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by Vasudha » Mon May 07, 2007 6:24 am
A works 4 times as fast as B. Together they can do all the work in 4 days. How many days would each require alone?
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by f2001290 » Mon May 07, 2007 6:32 am
Hi Vasudha

A alone can complete the work in 5 days
B alone can complete the work in 20 days.

Is this the right answer. I can explain this if u can confirm the solution.
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by Vasudha » Mon May 07, 2007 6:36 am
yes it is the right answer... how did u set up the equation...thnks....
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by f2001290 » Mon May 07, 2007 7:51 am
(a+b)'s one day work = 1/4

Assume B can complete the work in x days.
Then B's one day work = 1/x

Since A is 4 times as fast as B,
Then A's one day work = 4(1/x)

4(1/X) + (1/X) = 1/4

Solve for x... That is the answer
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by Vasudha » Mon May 07, 2007 8:03 am
Thanks.. :)
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by Cybermusings » Mon May 07, 2007 9:08 am
A works 4 times as fast as B. Together they can do all the work in 4 days. How many days would each require alone

If A works 4 times as fast as B--- he would take 1/4 of the time taken by B

So 1/x + 1/4x = 1/4 days

5/4x = 1/4
4x = 20
x = 20/4 = 5 days

A completes task in 5 days and B completes task in 4*5 = 20 days
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