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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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+3x=60

4x=60

x=15 hours together

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.

Sincerely,

Jose

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