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How far is it to his office?

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    Please I need help with this question:
    A man drives to a car park at an average speed of 40 km/hr and then walks to his office at average speed of 6km/h. The total journey takes him 25 minutes. One day his car breaks down and he has to walk three times as far, so arriving at his office 17 minutes late. How far is it to his office?
    2. Relevant equations

    time = distance /speed

    3. The attempt at a solution
    x = distance driven (normally)
    > y = distance walked (normally)
    For the normal situation,
    5/12 = x/40 + y/6
    When the car breaks down, he walked three times as far.That means he walked three times the distance he used to cover when his car had a mechanical fault. This distance include (3x) 3 times the distance to the car park and (3y) 3 times the distance to the office. Am saying this because on that the day he had no car at his disposal, so he made the journey with his leg i.e is the journey to the car park and to the office. So because of this reason the second equation becomes
    3x/40 + 3y/6 = 7/10h

    Am I equations correct?
  2. jcsd
  3. Apr 2, 2013 #2
    2nd equation isn't quite right. The car broke down somewhere along the way. If he walked a greater distance, then obviously the distance he drove isn't going to go up as well.
  4. Apr 3, 2013 #3
    You can think of it this way:

    If the distance walked (x) + distance driven (y) = the distance to his office (z)


    The distance to his office doesn't change and if his car is broken down, he can't use it anymore. which means the distance driven =?
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