Question: 20. The mass of a container is $116$ kg when it is $\frac{5}{6}$ filled with sand.
When the same container is half filled with sand, its mass is $78$ kg.
Find the mass of the empty container.
$78 \div 3 = 26$
$26 = \frac{1}{6}$
1. **State the problem:** We have a container partially filled with sand. When it is $\frac{5}{6}$ full, the total mass is $116$ kg. When it is half full, the total mass is $78$ kg. We want to find the mass of the empty container.
2. **Define variables:** Let $m$ be the mass of the empty container, and $s$ be the mass of sand that fills the container completely.
3. **Write equations:**
- When $\frac{5}{6}$ full: $$m + \frac{5}{6}s = 116$$
- When $\frac{1}{2}$ full: $$m + \frac{1}{2}s = 78$$
4. **Subtract the second equation from the first:**
$$\left(m + \frac{5}{6}s\right) - \left(m + \frac{1}{2}s\right) = 116 - 78$$
$$\cancel{m} + \frac{5}{6}s - \cancel{m} - \frac{1}{2}s = 38$$
5. **Simplify the left side:**
$$\frac{5}{6}s - \frac{1}{2}s = 38$$
Find common denominator $6$:
$$\frac{5}{6}s - \frac{3}{6}s = 38$$
$$\frac{2}{6}s = 38$$
6. **Simplify fraction:**
$$\frac{\cancel{2}}{\cancel{6}}s = 38 \Rightarrow \frac{1}{3}s = 38$$
7. **Solve for $s$:**
$$s = 38 \times 3 = 114$$
8. **Substitute $s$ back into one equation to find $m$:**
Using $$m + \frac{1}{2}s = 78$$
$$m + \frac{1}{2} \times 114 = 78$$
$$m + 57 = 78$$
9. **Solve for $m$:**
$$m = 78 - 57 = 21$$
**Final answer:** The mass of the empty container is $21$ kg.