Question 178779
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Let <i>x</i> be the number of hours it takes thing A to do one job.  Then thing A can do *[tex \Large \frac {1}{x}] of the job in 1 hour.  If thing B takes twice as long, then 2<i>x</i> represents the number of hours it takes thing B to do the same job, and thing B can do *[tex \Large \frac {1}{2x}] of the job in 1 hour.


Together, thing A and thing B can do *[tex \Large \frac {1}{x} + \frac {1}{2x} = \frac {2 + 1}{2x} = \frac {3}{2x}] of the job in 1 hour.


That means that thing A and thing B can do the entire job in *[tex \Large \frac {2x}{3}] hours.


But we are given that thing A and B can do the entire job in 4 hours.


Solve for <i>x</i> in *[tex \Large \frac {2x}{3} = 4] to get the time it takes A to do the job, and double it to get the time for thing B.


Notice I described this as 'thing A' and 'thing B' and just called what they were doing a 'job'  That's because this method works for 'painting a room' questions, 'two different sized pipes filling or emptying a tank' problems, and anything else like it.



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