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The Baynard High School Boosters are selling hot dogs and pretzels at the concession stand. They have 300 hot dogs and pretzels...Asked by .....
The Baynard High School Boosters are selling hot dogs and pretzels at the concession stand. They have 300 hot dogs and pretzels in stock and want to earn an income of $800. They plan to sell the hot dogs for $4 each and the pretzels for $2 each. How many hot dogs and how many pretzels do the boosters need to sell to reach their goal of $800?(1 point)
Responses
50 hot dogs and 250 pretzels
50 hot dogs and 250 pretzels
100 hot dogs and 200 pretzels
100 hot dogs and 200 pretzels
250 hot dogs and 50 pretzels
250 hot dogs and 50 pretzels
200 hot dogs and 100 pretzels
Responses
50 hot dogs and 250 pretzels
50 hot dogs and 250 pretzels
100 hot dogs and 200 pretzels
100 hot dogs and 200 pretzels
250 hot dogs and 50 pretzels
250 hot dogs and 50 pretzels
200 hot dogs and 100 pretzels
Answers
Answered by
GPT-4o mini
Let's denote the number of hot dogs sold as \( h \) and the number of pretzels sold as \( p \).
We have two conditions:
1. The total number of items sold (hot dogs and pretzels) must equal 300:
\[
h + p = 300
\]
2. The total income must be $800. The income from hot dogs sold at $4 each and pretzels sold at $2 each gives the equation:
\[
4h + 2p = 800
\]
Now we can solve these equations simultaneously.
From the first equation, we can express \( p \) in terms of \( h \):
\[
p = 300 - h
\]
Next, substitute \( p \) in the second equation:
\[
4h + 2(300 - h) = 800
\]
Expanding this:
\[
4h + 600 - 2h = 800
\]
Combining like terms:
\[
2h + 600 = 800
\]
Now subtract 600 from both sides:
\[
2h = 200
\]
Dividing by 2:
\[
h = 100
\]
Now substitute \( h \) back into the equation for \( p \):
\[
p = 300 - 100 = 200
\]
So the solution is:
- Hot dogs: 100
- Pretzels: 200
Thus, the boosters need to sell **100 hot dogs and 200 pretzels**.
The correct response is:
**100 hot dogs and 200 pretzels**.
We have two conditions:
1. The total number of items sold (hot dogs and pretzels) must equal 300:
\[
h + p = 300
\]
2. The total income must be $800. The income from hot dogs sold at $4 each and pretzels sold at $2 each gives the equation:
\[
4h + 2p = 800
\]
Now we can solve these equations simultaneously.
From the first equation, we can express \( p \) in terms of \( h \):
\[
p = 300 - h
\]
Next, substitute \( p \) in the second equation:
\[
4h + 2(300 - h) = 800
\]
Expanding this:
\[
4h + 600 - 2h = 800
\]
Combining like terms:
\[
2h + 600 = 800
\]
Now subtract 600 from both sides:
\[
2h = 200
\]
Dividing by 2:
\[
h = 100
\]
Now substitute \( h \) back into the equation for \( p \):
\[
p = 300 - 100 = 200
\]
So the solution is:
- Hot dogs: 100
- Pretzels: 200
Thus, the boosters need to sell **100 hot dogs and 200 pretzels**.
The correct response is:
**100 hot dogs and 200 pretzels**.
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