Question

Which of the following can be the order of lectures and lunch at the conference?

• A. W, X, Lunch, Y, S, T, Z
• B. X, Y, T, Lunch, S, Z, W
• C. Y, T, Lunch, S, W, X, Z
• D. Z, T, W, S, Lunch, Y, X
• E. Z, W, Y, S, Lunch, X, T
• A. X and T
• B. Y and T
• C. Z and T
• D. Z and W
• J. Z and Y
• A. T is the second lecture.
• B. T is the fifth lecture.
• C. W is the third lecture.
• D. X is the first lecture.
• O. X is the third lecture.
• A. X is the first lecture.
• B. X is the second lecture.
• C. Exactly two lectures are given before lunch.
• D. Exactly three lectures are given before lunch.
• T. Exactly four lectures are given before lunch.
• A. X is given before lunch.
• B. X is given after lunch.
• C. Y is given before lunch.
• D. T is the third lecture.
• Y. Z is the fifth lecture.
• A. S
• B. T
• C. X
• D. Y
• ^. Z

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question asks for a valid sequence of lectures that satisfies all the given rules. We can test each option against the rules. Rules Summary: \begin{enumerate} \item S is 4th. \item Exactly one resident (S or T) is before lunch. \item Y occurs before T (Y ... T). \item If W is before lunch, Y is after lunch (W\textsubscript{BL} \(\rightarrow\) Y\textsubscript{AL}). Contrapositive: If Y is before lunch, W is after lunch (Y\textsubscript{BL} \(\rightarrow\) W\textsubscript{AL}). \end{enumerate} Step 2: Detailed Explanation:
Let's check each option: \begin{itemize} \item (A) W, X, Lunch, Y, S, T, Z. Lectures are W(1), X(2), Y(3), S(4), T(5), Z(6). Rule 1: S must be 4th. This is satisfied. Rule 2: T and S are residents. Both are after lunch. This violates the rule that exactly one resident is before lunch. Invalid. \item (B) X, Y, T, Lunch, S, Z, W. Lectures are X(1), Y(2), T(3), S(4), Z(5), W(6). \begin{itemize} \item Rule 1: S is 4th. OK. \item Rule 2: T is before lunch, S is after lunch. Exactly one resident is before lunch. OK. \item Rule 3: Y(2) is before T(3). OK. \item Rule 4: Y is before lunch. The contrapositive requires W to be after lunch. W is 6th (after lunch). OK. \end{itemize} All rules are satisfied. Valid. \item (C) Y, T, Lunch, S, W, X, Z. Lectures are Y(1), T(2), S(3), W(4), X(5), Z(6). Rule 1: S must be 4th. Here S is 3rd. Invalid. \item (D) Z, T, W, S, Lunch, Y, X. Lectures are Z(1), T(2), W(3), S(4), Y(5), X(6). Rule 1: S is 4th. OK. Rule 3: Y must be before T. Here Y(5) is after T(2). Invalid. \item (E) Z, W, Y, S, Lunch, X, T. Lectures are Z(1), W(2), Y(3), S(4), X(5), T(6). Rule 1: S is 4th. OK. Rule 4: W is before lunch (2nd) and Y is also before lunch (3rd). This violates the rule "If W is given before lunch Y will be given after lunch". Invalid. \end{itemize} Step 3: Final Answer:
The only sequence that satisfies all the conditions is (B).

Answer 2

Step 1: Understanding the Concept:
This question adds a condition and asks what must be true as a result. We need to combine the new condition with the original rules to deduce the identities of the first two lectures.
Step 2: Detailed Explanation:
\begin{enumerate} \item Condition: Exactly two lectures are before lunch. This means lunch is between lecture 2 and 3. \item Apply Rule 1: S is the 4th lecture. This means S is after lunch. \item Apply Rule 2: Exactly one resident (S or T) must be before lunch. Since S is after lunch, T must be before lunch. The lectures before lunch are in positions 1 and 2, so T must be in position 1 or 2. \item Apply Rule 3: Y must be given at some time before T. \begin{itemize} \item If T were in position 1, no lecture could be given before it. This would violate Rule 3. \item Therefore, T cannot be in position 1. T must be in position 2. \end{itemize} \item Deduce the first lecture: Since T is in position 2, and Y must be before T, Y must be in position 1. \item Conclusion: The two lectures given before lunch must be Y (in position 1) and T (in position 2). \item Final Check: Does this arrangement conflict with other rules? With Y before lunch, the contrapositive of Rule 4 applies (Y\textsubscript{BL} \(\rightarrow\) W\textsubscript{AL}), meaning W must be after lunch. This is perfectly consistent. \end{enumerate} Step 3: Final Answer:
Based on the logical deductions, if exactly two lectures are given before lunch, they must be Y and T.

