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标题: 问三道OG11th上的题目，no217.no224,no248 [打印本页]

作者: mayizhong 时间: 2007-10-18 04:09
标题: 问三道OG11th上的题目，no217.no224,no248

no217.

a certain junior class has 1000 students and a senior class has 800 students,there are 60sibling pairs, each consisting of 1 junior and 1 senior. If a student is to be selected at random from each class, what is the probability that the two students selected will be a sibling pair

A. 3/40000, B. 1/3600, C. 9/2000, D.1/60, E.1/15 (key:A)

no224

For the past n days, the average (arithmetic mean) daily product at a company was 50 units. If today's production of 90 units raises the average to 55 units per day, what is the value of n?

A. 30, B. 18, C.10, D.9, E.7 (key: E)

no248.

Right triangle PQR is to be constructed in the xy-plane so that the right angles is at P and PR is parrallel to the x-axis.The x-and y-coordinates of P , Q and R are to be integers that satisfy the inequalities -4<=x<=5 and 6<=y<=16,how many different triangles with these properties could be constructed?

A. 110, B. 1100, C.9900, D.10000, E.12100 (key:C)

请好心人帮助。不胜感激！！！

作者: celine2007 时间: 2007-10-18 05:17

217 60/1000)*(1/800)=3/40000

224: (50*N+90)/(N+1) = 7

248: 排列问题

设P (X,Y), 因为 (-4<=X<=5), (6<=Y<=16), 所以P 有 10*11=110 种可能的位置(在限定区域)

根据条件知道, R 与P的Y 值总是一致(水平线), 所以 R 只有 11-1 = 10 的可能性.

通理, Q 与P的X值一致, 所以只有 10-1=9 的可能性

所有的排列可能 P*R*Q =110*10*9=9900

作者: mayizhong 时间: 2007-10-18 09:26
谢谢，不理解的是为什么217题，*1/800？能否讲一下题目？

作者: felix_kobe 时间: 2007-10-18 13:35

呵呵，我刚刚接触GMAT。

以我的理解：初级班1000人，高级班800人，这其中有60对儿双胞胎，并且每对都是分到初级班和高级班各一个。问：如果从每个班都随机抽出一个学生，这两个学生恰是一对双胞胎的概率是多少。

作者: felix_kobe 时间: 2007-10-18 13:36
解题思路，2楼给的可以啊

作者: grace_cd 时间: 2008-1-23 19:31

我觉得应该是(30/1000)*(1/800),不是一个在JUINOR,一个在SENIOR吗?

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