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Working simultaneously at their respective constant rates, Machines A
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02 Nov 2021, 07:49
02 Nov 2021, 07:49
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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) \(\frac{x}{x+y}\)
(B) \(\frac{y}{x+y}\)
(C) \(\frac{xy}{x+y}\)
(D) \(\frac{xy}{x-y}\)
(E) \(\frac{xy}{y-x}\)
(A) \(\frac{x}{x+y}\)
(B) \(\frac{y}{x+y}\)
(C) \(\frac{xy}{x+y}\)
(D) \(\frac{xy}{x-y}\)
(E) \(\frac{xy}{y-x}\)
More problems on work/rate https://gre.myprepclub.com/forum/multiple- ... %5B%5D=356
And our GRE Work/Rate Problems - Tips and Tricks you must know with examples
And our GRE Work/Rate Problems - Tips and Tricks you must know with examples
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Re: Working simultaneously at their respective constant rates, Machines A
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02 Nov 2021, 08:39
02 Nov 2021, 08:39
3
Carcass 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) \(\frac{x}{x+y}\)
(B) \(\frac{y}{x+y}\)
(C) \(\frac{xy}{x+y}\)
(D) \(\frac{xy}{x-y}\)
(E) \(\frac{xy}{y-x}\)
(A) \(\frac{x}{x+y}\)
(B) \(\frac{y}{x+y}\)
(C) \(\frac{xy}{x+y}\)
(D) \(\frac{xy}{x-y}\)
(E) \(\frac{xy}{y-x}\)
Machines A and B produce 800 nails in x hours.
Rate = ouput/time
So, their COMBINED rate = 800/x nails per hour
Working alone at its constant rate, Machine A produces 800 nails in y hours
Rate = ouput/time
So, Machine A's rate = 800/y nails per hour
We know that: (Machine A's rate) + (Machine B's rate) = (their combined rate)
Substitute to get: 800/y + (Machine B's rate) = 800/x
Rewrite as follows: Machine B's rate = 800/x - 800/y
Rewrite with common denominators: Machine B's rate = 800y/xy - 800x/yx
Combine: Machine B's rate = (800y - 800x)/xy
How many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?
Time = output/rate
= 800/((800y - 800x)/xy)
= (800)(xy/800y - 800x)
= 800xy/(800y - 800x)
= xy/(y-x)
Answer: E
General Discussion
Re: Working simultaneously at their respective constant rates, Machines A
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03 Nov 2021, 08:53
03 Nov 2021, 08:53
1
Rate per hour of both A and B together = \(\frac{800}{x}\)
Rate per hour of both A alone = \(\frac{800}{y}\)
We can deduce that,
Rate per hour of both A and B together = Rate per hour of both A alone + Rate per hour of both B alone
Let B take \(z\) hours to produce 800 nails, then rate per hour of B alone = \(\frac{800}{z}\)
We have to find \(z\) in terms of \(x\) and \(y\)
Using the deduction above,
\(\frac{800}{x} = \frac{800}{y} + \frac{800}{z}\)
\(\frac{1}{x} = \frac{1}{y} + \frac{1}{z}\)
\(\frac{1}{x} - \frac{1}{y} = \frac{1}{z}\)
\(\frac{y-x}{xy} = \frac{1}{z}\)
\(z = \frac{xy}{y-x}\)
Hence, the answer is E
Rate per hour of both A alone = \(\frac{800}{y}\)
We can deduce that,
Rate per hour of both A and B together = Rate per hour of both A alone + Rate per hour of both B alone
Let B take \(z\) hours to produce 800 nails, then rate per hour of B alone = \(\frac{800}{z}\)
We have to find \(z\) in terms of \(x\) and \(y\)
Using the deduction above,
\(\frac{800}{x} = \frac{800}{y} + \frac{800}{z}\)
\(\frac{1}{x} = \frac{1}{y} + \frac{1}{z}\)
\(\frac{1}{x} - \frac{1}{y} = \frac{1}{z}\)
\(\frac{y-x}{xy} = \frac{1}{z}\)
\(z = \frac{xy}{y-x}\)
Hence, the answer is E
