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Homework Help: Word Problem. Am I wrong, or is my Algebra book?

  1. Jan 27, 2005 #1
    In my algebra textbook the following problem is given:

    Ken and Bettina Wikendt live in Minneapolis, Minn. Following a severe snowstorm, Ken and Bettina must Clear the driveway and sidewalk. Ken can clear the snow by himself in 4 hours. Bettina can clear the snow by herself in 6 hours. After Bettina has been working for 3 hours, Ken is able to join her. How much longer will it take them working together to move the rest of the snow.

    The book gives the answer as:

    t/6 + (t+3)/4 = 1,
    thus t = 3/5 of an hour to complete the rest of the job.

    When I worked the problem I set it up as follows:

    (t+3)/6 + t/4 = 1.
    Which works out as t = 1 and 1/5 hours to complete the rest of the job.

    At first I figured I just worked the problem wrong, but after reviewing it, I'm not sure if I'm wrong or the book is wrong.

    I used "t" as the time they have been working together. So "t+3" equals the total amount of time Bettina has been working.

    So using the formula "amount of work = rate * time" with Bettina's rate of 1/6 the total job per hour, and a working time of "t+3" that means the amount of work she did was "(t+3)* 1/6" or simply "(t+3)/6"
    For Ken's the amount of work equals "t * 1/4" or simply "t/4"

    The total job, "1" should be Ken's work plus Bettina's work, which is "(t+3)/6 + (t/4) = 1"
    Working this I get 1 and 1/5 hours for "t"

    I even tried working the problem from a diffent angle.
    Since after 3 hours Bettina will have completed half the job ("3 * 1/6 = 3/6")
    I can figure out the time left by simply firguring out how much time it takes Ken and Bettina to finish the other half of the job. Or simply: "t/6 + t/4 = 1/2". Again, I get 1 and 1/5 hours, NOT 3/5 of an hour.

    Is the book wrong or am I? :confused:

  2. jcsd
  3. Jan 27, 2005 #2


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    P.S.Both lines of logic are correct.
  4. Jan 27, 2005 #3


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    The book (or you in copying the problem) has mixed up Ken and Bettina.
    The equation t/6+ (t+3)/4= 1 would be correct if it were Ken working for the extra 3 hours. If we use the equation (t+3)/6+ t/4= 1. Multiplying through by 12, 2(t+3)+ 3t= 5t+ 6= 12 so 5t= 6 and t= 6/5.

    Here's how I would analyse this problem:

    It would take Ken 4 hours to clear the sidewalk by himself so he works at the rate of 1/4 "sidewalk per hour".

    It would take Bettina 6 hours to clear the sidewalk by herself so she works at the rate of 1/6 "sidewalk per hour". She works by herself for 3 hours so she will have cleared half the sidewalk.

    Now Ken and Bettina work together to clear the rest of the sidewalk. Their rates add so they work at 1/4+ 1/6= 3/12+ 2/12= 5/12 "sidewalk per hour". To clear 1/2 a sidewalk at the rate of 5/12 sidewalk per hour requires (1/2)/(5/12)= (1/2)(12/5)=
    6/5= 1 and 1/5 hour: 1 hour and 12 minutes.

    Hmmm, "two great minds"...
    Last edited by a moderator: Jan 27, 2005
  5. Jan 27, 2005 #4
    The problem as shown above is word for word from the book, as is the answer they gave.

    I knew something was screwy. I just didn't know if it was me or the book. :tongue2:

    Last edited: Jan 27, 2005
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