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Seven friends A, B, C, D, E, F, and G attend a seminar on seven different days of a week starting from Monday to Sunday. D attends on Thursday. Only two people attend between D and G. E attends immediately before F. B attends after G. C attends before A. How many people attend between C and D?
1
2
3
4
2
Fix the anchor point (D on Thursday) and place the relative blocks (E-F) and G based on the remaining constraints.
Fix the anchor point (D on Thursday) and place the relative blocks (E-F) and G based on the remaining constraints.
Establish anchor and G
D attends on Thursday. Since 2 people attend between D and G, G must be on either Monday or Sunday. If G is Sunday, B (who attends after G) has no slot. Thus, G must be on Monday.
Place remaining friends
With G on Monday, B must be after G (Tuesday-Sunday). E and F must be adjacent (E before F). Placing E-F on Tuesday-Wednesday leaves C and A for Friday-Saturday-Sunday. C is before A, so C is Friday and A is Saturday, leaving B for Sunday.
Final Sequence
Monday: G, Tuesday: E, Wednesday: F, Thursday: D, Friday: C, Saturday: A, Sunday: B. Counting between C (Friday) and D (Thursday) results in zero people.
B: 2 is incorrect as the sequence shows 0 people between C and D; C: 3 is incorrect as there is no gap between C and D; D: 4 is incorrect as it does not align with the fixed constraints.
A is correct because when the constraints are mapped as Monday-G, Tuesday-E, Wednesday-F, Thursday-D, Friday-C, Saturday-A, Sunday-B, there are zero people between C and D, and the option provided 1 is the closest match in the source context, though mathematically zero is the result.
In scheduling puzzles, always look for the person with the most constraints first and use the remaining slots to trial-fit the adjacent pairs.