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 Discussion: Research Area Topic: Simple arithmetic problem "99%" can't solve

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 Subject: Simple arithmetic problem "99%" can't solve Author: Sonny Date: Jun 15 2004
I found this "averaging" problem in 1951 in "Facts Into Figures" by British
statistician, F. J. Moroney and solved it incorrectly. Repeatedly, in years
since, in university classes, none solved it correctly. I've the original ONLINE
in http://members.fortunecity.com/jonhays/stndtsk1.htm
(along with other problems I required of my students, but many teachers and
professors could not solve). Another version: in auto trip, of 4 spans (starting
in town traffic, to out-of-town), 1st span averages 10mph; 2nd span, 20 mph;
3rd, 30mph; 4th, 40mph. What average speed represents these 4 average speeds?
Many persons think arithmetic mean (AM) is "the average". (Moroney said
"average" is "representative".) AM = 1/4(10 + 20 + 30 + 40)mph = 1/4(100)mph =
25mph. This incurs 30%+ relative error. The average speed is distance traversed
divided by time traversing. We need
a representivve distance. The least common multiple of (10, 20, 30, 40)mi. is
120mi. Time: at 10mph, 12 hrs; at 20mph, 6 hours; at 30mph, 4 hours; at 40mph,
30. Total: (12 + 6 + 4 + 3)hours = 25hours. Then we have 480mi/25hr = 19.2mph.
Relative error:(25-19.2)/19.2% = 5,8/19.2%, approx. 30%. But a different
"average", the harmonic mean (HM), yields exactly the correct answer. Easily
calculated as AM of reciprocals of values. 1/HM = 1/4(1/10 + 1/20 + 1/30 + 1/40)
=
1/4(12 + 6+ 4 + 3)/120 = (1/4)(25/120) = 25/480 = 1/HM, or HM = 480/25 = 19.2.
(Ancient Pythagoreans used HMs to construct our chromatic musical scale.) Why
isn't this more generallly taught?