SAT Math Friday: Work Rate Questions

Paul and Mike are painting their backyard fence. If Paul takes 6 hours working at a constant rate to finish the fence, and Mike takes 4 hours, how long would it take them working together to finish painting the fence?

This question is called a work rate question and it actually, once in awhile, comes up on the SAT. So how do you approach it? Well, one way is to find the hourly rate of each person or thing involved (in this case, Paul and Mike). That is, in one hour, how much of the fence does each paint? In the case of Paul, he paints 1/6 of the fence in 6 hours. (6 x 1/6 = 1, where ‘1’ represents a complete job, i.e., you’ve painted 1 fence).

Mike would paint ¼ of a fence in an hour. Now that we’ve found how much of the fence each can paint in an hour, we just need to add those two rates together to see how much of the fence the two working together finish: ¼ + 1/6 = 10/24. Again, this is not a measure of time, but a measure of how much of the fence the two working together complete in one hour. In work rate questions, all you have to do is to take this “working together” hourly rate, and “flip it”: so 10/24 becomes 24/10 hrs, of 2 hours and 24 minutes.

Now let’s try a more difficult question:

Working alone and at a constant rate, Machine A takes 3 hours to create a widget. Machine B, working alone and at a constant rate, takes x hours to create a widget. Working together, Machine A and B take 12 hours to build two widgets. What is the value of x?

    (A) 3
    (B) 4
    (C) 17/3
    (D) 6
    (E) 34/3

Author

  • Chris Lele

    Chris Lele is the Principal Curriculum Manager (and vocabulary wizard) at Magoosh. Chris graduated from UCLA with a BA in Psychology and has 20 years of experience in the test prep industry. He's been quoted as a subject expert in many publications, including US News, GMAC, and Business Because. In his time at Magoosh, Chris has taught countless students how to tackle the GRE, GMAT, SAT, ACT, MCAT (CARS), and LSAT exams with confidence. Some of his students have even gone on to get near-perfect scores. You can find Chris on YouTube, LinkedIn, Twitter and Facebook!

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