Solve the following Word Problems. Write equations and then solve the equations. Some of the questions were adapted from (Seeliger \& Mouton, n.d.-a, p. 99) and (Loots et al., 2011, p. 55) and (Seeliger \& Mouton, n.d.-b, p. 34) and (Bothma, 2015, p. 175). a. A tank is \( \frac{2}{3} \) full of water. If 35 litres are tapped off, the tank is \( \frac{1}{6} \) full. Determine the volume of water in a full tank.

Sure, let's solve the problem step by step by setting up and solving the appropriate equations. ### **Problem Statement** A tank is \( \frac{2}{3} \) full of water. If 35 liters are tapped off, the tank is \( \frac{1}{6} \) full. Determine the volume of water in a full tank. ### **Step 1: Define the Variable** Let: - \( V \) = Total volume of the tank when it's completely full (in liters). ### **Step 2: Express the Given Information in Terms of \( V \)** 1. **Initial Volume of Water:** - The tank is \( \frac{2}{3} \) full. - Volume of water initially = \( \frac{2}{3}V \). 2. **Volume After Tapping Off 35 Liters:** - After removing 35 liters, the tank becomes \( \frac{1}{6} \) full. - Volume of water after tapping off = \( \frac{1}{6}V \). ### **Step 3: Set Up the Equation** The difference between the initial volume and the volume after tapping off 35 liters should equal 35 liters: \[ \frac{2}{3}V - 35 = \frac{1}{6}V \] ### **Step 4: Solve the Equation for \( V \)** 1. **Subtract \( \frac{1}{6}V \) from both sides to isolate terms with \( V \):** \[ \frac{2}{3}V - \frac{1}{6}V = 35 \] 2. **Find a common denominator to combine the terms:** \[ \frac{4}{6}V - \frac{1}{6}V = 35 \] \[ \frac{3}{6}V = 35 \] \[ \frac{1}{2}V = 35 \] 3. **Multiply both sides by 2 to solve for \( V \):** \[ V = 35 \times 2 \] \[ V = 70 \, \text{liters} \] ### **Conclusion** The volume of water in a full tank is **70 liters**.