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Working simultaneously at their respective constant rates, machine A a
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02 Jun 2016, 15:32
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74% (02:30) correct 26% (02:25) wrong based on 212 sessions
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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/2a2c (E)ac/2c2a
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Re: Working simultaneously at their respective constant rates, machine A a
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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(ac)/ac = 20/(ac/(ac)), which basically means that b produces 20 widgets in ac/ac 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(ac).
Hence the answer is D.



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Re: Working simultaneously at their respective constant rates, machine A a
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03 Aug 2017, 06:59
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 plugin 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/2a2c => 2/2, yes! (E)ac/2c2a => 2/2, no
Answer (D)



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Re: Working simultaneously at their respective constant rates, machine A a
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09 Aug 2017, 12:47
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/2a2c (E)ac/2c2a 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
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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(ac)/ac = 20/(ac/(ac)), which basically means that b produces 20 widgets in ac/ac 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(ac).
Hence the answer is D. How do you get "20/(ac/(ac))"? I got to 20(ac)/ac, then I don't understand the rest.. Thank you!



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Re: Working simultaneously at their respective constant rates, machine A a
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05 Jan 2019, 09:58
pclawong wrote: 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(ac)/ac = 20/(ac/(ac)), which basically means that b produces 20 widgets in ac/ac 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(ac).
Hence the answer is D. How do you get "20/(ac/(ac))"? I got to 20(ac)/ac, then I don't understand the rest.. Thank you! understand simple: a+b=20/c in 1 hour a produce 20/a in 1 hour so a+b=20/c 20/a+b=20/c b=20/c20/a b=(20a20c)/ac b=[20(ac)]/ac now : B [20(ac)]/ac widget in=1 hr so b will produce 10 widget in=10/[20(ac)]/ac=ac/2(ac).



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Working simultaneously at their respective constant rates, machine A a
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06 Jan 2019, 19:12
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/2a2c (E)ac/2c2a (A+B)"s 1 hr work = 1/c hrs.
A's 1 hr work = 1/a
B's 1 hr work = 1/c  1/a = ac/ca.
Total time needed for B to finish this work = ca/ca.
ca/ac is basically the time for 20 widgets. Now we have to know time for 10 widgets.
ca / ca *1/2 = ca / 2a  2c.
D is the correct answer.




Working simultaneously at their respective constant rates, machine A a
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