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Working simultaneously at their respective constant rates, M

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

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New post 17 Oct 2018, 18:27
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, 10:55
1
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, 19:03
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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Re: Working simultaneously at their respective constant rates, M  [#permalink]

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New post 17 Apr 2020, 03:25
Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)



Machine, R, T, T
A, 800/y, y, 800
B, 800/B, B, 800
A+B, 800/x, x, 800

Method 1: Using Time


Time = T1*T2/(T1+T2)
x = yB/(y+B) => xy + xB = yB => xy = yb - xB => y = xy/(y-x)

Method 2: Using Rate


800/B = 800/x - 800/y
1/B = 1/x - 1/y
1/B = (y - x)/xy
B = xy/(y-x)
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Re: Working simultaneously at their respective constant rates, M  [#permalink]

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New post 07 Jul 2020, 04:01
Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


Clearly, x < y and since A and B together produces 800/x nails in 1 hr, (x+y) can never be an option. A,B,C goes off. and since x< y and time > 0, E is the option.
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Re: Working simultaneously at their respective constant rates, M   [#permalink] 07 Jul 2020, 04:01

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