Question
If the birthday party at the science center is $150, and it is $10.50 per child and $5.50 per adult, how many children can attend with $300, assuming there is at least one adult?
Answers
GPT-4o mini
To solve the problem, we first establish that the total cost of the birthday party is $150, plus additional costs for children and adults.
Let \( c \) represent the number of children and \( a \) represent the number of adults.
The costs can be expressed with the following equation:
\[
\text{Total cost} = 150 + 10.50c + 5.50a
\]
Given that the total amount available is $300, we can write:
\[
150 + 10.50c + 5.50a \leq 300.
\]
Subtracting 150 from both sides gives us:
\[
10.50c + 5.50a \leq 150.
\]
This can be simplified for easier calculations:
\[
10.50c + 5.50a = 150.
\]
Next, we will explore how many children can attend by assigning a value to \( a \), making sure to keep \( a \) at least 1 (since there must be at least one adult).
First, let's express \( a \) in terms of \( c \):
\[
5.50a = 150 - 10.50c \quad \Rightarrow \quad a = \frac{150 - 10.50c}{5.50}.
\]
We need to ensure \( a \) is a positive integer, so we have:
\[
150 - 10.50c > 0 \quad \Rightarrow \quad 150 > 10.50c \quad \Rightarrow \quad c < \frac{150}{10.50} \approx 14.2857.
\]
Since \( c \) must be an integer, the maximum number of children that can attend is \( 14 \).
Now we will try the maximum number of children, \( c = 14 \):
\[
a = \frac{150 - 10.50 \cdot 14}{5.50} = \frac{150 - 147}{5.50} = \frac{3}{5.50} \approx 0.545,
\]
which is not valid since \( a \) must be at least 1.
Next, we try \( c = 13 \):
\[
a = \frac{150 - 10.50 \cdot 13}{5.50} = \frac{150 - 136.5}{5.50} = \frac{13.5}{5.50} \approx 2.4545,
\]
still not an integer.
Next, we try \( c = 12 \):
\[
a = \frac{150 - 10.50 \cdot 12}{5.50} = \frac{150 - 126}{5.50} = \frac{24}{5.50} \approx 4.3636,
\]
still not an integer.
Next, we try \( c = 11 \):
\[
a = \frac{150 - 10.50 \cdot 11}{5.50} = \frac{150 - 115.5}{5.50} = \frac{34.5}{5.50} = 6.2727,
\]
still not an integer.
Next, we try \( c = 10 \):
\[
a = \frac{150 - 10.50 \cdot 10}{5.50} = \frac{150 - 105}{5.50} = \frac{45}{5.50} \approx 8.1818,
\]
still not an integer.
Next, we try \( c = 9 \):
\[
a = \frac{150 - 10.50 \cdot 9}{5.50} = \frac{150 - 94.5}{5.50} = \frac{55.5}{5.50} = 10.0909,
\]
still not an integer.
Next, we try \( c = 8 \):
\[
a = \frac{150 - 10.50 \cdot 8}{5.50} = \frac{150 - 84}{5.50} = \frac{66}{5.50} \approx 12,
\]
which is a valid integer.
Now, checking for \( c = 8 \) and \( a = 12 \):
- Calculate total cost:
\[
\text{Total cost} = 150 + 10.50 \cdot 8 + 5.50 \cdot 12 = 150 + 84 + 66 = 300,
\]
and it fits within our budget with exactly one adult.
Continuing downwards for even more children, we can check for \( c = 7 \):
\[
a = \frac{150 - 10.50 \cdot 7}{5.50} = \frac{150 - 73.5}{5.50} = \frac{76.5}{5.50} \approx 13.9091,
\]
invalid.
Thus, and up to now, the best possible solution while keeping adults is:
- At least 1 adult and \( 8 \) children.
So, the maximum number of children that can attend with $300, while having at least one adult is:
\[
\boxed{8}.
\]
Let \( c \) represent the number of children and \( a \) represent the number of adults.
The costs can be expressed with the following equation:
\[
\text{Total cost} = 150 + 10.50c + 5.50a
\]
Given that the total amount available is $300, we can write:
\[
150 + 10.50c + 5.50a \leq 300.
\]
Subtracting 150 from both sides gives us:
\[
10.50c + 5.50a \leq 150.
\]
This can be simplified for easier calculations:
\[
10.50c + 5.50a = 150.
\]
Next, we will explore how many children can attend by assigning a value to \( a \), making sure to keep \( a \) at least 1 (since there must be at least one adult).
First, let's express \( a \) in terms of \( c \):
\[
5.50a = 150 - 10.50c \quad \Rightarrow \quad a = \frac{150 - 10.50c}{5.50}.
\]
We need to ensure \( a \) is a positive integer, so we have:
\[
150 - 10.50c > 0 \quad \Rightarrow \quad 150 > 10.50c \quad \Rightarrow \quad c < \frac{150}{10.50} \approx 14.2857.
\]
Since \( c \) must be an integer, the maximum number of children that can attend is \( 14 \).
Now we will try the maximum number of children, \( c = 14 \):
\[
a = \frac{150 - 10.50 \cdot 14}{5.50} = \frac{150 - 147}{5.50} = \frac{3}{5.50} \approx 0.545,
\]
which is not valid since \( a \) must be at least 1.
Next, we try \( c = 13 \):
\[
a = \frac{150 - 10.50 \cdot 13}{5.50} = \frac{150 - 136.5}{5.50} = \frac{13.5}{5.50} \approx 2.4545,
\]
still not an integer.
Next, we try \( c = 12 \):
\[
a = \frac{150 - 10.50 \cdot 12}{5.50} = \frac{150 - 126}{5.50} = \frac{24}{5.50} \approx 4.3636,
\]
still not an integer.
Next, we try \( c = 11 \):
\[
a = \frac{150 - 10.50 \cdot 11}{5.50} = \frac{150 - 115.5}{5.50} = \frac{34.5}{5.50} = 6.2727,
\]
still not an integer.
Next, we try \( c = 10 \):
\[
a = \frac{150 - 10.50 \cdot 10}{5.50} = \frac{150 - 105}{5.50} = \frac{45}{5.50} \approx 8.1818,
\]
still not an integer.
Next, we try \( c = 9 \):
\[
a = \frac{150 - 10.50 \cdot 9}{5.50} = \frac{150 - 94.5}{5.50} = \frac{55.5}{5.50} = 10.0909,
\]
still not an integer.
Next, we try \( c = 8 \):
\[
a = \frac{150 - 10.50 \cdot 8}{5.50} = \frac{150 - 84}{5.50} = \frac{66}{5.50} \approx 12,
\]
which is a valid integer.
Now, checking for \( c = 8 \) and \( a = 12 \):
- Calculate total cost:
\[
\text{Total cost} = 150 + 10.50 \cdot 8 + 5.50 \cdot 12 = 150 + 84 + 66 = 300,
\]
and it fits within our budget with exactly one adult.
Continuing downwards for even more children, we can check for \( c = 7 \):
\[
a = \frac{150 - 10.50 \cdot 7}{5.50} = \frac{150 - 73.5}{5.50} = \frac{76.5}{5.50} \approx 13.9091,
\]
invalid.
Thus, and up to now, the best possible solution while keeping adults is:
- At least 1 adult and \( 8 \) children.
So, the maximum number of children that can attend with $300, while having at least one adult is:
\[
\boxed{8}.
\]