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

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It takes 3 workers and one apprentice who works 3 times as [#permalink] New post 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?
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 [#permalink] New post 01 Aug 2003, 11:49
Is it 45 minutes?
It took almost 45 minutes to calculate it:) :roll:
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 [#permalink] New post 01 Aug 2003, 11:53
kpadma wrote:
Is it 45 minutes?
It took almost 45 minutes to calculate it:) :roll:


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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 [#permalink] New post 01 Aug 2003, 11:59
AkamaiBrah wrote:
kpadma wrote:
Is it 45 minutes?
It took almost 45 minutes to calculate it:) :roll:


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.
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 [#permalink] New post 01 Aug 2003, 12:04
I got 5/4 as much time or 20min longer
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 [#permalink] New post 01 Aug 2003, 12:06
I mean 15min. I probably would have gotten caught on the test
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 [#permalink] New post 01 Aug 2003, 12:43
It will take the new group about 3 hours and 45 minutes, so it is 45 minutes more than the first group
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 [#permalink] New post 02 Aug 2003, 08:21
Good work Stolyar, I heard they still test this stuff and it will never change.

:wink:

I will show you my way soon. I don't know if it is good. As I am a piece of kasheska on math. :wink:
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Good Idea [#permalink] New post 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.

Please, bear with me.

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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KL

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 [#permalink] New post 06 Aug 2003, 09:13
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)
  [#permalink] 06 Aug 2003, 09:13
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