Question
The table shows the total number of hamburgers and of hot dogs sold at a food stand at a local fair on two separate days. It also shows the dollar amount taken in each day.
Hamburgers Hot dogs Total
Day 1 200 150 $1,450
Day 2 200 250 $1,750
What is the cost of a hamburger, and what is the cost of a hot dog?
Enter your answers in the boxes.
Hamburger: $
Hot dog: $
Hamburgers Hot dogs Total
Day 1 200 150 $1,450
Day 2 200 250 $1,750
What is the cost of a hamburger, and what is the cost of a hot dog?
Enter your answers in the boxes.
Hamburger: $
Hot dog: $
Answers
GPT-4o mini
To find the cost of a hamburger and a hot dog, we can set up a system of equations based on the information provided.
Let \( H \) be the cost of a hamburger and \( D \) be the cost of a hot dog.
From Day 1, the equations are:
\[
200H + 150D = 1450
\]
From Day 2, the equations are:
\[
200H + 250D = 1750
\]
Now we can solve this system of equations. First, we can simplify both equations.
Starting from Day 1:
\[
200H + 150D = 1450 \quad \text{(1)}
\]
From Day 2:
\[
200H + 250D = 1750 \quad \text{(2)}
\]
Next, we can eliminate \( 200H \) by subtracting equation (1) from equation (2):
\[
(200H + 250D) - (200H + 150D) = 1750 - 1450
\]
This simplifies to:
\[
250D - 150D = 300
\]
\[
100D = 300
\]
So, we find:
\[
D = \frac{300}{100} = 3
\]
Now that we have the cost of a hot dog \( D = 3 \), we can substitute \( D \) back into one of the original equations to find \( H \). We can use equation (1):
\[
200H + 150(3) = 1450
\]
This simplifies to:
\[
200H + 450 = 1450
\]
Subtracting 450 from both sides:
\[
200H = 1450 - 450
\]
\[
200H = 1000
\]
Dividing both sides by 200:
\[
H = \frac{1000}{200} = 5
\]
Thus, the cost of a hamburger is $5 and the cost of a hot dog is $3.
So, the final answers are:
- Hamburger: $5
- Hot dog: $3
Let \( H \) be the cost of a hamburger and \( D \) be the cost of a hot dog.
From Day 1, the equations are:
\[
200H + 150D = 1450
\]
From Day 2, the equations are:
\[
200H + 250D = 1750
\]
Now we can solve this system of equations. First, we can simplify both equations.
Starting from Day 1:
\[
200H + 150D = 1450 \quad \text{(1)}
\]
From Day 2:
\[
200H + 250D = 1750 \quad \text{(2)}
\]
Next, we can eliminate \( 200H \) by subtracting equation (1) from equation (2):
\[
(200H + 250D) - (200H + 150D) = 1750 - 1450
\]
This simplifies to:
\[
250D - 150D = 300
\]
\[
100D = 300
\]
So, we find:
\[
D = \frac{300}{100} = 3
\]
Now that we have the cost of a hot dog \( D = 3 \), we can substitute \( D \) back into one of the original equations to find \( H \). We can use equation (1):
\[
200H + 150(3) = 1450
\]
This simplifies to:
\[
200H + 450 = 1450
\]
Subtracting 450 from both sides:
\[
200H = 1450 - 450
\]
\[
200H = 1000
\]
Dividing both sides by 200:
\[
H = \frac{1000}{200} = 5
\]
Thus, the cost of a hamburger is $5 and the cost of a hot dog is $3.
So, the final answers are:
- Hamburger: $5
- Hot dog: $3