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

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Working alone at its own constant rate, a machine seals k [#permalink]  17 Nov 2010, 07:35
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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%
[Reveal] Spoiler: OA
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Re: working alone at its own constant rate [#permalink]  17 Nov 2010, 07:38
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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%

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).

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Re: working alone at its own constant rate [#permalink]  17 Nov 2010, 08:03
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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%

or think of it this way:
First machine seals k cartons in 8 hours.
Second machine seals k cartons in 4 hours i.e. 2k cartons in 8 hours.

If they both are working for the same period of time i.e. 8 hours, they together seal 3k cartons. Faster machine (second one) seals 2k out of these 3k so it seals 2/3 = 66.66% of the total cartons.
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Re: working alone at its own constant rate [#permalink]  23 May 2011, 19:52
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I solved it this way:

Step 1:

In one hour, the amount of work completed by Machine 1 is 1/8 and the amount of work completed by Machine 2 is ¼ i.e. Machine 2 works twice as fast as Machine 1.

(You can get the answer in Step 1 only. Anyhow, you will understand how to get the answer in the Step 1 by using a shortcut in Step 3)

Step 2:

In two hours, the amount of work completed by Machine 1 is 2/8 and the amount of work completed by Machine 2 is 2/4.

⇨ Total work completed in 2 hours by 2 machines = (2/8)+(2/4) => (1/4) + (1/2) => 0.25+0.50 = 0.75.
⇨ So the total amount of work completed by both the machines is 75 %. Out of the 75%, Machine 1 completes 50% of the work and Machine 2 completes 25% of the work.

Step 3:

Only 25% of the work is left. So the remaining work has to be divided between Machine 1 and Machine 2. As we know, Machine 2 can do the work twice as fast as Machine 1.

For example, if Machine 1 can do 10% of the work, Machine 2 can do 20% of the work in the same time. So, together they will complete 30% of the work. So from this example, we can understand that any work can be divided into 3 parts, and if Machine 1 completes 1 part of the work, and Machine 2 completes 2 parts of the work in the same time, irrespective of the quantum of work.

Now, the pending work is 25%. Let’s divide the work into 3 equal parts i.e. 8.33% per part. So, Machine 1 will complete 8.33% of the work whereas Machine 2 will complete 16.66% of the work, which is remaining.

So the total work completed by Machine 1 = 25%+8.33% = 33.33%
The total work completed by Machine 2 = 50% + 16.66% = 66.66%

We can apply Step 3 in Step 1 to get the solution faster, avoiding all the unnecessary calculations of Step 2 and Step 3.
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Re: working alone at its own constant rate [#permalink]  23 May 2011, 20:34
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if each one has worked for 8 hrs then number of cartoons = k + 2k = 3k.
Mac B produces 2k of these.

hence % = 2k/3k = 66.66%
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Re: working alone at its own constant rate [#permalink]  30 Oct 2011, 09:13
Prefer Bunuel's approach
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Re: working alone at its own constant rate [#permalink]  18 Dec 2011, 09:59
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I solved the problem directly
Rate one machine = k/8
Rate of the other machine = k/4
Total k/8+ k/4 = 3k/8
Faster seal to total seal = k/4 x 8/k = 66.33
Ans. D
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Re: working alone at its own constant rate [#permalink]  05 Jan 2012, 23:17
As the second machine works twice as fast as does the first one, the second machine will do twice as much work as does the first one. Together, if they complete a job, the second machine has worked twice as much as has the first machine.
Thus, work done by second machine relative to the total work done is 2/3 => 66.67%
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  18 Sep 2013, 17:07
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  28 Dec 2014, 10:04
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  28 Dec 2014, 14:10
Expert's post
Hi All,

Even if you don't see the immediate ratios involved, you can still answer this question rather easily by TESTing VALUES. Here's how:

Machine A can produce K cartons in 8 hours
Machine B can produce K cartons in 4 hours

Let's TEST K = 2

So....
Machine A = 2 cartons every 8 hours
Machine B = 2 cartons every 4 hours

We're told that each machine works on its own for the SAME amount of time.

Let's say they both work for 8 hours. This means...

Machine A seals 2 cartons
Machine B seals 4 cartons
Total = 6 cartons

The question asks what ratio of the cartons the faster machine sealed. Machine B is the faster machine, and it sealed 4/6 of the cartons.

4/6 = 2/3 = 66 2/3%

[Reveal] Spoiler:
D

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Re: Working alone at its own constant rate, a machine seals k [#permalink]  31 Dec 2014, 00:23
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Rate of Machine A $$= \frac{k}{8}$$

Rate of Machine B $$= \frac{k}{4}$$

Combined rate of A & B $$= \frac{k}{8} + \frac{k}{4} = \frac{3k}{8}$$

Say they work for 1 hour

Work done by faster machine (Machine B) $$= \frac{k}{4} * 1 = \frac{k}{4}$$

Combined work done in 1 hour $$= \frac{3k}{8} * 1 = \frac{3k}{8}$$

Percentage work of faster machine $$= \frac{\frac{k}{4}}{\frac{3k}{8}} * 100 = \frac{200}{3} = 66.66%$$

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Re: Working alone at its own constant rate, a machine seals k [#permalink]  10 Jan 2015, 11:51
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?
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Re: Working alone at its own constant rate, a machine seals k [#permalink]  10 Jan 2015, 12:02
Expert's post
Hi pacifist85,

You are correct - we could use ANY number of hours when TESTing VALUES. However, since the prompt offers us two rates that are based on 8 hours of work and 4 hours of work, respectively, choosing 8 hours allows us to make the calculations as simple as possible. In many Quant questions on Test Day, the specific numbers that appear in the prompt (and in the Answer choices) can help you to make choices that will get you to the correct answer in the fastest/easiest way possible.

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Re: Working alone at its own constant rate, a machine seals k [#permalink]  10 Jan 2015, 12:31
Thank you Rich.

I was just wondering the reason why, since - as they were supposed to be working for some time - my intuition would be to use 1 hour, so I wouldn't have to calculate anything.

As I am not used to doing quick calculations and haven't achieved to immediately see the connection between numbers yet (except for finding the LCM which I have managed to do almost instantly!!!!!) I was wondering if there was a reasoning I should know about and can use during the actual test.

Thank god, I only need a 650, which is not really a low score for a psychologist to achieve... I am around the 600ts at the moment and can do most of the problems. However, calculations and time are the problem with me...

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Re: Working alone at its own constant rate, a machine seals k [#permalink]  11 Jan 2015, 06:25
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]  11 Jan 2015, 15:09
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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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Re: Working alone at its own constant rate, a machine seals k   [#permalink] 11 Jan 2015, 15:09
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