A hot dog cart at the entrance of the Stadium sells hot dogs ($3.50) and hamburger ($5.25) for a total of $1,225.00. If it sells twice as many hot dogs as hamburgers, how many of each did it sell?
3.50x + 5.25y =2(3.50x)
3.50x + 5.25y = 1,225.00
Use Elimination Method
-3.50( 3.50x + 5.25y ) = -3.50(7)
-12.25x - 18.375y = -24.5
3.50 + 5.25y = 1,225.00
Hmm. I get
x = 2y
3.50x + 5.25y = 1,225.00
Having twice as many dogs as burgers has nothing to do with their prices.
To solve this problem, we can use a system of equations. Let's start by defining some variables:
Let H represent the number of hamburgers sold.
Let D represent the number of hot dogs sold.
According to the problem statement, the hot dog cart sells hot dogs for $3.50 and hamburgers for $5.25, with a total revenue of $1,225.00. We can express this information in the form of an equation:
3.50D + 5.25H = 1225.00
Additionally, the problem states that the cart sells twice as many hot dogs as hamburgers:
D = 2H
Now we have a system of two equations. We can substitute the value of D from the second equation into the first equation:
3.50(2H) + 5.25H = 1225.00
Expanding and simplifying:
7H + 5.25H = 1225.00
12.25H = 1225.00
H = 1225.00 / 12.25
H ≈ 100
So, the hot dog cart sold approximately 100 hamburgers. Since the number of hot dogs sold is twice the number of hamburgers, there were 2 × 100 = 200 hot dogs sold.
Therefore, the hot dog cart sold 100 hamburgers and 200 hot dogs.