The concession stand is selling hot dogs and hamburgers during a game. At halftime, they sold a total of 78 hot dogs and hamburgers and brought in $105.50. how many of each item did they sell if hamburgers sold for $1.50 and hot dogs sold for $1.25?

1 Answer
Jul 5, 2016

The concession stand sold 4646 hot dogs and 3232 hamburgers.

Explanation:

The first thing to do in algebraic problems is assign variables to things we don't know, so let's start there:

  • We don't know how many hot dogs the concession stand sold, so we will call that number dd.
  • We don't know how many hamburgers the concession stand sold, so we will call that number hh.

Now we translate the statements into algebraic equations:

  • The number of hot dogs and hamburgers that were sold is 7878, so d+h=78d+h=78.
  • If each hot dog is sold for 1.251.25, then the total revenue from hot dogs is given by 1.25d1.25d. In the same way, the total revenue from hamburgers is 1.50h1.50h. The total revenue from both hot dogs and hamburgers should be the sum of these, and since we are told the total revenue is 105.50105.50, we can say 1.25d+1.5h=105.51.25d+1.5h=105.5.

We now have a system of two linear equations:
d+h=78d+h=78
1.25d+1.5h=105.51.25d+1.5h=105.5

We can solve it using several methods, though I'm going to go with substitution. Use the first equation to solve for dd:
d+h=78d+h=78
->d=78-hd=78h

Now plug this in for dd in the second equation:
1.25d+1.5h=105.51.25d+1.5h=105.5
->1.25(78-h)+1.5h=105.51.25(78h)+1.5h=105.5

Solving for hh, we have:
97.5-1.25h+1.5h=105.597.51.25h+1.5h=105.5
0.25h=80.25h=8
h=8/.25->h=32h=8.25h=32

Since h+d=78h+d=78,,
32+d=78->d=4632+d=78d=46

The concession stand therefore sold 4646 hot dogs and 3232 hamburgers.