gocanucks

48．ds 以速度p1完成路程s1所需时间比以速度p2完成路程s2所需时间长？
(1)p1比p2大30
(2)s1比s2大30
[确认]：C
[思路]：以速度p1完成路程s1所需时间:S1/P1=(S2+30)/(P2+30); 以速度p2完成路程s2所需时间:S2/P2
真分数分子分母同加上相等的数，分数的值越来越大。假分数相反。所以，可以判断。

How do you know S2/P2 is 真分数?
s2=10, p2= 30 => s2/p2 < s1/p1
s2=60, p2=30 => s2/p2 > s1/p1
should be E instead.

gocanucks

Question about #47.

47．用C.I.R.L.E 五个字母组成六个字母的词，C和C之间必须隔开，多少种？
[确认]：5P(6,6)-P(5,5)
[思路]：先不考虑C和C相邻的情况，按照六个位置进行全排列为5P(6,6)
再减去两个C相邻的情况，在这种情况下，把两个C看成一个元素，有P(5,5)中情况，所以答案是5P(6,6)-P(5,5)

I understand the part of P(5,5), but what about 5P(6,6)? I know 5 is for picking the extra one out of 5 characters. But is P(6,6) right? Since there are two same characters, we might be counting a same permutation twice.

Or look at it this way, P(6,6) means we have 6 choices for position 1, 5 for pos 2, etc. But since we have two same characters, we really have only 5 choices for position 1.

oddyuibe

Hi,
the numbers for AWA part are based on a book about AWA written by Sun Yuan, a teacher in New Oriental School. The book has a green cover (the new version has a red cover).

I don't know whether you can get the AWA topic numbers from that book on the net. you can check the download center of Topway.com.

Good luck~

1036

Hi, oddyuibe,
Thank you very much. YOu make it even more clear. BTW, it looks like you are in US, right? I am in San Jose. Pls contact me at : 408-243-2865. JUst want to say hi to you.

oddyuibe

to 1036

well, i think we can tackle this problem in this way:

from S1, we should subtract those who subscribe S or T, thus S1-B1-B3, however in this process, one more "a" is subtracted, so we should add "a" afterwards: S1-B1-B3+a

similarly, from S2, we subtract those who subscribe T or R, and add the "a" which we unnecessarily subtract, thus S2-B1-B2+a

accordingly, the third one is: S3-B2-B3+a

so, the answer is S1+S2+S3-2B1-2B2-2B3+3a

If you draw venn diagram, the relationship will be very clear.

I wish this is helpful~

1036

SOrry, I think I am wrong. I notice
"Both R and S" is different from " only subscribe R and S".

math_jj_4-28-5-04   13 th T

1036

IS the answeer wrong ?
math JJ 4-28 to 5-04 the 13 th T: ( 45 people have newspaper)
13. 一社区有45户, 订报纸, 有 R, S, T 三种报,给出了三种报的订户人数分别是, S1=28, S2=17, S3=12,then, subscribe, both R&S, b1=9, both S&T b2=8, both R&T b3=7; Now 给出了一个重要条件, 订了三种的和什么都没有订的人,一样. 然后说, at least subscribe one kind of newspaper,29.
问: Subscribe only one kind Newspaper 的人数 X=?
[讨论]：X=S1+S2+S3-2B1-2B2-2B3+3a, 其中a是三种报纸都订了的户数 a=45-29=16

Since this problem asks : Subscribe only one kind Newspaper 的人数 X=?
I think X= S1+S2+S3-2B1-2B2-2B3-3a,( substract 3a) 其中a是三种报纸都订了的户数 a=45-29=16  Pls help me to confirm!

bluesky

oddyuibe

嗯，Mars_gmat 说的对，

相应题目已经改正~

Mars_gmat

17 .38 的答案好象错了吧?
17.有6个组,每组出三个代表, 分别与其他组的代表握手,但自己人之间不握手. 问一共有多少次握手?
[确认]：(15+12+9+6+3)*3=135
38.一个人持有25分硬币的个数（coins）是10分硬币个数的2倍。持有10分硬币的个数（coins）是5分硬币个数的2倍。问 此人所持25分硬币的金额（value）是5分硬币金额的多少倍。
[确认]：4*5=20倍

oddyuibe

to fen797

谢谢你的补充，没关系的，先好好休息吧，考试很辛苦的~

祝今后申请顺利~