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# It takes 3 workers and one apprentice who works 3 times as

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Joined: 03 Feb 2003
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It takes 3 workers and one apprentice who works 3 times as [#permalink]  01 Aug 2003, 11:31
It takes 3 workers and one apprentice who works 3 times as slowly as does a worker 3 hours to dig a certain ditch. If one worker is sick and supplanted by one apprentice, then how much longer a new group will perform the same work?
Director
Joined: 03 Jul 2003
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Is it 45 minutes?
It took almost 45 minutes to calculate it:)
GMAT Instructor
Joined: 07 Jul 2003
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Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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Is it 45 minutes?
It took almost 45 minutes to calculate it:)

Think logically. You replace a good worker with an inferior one. It took 3 hours before. Why would it take 45 minutes now?
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

SVP
Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

AkamaiBrah wrote:
Is it 45 minutes?
It took almost 45 minutes to calculate it:)

Think logically. You replace a good worker with an inferior one. It took 3 hours before. Why would it take 45 minutes now?

it would take a new group 45 minutes more that it would a usual one.
Intern
Joined: 04 Jun 2003
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I got 5/4 as much time or 20min longer
Intern
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I mean 15min. I probably would have gotten caught on the test
Manager
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It will take the new group about 3 hours and 45 minutes, so it is 45 minutes more than the first group
Eternal Intern
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Good work Stolyar, I heard they still test this stuff and it will never change.

I will show you my way soon. I don't know if it is good. As I am a piece of kasheska on math.
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Ride em cowboy

Manager
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Good Idea [#permalink]  06 Aug 2003, 08:31
Although I used to think I knew everything about work problems, I must admit it took me a while before reaching the answer. Therefore, let me show my approach for those who still have problems with this type of questions.

Let X be the time it takes a worker to perform a task alone. Then, 3X - time of an apprentice. 1/X - worker's hourly rate and 1/3X - that of an apprentice.

Substitute variable into an equation:

1/X+1/X+1/X+1/3X=1/3 (Watch it!, it's not 3 as I did originally)
10/3X=1/3
X=10 => time of an apprentice = 30 hours.

Now, new conditions are 2 workers and 2 apprentices.
A new equation looks as follows:

1/10+1/10+1/30+1/30=Y
8/30=Y (this is the amount of work a given group can complete in an hour's time). => reverse the ratio to get the total amount of time.
30/8=3and6/8 hours

3 and 6/8 hours - 3 hours = 6/8 hours (or 45 minutes)

Special thanks to Stolyar. Beautiful questions!
It's a pleasure to seeing Him here.
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Respect,

KL

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The best approach for such problems and many others is using a basic rule:

Work=Rate*Time

Even the velocity/time problem is the same.

Distance is work
Velocity is a rate
Time is time (even in Africa)
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