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FormosaMBA 傷心咖啡店 • 查看主题 - GWD 12-4 13-7 13-23

GWD 12-4 13-7 13-23

關於 Problem Solving 和 Data Sufficiency 的問題都可以在這邊發表

版主: shpassion, Traver0818

帖子yaokk » 2005-11-13 01:52

吳志道 \$m[1]:再補充一下 因為16X要為整數 當0<x<1 所以X最小為1/16...再以此類推
==>x=1/16 十分位為0
[tab]之後x=1/8...1/4 十分位都不為0 所以不充分


不知是否解決了M大的困惑 ^Q^


(2)8x為整數
還有一個x=1/2哦~~別忘了

不過不影響答案就是了
報告完畢
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Re: 問幾題數學題....GWD 12和13回裡的

帖子EmmaCYC » 2006-06-09 09:55

asicschu \$m[1]:
homaru \$m[1]:GWD 13-7
For a certain race, 3 teams were allowed to enter 3 members each. A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5. There were no ties, disqualifications, or withdrawals. If no team earned more than 6 points, what is the least possible score a team could have earned?

A. 0
B. 1
C. 2
D. 3
E. 4

3個隊伍,每一隊3個人,每一名得6-n分,n是名次
第一名得6-1=5分,第二名4分,第三名3分,第四名2分,第五名1分,沒有平手棄權不合資格...
如果沒有隊伍得分超過6分
(5+1)(4+2)(3)-->每對最低得分為3
最低得分2 or 1都不可能,因為就會有隊伍得分超過6分


這題OG裡面有吧... :P


請問市第10版OG嗎 我的是第11版
印象中沒看過...

對不起...我還是搞不清楚
是說只有3個隊伍比嗎,如果這樣不是只有前3名而已
可以再詳細解釋ㄧ下
"(5+1)(4+2)(3)-->每對最低得分為3 [color=blue]這是什麼意思

最低得分2 or 1都不可能,因為就會有隊伍得分超過6分"[/color]

謝謝各位 真的搞不太清楚...
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帖子scujean » 2006-06-09 16:53

見圖.......
感謝BULL的熱心幫忙~
--------------------------
MS Accounting 2007 spring at UT Dallas
----------------------------
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帖子yellowten » 2006-08-21 16:23

scujean \$m[1]:見圖.......

還是不太懂耶T T
一隊有三個人 所以是和其他隊伍的人比賽囉?
是說每一個隊伍的"累積得分"不可以超過六分嗎?
好怪喔><

那最低得分三分是說 tem裡面的每一個人都拿最後一名 所以 1+1+1=3降嗎?可是為什麼可以有這麼多的最後一名呢??

看不太懂 5+1 4+2 3+3 那邊的意思

拜託可以再講解一次嗎?
:sad
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帖子dorisdi » 2006-09-04 15:17

頂一下,我和樓上的大大不懂的地方一樣,有人可以說明嗎?感恩
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Re: 問幾題數學題....GWD 12和13回裡的

帖子skull » 2006-09-25 00:35

homaru \$m[1]:GWD 13-7
For a certain race, 3 teams were allowed to enter 3 members each. A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5. There were no ties, disqualifications, or withdrawals. If no team earned more than 6 points, what is the least possible score a team could have earned?

A. 0
B. 1
C. 2
D. 3
E. 4

3個隊伍,每一隊3個人,每一名得6-n分,n是名次
第一名得6-1=5分,第二名4分,第三名3分,第四名2分,第五名1分,沒有平手棄權不合資格...
如果沒有隊伍得分超過6分
(5+1)(4+2)(3)-->每對最低得分為3
最低得分2 or 1都不可能,因為就會有隊伍得分超過6分



看了樓樓上的圖,我想了解了,謝謝~~
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帖子sugarcane » 2006-12-26 14:13

13 Q7還是不太懂
因為看不到圖了
請幫忙

謝謝
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Re: 問幾題數學題....GWD 12和13回裡的

帖子vera1010 » 2007-08-16 15:11

homaru \$m[1]:GWD 13-7
For a certain race, 3 teams were allowed to enter 3 members each. A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5. There were no ties, disqualifications, or withdrawals. If no team earned more than 6 points, what is the least possible score a team could have earned?

A. 0
B. 1
C. 2
D. 3
E. 4

3個隊伍,每一隊3個人,每一名得6-n分,n是名次
第一名得6-1=5分,第二名4分,第三名3分,第四名2分,第五名1分,沒有平手棄權不合資格...
如果沒有隊伍得分超過6分
(5+1)(4+2)(3)-->每對最低得分為3
最低得分2 or 1都不可能,因為就會有隊伍得分超過6分



3team 都不能超過6分,所以把最高分先抓出來:
the first team: 5(第一名)+1(第五名)=6 score
the second team:4(第二名)+2(第四名)=6
the third team:3(第三名)
第一名~第五名中,除了第三名以外,全都被前兩個 team 包辦了,所以the third team=3 score only

我的理解對嗎?
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帖子chris8888 » 2007-11-15 12:01

In the decimal representation of x, where 0 < x < 1, is the tenths digit of x nonzero?
(1) 16x is an integer.
(2) 8x is an integer.
同協, 題目都貼錯, 讓後人一頭霧水, 花掉很多時間在了解題目的文法好像有錯, 不好喔
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帖子chris8888 » 2007-11-15 17:24

GWD 13-7
For a certain race, 3 teams were allowed to enter 3 members each. A team earned 6 – n points whenever one of its members finished in nth place, where 1 ≤ n ≤ 5. There were no ties, disqualifications, or withdrawals. If no team earned more than 6 points, what is the least possible score a team could have earned?

這一題我做兩次了. 我發現一個問題, 大家都認為是3, 我認為是4

因為There were no ties <== 沒有平手的
(3,0,0)(5,1,0)(4,2,0) <== 雖然最低分數可以是3, 但是會出現平手的局面Team1=3分, Team2=6分,Team3=6分, 違反題意

所以應該是
(3,2,0)(5,1,0)(4,0,0) 分數比為5:6:4, 最低分應該是4分, 答案為E

大家有何看法?
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