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Junior | Next Rank: 30 Posts
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by muna_m » Mon Aug 03, 2009 5:17 am
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?

Can somebody plz post the explanation for this one... Appreciate it!

OA 12
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by truplayer256 » Mon Aug 03, 2009 6:07 am
Check this thread out:

https://www.beatthegmat.com/widgets-t40224.html#166925
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by m&m » Mon Aug 03, 2009 6:22 am
I typically solve this algebraically, but it may not be the fastest way

let d be the num days machine Y takes to build w widgets

Machine X
days= d+2
widgets= w
rate = w/(d+2)

Machine Y
days= d
widgets= w
rate = w/(d)

Machine X & Y together
days= 3
widgets= 5/4w
rate = (5/4w)/3 = 5/12w

so rate of X & Y = w/(d+2) + w/d = 5/12w
1/(d+2) + 1/d = 5/12
(2d+2)/(d^2+2d) = 5/12
24d+24=5d^2+10d
0=5d^2-14d-24
0=(5d+6)(d-4)
d=-ve or 4

so X take d+2=4+2=6 to make w widgets, so to make 2w widgets it takes 12 days

hope that helps


solving for d
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by shahdevine » Mon Aug 03, 2009 7:34 am
x=y+2

w/x+w/y=5/4 * 1/3 days

rate then is:

w/(y+2)+w/y=5/12

solve for y. the w's will cancel out.

0=5y^2-14y-24
0=(5y+6)(y-4)

y has to equal 4.

x=4+2=6

w/6=2w/x
x=12

cheers.
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by muna_m » Mon Aug 03, 2009 8:04 am
Thank u
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