Most uninformed students, when asked to calculate
the average speed for a round trip with a “going” average speed of 30 miles per
hour and a “returning” average speed of 60 miles per hour, would think that
their average speed for the entire trip is 45 miles per hour [calculated as (30+60)/2 = 45].
The first task is to convince the students that this is the wrong answer.
For starters, you might ask the students if they
believe it is fair to consider the two speeds with equal “weight.” Some may
realize that the two speeds were achieved for different lengths of time and
therefore cannot get the same weight. This might lead someone to offer that the
trip at the slower speed, 30 mph, took twice as long and, therefore, ought to
get twice the weight in the calculation of the average round-trip speed. This
would then bring the calculation to the following:
(30+30+60)/3 = 40,
which happens to be the correct average speed.
For those not convinced by this argument, try
something a bit closer to “home.” A question can be posed about the grade a
student deserves who scored 100% on nine of ten tests in a semester and on one
test scored only 50%. Would it be fair to assume that this student’s
performance for the term was 75% (i.e., 100+50/2)?
The reaction to this suggestion will tend toward applying appropriate weight to
the two scores in consideration. The 100% was achieved nine times as often as
the 50% and therefore ought to get the appropriate weight. Thus, a proper
calculation of the student’s average ought to be
(9×100)+50/10 = 95
This clearly appears more just!
An astute student may now ask, “What happens if the
rates to be averaged are not multiples of one another?” For the speed problem
above, one could find the time “going” and the time “returning” to get the
total time, and then, with the total distance, calculate the “total rate,”
which is, in fact, the average rate.
There is a more efficient way; a concept called the
harmonic mean, which is the mean of a harmonic sequence. This frequently
misunderstood mean (or average) usually causes confusion, but to avoid this,
once we identify that we are to find the average of rates (i.e., the harmonic
mean), then we have a lovely formula for calculating the harmonic mean for
rates over the same base. In the above situation, the rates were for the same
distance (round-trip legs).
The harmonic mean for two rates, a and
b,
is 2ab/(a+b); and for
three rates, a, b, and c, the harmonic mean is 3abc/(ab+bc+ac).
You can see the pattern evolving, so that for 4
rates the harmonic mean is
4abcd/(abc+abd+acd+bcd).
Applying this to the above speed problem gives us
(2×30×60)/(30+60) = 3600/90
= 40
Begin by posing the following problem:
On Monday,
a plane makes a round-trip flight from New York City to Washington with an
average speed of 300 miles per hour. The next day, Tuesday, there is a wind of
constant speed (50 miles per hour) and direction (blowing from New York City to
Washington). With the same speed setting as on Monday, this same plane makes
the same round trip on Tuesday. Will the Tuesday trip require more time, less
time, or the same time as the Monday trip?
This problem should be slowly and carefully posed,
so that students notice that the only thing that has changed is the “help and
hindrance of the wind.” All other controllable factors are the same: distances,
speed regulation, airplane’s conditions, etc. An expected response is that the
two round-trip flights ought to be the same, especially since the same wind is
helping and hindering two equal legs of a round-trip flight.
Realization that the two legs of the “wind trip”
require different amounts of time should lead to the notion that the two speeds
of this trip cannot be weighted equally as they were done for different lengths
of time. Therefore, the time for each leg should be calculated and then
appropriately apportioned to the related speeds.
We can use the harmonic mean formula to find the
average speed for the “windy trip.” The harmonic mean is 2(350)(250)/(250+350) = 291.667, which is
slower than the no-wind trip.
What a surprise!
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