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标题: PREP中三题,请教高手。。附答案 [打印本页]

作者: XIAOYUEHAN    时间: 2009-6-13 10:07     标题: PREP中三题,请教高手。。附答案

166.

70, 75, 80, 85, 90, 105, 105, 130, 130, 130

The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 meters.  If the standard deviation of the 10 running times is 22.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?

 

(A) One

(B) Two

(C) Three

(D) Four

(E) Five

选B,看不懂,也不会做。。

3   A manufacturer produced x percent more video cameras in 1994 than in 1993 and y percent more video cameras in 1995 than in 1994.  If the manufacturer produced 1,000 video cameras in 1993, how many video cameras did the manufacturer produce in 1995 ?

1 xy=20

2 X+Y+XY/100=9.2

答案选B

请问2为什么可以求出?


4. Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours.  If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

(A) 25%
(B)
(C) 50%
(D) 66*2/3%
(E) 75%

选D, 看不懂题目也不会解答


作者: BruceNornia    时间: 2009-6-13 18:24

166 题目的意思:

70 75 80 85 90 105 105 130 130 130

上述列表所示为10个学生跑400米跑所用的时间。该10个数的标准差为22.4,四舍五入到10分位,有多少学生的跑步成绩比上述列表的平均数低出一个标准差以上?

平均数为: (70+75+80+85+90+105+105+130+130+130)/10 = 100

比平均数低1个标准差的数字 必须 小于 100 - 22.4 = 77.6

所有数字中 比77.6 小的只有 70和75 两个 所以B Tow


作者: BruceNornia    时间: 2009-6-13 18:27

3 1993的产量为 1000

1994的产量为  1000(1+y%)

1995的产量为  1000(1+y%)(1+x%) = 1000(1+x%+y%+x%y%)

 

如果知道 X+Y+XY/100=9.2 则 1+x%+y%+x%y% = 9.2%

从而 1000(1+x%+y%+x%y%) 可以计算

所以条件2 是可以的


作者: BruceNornia    时间: 2009-6-13 18:31

4 比较奇怪

 

自愧资质有限  难以领悟题目要旨


作者: yahooww    时间: 2009-6-13 19:25

1,额..要找10个数中,在比MEAN小的数中,与MEAN的差比标准方差小的数...

只有70,75

2,X+Y+XY/100=9.2

看到这,我们准备求1+X/100)(1+Y/100)=1+x/100+y/100+xy/10000

到这,应该明白了吧..

3,V1=K/8

V2=K/4

面对K CARTONS

SET THE NEW ONE COST AS Y

1/Y=1/8+1/4

Y=8/3

V3=3K/8

额,后来看到问题,发现上面不要算的...随便写出来...权当偶的思路

问最快机器在相同时间内生产的产品在所有中的比率.

一看就知道.8小时,4小时.肯定是2/3的产品...所以是66.67%


作者: XIAOYUEHAN    时间: 2009-6-14 09:05

what percent of the cartons were sealed by the machine working at the faster rate? 这句啥意思,要求啥,我也当时算到V3=3K/8
作者: BruceNornia    时间: 2009-6-14 18:44

... 我把第三个题目翻译一遍吧 我困惑的是  66*2/3%这个数字是什么意思

 

Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours.  If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

 

当单独工作于其自身的稳定工作率时,某机器密封k个纸箱需8小时;同样当单独工作于自身的稳定的工作频率时,另一机器密封k个纸箱需4个小时。如果两台机器(每个都工作在与上述相同的工作率)一起密封一定数量的纸箱时,那么工作率较高的那台机器密封了百分之多少的纸箱?

 

这个题目其实很容易理解,题目的第一句话描述了两个机器其中一个的工作率为另一个的两倍

所以一起工作的话 必然其中快的那个密封的纸箱书为另一个的两倍 所以占总量的66%


作者: wumeier    时间: 2009-6-15 06:43

第一题:

十个小孩儿跑400米所需时间分别是以上十个数,这十个时间的标准方差是22.4,精确到小数点后第一位。这十个数中,有几个数比这十个数的算术平均数小一个以上(more than one)标准方差(22.4)?

答案:两个,分别是70,75。求出算数平均数为100,then100-22.4=77.6,因为是more than one standard deviation,所以结果肯定该比77.6小,那就是前两个喽。

第二题:

(1+X%)x1000x(1+Y%)=一九九五年的产量=(1+X%+Y%+XY%%)x1000

                                      =1000+10x+10y+0.1xy

                                      =1000+92

                                      =1092






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