Question

Then, all the following statements could be true EXCEPT :

• A. Twice as many experts take sessions after the school as before the school
• B. J will not take sessions during lunch period
• C. The same number of experts take sessions before school as during lunch
• D. The same number of experts take sessions after school as during lunch
• E. The same number of experts take sessions before school as after school

Hint:

For scheduling EXCEPT questions, check each option by trying to construct a valid schedule. The option that violates at least one constraint or is arithmetically impossible is the answer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the constraints to determine which statement CANNOT be true.

Step 2: Detailed Explanation:
J is only after school; W always during lunch; M is not after school. Minimum: 3 after school (includes J). At least 2 before school. W is fixed at lunch. If after school = lunch count: after school $\geq$ 3, lunch includes at least W = 1. For them to be equal, lunch $\geq$ 3 also. But W is the only guaranteed lunch person; if lunch = 3, after school = 3: total used = 6 slots for 7 experts + W already in lunch. Testing: after school = lunch = 3 means after school has J + 2 others, lunch has W + 2 others, before school has 2 remaining. That uses $3+3+2=8 > 7$. Impossible since only 7 experts. So ``after school = lunch'' cannot hold given $\text{after} \geq 3 + \text{lunch} \geq 1$ (W) and total = 7. Statement (D) is impossible (except in a degenerate case) -- official answer is (D).

Step 3: Final Answer:
Statement (D) cannot be true: The same number of experts take sessions after school as during lunch.