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

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Re: Working alone at its own constant rate, a machine seals k [#permalink]

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New post 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 [#permalink]

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New post 11 Jan 2015, 16:09
Hi pacifist85,

Here's an interesting "drill" that you can do to help develop this skill. Go back and review all of the past Quant questions that you've done and look for situations in which TESTing VALUES is applicable. When you come across those questions, look for the various "clues" in the prompt that would help you make a smart choice for the value(s) that you'd pick. Look for what's listed in the answers, the wording/descriptions in the prompt, any fractions that you're given, etc. For example, when I see the % sign in the answers, my first thought is that I might be able to TEST the number 100 in this question... In this way, you'll be training to not only spot the potential uses for this tactic, but you'll also be training to test the values that will make solving the problem most efficient.

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Rich
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Re: Working alone at its own constant rate, a machine seals k [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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New post 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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Re: Working alone at its own constant rate, a machine seals k   [#permalink] 10 Apr 2018, 11:03

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