Nonsense Solutions to Word Problems

Date: 04/28/2003 at 16:13:42
From: Jason
Subject: How do you solve this problem?

Bill, Simon, and John are brothers. Bill is as many years younger than
one brother as he is older than the other. Simon is 7 years younger 
than twice the age of John. John is 5 years older than half the age 
of one of his brothers. How old is each brother?

Here's what I tried:

  Let Bill  = B
      John  = J
      Simon = S

  The equations are:  Simon's age is S=2(J) -7
                      John'age is    J=1/2(S) + 5
                      Bill'age is between that of Simon and John.

  Then 

    S = 2((1/2)S + 5) - 7

     = 2S + 10 - 7

     = 2S + 3
 
But this gives me S = -3, which I find misleading.  What am I doing
wrong? 

Date: 04/28/2003 at 17:12:39
From: Doctor Ian
Subject: Re: How do you solve this problem?

Hi Jason, 

It's easy to get lost in problems like this... which is, of course,
exactly why people make them up.  :^D

I've always found that one key to solving them is to avoid the
temptation to try to do too much in one step.  Another key is to avoid
making assumptions, at least without keeping track of them. 

You started well by assigning a variable to each age, instead of
trying to express everything in terms of one variable.  However, you
made an assumption about the order of their ages.  

The problem merely says that John is "five years older than half the
age of one of his brothers".  We don't know which brother that is. 
And it's possible that there are multiple solutions to the problem,
depending on which order you choose.  

Here's what we know:

  1) Bill is as many years younger than one brother as he is 
     older than the other. 

  2) Simon is 7 years younger than twice the age of John. 

  3) John is 5 years older than half the age of one of his brothers. 

From (1), we can conclude that Bill is in the middle.  But we can't
yet conclude anything about who is oldest.  Maybe we don't have to! 
Can we convert facts (1-3) into equations? 

  1a)  J - B = B - S     or      1b)  S - B = B - J

  2)   S = 2J - 7 

  3a)  J = B/2 + 5       or      3b)  J = S/2 + 5

As you can see, we get different equations depending on how we
interpret the problem. 

So this gives us four distinct possibilities:

  i)   1a, 2, 3a
  ii)  1b, 2, 3a
  iii) 1a, 2, 3b
  iv)  1b, 2, 3b

And some of them may lead to inconsistent results.  Let's examine the
first possibility:

    J - B = B - S              (1a)

        S = 2J - 7             (2)

        J = B/2 + 5            (3a)

We can substitute the second equation into the first to get

    J - B = B - (2J - 7)

        J = B/2 + 5

[Were you able to follow that?  Wherever there was an 'S' in the first
equation, we replaced it with (2J - 7).  We can do that, because what
it _means_ for two expressions to be equal is that they can be
interchanged in this way.]

And we can do a second substitution for J to get an equation
containing only B's:

    (B/2 + 5) - B = B - (2(B/2 + 5) - 7)

This yields sensible values for the ages of all three brothers.  But
it doesn't mean we're done!  We have to go through _all_ of the
possible sets of equations.  Otherwise, there may be solutions that
we've ignored. 

Note that in trying the other cases, you may in fact end up with a
result like S = -3.  What does it mean when that happens? The 'result'
represents a _mathematical_ solution to the equations which happens
not to correspond to a _physical_ solution to the problem.  

This happens sometimes, because any given set of equations can
represent an infinite number of possible situations.  (This is exactly
what makes equations so useful!)  In a situation where S is supposed
to represent an age, or a length, we throw negative values away. But
there could be other situations where S represents a temperature, or
an amount of money, or something else where a negative value makes
perfect sense.  

This is why it's important not to blindly solve equations without
understanding what the equations are supposed to be telling you.  

For another very common context where you need to keep this in mind, see

  Negative Square Roots
  http://mathforum.org/library/drmath/view/62047.html 

I hope this helps!  Write back if you'd like to talk more about this,
or anything else. 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/