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Working alone at its own constant rate, a machine seals k [#permalink]
17 Nov 2010, 07:35

00:00

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Difficulty:

35% (medium)

Question Stats:

76% (01:49) correct
24% (01:46) wrong based on 137 sessions

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?

Re: working alone at its own constant rate [#permalink]
17 Nov 2010, 07:38

2

This post received KUDOS

Expert's post

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).

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

4

This post received KUDOS

Expert's post

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

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.

Re: working alone at its own constant rate [#permalink]
23 May 2011, 19:52

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%

So, the answer is (D)

We can apply Step 3 in Step 1 to get the solution faster, avoiding all the unnecessary calculations of Step 2 and Step 3.

Re: working alone at its own constant rate [#permalink]
18 Dec 2011, 09:59

1

This post received KUDOS

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

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% Answer: D

Re: Working alone at its own constant rate, a machine seals k [#permalink]
18 Sep 2013, 17:07

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