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Longfellow's Bees

Date: 08/01/2002 at 05:57:08
From: Eliza
Subject: Henry Wadsworth Longfellow

Bees are let out of a hive. One-fifth of the bees fly to the rosebush, 
one-third fly to the apple tree, and three times the difference
fly to the acorn tree. One bee is left flying around. How many bees
are there altogether? 

I tried answering this by saying that the total must be one plus a 
number that both 3 and 5 can go into, because we can't have half-bees, 
but that's as far as I got. 

Can you please help? I'm finding it really challenging.

Thank you, 
Eliza


Date: 08/01/2002 at 10:03:18
From: Doctor Ian
Subject: Re: Henry Wadsworth Longfellow

Hi Eliza,

Often, when you don't know what else to do, a good place to start is 
to make a guess, and see if you can figure out how to check whether 
it's correct.  

You made a good start, noting that the number of bees will have to be 
divisible by both 3 and 5. 30 is divisible by both 3 and 5, so could 
that be the answer? Let's see:

  30/5 = 6 bees fly to the rose bush. 

  30/3 = 10 bees fly to the apple tree

  3*(10 - 6) = 12 bees fly to the acorn tree

  1 just flies around

  Total = 6 + 10 + 12 + 1 = 29 bees

which is close, but not quite right, since we need to end up with 30.  
So we didn't get the right answer. 

However, we got something almost as good!  Note that if we write 
our sum this way, 

  30/5 + 30/3 + 3(30/3 - 30/5) + 1 = ?

we can just substitute other values for 30 to check them.  For 
example, 60 is also divisible by both 3 and 5, so we can test 
that value:

  60/5 + 60/3 + 3(60/5 - 60/3) + 1 = 

          12 + 20 + 3(20 - 12) + 1 = 

                       32 + 24 + 1 = 57

So that doesn't work either, but we can use this expression to check 
answers very quickly. So at this point, we might start examining the 
possibilities systematically, starting with 15, 30, 45, 60, and so on, 
until we come across the answer. 
             
But we can be a little more clever than that. Note that we'd like to 
end up with our guess on the right side of the equals sign, e.g., 

  30/5 + 30/3 + 3(30/3 - 30/5) + 1 =? 30
 
Suppose we replace the guess, wherever it appears, with an unknown 
quantity, say, 'U'.  Then we end up with this:

  U/5 + U/3 + 3(U/3 - U/5) + 1 = U

Now, instead of using arithmetic to check guesses, we can use algebra 
to find the correct 'guess' directly. 

But the important lesson here isn't that this equation solves this 
particular problem. The important lesson is that this is how you 
_find_ an equation to solve a problem like this!  

One of the most damaging ideas that students seem to pick up in math 
class is that you should be able to read a word problem and just write 
down an equation that will let you solve it. In practice, it only 
works like that for problems that you've already solved in the past!  
For problems you've never seen before, you're much better off figuring 
out what steps you'd need to take to _check_ an answer, and then 
trying to turn those steps into the equation (or equations - sometimes 
you need more than one) that you need. 

Can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Elementary Fractions
Elementary Word Problems
Middle School Fractions
Middle School Word Problems

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