Santiago is working at his mom’s daycare this summer. His mom has asked him to check out the library book sale to look for books they might be able to use in the daycare. When he gets to the book sale, he sees that board books are $1.50 and soft cover books are $0.50. He picks out a variety of books and spends $20 find all viable solutions
Let's denote the number of board books Santiago buys as \( B \) and the number of softcover books as \( S \). Given the prices, we have the equations:
\[ 1.50B + 0.50S = 20 \]
Let's find the all viable solutions (all pairs of \( B \) and \( S \) that are whole numbers) because we can't have a fraction of a book.
First, let's express one variable in terms of the other. We can express \( S \) in terms of \( B \):
\[ S = \frac{20 - 1.50B}{0.50} \]
\[ S = 40 - 3B \]
The solutions will be valid when \( S \) and \( B \) are non-negative integers because you can’t buy a negative number of books.
Let's start by finding the maximum number of board books \( B \) that can be purchased without exceeding $20. Since board books are more expensive, we divide our total budget by the cost of one board book:
\[ B_{max} = \frac{20}{1.50} \]
\[ B_{max} \approx 13.33 \]
Since we can't buy a fraction of a book, the maximum number of board books Santiago can buy is 13.
Now, we’ll need to increment down from 13 and find the corresponding number of softcover books \( S \) using the equation \( S = 40 - 3B \) until \( S \) is no longer a positive integer.
Let’s start:
For \( B = 13 \):
\[ S = 40 - 3 \times 13 = 40 - 39 = 1 \]
Which is a viable solution: 13 board books and 1 softcover book.
For \( B = 12 \):
\[ S = 40 - 3 \times 12 = 40 - 36 = 4 \]
Another solution: 12 board books and 4 softcover books.
We continue this process decrementing \( B \) by 1 each time until \( S \) or \( B \) becomes negative:
For \( B = 11 \):
\[ S = 40 - 3 \times 11 = 40 - 33 = 7 \]
For \( B = 10 \):
\[ S = 40 - 3 \times 10 = 40 - 30 = 10 \]
We could continue this, but let's find a quicker way since we know the structure of our equation. Since we subtract three times the number of board books from 40 to get the number of softcover books, we are looking for multiples of 3 between 0 and 40. These are 0, 3, 6, 9, ... up to 39.
So we can continue:
For \( B = 9 \):
\[ S = 40 - 3 \times 9 = 40 - 27 = 13 \]
For \( B = 8 \):
\[ S = 40 - 3 \times 8 = 40 - 24 = 16 \]
And so on, all the way until:
For \( B = 0 \):
\[ S = 40 - 3 \times 0 = 40 \]
Each pair ( \( B, S \) ) we calculated is a viable solution as long as both \( B \) and \( S \) are non-negative integers and their total cost equals $20. You can use the equation to calculate the rest of the values or follow the pattern we identified:
Every time \( B \) decreases by 1, \( S \) increases by 3. This will continue until \( S \) exceeds 40 (which it will not in this scenario), or \( B \) becomes negative (which is our stopping criterion). The solutions form a straight line on a graph with \( B \) on the x-axis and \( S \) on the y-axis, where \( S \) decreases by 3 as \( B \) increases by 1.
The full list of viable solutions:
- \( B = 13 \), \( S = 1 \)
- \( B = 12 \), \( S = 4 \)
- \( B = 11 \), \( S = 7 \)
- \( B = 10 \), \( S = 10 \)
- \( B = 9 \), \( S = 13 \)
- \( B = 8 \), \( S = 16 \)
- \( B = 7 \), \( S = 19 \)
- \( B = 6 \), \( S = 22 \)
- \( B = 5 \), \( S = 25 \)
- \( B = 4 \), \( S = 28 \)
- \( B = 3 \), \( S = 31 \)
- \( B = 2 \), \( S = 34 \)
- \( B = 1 \), \( S = 37 \)
- \( B = 0 \), \( S = 40 \)
These pairs all meet the criteria, and Santiago can choose any of these combinations based on the daycare's needs.
What are all viable solutions
Apologies for the oversight. To find all viable solutions where the total cost does not surpass $20, purchasing combinations of board books at $1.50 each and softcover books at $0.50 each, you would need to express one of the variables as a function of the other and check for non-negative integer solutions. The equation to use would be:
\[ 1.50B + 0.50S = 20 \]
Let's simplify this equation by multiplying through by 2 to get rid of the decimal points:
\[ 3B + S = 40 \]
Now \( S = 40 - 3B \). We need to find values of \( B \) (the number of board books) where \( S \) (the number of softcover books) is a non-negative integer, and \( B \) is also an integer.
For \( B = 0 \):
\[ S = 40 - 3(0) = 40 \]
Solution: 0 board books, 40 softcover books.
For \( B = 1 \):
\[ S = 40 - 3(1) = 37 \]
Solution: 1 board book, 37 softcover books.
This pattern will continue, and each time \( B \) increases by 1, \( S \) will decrease by 3, as long as \( S \) remains non-negative. Let’s list all solutions:
1. \( B = 0, S = 40 \)
2. \( B = 1, S = 37 \)
3. \( B = 2, S = 34 \)
4. \( B = 3, S = 31 \)
5. \( B = 4, S = 28 \)
6. \( B = 5, S = 25 \)
7. \( B = 6, S = 22 \)
8. \( B = 7, S = 19 \)
9. \( B = 8, S = 16 \)
10. \( B = 9, S = 13 \)
11. \( B = 10, S = 10 \)
12. \( B = 11, S = 7 \)
13. \( B = 12, S = 4 \)
14. \( B = 13, S = 1 \)
The process ends once \( S \) will result in a negative number, which is when \( B \) exceeds 13.
So the viable solutions for Santiago to buy board books and softcover books for $20 are all the pairs listed above.
To find all the viable solutions, we need to determine the combinations of board books and softcover books that Santiago can purchase within his budget of $20.
Let's start by assuming that Santiago will buy a certain number of board books and a certain number of softcover books. We can represent this with two variables:
- Let's say 'x' represents the number of board books he buys.
- And 'y' represents the number of softcover books he buys.
Now, we need to set up an equation to represent the total cost of Santiago's purchase. Since board books cost $1.50 each and softcover books cost $0.50 each, the equation becomes:
1.50x + 0.50y = 20
To find the viable solutions, we need to determine the integer values of x and y that satisfy this equation, as Santiago cannot purchase a fraction of a book.
One way to find the viable solutions is by using a technique called trial and error. We can try different values for x and then calculate the corresponding value of y that satisfies the equation.
Let's try some values of x and calculate the corresponding y:
For x = 1:
1.50(1) + 0.50y = 20
1.50 + 0.50y = 20
0.50y = 18.50
y = 37 (which is not an integer, so this solution is not viable)
For x = 2:
1.50(2) + 0.50y = 20
3.00 + 0.50y = 20
0.50y = 17.00
y = 34 (which is an integer, so this solution is viable)
For x = 3:
1.50(3) + 0.50y = 20
4.50 + 0.50y = 20
0.50y = 15.50
y = 31 (which is an integer, so this solution is viable)
We can continue this process to find more viable solutions by trying different values of x and solving for y.
By following this method, Santiago can find all the viable solutions by trying different values of x and solving for y in the equation 1.50x + 0.50y = 20, making sure that y results in an integer value.