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Shortcut ???

by arorag » Mon Sep 01, 2008 11:45 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?
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Re: Shortcut ???

by sudhir3127 » Mon Sep 01, 2008 12:22 pm
arorag wrote: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?
IMO 12..do let me know if its right to post my solution...
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Re: Shortcut ???

by anju » Mon Sep 01, 2008 12:49 pm
IMO 12..do let me know if its right to post my solution...
My ans is 12 too
Here's my solution:
Let's Y's time be t then X's time is t+2
Rate of Y is w/t
Rate of X is w/t+2

w/t + w/t+2 = 5/12w (5/4w in 3 days so 5/12w in 1 day)
solving we get t = 4
so for x t+2 = 4+2 = 6 for 1 widget
for 2w 6*2 = 12 days

Sudhir, let us know if you have a faster solution
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by rosh26 » Tue Sep 02, 2008 1:00 pm
w/t + w/t+2 = 5/12w (5/4w in 3 days so 5/12w in 1 day)

can u please explain how u solved for t=4??? I dont see how w cancels out...thanks!
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by anju » Tue Sep 02, 2008 1:04 pm
rosh26 wrote:w/t + w/t+2 = 5/12w (5/4w in 3 days so 5/12w in 1 day)

can u please explain how u solved for t=4??? I dont see how w cancels out...thanks!
w/t + w/t+2 = 5/12w
w[1/t + 1/(t+2)] = 5/12w --- in this step w is common
1/t + 1/(t+2) =(5/12)*(w/w)
1/t + 1/(t+2) = (5/12) as w/w is 1
i hope it's clear now
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