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Quant Rev. #173 PS

by tonebeeze » Wed Jun 15, 2011 8:46 pm
Does someone have an efficient method of solving this problem?

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 5/4w 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

OA = e
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by Frankenstein » Wed Jun 15, 2011 9:32 pm
Hi,
Let Y take 'n' days to produce 'w' widgets. So, it produces (w/n) widgets per day
So X takes (n+2) days to produce 'w' widgets. So, it produces (w/n+2) widgets per day
Together they produce (w/n) + (w/n+2) in a day
In 3 days, [w/n + w/(n+2)]*3 = 5w/4 => 5n^2 - 14n -24 = 0
So, (5n+6)(n-4) = 0.
n=4
So, X produces w/(4+2)=w/6 widgets in a day.
To produce 2w widgets it takes (2w)/(w/6) = 12 days

Hence, E
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by Mom4MBA » Wed Jun 15, 2011 9:36 pm
Let Y take 'a' days to complete 'w' widgets
then X will take 'a+2' days to complete 'w' widgets

therefore the rate for Y will be w/a widgets per day
and the rate for X will be w/(a+2) widgets per day

together their rate of completing 'w' widgets will be w/a + w/(a+2) = (2a+2)w/(a(a+2))...........(i)

given: they take 3 days to complete (5/4)w widgets
so the rate together will be 5w/(4x3) widgets per day .................(ii)

on equating (i) and (ii)

(2a+2)w/(a(a+2)) = 5w/(4x3)

on solving we get
5a² - 14 a - 24 = 0

a= -6/5 not possible ; and a=4

hence X will make 'w' widgets in (a+2)=6 days

so to make '2w' widgets X will take 12 days
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