Question

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?

Answer 1

The concession stand sold `46` hot dogs and `32` 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 `d`.
• We don't know how many hamburgers the concession stand sold, so we will call that number `h`.

Now we translate the statements into algebraic equations:

• The number of hot dogs and hamburgers that were sold is `78`, so `d+h=78`.
• If each hot dog is sold for `1.25`, then the total revenue from hot dogs is given by `1.25d`. In the same way, the total revenue from hamburgers is `1.50h`. The total revenue from both hot dogs and hamburgers should be the sum of these, and since we are told the total revenue is `105.50`, we can say `1.25d+1.5h=105.5`.

We now have a system of two linear equations:
`d+h=78`
`1.25d+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 `d`:
`d+h=78`
`->d=78-h`

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

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

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

The concession stand therefore sold `46` hot dogs and `32` hamburgers.