Question: DATE: 8/7/26
2. A farmer has $3 \frac{3}{5}$ liters of organic fertilizer he pours it into containers holding $\frac{3}{10}$ liter each. He uses $\frac{1}{3}$ of the containers. How many pieces remain?
1. **State the problem:**
A farmer has $3 \frac{3}{5}$ liters of fertilizer and pours it into containers each holding $\frac{3}{10}$ liter. He uses $\frac{1}{3}$ of the containers. We need to find how many containers remain unused.
2. **Convert mixed number to improper fraction:**
$$3 \frac{3}{5} = \frac{3 \times 5 + 3}{5} = \frac{15 + 3}{5} = \frac{18}{5}$$
3. **Find total number of containers:**
Divide total fertilizer by container size:
$$\frac{18}{5} \div \frac{3}{10} = \frac{18}{5} \times \frac{10}{3} = \frac{18 \times 10}{5 \times 3} = \frac{180}{15}$$
4. **Simplify the fraction:**
$$\frac{180}{15} = \cancel{\frac{180}{15}} = 12$$
So, there are $12$ containers in total.
5. **Calculate number of containers used:**
He uses $\frac{1}{3}$ of the containers:
$$12 \times \frac{1}{3} = 4$$
6. **Calculate remaining containers:**
$$12 - 4 = 8$$
**Answer:** There are $8$ containers remaining unused.