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NYT math problem

  1. Mar 13, 2007 #1
    Hi. Has anyone seen this?
    http://www.nytimes.com/ref/science/20070313_PROF_GRAPHIC.html

    I got all the problems except the one that says

    "I walk from home to school in 30 minutes and my brother takes 40 minutes. My brother left 5 minutes before I did. In how many minutes will I overtake him?"

    I said we need to know how fast each brother is walking.
    apparently the answer is 15 minutes??
    Is it because their difference in 10 minutes + 5 minutes head start?
    if so, that makes no sense to me.

    why is the answer 15 minutes?


    thanks
     
  2. jcsd
  3. Mar 13, 2007 #2
    They are walking at a constant speed apparently. If it takes you 30 minutes to walk home, then you will be halfway there in 15 minutes. If it takes your brother 40 minutes to walk home then he will be halfway there in 20 minutes. So if your brother starts walking home, and then you start walking home five minutes later, you and your brother will meet at the halfway mark. If that does not make sense, consider a concrete example. Suppose your brother starts walking home at 1. Then he will be halfway home at 1:20 (1:00 + 20 minutes). If you start walking at 1:05 (5 minutes after your brother), then you will be halfway home also at 1:20 (1:05 + 15 minutes).

    You could also solve it algebraically.
     
  4. Mar 13, 2007 #3
    You go from point A to point B in 30 minutes, and your brother does it in 40 minutes. You want to find out at what time are your positions equal.
    If you say that the distance from home to school is d meters, then your speed is d/30 m/min. Your brother's speed is d/40 m/min. You might know that your position is your speed times your time, or x = vt. If we say you leave at time t=0 min, then your brother left at time t=-5 min (five minutes earlier). So,

    xyou = vyout = d/30 * t
    and
    xbrother = vbrothert = d/40 * (t+5) (since your brother has been traveling for five more minutes than you have)

    At the point where you overtake your brother, your x positions are equal. So, set the two equations equal to one another to obtain

    d/30 * t = d/40 * (t+5)

    Now solve for t. (d's on both sides cancel)

    t/30 = (t+5)/40
    t/(t+5) = 30/40
    (t+5)/t = 40/30
    1 + 5/t = 40/30
    5/t = 4/3 - 1
    t = 5/(4/3 - 1) = 15 min.
     
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