enclosure 5? Yes, if enclosure 3 is adjacent to enclosure 5 and enclosure 5 is empty. Both are reasonable assumptions in this type of puzzle. The move would be T: 3 \(\rightarrow\) 5.
State after Move 2: Lion is at 2, Tiger is at 5. (L@2, T@5). \end{itemize} This sequence is logically sound. The tiger has now reached its final destination. The problem is simplified to moving the lion from enclosure 2 to enclosure 6. This is a very efficient sequence.
Why other options are less likely:
While other options like (A) and (C) can also be part of a valid two-move start (e.g., Move 1: T: 3\(\rightarrow\)4, Move 2: L: 1\(\rightarrow\)2), the sequence leading from option (D) is strategically superior. By moving the tiger directly to its destination in the second step, the problem becomes much simpler. Logic puzzles often imply an efficient or optimal path. The move T: 3 \(\rightarrow\) 5 is a key move that resolves half of the puzzle very early. Therefore, it is a very strong candidate for a "could be" scenario.
Step 3: Final Answer:
A possible and efficient sequence of moves exists where the second transfer is the tiger moving to enclosure 5. The sequence is: (1) Lion moves from 1 to 2. (2) Tiger moves from 3 to 5. This makes option (D) a valid possibility.

Answer 7

Step 1: Understanding the Concept:
This puzzle describes a classic "swap" problem. Two items (the lion and the tiger) need to exchange their positions (enclosures 3 and 6). To do this, they cannot simply pass through each other. They must use intermediate empty enclosures to maneuver around each other. The question asks what must be true for any valid sequence of moves that accomplishes this swap.
Step 2: Detailed Explanation:
Let's analyze the setup:
\begin{itemize} \item Initial State: Lion (L) at 3, Tiger (T) at 6. \item Final State: Lion (L) at 6, Tiger (T) at 3. \end{itemize} The direct path between 3 and 6 would likely involve enclosures 4 and 5. For the animals to swap places, they cannot both be on this direct path at the same time moving towards each other. One must move aside to let the other pass.
Let's evaluate each option to see if it's a necessity:
\begin{itemize} \item (A) Exactly five enclosures are used in the move: The number of enclosures used depends on the specific layout and path taken. It's possible to construct a path that uses five enclosures (e.g., 2, 3, 4, 5, 6), but another path might use more (e.g., if a detour through 7 is needed). Since the exact layout isn't given, we cannot say this must be true. \item (B) One animal is transferred exactly twice as many times as the other: The number of transfers for each animal will depend on the path chosen. This is highly unlikely to be a necessary condition for all possible solutions. \item (C) All of the transfers of the lion are completed before any transfer of the tiger occurs: This is impossible. The lion needs to move to enclosure 6, but the tiger starts there. The tiger must be moved out of enclosure 6 before the lion can complete its journey. \item (D) At one point one of the animals is transferred to either enclosure 2 or enclosure 4: This is the most logical necessity. Consider the Lion at enclosure 3. To begin its journey to 6, it must move to an adjacent enclosure. In a standard linear or grid layout, the neighbors of 3 are 2 and 4. So the lion's first move will almost certainly be to 2 or 4. Now, consider the Tiger at enclosure 6. Its goal is enclosure 3. Its path must eventually lead it through the enclosures preceding 3, which would be 4 and 5. Therefore, the tiger will inevitably be transferred to enclosure 4 (and 5) to reach 3. Since either the lion must move to 2 or 4 to start, or the tiger must move to 4 to finish, it is unavoidable that at some point, an animal is transferred to enclosure 2 or 4. \item (E) At one point neither the lion nor the tiger is in enclosure 3, enclosure 5, or enclosure 6: While it's possible to construct a scenario where this happens (e.g., L moves to 2, T moves to 4), we can't be certain it must happen. For example, consider the sequence: L(3\(\rightarrow\)4), T(6\(\rightarrow\)5). At this point, L is at 4 and T is at 5. The condition is false because T is in 5. From here, one animal has to move aside. If L moves (4\(\rightarrow\)2), the state is L@2, T@5. The condition is still false. It's not a guaranteed state in every possible solution. \end{itemize} Step 3: Final Answer:
The most robust conclusion that holds regardless of the specific path is (D). The geometry of the problem (swapping from 3 to 6) necessitates movement through the intermediate and adjacent enclosures. The enclosures adjacent to 3 are likely 2 and 4, and the path from 6 to 3 must pass through 4. Therefore, an animal being transferred to 2 or 4 is a necessary step in the process.