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Word problem about distance, rate and time

  1. Jun 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Q35hall_XXIV.jpg
    2. Relevant equations
    D = rt

    3. The attempt at a solution

    I let d to be the total distance

    d/3a = t1 ----- time taken for 1st part of the trip

    d/3b = t2 ----- time taken for 2nd part of the trip

    d/3c = t3 ------ time taken for the whole trip with uniform speed

    Since t1+t2 =t3

    d/3a + d/3b = d/3c

    Multiplying both sides by 3abc

    bcd + acd = abd

    Dividing both sides by d

    bc + ac = ab

    Factoring out c and diving both sides by it.

    1/c = (a + b)/ab = 1/a + 1/b

    1/c = 1/a + 1/b not the same as 2/c =1/a + 1/b

    Do you think it is just a typo?
     
  2. jcsd
  3. Jun 8, 2016 #2
    Hi Paul:

    The problems statement says:
    "He could have ridden from A to B and back again in the same time."​
    Therefore if t3 = time to go from A to B, then t3 = ( t1 + t2) / 2.

    Regards,
    Buzz
     
  4. Jun 9, 2016 #3

    ehild

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    2/c = 1/a +1/b is a typo. It should be 2/c = 1/a +2/b.
     
  5. Jun 9, 2016 #4

    SammyS

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    No, it's not a typo.

    See what Buzz said.
     
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