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. 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

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.