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Alpine's street sweeper cleans downtown in 60 hrs. A newer model cleans in 20 hrs. Together, how long to sweep the streets?

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Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (7 lesson ratings) (7)
x=# of hours working together
old model does 1/60 of job alone per hour
new model does 1/20 of job alone per hour
(1/60)x+(1/20)x=1 (job done)
multiply both sides by 60
x=15 hours together
Jose S. | Friendly Mathematics, Philosophy, and Computer Science tutorFriendly Mathematics, Philosophy, and Co...
Hello Noel,
The most important fact to gleam from the problem is that the faster sweeper is 3 times faster than the slower one.  Ideally if the two are working together they work for the same amount of time(otherwise there is a time when one is working and the other is not).  You want to divide the work into fractions s and f such that s+f =1, where s is the amount of work done by the slower sweeper and f is the amount of work done by the faster one.  Here we use the fact that the faster one should do three times as much work as the slower one: 3s=f, substitute this into the equation above and you have 3s+s=1 or 4s =1 so s = 1/4.  Hence the slower sweeper should do only 1/4 of the work.  How long does this take?  well 1/4 of 60 is 15.  And how long does it take the faster one to do 3/4s of the work?  Again 3/4 of 20 is 15.  Hence together working for 15 hours they can clean the whole of downtown. 
This is not really how I initially thought about the problem.  
For me once it was clear that the slower sweeper was 3 times slower I couldn't help but think 15 and quickly check that they could do all the work in just 15 hours.  I hope the above approach is helpful though since there is no way to teach the intuitive leap other than by practicing a lot of math. I had to think for a bit about how to justify my answer.