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.

Camitava, thank you for explanation. I got it.

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!

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?

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.