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Working simultaneously at their respective constant rates, M

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Senior Manager
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Working simultaneously at their respective constant rates, M  [#permalink]

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New post 17 Oct 2018, 19:27
We know that work done is 800 nails and we are asked to find that how many hours B take to produce same 800 nails.

Since work done is uniform , we take it as 1 instead of 800 to avoid heavy calculation.

Both A and B complete work in x hours.

1/A+1/B=1/x

Given A=y

So 1/B=1/x-1/y

Get B from here =(xy)/(y-x)

IMO E.
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Re: Working simultaneously at their respective constant rates, M  [#permalink]

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New post 14 Mar 2019, 11:55
As the work is same (800 nails) let's not think about it.
If they can work together in x hours then their combined rate is 1/x
The rate of A is 1/y
so rate of b is = 1/x-1/y
= (y-x)/xy
so time will be the reverse of rate = xy/(y-x)
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Working simultaneously at their respective constant rates, M  [#permalink]

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New post 06 Nov 2019, 20:03
since the work units are same i.e. 800 so we can take it as 1 unit as well.
Now work done by A + B = 1/x
work done by A alone = 1/y

\(\frac{1}{y}\) + \(\frac{1}{B}\) = \(\frac{1}{x}\)
On solving, we get B = \(\frac{xy}{(y-x)}\)

Answer - Option E
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Working simultaneously at their respective constant rates, M   [#permalink] 06 Nov 2019, 20:03

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