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D, E, F, G, H, I, and J, each has an exam on a different day of a week starting from Monday and ending on Sunday of the same week. E has the exam on Friday. Only one person has the exam between E and D. J has the exam on the day immediately after E. H has the exam on the day immediately after D. Only two people have exams between H and G. F has the exam on the day immediately before D. I does not have the exam on Monday. How many people have exams between J and F?
Five
Four
Three
Two
Three
Fix the anchor point (Friday) and list days immediately to deduce the fixed sequence (F, D, H, G).
Fix the anchor point (Friday) and list days immediately to deduce the fixed sequence (F, D, H, G).
Establish the anchor points
Days: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. E is on Friday. J is immediately after E, so J is on Saturday. Only one person is between E and D; D must be on Wednesday (cannot be Sunday since H is after D).
Fill remaining positions
F is immediately before D, so F is on Tuesday. H is immediately after D, so H is on Thursday. Only two people are between H (Thursday) and G; counting from Thursday, G must be on Sunday (H, _, _, G). The remaining person I is on Monday.
Final Sequence Check
Monday: I, Tuesday: F, Wednesday: D, Thursday: H, Friday: E, Saturday: J, Sunday: G. People between J (Saturday) and F (Tuesday) are D, H, and E.
A: Five people do not exist between J and F. B: Four people do not exist between J and F. D: Two people is an incorrect count as there are exactly three (D, H, E).
C is correct because the people placed between J (Saturday) and F (Tuesday) are D (Wednesday), H (Thursday), and E (Friday), totaling three people.
In scheduling puzzles, always prioritize the fixed anchor points provided and build the timeline step-by-step to avoid errors.
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