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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
Machine A produces pencils at a constant rate of 9000 pencils per hour
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22 Dec 2022, 07:00
22 Dec 2022, 07:00
Expert Reply
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Question Stats:
66% (01:41) correct
33% (02:19) wrong based on 6 sessions
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Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?
A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4
A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
Re: Machine A produces pencils at a constant rate of 9000 pencils per hour
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24 Dec 2022, 02:15
24 Dec 2022, 02:15
1
Expert Reply
To minimize the time that machine B must operate we must maximize the time machine A can operate, so make it operate 8 hours. In 8 hours machine A will produce \(8*9,000 = 72,000\) pencils, so \(100,000 - 72,000 = 28,000\) pencils are left to produce, which can be produced by machine B in \(\frac{28,000}{7,000}=4\) hours.
Answer: A
Answer: A
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Machine A produces pencils at a constant rate of 9000 pencils per hour
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02 Jan 2023, 13:03
02 Jan 2023, 13:03
1
GeminiHeat wrote:
Machine A produces pencils at a constant rate of 9000 pencils per hour, and machine B produces pencils at a constant rate of 7000 pencils per hour. If the two machines together must produce 100,000 pencils and if each machine can operate for at most 8 hours, what is the least amount of time, in hours, that machine B must operate?
A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4
A. 4
B. 4 2/3
C. 5 1/3
D. 6
E. 6 1/4
To MINIMIZE machine B's operating time, we must MAXIMIZE the time machine A's operating time.
So, let machine A operate for the full 8 hours.
In 8 hours, machine A produces 72,000 pencils [8 hours x 9,000 pencils/hour =72,000 pencils]
So, the number of pencils machine B must make = 100,000 - 72,000 = 28,000
Time = output/rate
So, operating time = 28,000/7000 = 4 hours
Answer: A
Cheers,
Brent
