Five members A, B, C, D and E of a family together eat five different fruits -…
2014
Five members A, B, C, D and E of a family together eat five different fruits - Grapes, Apple, Watermelon, Pomegranate and Pineapple - after their lunch, one fruit each per day, from Tuesday to Saturday. Each member eats a different one of these five fruits on each of the five days, so every member eats all five fruits exactly once during the week, and no two members eat the same fruit on the same day. Neither B nor E eats Watermelon or Grapes on Wednesday. D eats Apple on Tuesday. E does not eat Pineapple on Tuesday. B eats Pomegranate on Friday. C eats Grapes on Saturday. A eats Watermelon on Tuesday. D eats Pineapple on Wednesday.
Which fruit does E eat on Friday?
- A.
Grapes
- B.
Watermelon
- C.
Apple
- D.
Pomegranate
Attempted by 3 students.
Show answer & explanation
Correct answer: B
This is a Latin-square style arrangement puzzle: across the five days from Tuesday to Saturday, the five family members together eat all five different fruits with no repeats, so every day's five fruits are all different and every member's five days are all different fruits. It is solved by first placing every explicitly stated fact onto a 5-by-5 grid of member versus day, and then repeatedly applying the two "no-repeat" rules - one fruit per member across the week, one fruit per day across the members - to force each remaining, unstated cell by elimination.
Place the seven given facts directly: D eats Apple on Tuesday and Pineapple on Wednesday; A eats Watermelon on Tuesday; B eats Pomegranate on Friday; C eats Grapes on Saturday.
On Wednesday, with D's Pineapple fixed, A, B, C and E must share Grapes, Apple, Watermelon and Pomegranate. Neither B nor E may eat Watermelon or Grapes that day, so between them B and E take Apple and Pomegranate, leaving Grapes and Watermelon for A and C.
A already ate Watermelon on Tuesday, so it cannot repeat Watermelon on Wednesday - so A eats Grapes on Wednesday and C eats Watermelon on Wednesday.
B already has Pomegranate booked for Friday, so it cannot repeat Pomegranate on Wednesday - so B eats Apple on Wednesday, leaving Pomegranate for E on Wednesday.
On Tuesday, with A's Watermelon and D's Apple fixed, B, C and E must share Grapes, Pomegranate and Pineapple. E cannot eat Pineapple that day (given) and cannot repeat the Pomegranate it now has on Wednesday, so E eats Grapes on Tuesday. C cannot repeat the Grapes it has booked for Saturday, so C eats Pomegranate on Tuesday, leaving Pineapple for B on Tuesday.
B's fruits so far are Pineapple (Tuesday), Apple (Wednesday) and Pomegranate (Friday); only Grapes and Watermelon remain, for Thursday and Saturday. C already has Grapes on Saturday, so B cannot repeat that fruit on Saturday - B eats Watermelon on Saturday and Grapes on Thursday.
D's fruits so far are Apple (Tuesday) and Pineapple (Wednesday); only Grapes, Watermelon and Pomegranate remain, for Thursday, Friday and Saturday. Saturday cannot be Grapes (C already has it) or Watermelon (B already has it), so D eats Pomegranate on Saturday; Thursday cannot be Grapes (B already has it), so D eats Watermelon on Thursday, leaving Grapes for D on Friday.
On Friday, Pomegranate belongs to B and Grapes belongs to D, leaving Apple, Watermelon and Pineapple for A, C and E. Watermelon cannot go to A (already used on Tuesday) or to C (already used on Wednesday) - so Watermelon must go to E.
Member | Friday's fruit |
|---|---|
A | Apple or Pineapple |
B | Pomegranate |
C | Whichever of Apple / Pineapple A does not take |
D | Grapes |
E | Watermelon |
Every one of the seven stated clues, and the rule that no two members share a fruit on the same day, holds throughout this arrangement, and Watermelon is the only fruit left once Pomegranate and Grapes are fixed to B and D and Watermelon itself is ruled out for both A and C by their own earlier days. So E eats Watermelon on Friday.