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GWD-18-13

Q13:

Each of 20 parents chose one of five days from Monday through Friday to attend parent-teacher conferences. If more parents chose Monday than Tuesday, did at least one of the parents choose Friday?

(1) None of the five days was chosen by more than 5 parents.

(2) More parents chose Monday than Wednesday.

Why not A?

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1. 5x4 left none chooses Friday. 没有一天有超过5人,那么5人就不能算超过5人.

2. gone

ft, E is wrong, C: MTW = 5+4+4 = 13, ThF = 20-13=7, but Th<=5, so F>=2

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我也支持A.

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more parents chose Monday than Tuesday A

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(1)the maximum number of parents who chose one of five days from Monday through Friday is 5;

at the same time, more parents chose Monday than Tuesday, Tuesday<=4;

so the result is  at least one of the parents choose Friday.

A is correct

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偶写错了,其实偶选C,偶是想问为 WHY A?

星期五在A下,可以有0也可以有>0的情况啊

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C.

According to (1), since Friday could be 0 or 1,2,3,4,5, A is unsufficent

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I think is A, since the questions states: More parents choose Monday than choose Tuesday"

1) None of the five days was chosen by more than 5 parents

So it is impossible have 0 for Friday since it would go against the statement above (M 5 T 5 W 5 Th 5 F 0)

Any combination would show Friday has at least one.

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a是对的。

没有一天的数字超过5,那就是20个人,每天四个人去。

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1) None of the five days was chosen by more than 5 parents  指的是一天最多5 parents

如果按照樓上的意思, 就變成一天最多4 patents, 這樣理解是錯誤的, 原因如下:

若一天最多4 parents, 而且按照題目內含的條件Mon.>Tue.

則剩下的3天要分配13 parents(20-4-3=13), 就算每天達到最大量4 parents, 這樣還是會剩下1 parents沒有排到, 這樣不合理

所以條件(1)的正確理解是每天最多5 parents

如果按照樓上的意思, 就變成一天最多4 patents, 這樣理解是錯誤的, 原因如下:

若一天最多4 parents, 而且按照題目內含的條件Mon.>Tue.

則剩下的3天要分配13 parents(20-4-3=13), 就算每天達到最大量4 parents, 這樣還是會剩下1 parents沒有排到, 這樣不合理

所以條件(1)的正確理解是每天最多5 parents

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