SOLUTION: Jimmy has $200 to buy dessert. Pies cost $10 each, cakes $20. 1 pie feeds 12 people, 1 cake feeds 24. There are 210 people coming. Right 4 inequalities/equations. - This isn't t

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Jimmy has $200 to buy dessert. Pies cost $10 each, cakes $20. 1 pie feeds 12 people, 1 cake feeds 24. There are 210 people coming. Right 4 inequalities/equations. - This isn't t      Log On


   

Question 1150683: Jimmy has $200 to buy dessert. Pies cost $10 each, cakes $20. 1 pie feeds 12 people, 1 cake feeds 24. There are 210 people coming. Right 4 inequalities/equations.
- This isn't the exact problem, but all of the values and variables are correct, but I am pretty sure it is 4 equations/inequalities, and it needs to be solved.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There are only two unknowns -- the number of pies and the number of cakes. With two unknowns you only need two equations or inequalities.

The statement of the problem doesn't make it clear whether we are to use equations or inequalities; so let's try equations first.

Let p = number of pies
Let c = number of cakes

The total cost of the desserts, at $10 per pie and $20 per cake, must be $200:

10p%2B20c+=+200

The number of people to be served, at 12 per pie and 24 per cake, is 210:

12p%2B24c+=+210

The first equation is equivalent to c%2B2p+=+20; the second equation is equivalent to c%2B2p+=+17.5

Obviously there is no common solution to those two equations.

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.

Then the inequalities are equivalent to c%2B2p+%3C=+20 and c%2B2p+%3E=+17.5.

And that pair of inequalities is equivalent to the compound inequality

17.5+%3C=+c%2B2p+%3C=+20

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.

There are numerous such pairs; here are a few of them....
   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
     ...