8.17换题后JJ分类汇总（有补充整理）

prllrpnt

1.        X 被5除余3，被7除余4，y也是被5除余3，被7除余4，问一下哪个是x-y的factor

我凑数做出来35，不晓得对不对

 对的，根据同余的性质，两数同余，则x-y得出来的数要即能被5除也能被7整除

duckzxcv

以下是引用emmadon在2007-8-26 18:25:00的发言：

79.   某人有一本１到１００页的书，她往第一页放一个红色的ＳＴＩＣＫ，然后每隔６页再放一个，然后在第１页又放一个蓝色的ＳＴＩＣＫ，然后每隔８页又放一个，问一共有多少页既没有红色ＳＴＩＣＫ也没有蓝色的。

我觉得这题不是问“同时没有红和蓝”，而是问“没有红”+“没有蓝”，就是说排除了3种情况，1、有红，2、有蓝，3、有红和蓝。所以是100-17-13=70

open to discussion..

你應該是沒扣掉重複的部份

第1頁重複先算1 +6page 99/6=16... +8page 99/8=12....-重複99/24=4...

100-1-16-12+4=75

to be discuss

iflyagain

LZ SC补充上下面这题。GWD答案选A,讨论主潮流选E

GWD 2-33.

Q33:

When an active tooth in the shark’s jaws is lost or worn down, many spare teeth lie in seemingly limitless reserve, each of which are ready to slide into the appropriate position.

 

1. When an active tooth in the shark’s jaws is lost or worn down, many spare teeth lie in seemingly limitless reserve, each of which are ready to slide into the appropriate position.
           
2. Whenever an active tooth is lost or worn down, many spare teeth lie in seemingly limitless reserve in the shark’s jaws, which are each ready to slide into the appropriate position.
           
3. Many spare teeth lie in seemingly limitless reserve in the shark’s jaws, each one of which are ready to slide into the appropriate position when an active tooth is lost or worn down.
           
4. The many spare teeth lying in seemingly limitless reserve in the shark’s jaws, each one of which is ready to slide into the appropriate position whenever an active tooth is lost or worn down.
           
5. In the shark’s jaws, many spare teeth lie in seemingly limitless reserve, each one ready to slide into the appropriate position whenever an active tooth is lost or worn down.

prllrpnt

1.        一对夫妻准备生4个孩子，问生出来2男2女几率是多少？

答案说3/8

C(4,2)*(1/2)^4=3/8

想问问大家，这个思路是怎么样的啊为什么要乘 C(4,2)， 谢谢！

sdydean

感谢楼主

prllrpnt

1.        一些学生放假，75%放了寒假，34%（这个数字不准）又放寒假又放暑假，20%没放假，问只放了暑假没放寒假的有百分之几？
                

0.39

1- 0.2-（0.75-0.34）=0.39 

 

我感觉这道题的答案应该是0.05，我是这样算的：设X为放了暑假的
        则1=0.75+X-0.34+0.20 则X为0.39，然后减去0.34 为0.05

waninayang

MM实在是太好了~！！！o(∩_∩)o...

还特意给偶发短信说更新了~

好感动``````````````````

prllrpnt

 

1.        从n个东西取出三个(0<n<=60)，问n(n+1)(n+2)可以被6整除的概率。

这道题的概率是不是1啊，根据吴强GMAT的数学的第38页的性质6.任何三个连续整数中，恰好有一个数是3的倍数，并且 这三个连续整数之积能够被6整除。

illumination

thanks

kevin_lyx

1.        六个不同的公司,每个公司派3人,每2个不同的人握手(自己公司的人不用握手),最后求握手的次数. 135次

怎么算出来的?算了几次都不是这个结果.