Answer 3

Step 1: Understanding the Concept:
This question gives a new condition and asks what is possible ("can be true"). We need to find an option that is consistent with the new condition and all original rules.
Step 2: Detailed Explanation:
\begin{enumerate} \item Condition: Exactly three lectures before lunch, and they include Y and Z. Lunch is between 3 and 4. \item Apply Rule 1: S is the 4th lecture, so S is after lunch. \item Apply Rule 2: Exactly one resident is before lunch. Since S is after lunch, T must be before lunch. \item Deduce the before-lunch group: The three lectures before lunch are given in positions 1, 2, and 3. We know they include Y and Z (from the condition) and T (from our deduction). So, the set of lectures before lunch is \{Y, Z, T\}. \item Apply Rule 4: Since Y is before lunch, the contrapositive (Y\textsubscript{BL} \(\rightarrow\) W\textsubscript{AL}) means W must be after lunch. \item Deduce the after-lunch group: The lectures after lunch are in positions 4, 5, and 6. We know S is 4th. We deduced W must be after lunch. The only remaining lecture is X. So, the set of lectures after lunch is \{S, W, X\}. \item Apply Rule 3: Y must be before T. Since both are in the \{1, 2, 3\} group, T cannot be in position 1. T can be in position 2 or 3. \end{enumerate} Now let's check the options to see what can be true. \begin{itemize} \item (A) T is the second lecture. Is this possible? Yes. If T is 2nd, Y must be 1st to satisfy Y...T. Z could be 3rd. The order `Y, T, Z` before lunch is possible. \item (B) T is the fifth lecture. False. We deduced T must be before lunch (in slots 1, 2, or 3). \item (C) W is the third lecture. False. We deduced W must be after lunch. \item (D) X is the first lecture. False. We deduced X must be after lunch. \item (E) X is the third lecture. False. We deduced X must be after lunch. \end{itemize} Step 3: Final Answer:
It is possible for T to be the second lecture. Therefore, option (A) can be true.

Answer 4

Step 1: Understanding the Concept:
This is a conditional question that asks for a necessary conclusion ("must be true"). The condition fixes T to the last position.
Step 2: Detailed Explanation:
\begin{enumerate} \item Condition: T is the 6th lecture. \item Analyze with Resident rule (Rule 2): We have two residents, S and T. Rule 2 states that exactly one of them must be before lunch. \item Position S and T relative to lunch: S is the 4th lecture. T is the 6th lecture. Since T is 6th, it is definitely after lunch, regardless of whether lunch is after 2, 3, or 4 lectures. \item Deduce S's position relative to lunch: For Rule 2 to be satisfied, S must be the resident who is before lunch. \item Deduce the number of lectures before lunch: For S, the 4th lecture, to be before lunch, lunch must take place after the 4th lecture. The number of lectures before lunch cannot be 2 or 3. It must be 4. \item Conclusion: The statement "Exactly four lectures are given before lunch" is a necessary consequence of T being the sixth lecture. \end{enumerate} Let's check the other options. \begin{itemize} \item (A) X is the first lecture. Not necessarily. The before-lunch lectures are 1, 2, 3, and S(4). We know Y...T, so Y is one of these. The remaining slots are filled by visitors. We also have the rule W\textsubscript{BL} \(\rightarrow\) Y\textsubscript{AL}. Since Y cannot be after lunch (T is last), Y must be before lunch. This means W must be after lunch. The only spot after lunch besides T(6) is 5. So W must be 5th. The lectures before lunch are \{1, 2, 3, S\}. They must be \{Y, X, Z, S\}. The order is not fixed, so X doesn't have to be first. \item (B) X is the second lecture. Not necessarily, for the same reason. \item (C) Exactly two lectures are given before lunch. False. We proved it must be four. \item (D) Exactly three lectures are given before lunch. False. We proved it must be four. \end{itemize} Step 3: Final Answer:
The condition that T is sixth forces the conclusion that there must be exactly four lectures before lunch to satisfy the rule about resident speakers.

