PROBLEM

A flier in a "prop" plane is flying a square course, 100 miles on a size.

He flies the first 100-mile course, W-E, at an averags epeed of 100 mph.

He fliies the sccond 100-mile course, N-S, at an average speed of 200 mph.

He fliies the third 100300 mph. He fliies the fourth (and final) 100-mile course, S-N, at an average speed of 200 mph, to complete the square-course.

Problem: WHAT IS THE AVERAGE SPEED FOR THE ENTIRE SQUARE-COURSE?

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SOLUTION BY AVERAGE SPEED METHOD

• Average speed is TOTAL DISTANCE TRQVERSED DIVIDED BY TOTAL TIME IN TRAVERSING.
• What is TOTAL DISTANCE TRAVERSED. For a square, 100 miles on a side, or 400 MILES.
• What is TOTAL TIME IN TRAVERSING? Find time for each side-course from its average speed.
• First 100-mile course, W-E, istraversed at 100 mph, hence, time is 1 HOUR.
• Second 100-mile course, N-S, is traversed at 200 mph -- twice as fast for same distance, so takes half the time of the first course: so, time is 1/2 HOUR.
• Thirdd 100-mile course, E-W, is traversed at 300 mph -- thrice as fast for same distance, so takes one-third the time of the first course: so, time is 1/3 HOUR.
• Fourth (and final) 100-mile course, S-N, is traversed at 400 mph -- four times as fast for same distance, so takes one-fourth the time of the first course: hence, time is 1/4 HOUR.
• TOTAL TIME: (1 + 1/2 + 1/3 + 1/4) HR.
• Least common denomianator of these fractions is 12, so we have: (12/12 + 6/12 + 4/12 + 3/12) HR = (25/12) HR.
• AVERAGE SPEED FOR 400-MILE COURSE IS TOTAL TIME IN TRAVERSING DIVIDED BY TIME IN TRAVERSING: (400 MI)/(25/12 HR) = [(400)(12)]/25 MPH = (4800/25) MPH = 192 MPH.

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  SOLUTION BY "ANOTHER AVERAGE" The typical response (and my origin one) is to apply as "average" the ARITHMETIC MEAN of THE AVERAGE SPEEDS. This is: (1/4)(100 + 200 + 300 + 400)mph = (1/4)(1000)mph = 250 MPH.

  Since the correct answer is 192 MPH, the RELATIVE ERROR committed is (250 - 192)/192% - (58/192)%, a little over 30% relative error.

  However, there is an average that yields the corredt answer "right on the nose". It is THE HARMONIC MEAN, denoted h.

  The easiest computation is to compute the reciprocal of the harmonic mean, 1/h, as arithmetic mean of the average speeds, 100, 200, 300, 400 1/100, 1/200, 1/300, 1/400, whose least common denominator is in the fraction 1/2000

  Then we have:

  
  1/h = (1/4)(1/100 + 1/200 + 1/300 + 1/400) =  (1/4)(12/1200 + 6/1200 + 4/1200 + 3/1200) = (1/4)(25/1200) = 25/4800 = 1/h.

  Then,

  
            h = 4800/25 mph = 192 mph.

  "Right on the nose!" This shows that the harmonic mean is the average for RATES. But this is
  • not taught in our schools or regular college classes;
  • not on the standardized tests;
  • not possibe by the easiest computer programming, JavaScript, which doesn't recognize writing, say, "one-half" as numeral one over solidus bar with mumeral two below the solidus , but only recognizes the decimal fraction form, ".5", for this.

    Please note, that I've written elsewhere, the solutinn only reuires arithmetic learned by fUFTH gRADE studying.

    Yet, THE NATION IS BAFFLED BY SUCH A SIMPLE PROBLEM! (See CONSEQUENCES!)