Question

Santiago is working at his mom’s day care 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 day care. 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. how many viable solutions that include both board books and soft cover books are there?

Answer 1

To determine the number of viable solutions for Santiago's purchase of board books and soft cover books, we can set up a mathematical equation based on the problem.

Let's define:
- \( b \) = number of board books
- \( s \) = number of soft cover books

From the information given, we know that:
- Each board book costs $1.50
- Each soft cover book costs $0.50
- Santiago spends a total of $20

We can set up the following equation based on the prices and total spent:

\[
1.5b + 0.5s = 20
\]

To simplify calculations, multiply the entire equation by 2 to eliminate decimal fractions:

\[
3b + s = 40
\]

Now, we need to express the number of solutions involving both board books and soft cover books. First, we can express \( s \) in terms of \( b \):

\[
s = 40 - 3b
\]

Now we need to find the constraints on \( b \) and \( s \). Since both \( b \) and \( s \) must be non-negative integers, we can evaluate the conditions:

1. Since \( s \) must be non-negative:
\[
40 - 3b \geq 0 \implies 40 \geq 3b \implies \frac{40}{3} \geq b \implies b \leq 13.33
\]
Since \( b \) must be an integer, the maximum value for \( b \) is 13.

2. Since \( b \) must also be non-negative:
\[
b \geq 0
\]

Now we can find the possible integer values for \( b \):
- The possible integer values for \( b \) are \( 0, 1, 2, \ldots, 13 \).

Next, we find the corresponding \( s \) values to ensure \( s \) is also a non-negative integer:
- If \( b = 0 \), \( s = 40 - 3(0) = 40 \)
- If \( b = 1 \), \( s = 40 - 3(1) = 37 \)
- If \( b = 2 \), \( s = 40 - 3(2) = 34 \)
- If \( b = 3 \), \( s = 40 - 3(3) = 31 \)
- If \( b = 4 \), \( s = 40 - 3(4) = 28 \)
- If \( b = 5 \), \( s = 40 - 3(5) = 25 \)
- If \( b = 6 \), \( s = 40 - 3(6) = 22 \)
- If \( b = 7 \), \( s = 40 - 3(7) = 19 \)
- If \( b = 8 \), \( s = 40 - 3(8) = 16 \)
- If \( b = 9 \), \( s = 40 - 3(9) = 13 \)
- If \( b = 10 \), \( s = 40 - 3(10) = 10 \)
- If \( b = 11 \), \( s = 40 - 3(11) = 7 \)
- If \( b = 12 \), \( s = 40 - 3(12) = 4 \)
- If \( b = 13 \), \( s = 40 - 3(13) = 1 \)

Now we need to ensure that each solution includes at least one of each type of book (both board and soft cover):

1. **Board books requirement:** \( b \geq 1 \)
2. **Soft cover books requirement:** \( s \geq 1 \)

The viable values for \( b \) satisfying both conditions (and \( s \geq 1 \)) are:
- \( b = 1 \): \( s = 37 \) (valid)
- \( b = 2 \): \( s = 34 \) (valid)
- \( b = 3 \): \( s = 31 \) (valid)
- \( b = 4 \): \( s = 28 \) (valid)
- \( b = 5 \): \( s = 25 \) (valid)
- \( b = 6 \): \( s = 22 \) (valid)
- \( b = 7 \): \( s = 19 \) (valid)
- \( b = 8 \): \( s = 16 \) (valid)
- \( b = 9 \): \( s = 13 \) (valid)
- \( b = 10 \): \( s = 10 \) (valid)
- \( b = 11 \): \( s = 7 \) (valid)
- \( b = 12 \): \( s = 4 \) (valid)
- \( b = 13 \): \( s = 1 \) (valid)

Thus, the values of \( b \) from 1 to 13 give us:

- There are \( 13 - 1 + 1 = 13 \) choices for \( b \), all resulting in valid \( s \) satisfying both book requirements.

Therefore, the total number of viable solutions that include both board books and soft cover books is **13**.