Question

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

Answer 1

Let's denote the number of board books Santiago buys as \( b \) and the number of soft cover books as \( s \).

The cost of board books is $1.50 each, so the total cost for the board books is \( 1.50b \).
The cost of soft cover books is $0.50 each, so the total cost for the soft cover books is \( 0.50s \).

According to the problem, the total amount Santiago spends is $20. We can set up the following equation based on this information:

\[
1.50b + 0.50s = 20
\]

To simplify this equation, we can multiply everything by 2 to eliminate the decimals:

\[
3b + s = 40
\]

Now, we have a linear equation in two variables. To find possible solutions, we can express \( s \) in terms of \( b \):

\[
s = 40 - 3b
\]

Since both \( b \) and \( s \) represent quantities of books, they must both be non-negative integers.

Now we can find possible integer values for \( b \):

1. If \( b = 0 \):
\[
s = 40 - 3(0) = 40 \quad (\text{Total cost } = 0 \times 1.50 + 40 \times 0.50 = 20)
\]

2. If \( b = 1 \):
\[
s = 40 - 3(1) = 37 \quad (\text{Total cost } = 1 \times 1.50 + 37 \times 0.50 = 20)
\]

3. If \( b = 2 \):
\[
s = 40 - 3(2) = 34 \quad (\text{Total cost } = 2 \times 1.50 + 34 \times 0.50 = 20)
\]

4. If \( b = 3 \):
\[
s = 40 - 3(3) = 31 \quad (\text{Total cost } = 3 \times 1.50 + 31 \times 0.50 = 20)
\]

5. If \( b = 4 \):
\[
s = 40 - 3(4) = 28 \quad (\text{Total cost } = 4 \times 1.50 + 28 \times 0.50 = 20)
\]

6. If \( b = 5 \):
\[
s = 40 - 3(5) = 25 \quad (\text{Total cost } = 5 \times 1.50 + 25 \times 0.50 = 20)
\]

7. If \( b = 6 \):
\[
s = 40 - 3(6) = 22 \quad (\text{Total cost } = 6 \times 1.50 + 22 \times 0.50 = 20)
\]

8. If \( b = 7 \):
\[
s = 40 - 3(7) = 19 \quad (\text{Total cost } = 7 \times 1.50 + 19 \times 0.50 = 20)
\]

9. If \( b = 8 \):
\[
s = 40 - 3(8) = 16 \quad (\text{Total cost } = 8 \times 1.50 + 16 \times 0.50 = 20)
\]

10. If \( b = 9 \):
\[
s = 40 - 3(9) = 13 \quad (\text{Total cost } = 9 \times 1.50 + 13 \times 0.50 = 20)
\]

11. If \( b = 10 \):
\[
s = 40 - 3(10) = 10 \quad (\text{Total cost } = 10 \times 1.50 + 10 \times 0.50 = 20)
\]

12. If \( b = 11 \):
\[
s = 40 - 3(11) = 7 \quad (\text{Total cost } = 11 \times 1.50 + 7 \times 0.50 = 20)
\]

13. If \( b = 12 \):
\[
s = 40 - 3(12) = 4 \quad (\text{Total cost } = 12 \times 1.50 + 4 \times 0.50 = 20)
\]

14. If \( b = 13 \):
\[
s = 40 - 3(13) = 1 \quad (\text{Total cost } = 13 \times 1.50 + 1 \times 0.50 = 20)
\]

15. If \( b = 14 \):
\[
s = 40 - 3(14) = -2 \quad (\text{Not allowed since }\ s \text{ cannot be negative})
\]

Now, we can list the valid integer pairs \( (b, s) \):

- \( (0, 40) \)
- \( (1, 37) \)
- \( (2, 34) \)
- \( (3, 31) \)
- \( (4, 28) \)
- \( (5, 25) \)
- \( (6, 22) \)
- \( (7, 19) \)
- \( (8, 16) \)
- \( (9, 13) \)
- \( (10, 10) \)
- \( (11, 7) \)
- \( (12, 4) \)
- \( (13, 1) \)

These pairs represent the different quantities of board books and soft cover books Santiago could buy with $20.