Question 1150683
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There are only two unknowns -- the number of pies and the number of cakes.  With two unknowns you only need two equations or inequalities.<br>
The statement of the problem doesn't make it clear whether we are to use equations or inequalities; so let's try equations first.<br>
Let p = number of pies
Let c = number of cakes<br>
The total cost of the desserts, at $10 per pie and $20 per cake, must be $200:<br>
{{{10p+20c = 200}}}<br>
The number of people to be served, at 12 per pie and 24 per cake, is 210:<br>
{{{12p+24c = 210}}}<br>
The first equation is equivalent to {{{c+2p = 20}}}; the second equation is equivalent to {{{c+2p = 17.5}}}<br>
Obviously there is no common solution to those two equations.<br>
So apparently we are to use inequalities -- the total cost must be AT MOST $200; and the number of people to be fed must be AT LEAST 210.<br>
Then the inequalities are equivalent to {{{c+2p <= 20}}} and {{{c+2p >= 17.5}}}.<br>
And that pair of inequalities is equivalent to the compound inequality<br>
{{{17.5 <= c+2p <= 20}}}<br>
We can assume that the numbers of pies and cakes must be whole numbers.  So any pair of whole numbers c and p that satisfy that inequality are the possible numbers of pies and cakes.<br>
There are numerous such pairs; here are a few of them....<br><pre>
   pies  cakes        cost          people served
  ----------------------------------------------------
     0     9     0(10)+ 9(20) = 180   0(12)+ 9(24) = 216
     0    10     0(10)+10(20) = 200   0(12)+10(24) = 240
     1     9     1(10)+ 9(20) = 190   1(12)+ 9(24) = 228
     2     8     2(10)+ 8(20) = 180   2(12)+ 8(24) = 216
     2     9     2(10)+ 9(20) = 200   2(12)+10(24) = 264
     ...
    18     0    18(10)+ 0(20) = 180  18(12)+ 0(24) = 216
    18     1    18(10)+ 1(20) = 200  18(12)+ 1(24) = 240
     ...</pre>