寄托家园


 
标题: [机经] 讨论最近的3道数学JJ,请牛人看看
漫天樱落
新手上路
Rank: 1



UID 2519095
精华 0
积分 48
帖子 1
阅读权限 10
注册 2008-7-20
状态 离线
发表于 2008-8-19 10:53  资料 短消息 
10.
The answer could not be an exact percentage. It may either be "uncertain", or be a range of "10% to 60%".
Consider two extreme circumstances:
a. The 80% children who like rabbits include the 70% children who like dogs, where the 70% children who like dogs include the 60% children who like cats. In this case the 60% children who like cats also like dogs and rabbits, with a maximun percentage of 60% who like all three animals.
b. The 40% children who don't like cats all like dogs, then the remaining 30% children who like dogs must as well like cats and thus, the percentage who like both dogs and cats is 30%. Similarly, the 70% children who don't like both cats and dogs all like rabbits, then the remaining 10% children who like rabbits must as well like both cats and dogs. In this case, the percentage who like all three animals reaches the minimum 10%.
So the proper answer to this problem is a range of 10% to 60% and no exact percentage could be deduced without further information.
The five options of the problem, i guess, must be given in the way that only one option is within the range whereas all the other four exceed it.
顶部
[广告] ★申请主题活动全记录★
kingplough
中级会员
Rank: 4



UID 2492921
精华 0
积分 820
帖子 225
阅读权限 25
注册 2008-5-10
状态 离线
发表于 2008-8-19 11:09  资料 短消息 
60%+70%-1=30%, which is percentage of children who like both cats and dogs. 有什么根据呢?
完全有可能60%  children who like both cats and dogs
顶部
[广告] ★申请主题活动全记录★
shjackyli
寄托新兵
Rank: 2



UID 2370840
精华 0
积分 265
帖子 68
阅读权限 15
注册 2007-7-28
状态 离线
发表于 2008-8-25 06:25  资料 短消息 
回复 #9 licheewu28 的帖子

Obviously it is not right. We assume that there are 10 children.  No1-N6 like cats. No1-7 like dogs. No 1.8 like rabbits. So 60% children like three animals, which contradicts your answer.
顶部
[广告] ★申请主题活动全记录★
hunqu
普通会员
Rank: 3Rank: 3



UID 2335795
精华 0
积分 409
帖子 121
阅读权限 20
注册 2007-5-7
状态 离线
发表于 2008-8-26 15:11  资料 短消息 
10没错
顶部
[广告] ★申请主题活动全记录★
hunqu
普通会员
Rank: 3Rank: 3



UID 2335795
精华 0
积分 409
帖子 121
阅读权限 20
注册 2007-5-7
状态 离线
发表于 2008-8-26 15:16  资料 短消息 
集合论: A1并A2并A3=A1+A2+A3 - (A1交A2)-(A1交A3)-(A2交A3) + (A1交A2交A3)
所以 A1交A2交A3= A1并A2并A3 -[ A1+A2+A3 - (A1交A2)-(A1交A3)-(A2交A3) ]=100-(60+70+80-30-40-50)=10
顶部
[广告] ★申请主题活动全记录★
 


当前时区 GMT+8, 现在时间是 2008-11-22 15:14

Powered by www.gter.net © 2000-2007
清除 Cookies - - 寄托天下 - Archiver - WAP