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Working simultaneously at their respective constant rates, machine A a

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Working simultaneously at their respective constant rates, machine A a  [#permalink]

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New post 02 Jun 2016, 15:32
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Working simultaneously at their respective constant rates, machine A and B produces 20 widgets in c hours. Working alone at its constant rate, Machine A produces 20 widgets in 'a' hours.In terms of a and c, how many hours does it take Machine B, working alone at its constant rate, to produce 10 widgets

(A)ac/a+c
(B)2ac/a+c
(C)ac/2a+2c
(D)ac/2a-2c
(E)ac/2c-2a

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Re: Working simultaneously at their respective constant rates, machine A a  [#permalink]

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New post 02 Jun 2016, 20:13
Let the individual rates be R(a) and R(b) for machines a and b respectively. Given than combined rate is 20 widgets in c hours => 20/c

R(a) + R(b) = 20/c

Also given is the individual rate for machine a = R(a) = 20/a

Therefore, R(a) + R(b) = 20/c => 20/a + R(b) = 20/c => R(b) = 20/c - 20/a = 20(a-c)/ac = 20/(ac/(a-c)), which basically means that b produces 20 widgets in ac/a-c Hrs. We have to calculate how much does it take to produce 10 widgets. So divide the rate by 2 which gives us ac/2(a-c).

Hence the answer is D.
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Re: Working simultaneously at their respective constant rates, machine A a  [#permalink]

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New post 03 Aug 2017, 06:59
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Since there is no restriction given that a and b must be different, let's assume that Machine A and Machine B have the same rates. => a=b.

Let's take some small numbers and see how that works out:

Let a = b = 2, which means both Machines A and B need 2 hours each independently to produce 20 widgets.
Combined, they'll need half the time, namely 1 hour to produce 20 widgets => c = 1

What happens if we plug-in these numbers now?
Before we do that we have to keep in mind that we are looking for the time that b needs to produce 10 widgets.
Therefore our target value = b/2 = 1.

First we see that we can immediately rule out answers A) and B) as we'd divide an even number ac or 2ac with an uneven number a+c. That would not result in an integer value.

(A)ac/a+c =>2/3, no
(B)2ac/a+c =>4/3, no
(C)ac/2a+2c => 2/6, no
(D)ac/2a-2c => 2/2, yes!
(E)ac/2c-2a => 2/-2, no

Answer (D)
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Re: Working simultaneously at their respective constant rates, machine A a  [#permalink]

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New post 09 Aug 2017, 12:47
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snorkeler wrote:
Working simultaneously at their respective constant rates, machine A and B produces 20 widgets in c hours. Working alone at its constant rate, Machine A produces 20 widgets in 'a' hours.In terms of a and c, how many hours does it take Machine B, working alone at its constant rate, to produce 10 widgets

(A)ac/a+c
(B)2ac/a+c
(C)ac/2a+2c
(D)ac/2a-2c
(E)ac/2c-2a


Since machines A and B produce 20 widgets in c hours, their combined rate = 20/c. Since machine A produces 20 widgets in a hours, its rate = 20/a. We can let the time it takes machine B to produce 20 widgets = b; thus, its rate = 20/b and we can create the following equation:

20/a + 20/b = 20/c

Multiplying by abc, we have:

20bc + 20ac = 20ab

bc + ac = ab

ac = ab - bc

ac = b(a - c)

ac/(a - c) = b

Since the time it takes machine B to produce 20 widgets is ac/(a - c), the time it takes machine B to produce 10 widgets must be half as much, or:

ac/(a - c) x 1/2 = ac/(2a - 2c)

Answer: D
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Re: Working simultaneously at their respective constant rates, machine A a  [#permalink]

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New post 09 Aug 2017, 19:23
FacelessMan wrote:
Let the individual rates be R(a) and R(b) for machines a and b respectively. Given than combined rate is 20 widgets in c hours => 20/c

R(a) + R(b) = 20/c

Also given is the individual rate for machine a = R(a) = 20/a

Therefore, R(a) + R(b) = 20/c => 20/a + R(b) = 20/c => R(b) = 20/c - 20/a = 20(a-c)/ac = 20/(ac/(a-c)), which basically means that b produces 20 widgets in ac/a-c Hrs. We have to calculate how much does it take to produce 10 widgets. So divide the rate by 2 which gives us ac/2(a-c).

Hence the answer is D.


How do you get "20/(ac/(a-c))"?
I got to 20(a-c)/ac, then I don't understand the rest..
Thank you!
Re: Working simultaneously at their respective constant rates, machine A a &nbs [#permalink] 09 Aug 2017, 19:23
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