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Re: Working alone at its own constant rate, a machine seals k
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11 Jan 2015, 07:25
pacifist85 wrote: I used the same aproach as Paresh. So, after finding the comdined rate as 3k/8, I assumed that they work for 1 hour. So, the individual work should be the same as the individual rates, i.e. k/8 for the slower one and k/4 for the faster one. It is then (k/4)/(3k/8).
What I didn't understand in the other approaches is why we used 8 hours as the amount of hours they worked for. I guess, we could also have chosen 4 hours or any other amount of hours? LCM of 4 & 8 is 8..... Its just for ease of calculation......
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Re: Working alone at its own constant rate, a machine seals k
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04 Jul 2015, 23:07
In 8 hours, first machine will seal k cartons and second machine, the faster one, will seal 2k cartons. Total work done = 3k 2k/3k = 66 2/3%, so D



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Working alone at its own constant rate, a machine seals k
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25 Nov 2015, 13:30
anilnandyala wrote: Working alone at its own constant rate, a machine seals k cartoons in 8 hours, and working at its own constant rate a second machine seals k cartoons in 4 hours. if 2 machines each working at its own constant rate and for the same period of time together sealed a certain no of cartoons. what percentage of the cartoons were sealed by the machine working at the faster rate?
A. 25% B. 33 1/3% C. 50% D. 66 2/3% E. 75% Rate A = \(\frac{1}{8}\), Rate B =\(\frac{1}{4}\) > \(\frac{Rate A}{Rate B}=\frac{1}{8}*\frac{4}{1} = \frac{1}{2}\) So the faster rate has 2 of 3 > 66 2/3% Answer D
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Re: Working alone at its own constant rate, a machine seals k
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10 Apr 2018, 11:03
study wrote: Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?
A. \(25%\)
B. \(33\frac{1}{3}%\)
C. \(50%\)
D. \(66\frac{2}{3}%\)
E. \(75%\) Since the second machine can seal k cartons in 4 hours, it can seal 2k cartons in 8 hours, whereas the first machine can seal only k cartons in 8 hours. So the second machine is twice as fast as the first machine and thus it will seal twice as many cartons as the the first machine. Therefore, when working together, the second machine (the faster machine) will finish ⅔ or 66⅔% of the work whereas the first machine (the slower machine) will finish ⅓ or 33⅓% of the work. Alternate Solution: In 8 hours, the slower machine will seal k cartons, and the faster machine will seal 2k cartons. In total, k + 2k = 3k cartons will be sealed, and, of these cartons, the faster machine will seal 2k/3k = 2/3 = 66 ⅔% of them. Answer: D
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Working alone at its own constant rate, a machine seals k
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30 May 2018, 14:25
Bunuel wrote: anilnandyala wrote: working alone at its own constant rate, a machine seals k cartoons in 8 hours, and working at its own constant rate a second machine seals k cartoons in 4 hours. if 2 machines each working at its own constant rate and for the same period of time together sealed a certain no of cartoons. what percentage of the cartoons were sealed by the machine working at the faster rate? a 25% b 33 1/3% c 50% d 66 2/3% e 75%
i am missing something in this, please explain thanks in advance As second one works twice as fast as the first, working together second will do 2/3 of work and the first 1/3 of work (their ratio 1/2). Answer: D. Bunuel the question itself confused me  > what percent of the cartons were sealed by the machine working at the faster rate? what does it mean at faster rate ? I know there rates are 1/8 and 1/4 ….but at faster rate … what it means???




Working alone at its own constant rate, a machine seals k &nbs
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