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标题: [求助]GWD-1-26 涛涛 [打印本页]

作者: logowuyi 时间: 2010-9-30 06:52 标题: [求助]GWD-1-26 涛涛

Q26:

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y.  At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A.    4

B.    6

C.    8

D.    10

E.    12

我想到個比較煩但易明的方法,不知大家怎看?

先找c看,2wx=8,wx=4>wy=2, 不行,因x+y也要3days.

d:wx=5>wy=3,同樣不行,因x+y也要3days

e:wx=6>wy=4,在這題唯一可用答案.

作者: jingjane222 时间: 2010-9-30 22:11

设machine x和y的生产速度分别为x、y个widget/day,

根据题干有

w/x-w/y=2

3x+3y=5/4w

求2w/x

令m=w/x,n=w/y

转化为

m-n=2

3/m+3/n=5/4

将m=n+2代入3/m+3/n=5/4经化简得5n^2-14n-24=0,(n-4)(5n+6)=0,得n=4或-6(舍去）

所以m=n+2=6,2m=12

选E

作者: logowuyi 时间: 2010-10-1 09:30

謝謝

作者: amorer 时间: 2010-10-4 21:53

直接解是最准确的方法，但是很费时间，而且会因为粗心出现错误。
最好用代入法：
由题目知道，x,y一起用3/（5/4）=2.4天
x和y一起用 w/ (wx+wy)= xy/(x+y)
然后由B：2x=8得出x=4,y=2, 代入 xy/(x+y) 不等于2.4
C: 2x=10 得出x=5,y=3,代入 xy/(x+y) 不等于2.4
D: 2x=12; x=6,y=4, 代入xy/x+y 等于 24/10=2.4 正确答案

作者: windows10 时间: 2010-10-11 06:47

算起来更简单一点，它是假设x机器完成w件的时间是T

Let T total number of days taken by machine X to produce W widgets.Then Work done by x is 1 day is W/TWork done by machine y in 1 day = W/(T - 2).
Combined work done in 3 days = 3( W/T + W/(t-2) )= 3W( 2T-1 ) / (T- 2) which is equal to 5W/4So equating the two equations we get12( 2T - 1 ) = 5T( T - 2)24T - 12 = 5T2 - 10T5T2 - 32T + 12 = 0(T-6) (5T+2) = 0.
Which gives T = 6.So no. of days to produce 2W widgets = 2T = 12.Ans: E

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