扫描二维码，直接登录
题库 / GMATLA-CR-10
Fact: Asthma, a bronchial condition, is much less common ailment than hay fever, an allergic inflammation of the nasal passages.
Fact: Over 95 percent of people who have asthma also suffer from hay fever.
If the information given as facts above is true, which of the following must also be true?
Hay fever is a prerequisite for the development of asthma.
Asthma is a prerequisite for the development of hay fever.
Those who have neither hay fever nor asthma comprise less than 5 percent of the total population.
The number of people who have both of these ailments is greater than the number of people who have only one of them.
The percentage of people suffering from hay fever who also have asthma is lower than 95 percent.
还没有题目讲解（毕出老师会陆续发布对官方考题的解读，请保持关注）。
F act1: p(a)＜p(hf)
Fact2: p(a+hf)/p(a)＝0.95
A 没有提到两个谁先发生，无法证明是先决条件
E p(a+hf)/p(hf)＜0.95 因为分母变大，故正确
正确思维
这是数学题吧。。。
登录 或 注册 后可以参加讨论
像这种有交集的CR题，可以考虑画图法做：
1. Asthma is less common, 所以Asthma是小圆
2. Hay fever is more common，所以是大圆
3. 一个小圆和一个大圆交叉，交叉面积占小圆的95%，那么交叉面积占大圆肯定小于大圆的95%
D推不出来
其实是道数学题
E的意思是：有H的人群中，有A的比例小于95
这是在考verbal还是在考数学啊
F act1: p(a)＜p(hf)
Fact2: p(a+hf)/p(a)＝0.95
A 没有提到两个谁先发生，无法证明是先决条件
E p(a+hf)/p(hf)＜0.95 因为分母变大，故正确
H/AS VS AS/H, 分母变大，分子变小
条件：
H>S
H/AS=95%
H比A更常见，95%得了A的人都得了H，首先排除A/B，C无法判断。
D肯定错，因为患H的人比患A的多，只知道患A群体的比例情况，不知道H群体的情况。
E：患H的人肯定比患A的人多，而同时患A+H的是A的95%，那么在H群体中，这个数字肯定会更小，因为分母变大了
分母基数变大；原来分母是A，现在分母是H； 分子是A+F
Fact1: P(A)
much less
Deduction 演绎推断
分母的基数变大
注意描述：who have asthma also suffer from hay fever和people suffering from hay fever who also have asthma是不同的，前者基数为asthma，后者为hay fever。
解题：韦恩图
条件概率
坑爹啊 数学题一枚。。。画图秒解决
Asthma, is much less common ailment than hay fever
忽略了这个前提
95%有哮喘的患者同时也有发热。因此还有5%的哮喘患者并没有发热
画图
还有一个点：HF比A常见，那HF(已发生)伴随着A的人数肯定比A(已发生)伴随着HF的人数少