[求助] GMATPrep上的题

1 A certain city with population of132,000 is to be devided into 11 voting districts, and no district is to have population that is more than 10 percent greater than the population of any other district. What is the minimum possible population that the least population district could have?

A 10700    B 10800    C 10900     D 11000     E 11100    (answer: D)

主要是红色的这句话没看懂

2 A certain libary assesss fines for overdue books as follows. On the first day that a book is overdue, the total fine is $0.10. For each additional day that the book is overdue the total fine is etiher increased by $0.30 ro double. whichever results in the lesser amount. What is the total fine for a book on the fourth day it is overdue?

A $0.60          B $0.70        C $0.80            D $0.90         E $1.00 (answer: B)

我算得是E，不知道为什么是B，B应该表示的是第三天的吧？

3 If xy +z =x(y+z), which of the following must be true?

A x=0 and z=0

B x=1 and y=1

C y=1 and z=0

D x=1 or y=0

E x=1 or z=0   (answer: E)  这题实在搞不清楚，觉得好几个都可以满足，比如A，明显两边都等于零了。

4 On a certain tour bus, the ratio of the number of women and the number of childern is 5:2, what is the number of men?

(1)  Children: men = 5:11

(2)  women less than 30   (answer: C)

这几个题好像在汇总稿里都没看到，谢谢指教！！

1.红色部分是说任何一个区都不超过其他区10％。也就是最大和最小的范围差也就在10％之间

设最小的为X,令其他10个都取最大，为1.1X

则X+10*1.1X=132000

解得X=11000

2.For each additional day that the book is overdue the total fine is etiher increased by $0.30 ro double. 黄色部分是不是抄错了？不懂什么意思

3.

xy+z=x(y+z)

xy+z=xy+xz

z=xz

z(x-1)=0

所以x=1或z=0

E

4.题干给出women:children=5:2

(1)children:men=5:11,所以women:men=25:22

(2)women<30

综合上面，women只能取25,因为取其他数，men都不是正整数。所以men=22

我猜测了一下，那句话应该是either increased by $0.30 or double.那么第一天是0.1，第二、第三天显然是double比较合算，到第四天是+$0.3合算，也就是这样的一个罚款变化，0.1->0.2->0.4->0.7

支持这种看法

还是不明白为什么最大就是1.1X?

no district is to have population that is more than 10 percent greater than the population of any other district.

取1.2X就不符合条件了，最多取到1.1X

原来如此。...

我当从头到尾只能用一种方法的呢

以下是引用catcenter在2006-5-24 15:36:00的发言：

原来如此。...

我当从头到尾只能用一种方法的呢

我作的时候也这么以为～～