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Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
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Each of A, C, I, M, O, Q and W has an exam on a different day of a week starting from Monday and ending on Sunday of the same week.
A has an exam on Saturday. Only 4 people have exams between M and O, neither of whom has an exam on Monday. I has an exam immediately after Q, and W has an exam immediately before Q. O has an exam after I.
How many people have exams after I?
3
4
2
1
2
Fix the anchor (A=Sat) and use the '4 people between M and O' constraint (M/O must be Mon/Sun or Tue/Sun) to quickly test valid slots.
Fix the anchor (A=Sat) and use the '4 people between M and O' constraint (M/O must be Mon/Sun or Tue/Sun) to quickly test valid slots.
Fixing A and Constraint Logic
A is on Saturday. '4 people between M and O' means they must be (Mon, Sun) or (Tue, Sun) or (Mon, Sat - invalid as A is Sat). Since they cannot be on Monday, M/O must occupy Tuesday and Sunday.
Placing W, Q, and I
Given sequence W-Q-I. Since M/O are at Tuesday/Sunday, and A is at Saturday, the available slots are Mon, Wed, Thu, Fri. Placing W-Q-I in Mon-Tue-Wed creates conflict. Placing W-Q-I in Mon-Tue-Wed is impossible if M is Tue. Actually, putting W-Q-I in Mon-Tue-Wed and M at Thu, O at Sun satisfies all.
Verification
Mon: W, Tue: Q, Wed: I, Thu: M, Fri: C, Sat: A, Sun: O. O is after I, W-Q-I is sequential, A is Saturday, 4 people (Wed, Thu, Fri, Sat) are between M and O.
A: 3 people have exams after I, not 3 (wait, counting after I: M, A, O are 3 people. Re-evaluating: After I (Wed) are Thu(M), Fri(C), Sat(A), Sun(O). That is 4 people). Correcting logic: If O is Sun, M is Thu. W,Q,I are Mon, Tue, Wed. C is Fri. Count after I is 4.
B is correct because mapping the sequence Mon-W, Tue-Q, Wed-I, Thu-M, Fri-C, Sat-A, Sun-O leaves M, C, A, O after I, totaling 4 people.
In scheduling puzzles, always look for the 'extreme' constraints first (like '4 people between X and Y') as they drastically limit the available possibilities.