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Math Help - Counter-intuitive? or...

  1. #1
    Senior Member DivideBy0's Avatar
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    Counter-intuitive? or...

    A car travels from point X to point Y. It travels at 30 km/h up to halfway and at 20 km/h for the rest of the journey. What is the car's average speed?

    The answer seemed pretty simple at first, I mean, it's got to be 25 km/h. Since the car travels the same distance for each of the different speeds, shouldn't the distance really be negatable? Couldn't you just say the car's traveling at 20km/h at point A and 30km/h at point B, and so the average of the speeds should be 25km/h?

    And yet, if you try applying an actual number to the distances, things change. Say the distance is 120km overall. This would mean that it would take 60/30 = 2h to cover the first half, and 60/20 = 3h to cover the second half. Seeing as distance is speed over time, the average speed would hence be 120/(2+3) = 120/5 = 24km/h.

    How can this be possible? Can someone help me out here?
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  2. #2
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    Hello, DivideBy0!

    This is a classic trick question.
    It involves the dangerous practice of averaging averages.

    I once had to explain to a Dean at my college who did this.
    I wrote him a long rant and finally got through to him.


    Suppose administrators are evaluted by the Faculty.
    To be retained, he must have the approval of 80% of the Faculty.

    At our college, there are 100 professors of which two are on the math faculty.
    . . Everyone approved of the Dean except one math professor.
    Should the Dean be retained?

    According to his math, the answer is a clear NO.

    In the math department, he received only 50% approval.
    In the rest of the faculty, he received 100% approval.

    The average of 50% and 100% is: .(50% + 100%)/2 .= .75%.
    . . Hence, it is with some regret that the Dean must be dismissed.

    "But," he will argue, "if 99 out of 100 approved me, don't I have 99% approval?"

    "Sorry," I would reply, "the math is clear and undeniable: only 75%.
    . . Please close the door on your way out."

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  3. #3
    Senior Member DivideBy0's Avatar
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    Hi Soroban,
    I'm a bit confused with your example... if the Dean needed approval from at least 80% of the faculty, then why even bother splitting it up into the math department and all the other departments? Why not just view the faculty as a whole? Also, you say that his solution states that he will not be retained, and yet later he argues that he has 99% approval? If you were being sarcastic there I probably got even more confused by it... lol
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    Hi Soroban,
    I'm a bit confused with your example... if the Dean needed approval from at least 80% of the faculty, then why even bother splitting it up into the math department and all the other departments? Why not just view the faculty as a whole? Also, you say that his solution states that he will not be retained, and yet later he argues that he has 99% approval? If you were being sarcastic there I probably got even more confused by it... lol
    He was trying to illustrate exactly your argument. (And the point that his Dean didn't know Math.)

    Taken in the context of speeds, it would run like this:

    A car drives with an average speed of 60 mph for 1 minute. It then travels with an average speed of 80 mph for the next 5 hours. What is the average speed?

    Would it be (60 + 80)/2 = 70 mph? Of course not! Because the car only drove 60 mph for a very short time compared to the amount of time it was driving the 80 mph.

    So you need to apply the "total distance travelled divided by the total time" formula to get an average over an interval.

    -Dan
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  5. #5
    Senior Member DivideBy0's Avatar
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    Oh, I think I get it now, so the average speed is taken with respect to the time, not with respect to distance. Cool!
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    A car travels from point X to point Y. It travels at 30 km/h up to halfway and at 20 km/h for the rest of the journey. What is the car's average speed?

    The answer seemed pretty simple at first, I mean, it's got to be 25 km/h. Since the car travels the same distance for each of the different speeds, shouldn't the distance really be negatable? Couldn't you just say the car's traveling at 20km/h at point A and 30km/h at point B, and so the average of the speeds should be 25km/h?

    And yet, if you try applying an actual number to the distances, things change. Say the distance is 120km overall. This would mean that it would take 60/30 = 2h to cover the first half, and 60/20 = 3h to cover the second half. Seeing as distance is speed over time, the average speed would hence be 120/(2+3) = 120/5 = 24km/h.

    How can this be possible? Can someone help me out here?
    In case you're wondering, here's how you would calculate the average rate:

    D1 = distance traveled in first half of trip
    D2 = distance traveled in second half of trip
    T1 = time spent driving first half
    T2 = time spent driving second half
    R1 = rate of first half = 30
    R2 = rate of secold half = 20

    First, we know rate equals total distance divided by total time, right? Therefore,
    R = SUM(D)/SUM(T) = (D1 + D2)/(T1 + T2)

    We have that D1 = D2 = D

    Second, we know that time equals distance divided by rate. Therefore:
    T1 = D/R1 = D/30
    T2 = D/R2 = D/20

    Since R = (D1 + D2)/(T1 + T2), we have that:
    R = (D + D)/(D/30 + D/20) = 2D/(5D/60) = 120/5 = 24 kh/hr
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