Michael Paul Goldenberg wrote: > > Everything else aside, shouldn't the vos Savant (did she make up that last > name or what?) situation suggest that having "the world's highest IQ) is a > fairly worthless accomplishment? > > -----------------------------------------------------------------
According to her column of May 24, 1992, "vos Savant" is the last name of Marilyn's mother.
In recent years, Marilyn has simply stated the answers to several Algebra I problems that involved one variable. The following more complicated problem appeared in her column of Nov. 12, 1995 (p. 29). When she finally defined the variables and derived the equations, she made the solution extremely cumbersome. Even a person with "the world's highest IQ" can make a complete mess out of an Algebra I problem.
Problem: A ship is twice as old as its boiler was when the ship was as old as the boiler is. The sum of their ages is 49. How old is the ship, and how old is the boiler?
Marilyn's solution: Say that S is the age of the ship now; B is the age of the boiler now. Because S+B = 49, then S = 49-B. So, because S "is twice as old as its boiler was when the ship was as old as the boiler is," then 49-B is "twice as old as the boiler was when the ship was as old as" B. Continuing this substitution, 49-B is two times an earlier boiler age. How do we define this earlier age? The age difference in years between the ship and the boiler always remains the same. So, back at this earlier age, the boiler's age would have been the ship's earlier age of B ("when the ship was as old as the boiler is [now])" minus the number of years that the ship and the boiler are always apart. That "apart" number of years could be expressed as the ship's age minus the boiler's age, even now. So, because the ship is 49-B years old (now) and the boiler is B years old (now), that age difference can be expressed as (49-B)-(B). Therefore, the boiler's earlier age can be expressed as B minus (49-B)-(B). Rewritten, 49-B is 2 times as old as (B-[(49-B)-(B)]). Which means the same as 49-B = 2(B-[(49-B)-(B)]). Solving, we see that 49-B = 2(B-[49-2B]) Which is the same as 49-B = 2(B-49+2B) 49-B = 2(3B-49) 49-B = 6B-98 147 = 7B 21 = B And if the boiller (B) is 21 years old, the ship (S) is 28.
When Miss. Phoebe Fitzpatrick taught me Algebra I in 1962-63, she told my class that in many word problems the variables can be defined in several different ways. In particular, for age problems, the variables may be defined to denote either the present or the past/future ages; and we should choose the definitions that would yield the simplest solution.
This is my solution, based on the way I was taught by Miss. Fitzpatrick.
Let S = the original age of the ship, (these definitions will) and let B = the original age of the boiler. (avoid the fraction 1/2) Then, 2B = the present age of the ship, and S = the present age of the boiler.
The following equations must be satisfied:
(1) 2B + S = 49 (stated) (2) 2B - S = S - B (age differences are the same) or 3B - 2S = 0 (as observed by Marilyn )
By using the Sustitution or the Elimination Method, one easily finds that
B = 14 = the original age of the boiler. Thus: 2B = 28 = the present age of the ship, and S = 49 - 2B = 21 = the present age of the boiler.