Question

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 that means santiago spends all 20 dollars?

Answer 1

To solve this problem, let's define some variables:

- Let \( b \) be the number of board books Santiago buys.
- Let \( s \) be the number of soft cover books he buys.

From the problem, we know that:
- The cost of board books is $1.50 each.
- The cost of soft cover books is $0.50 each.
- Santiago spends a total of $20.

We can set up the following equation based on the costs:
\[
1.5b + 0.5s = 20
\]

To simplify the equation, we can multiply everything by 2 to eliminate the decimals:
\[
3b + s = 40
\]

Now, we want to find solutions for \( b \) and \( s \) where both values are non-negative integers (since they can't purchase a negative number of books). Additionally, we need to make sure both types of books are included, which means \( b \geq 1 \) and \( s \geq 1 \).

Next, we can express \( s \) in terms of \( b \):
\[
s = 40 - 3b
\]

To ensure that \( s \) is a non-negative integer, we set up the following condition:
\[
40 - 3b \geq 1
\]

This gives:
\[
39 \geq 3b
\]
\[
13 \geq b
\]

So, \( b \) can take any integer value from 1 to 13 (inclusive).

Now, let's calculate the corresponding values of \( s \) for each value of \( b \):

- 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 \( b = 13 \): \( s = 40 - 3(13) = 1 \)

All of these combinations of \( (b, s) \) yield valid non-negative integers with both \( b \geq 1 \) and \( s \geq 1 \).

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