I don't think you understood the question, sorry. Lets break up $9.90. With $9, you can buy 9 sets of 3 which is 27. You have 90 cents left, which you can buy one more 50 cent book.

Thank you for the clarification. Let's go through the solution with this approach.

### Step 1: Spend $9 on Sets of 3 Books
With $9, you can buy as many sets of 3 books as possible:
- **Cost per set of 3 books:** $1
- **Number of sets you can buy with $9:** 
  \[
  \frac{9}{1} = 9 \text{ sets}
  \]
- **Number of books in 9 sets:**
  \[
  9 \times 3 = 27 \text{ books}
  \]

### Step 2: Spend the Remaining $0.90
You have $0.90 left. Each book costs $0.50 if bought individually:
- **Number of books you can buy with $0.90:**
  \[
  \frac{0.90}{0.50} = 1 \text{ book (with $0.40 left over)}
  \]

### Total Number of Books
- **Total number of books:**
  \[
  27 + 1 = 28 \text{ books}
  \]

So, the greatest number of old books you can buy with $9.90 is **28 books**.

Would you like more details or have any other questions?

Here are some related questions:

1. What if you had $10 instead of $9.90? How many books could you buy?
2. How much would it cost to buy exactly 30 books?
3. What is the maximum number of books you could buy with $8?
4. If the deal was "5 books for $2," how would the number of books change with $9.90?
5. What is the minimum amount of money needed to buy 50 books?
6. How many books could you buy with $4.50 using both deals?
7. How much more would it cost to buy 35 books compared to 28 books?
8. What would happen if the deal was "2 for $1"?

**Tip:** When faced with a budget problem, break it down by using the best available deal first and then allocate any remaining funds.

Explore the solution to breaking down $9.90 to maximize book purchases. Learn how to allocate $9 to buy sets of 3 books and spend the remaining $0.90 effectively. Suitable for Grades 4-6, covering arithmetic, division, and decimal calculations.

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