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At Supersonic Corporation, the time required for a machine to complete
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21 Apr 2021, 16:03
21 Apr 2021, 16:03
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At Supersonic Corporation, the time required for a machine to complete a job is determined by the formula: \(t = \sqrt{w} + \sqrt{(w -1)}\), where w = the weight of the machine in pounds and t = the hours required to complete the job. If machine A weighs 8 pounds, and machine B weighs 7 pounds, how many hours will it take the two machines to finish one job if they work together?
A. \(\frac{6}{7-\sqrt{3}}\)
B. \(\frac{1}{2}(\sqrt{8}+\sqrt{6})\)
C. \(\frac{1}{3}(6-\sqrt{3})\)
D. \(3(\sqrt{3}+\sqrt{2})\)
E. \(\sqrt{8}+2\sqrt{7}+\sqrt{6}\)
A. \(\frac{6}{7-\sqrt{3}}\)
B. \(\frac{1}{2}(\sqrt{8}+\sqrt{6})\)
C. \(\frac{1}{3}(6-\sqrt{3})\)
D. \(3(\sqrt{3}+\sqrt{2})\)
E. \(\sqrt{8}+2\sqrt{7}+\sqrt{6}\)
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Re: At Supersonic Corporation, the time required for a machine to complete
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22 Apr 2021, 02:53
22 Apr 2021, 02:53
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GeminiHeat wrote:
At Supersonic Corporation, the time required for a machine to complete a job is determined by the formula: \(t = \sqrt{w} + \sqrt{(w -1)}\), where w = the weight of the machine in pounds and t = the hours required to complete the job. If machine A weighs 8 pounds, and machine B weighs 7 pounds, how many hours will it take the two machines to finish one job if they work together?
A. \(\frac{6}{7-\sqrt{3}}\)
B. \(\frac{1}{2}(\sqrt{8}+\sqrt{6})\)
C. \(\frac{1}{3}(6-\sqrt{3})\)
D. \(3(\sqrt{3}+\sqrt{2})\)
E. \(\sqrt{8}+2\sqrt{7}+\sqrt{6}\)
A. \(\frac{6}{7-\sqrt{3}}\)
B. \(\frac{1}{2}(\sqrt{8}+\sqrt{6})\)
C. \(\frac{1}{3}(6-\sqrt{3})\)
D. \(3(\sqrt{3}+\sqrt{2})\)
E. \(\sqrt{8}+2\sqrt{7}+\sqrt{6}\)
\(t_A = \sqrt{8} + \sqrt{7}\) and \(t_B = \sqrt{7} + \sqrt{6}\)
Let the time required to finish the job together be \(T\);
\(\frac{1}{T} = \frac{1}{t_A} + \frac{1}{t_B}\)
\(\frac{1}{T} = \frac{1}{(\sqrt{8} + \sqrt{7})} + \frac{1}{(\sqrt{7} + \sqrt{6})}\)
\(\frac{1}{T} = \frac{(\sqrt{8} - \sqrt{7})}{(\sqrt{8} + \sqrt{7})(\sqrt{8} - \sqrt{7})} + \frac{(\sqrt{7} - \sqrt{6})}{(\sqrt{7} + \sqrt{6})(\sqrt{7} - \sqrt{6})}\)
\(\frac{1}{T} = \frac{(\sqrt{8} - \sqrt{7})}{(8 - 7)} + \frac{(\sqrt{7} - \sqrt{6})}{(7 - 6)}\)
\(\frac{1}{T} = (\sqrt{8} - \sqrt{6})\)
\(T = \frac{1}{(\sqrt{8} - \sqrt{6})}\)
\(T = \frac{(\sqrt{8} + \sqrt{6})}{(\sqrt{8} - \sqrt{6})(\sqrt{8} + \sqrt{6})}\)
\(T = \frac{(\sqrt{8} + \sqrt{6})}{(8 - 6)}\)
\(T = \frac{(\sqrt{8} + \sqrt{6})}{2}\)
Hence, option B
Re: At Supersonic Corporation, the time required for a machine to complete
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10 Apr 2024, 10:01
10 Apr 2024, 10:01
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
