[求助]GWD数学！

aruora · 发表于 2009-8-16 22:17:00

数据充分题（选项省略）

In triangle ABC, point X is the midpoint of side AC and point Y is the midpoint of side BC. If point R is the midpoint of line segment XC and if point S is the midpoint of line segment YC. What is the area of triangular region RCS? （图麻烦各位画一下）

(1) The area of triangular region ABX is 32.
(2) The length of one of the altitudes of triangle ABC is 8.

我选的是E，也就是两个加一起也推不出来。而正确答案是A，认为第一个自己就充分。

我实在不明白给ABX的面积有啥用？！

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On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship's passengers held round-trip tickets?

A. 33又1/3 %
B. 40%
C. 50%
D. 60%
E. 66又2/3%

对这种整体到部分的比例题，我做一道错一道。。拜托各位给点技巧，好吗？

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If M is the least common multiple of 90, 196, and 300. Which of the following is NOT a factor of M?

A. 600
B. 700
C. 900
D. 2,100
E. 4,900

这道题的思路应该怎么想？

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数据充分题（选项省略）

If x and y are positive integers and r is the remainder when 3^(4x + 2) + y is divided by 10, what is the value of r?
（注： 3^(4x + 2)是指3的（4x + 2）次方）

(1) x = 25
(2) y = 1

我选的是C，正确答案是B。
这道题的思路是怎样的？

静夜 · 发表于 2009-8-16 23:43:00

1/给出ABX的面积,就可以知道AB*高,也就是XY到AB的距离,就是整个三角形ABC的高的一半.ABC的面积=64*2h*1/2=64

RCS=1/4CXY=1/16ABC=4

2/这种题我也不太会

3/90=2*3*3*5 ; 196=2*2*7*7 ; 300=2*2*3*5*5

least common multiple= 2*2*7*7*3*3*5*5

600=2*2*2*3*5*5

700=2*5*5*2*7

900=2*5*2*4*3*3

2100=3*7*2*2*5*5

4900=7*7*5*5*2*2

只有A多了一个2,选A

4/3的N次方,末尾数都是1,3,9,7循环,且当次幂为偶数时,为1,9;为积数时,为3,7

(4X+2)必为偶数,且末尾数字必为9(你找一下规律就知道了)

那么只要知道Y的值就可以了,所以选B

aruora · 发表于 2009-8-17 13:41:00

太谢谢啦！！~~

chenhaoming · 发表于 2009-8-18 00:12:00

问下这题答案是多少啊

On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship's passengers held round-trip tickets?

A. 33又1/3 %
B. 40%
C. 50%
D. 60%
E. 66又2/3%

bizgirl · 发表于 2009-8-18 01:46:00

I think it's 50%. 20%/(1-0.6)=50%

since round-trip no cars is 60%, so round-trip with cars are 40%. 0.2/0.4=0.5

chenhaoming · 发表于 2009-8-18 10:25:00

哦，它说有往返票的人中，百分60没车的，那百分之40就是有车的，那不就等于前面全部游客的百分之20么，两个假设前面是a，后面全体乘客为B 是不是就是 2a=b，就是百分之50，不知道我这么算对不对，高手指教

leosnoon · 发表于 2010-6-3 21:49:13

问下这题答案是多少啊On a certain transatlantic crossing, 20 percent of a ship's passengers held round-trip tickets and also took their cars abroad the ship. If 60 percent of the passengers with round-trip tickets did not take their cars abroad the ship, what percent of the ship's passengers held round-trip tickets?

A. 33又1/3 %
B. 40%
C. 50%
D. 60%
E. 66又2/3%

-- by 会员 chenhaoming (2009/8/18 0:12:00)

设S为ship's passengers 则有： 20% S = R —> C 40% R-> C
R为 Round trip 相等量是 C
C为 have cars 即： 20%S = 40%R R/S = 50%
此类题关键是找出相等的量，写出等式求解

rexxu · 发表于 2010-7-18 13:45:14

谢谢上面的大侠

[求助]GWD数学！

[求助]GWD数学！

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