Question 1151072
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I will show you three different ways of solving this problem.  Most problems of this type can be solved using any one of the three methods.  Look at all three of them; you might find one of them "works" for you much better than the others.<br>
(1) The traditional algebraic method, using the fraction of the job each worker or group of workers does in an hour.<br>
master and apprentice together take 15 hours --> in 1 hour the two together do 1/15 of the job.
master alone takes 20 hours --> in 1 hour the master does 1/20 of the job.
apprentice alone take x hours --> in 1 hour the fraction of the job the apprentice does is 1/x.<br>
The fraction of work they do together in 1 hour is the sum of the fractions they each do alone in 1 hour:<br>
{{{1/20+1/x = 1/15}}}<br>
Solve using basic algebra....<br>
(2) A very similar method that results is very similar calculations to get the answer.<br>
Consider the least common multiple of the two given times, which is 60 hours.<br>
In 60 hours, the two of them together could do the job 60/15 = 4 times.<br>
In 60 hours, the master alone could do the job 60/20 = 3 times.<br>
So in 60 hours the apprentice alone could do the job 4-3=1 time.<br>
So the time it would take the apprentice alone to do the job is 60 hours.<br>
(3) A very different approach to finding the answer....<br>
The master can do the job alone in 20 hours.  So when the two work together and complete the job in 15 hours, the fraction of the work the master does is 15/20 = 3/4.  That means the apprentice does 1/4 of the job in 15 hours; and that in turn means it take him 4*15 = 60 hours to do the job alone.<br>