document.write( "Question 159483: The worker on machine A can complete a certain job in 6 hours. The worker on machine B can do the same job in 4 hours. In a rush situation, how long would it take both machines to complete the job??? \n" ); document.write( "
Algebra.Com's Answer #117602 by Alan3354(69443)\"\" \"About 
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The worker on machine A can complete a certain job in 6 hours. The worker on machine B can do the same job in 4 hours. In a rush situation, how long would it take both machines to complete the job???
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\n" ); document.write( "This type of problem comes up in real life, and it's puzzling until you know how it works.
\n" ); document.write( "You have 2 machines, one takes 6 hours, one takes 4 hours. Obviously, working together will take less time than either of them alone, so you can't add the 4 and 6.
\n" ); document.write( "What's necessary is to look at how may jobs, or parts of the job, are completed per hour, not hours per job.
\n" ); document.write( "Machine A takes 6 hours/job, so that's 1/6 jobs/hour.
\n" ); document.write( "Machine B takes 4 hours/job, so that's 1/4 jobs/hour.
\n" ); document.write( "Now you can add those up, 1/6 + 1/4 jobs/hour working together.
\n" ); document.write( "= 5/12 jobs/hour.
\n" ); document.write( "Invert that, and you get 12/5 hours/job together, or 2.4 hours.
\n" ); document.write( "It looks right, it's less than 4, but more than half of 4. If both machines took 4 hours/job, it's obvious that together it would be half, or 2 hours, but machine B takes 6, so it's gonna be more than 2.
\n" ); document.write( "2.4 hours is it.
\n" ); document.write( "This is called \"the reciprocal of the sum of the reciprocals\", and is seen in physical phenomena.
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