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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Six machines at a certain factory operate at the same constant rate. I
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29 Nov 2022, 08:19
29 Nov 2022, 08:19
1
Expert Reply
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Question Stats:
53% (01:41) correct
46% (02:30) wrong based on 15 sessions
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Six machines at a certain factory operate at the same constant rate. If four of these machines, operating simultaneously, take 27 hours to fill a certain production order, how many fewer hours does it take all six machines, operating simultaneously, to fill the same production order?
A: 9
B: 12
C: 16
D: 18
E: 24
A: 9
B: 12
C: 16
D: 18
E: 24
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Six machines at a certain factory operate at the same constant rate. I
[#permalink]
30 Nov 2022, 22:41
30 Nov 2022, 22:41
1
Expert Reply
Suppose each machine finishes a job in x hours. This would mean that each machine has a rate of \(\frac{1}{x}\) jobs/hour. Using the equation R*T=Work, we can figure out the rest.
We know that four machines combined would work four times as fast as just one machine working alone. Thus, the combined rate of 4 machines is \(\frac{4}{x}\). Furthmore the time it takes to complete the order is 27 hours. Now we can put the complete order in terms of x:
\(R * T = Work\)
\(\frac{4}{x} * 27 = Work\)
\(Work = \frac{4*27}{x}\)
Note that I didn't do out the actual multiplication.
Next, let's look at 6 machines. Working together, they have a rate of \(\frac{6}{x}\). From before, we know the order size is \(\frac{4*27}{x}\). We need to figure out the new time.
\(\frac{6}{x} * T = \frac{4*27}{x}\)
\(T = \frac{4}{6}*27 = \frac{2}{3}*27 = 18\) (**Note that the x's cancel)
Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Answer: A
We know that four machines combined would work four times as fast as just one machine working alone. Thus, the combined rate of 4 machines is \(\frac{4}{x}\). Furthmore the time it takes to complete the order is 27 hours. Now we can put the complete order in terms of x:
\(R * T = Work\)
\(\frac{4}{x} * 27 = Work\)
\(Work = \frac{4*27}{x}\)
Note that I didn't do out the actual multiplication.
Next, let's look at 6 machines. Working together, they have a rate of \(\frac{6}{x}\). From before, we know the order size is \(\frac{4*27}{x}\). We need to figure out the new time.
\(\frac{6}{x} * T = \frac{4*27}{x}\)
\(T = \frac{4}{6}*27 = \frac{2}{3}*27 = 18\) (**Note that the x's cancel)
Thus, the difference is 27-18 = 9 hours, which is answer choice A.
Answer: A
Re: Six machines at a certain factory operate at the same constant rate. I
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19 Dec 2022, 06:33
19 Dec 2022, 06:33
GeminiHeat Hi Thanks for the detailed solution. Can we also solve it in the following way,
Given : Rate is constant, production order is same = work constant , then simply calculating by
M1 x T1 = M2 x T2 where M1= 4 machines , T1= 27 hrs , M2= 6 machines , substituting and calculating for T2 we get it as 18 hrs and T1-T2 = 9 hrs (answer)
Given : Rate is constant, production order is same = work constant , then simply calculating by
M1 x T1 = M2 x T2 where M1= 4 machines , T1= 27 hrs , M2= 6 machines , substituting and calculating for T2 we get it as 18 hrs and T1-T2 = 9 hrs (answer)
