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Machine : Rate Problem..

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Machine : Rate Problem..

by Thephu » Sun Jan 20, 2008 8:46 pm
Hi, I am always stuck in rate problem. Please help to explain how to solve this problem.. and any suggestion to solve this kind of problem is appreciated. Thank you very much.

:o


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

A : 4
B : 6
C : 8
D : 10
E : 12


OA : E
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by camitava » Sun Jan 20, 2008 9:04 pm
Thephu, for this type of work-rate problem, u can follow the following approach -

Let,
x = Days that machine X takes to complete W widgets
y = Days that machine Y takes to complete W widgets
Again x = y + 2
So
for x
------
y + 2 days to do W widgets
1 day to do W/(y + 2) widgets

for y
------
y days to do W widgets
1 day to do W/y widgets

So in combined, in 1 day they do = (2y + 2)W/(y^2 + 2y)
So they can do W widgets in = (y^2 + 2y)/(2y + 2)W
So they can do 5W/4 widgets in = 5 * (y^2 + 2y)/4 * (2y + 2) days.
So 5 * (y^2 + 2y)/4 * (2y + 2) = 3
---> y = 4
so x = 6
so to complete W widgets, X takes 6 days
so to complete 2W widgets, X takes 12 days

So IMO E.
Correct me If I am wrong


Regards,

Amitava
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by Thephu » Sun Jan 20, 2008 9:27 pm
Camitava, thank you for explanation. I got it. :)
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by cris » Wed Jan 23, 2008 8:27 pm
Camitiva, how did yoy resolve this ecuation:

"So 5 * (y^2 + 2y)/4 * (2y + 2) = 3
---> y = 4 "

I did correct until there....but I could not solve it in a quick way. I had to do a lot of calculations [-b+-squareroot(b^2-4ac)/2a] that I would definetly like to avoid in the exam....

Did you solve for "y" in a faster way?

Thanks!
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by camitava » Wed Jan 23, 2008 9:24 pm
Cris,
Why are you calculating by the formula - [-b+-squareroot(b^2-4ac)/2a] for all quadratic equ. This is basically time-consuming. Try to break the quadratic equ into the form of factors like - (x - a)(x - b) = 0 - That means either x - a = 0 or x - b = 0. Like this way, if u take the equ -
5 * (y^2 + 2y) = 12 (2y + 2)
or 5y^2 - 14y - 24 = 0
or (y - 4)(5y + 6) = 0
so y = 4 as y can not be negative. Got my point, Cris?
Correct me If I am wrong


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by cris » Thu Jan 24, 2008 12:08 pm
Is just that I did not know how to solve quadratic equ (ay^2+by+c=0) by fatoring when "a" was not 1.

...I checked Wikipedia and now I can do it (if the numbers are not too complicated).

Thanks Camitiva.
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