Answer 5

Step 1: Understanding the Concept:
This is a conditional question asking what must be true given a new constraint. We must combine the new information with the original rules to make a definitive deduction.
Rules Summary: \begin{enumerate} \item S is 4th. \item Exactly one resident (S or T) is before lunch. \item Y occurs before T (Y ... T). \item If W is before lunch, Y is after lunch (W\textsubscript{BL} \(\rightarrow\) Y\textsubscript{AL}). Contrapositive: If Y is before lunch, W is after lunch (Y\textsubscript{BL} \(\rightarrow\) W\textsubscript{AL}). \end{enumerate} Step 2: Detailed Explanation:
\begin{enumerate} \item Condition: S and Z are both given after lunch. \item Apply Rule 1 (S is 4th): Since S is 4th and is after lunch, lunch must take place before the 4th lecture. So, there are either 2 or 3 lectures before lunch. \item Apply Rule 2 (Resident rule): Exactly one resident must be before lunch. Since S is after lunch, T must be before lunch. T must therefore be in position 1, 2, or 3. \item Apply Rule 3 (Y...T): Y must be given before T. Since T is before lunch (in positions 1, 2, or 3), Y must also be given before lunch and in an earlier position than T. \item Conclusion: From step 4, it is a logical necessity that Y is given before lunch. This directly matches option (C). \item Checking other options for completeness: \begin{itemize} \item Since Y is before lunch, the contrapositive of Rule 4 applies: W must be after lunch. \item So, we know: Before Lunch includes \{Y, T\}, After Lunch includes \{S, Z, W\}. \item The last lecture, X, can be either before lunch (making it 3 before, 3 after) or after lunch (making it 2 before, 4 after). Both scenarios are possible. Therefore, we cannot say for sure whether X is before or after lunch, making options (A) and (B) not "must be true". \item T could be 2nd or 3rd, so (D) is not a must. \item Z is after lunch, but its specific position (5th or 6th) is not fixed, so (E) is not a must. \end{itemize} \end{enumerate} Step 3: Final Answer:
The condition that S is after lunch forces T to be before lunch. The rule that Y must come before T then forces Y to also be before lunch. This is a certain deduction.

Answer 6

Step 1: Understanding the Concept:
This question asks which lecture is impossible to place in the last slot before lunch. The last slot before lunch can be position 2, 3, or 4. We can test if each lecture could occupy this position.
Step 2: Detailed Explanation:
Let's test which lectures CAN be immediately before lunch. \begin{itemize} \item Can S be last before lunch? Yes. If there are 4 lectures before lunch, the order is 1, 2, 3, 4 | Lunch | 5, 6. Since S is 4th, S would be the lecture immediately before lunch. This is a valid scenario. (Eliminates A). \item Can T be last before lunch? Yes. If there are 2 lectures before lunch (1, 2 | Lunch), S is after lunch (at 4). Rule 2 requires T to be before lunch. Rule 3 (Y...T) requires Y to be before T. Thus, the order must be Y(1), T(2). Here, T is immediately before lunch. (Eliminates B). \item Can X or Z be last before lunch? Yes. If there are 3 lectures before lunch (1, 2, 3 | Lunch), S is after lunch (at 4). T must be before lunch. Y must be before T. This means Y and T occupy two of the first three spots. The third spot can be filled by a visiting speaker. It cannot be W (because Y is before lunch, so W must be after lunch). So, the third speaker could be X or Z. A possible order is Y(1), T(2), Z(3) | Lunch. In this case, Z is immediately before lunch. Another is Y(1), T(2), X(3) | Lunch. X is immediately before lunch. (Eliminates C and E). \end{itemize} Step 3: Proving Y Cannot be last before lunch:
By elimination, Y is the answer. Let's prove it directly. Assume Y is the lecture immediately before lunch. \begin{itemize} \item The rule Y...T means T must come after Y. Since Y is immediately before lunch, T must be after lunch. \item The rule states exactly one resident (S or T) is before lunch. Since T is after lunch, S must be before lunch. \item The rule states S is the 4th lecture. For S(4) to be before lunch, there must be exactly 4 lectures before lunch. \item This means the lecture immediately before lunch is the 4th lecture. \item We assumed Y is the lecture immediately before lunch. This would mean Y is the 4th lecture. \item This creates a contradiction: Y cannot be the 4th lecture because S is the 4th lecture. \item Therefore, the initial assumption that Y can be immediately before lunch is false. \end{itemize} Step 4: Final Answer:
S, T, X, and Z can all be placed in the last position before lunch under valid scenarios. Y cannot, because it leads to a direct contradiction with the rule that S is the fourth lecture.