Problem 174 — Million-dollar questions don’t always deserve answers, as this month’s problem by Ray Santillan Jr. of Euless, Texas, demonstrates.
We know that the human body is comprised of some percentage of water. Senator Rufus “Pork” Barrel has placed you in charge of a $10-million government study to find out how much water is contained in the all the people of the earth. You’ve acquired the following data, in addition to the distribution chart pictured:
• World population 55 × 109 people.
• Water weighs 8.345 lb per gal.
• Volume of water is 0.1337 ft3 per gal.
• Assume population is 50% men, 50% women.
• Assume women’s weight is 75% of men’s weight for each weight category.
• Person’s water content is 67% by weight.
How many cubic feet of water are contained in the world’s population?
Technical consultant, Jack Couillard, Menasha, Wis.
Solution to last month’s problem 173 — You rarely go around in circles if you answered no, the car would not stay on the track. Here’s how Lockjaw was in up to his neck:
dl = diameter of left front wheel, given as 27 in.
dr = diameter of right front wheel, given as 21 in.
R = radius of full circle described by left front wheel, ft
vr = angular velocity of right front wheel, rpm
vl = angular velocity of left front wheel, rpm
u = angle described by left and right front wheels during time, t, deg
Ar = arc described by right front wheel in time t, ft
Al = arc described by left front wheel in time t, ft
From the story we know the wheels were 4 ft apart. Using geometry:
Insert the values for (2) and (3) into (1):
From the story we know that the left front wheel rotates 28% faster than the right. So, vl 5 1.28vr. Eliminate common terms and substitute known values. The previous equation reduces to:
Solving for R, we get:
0.01R = 8.96 ft
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R = 896 ft. The radius described by the right front tire is R + 4 ft, or 900 ft. Thus the diameter described by the outside front wheel of the car is 1,800 ft.
Compare this circle to the track. The distance from the outer edge of one straight section to the outer edge of the other is 1,320 ft + (2 × 50 ft) = 1,420 ft. The distance from the inner edge of one semicircle to the inner edge of the other is (2 × 1/8 mile) 11/8 mile or 1,980 ft.
Rather than face a board of inquiry, Lockjaw decided to leave his car in the swamp.
Contest winner — Congratulations to Tom Amerine of Houston who won our May contest by having his name drawn from the 140 contestants who answered correctly out of a total of 216 for that month. A TI-68 calculator is in the mail to him.
The TI-68 Advanced Scientific Calculator by Texas Instruments can solve five simultaneous equations with real and complex coefficients and has 40 number functions that can be used in both the rectangular and polar coordinate systems. Other functions include formula programming, integration, and polynomial root finding. The calculator also features a last-equation replay function that lets you double-check your work.