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Tea and Cakes

Date: 11/13/2002 at 01:17:27
From: Jacquie Leferink
Subject: Tea and cakes

A cafe sold tea at 30 cents a cup and cakes at 50 cents each. Everyone 
in a group had the same number of cups of tea and the same number of 
cakes. The bill came to $13.30. How many cups of tea did each person 
have?

I set up the problem this way..

   x = cups of tea
   y = cakes

   .30x + .50y = $13.30

But when I try to solve it, it does not make sense.
Please help.


Date: 11/13/2002 at 12:51:53
From: Doctor Ian
Subject: Re: Tea and cakes

Hi Jacquie,

That equation is correct. The extra information that you need to solve 
the problem is that both x and y will have to be integers.  

You can solve the equation for either variable, e.g., 

       1330 - 30x
   y = ----------
           50

(Note that I'm using pennies instead of dollars to represent money,
because that lets me use integers instead of decimals.)

Now, for y to be an integer, the expression on the right must evaluate
to an integer. What you need to do is figure out for which value(s)
of x this occurs.  

One step in the right direction would be to simplify it:

       133 - 3x
   y = ----------
           5

Now it's clear that the numerator must be a multiple of 5, which is to
say, it has to end in a '0' or a '5'. 

Note that when x=1, we get

       133 - 3(1)
   y = ---------- = 130/5 = 26
           5

which says that (1 tea, 26 cakes) is a solution to the problem.  For
x = 6, we get

       133 - 3(6)
   y = ---------- = 115/5 = 23
           5
   
which says that (6 teas, 23 cakes) is also a solution. 

And there will be more solutions like these. How can we choose the
correct solution to the equation, i.e., the one that satisfies the
problem? 

The key is to look at the sentence

  Everyone in the group had the same number of cups of tea 
  and the same number of cakes.

If I'm reading this correctly, it tells me that the number of teas,
and the number of cakes, must be a multiple of the same number, which
is true for the solution

  (1 tea, 26 cakes)

but only if you count a 'group' as one person. (If you're willing to
argue for that interpretation, you're done with the problem.)

If you assume that a 'group' contains more than one person, then you
need a solution like 

  (T teas, C cakes) 

where T and C have a greatest common factor greater than one. 

Now, if you start listing these solutions in a table, 

   x    (133 - 3x)/5
  ---  --------------
    1   130/5 = 26 
    6   115/5 = 23
   11   100/5 = 20

you should quickly notice a pattern that you can use to add rows to
the table using only addition and subtraction...

Can you take it from here?

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