1. **State the problem.** Alan spent $\frac{7}{10}$ of his allowance on food, then spent half of what was left on stationery, and he had $3$ left. We need to find Alan’s original allowance. 2. **Set up the model.** Let the total allowance be $x$. After spending $\frac{7}{10}$ on food, the remaining amount is: $$x-\frac{7}{10}x=\frac{3}{10}x$$ 3. **Use the information about stationery.** He spent half of the remaining allowance on stationery, so he kept the other half. That means the final amount left is half of $\frac{3}{10}x$: $$\frac{1}{2}\cdot \frac{3}{10}x=3$$ 4. **Simplify the equation.** $$\frac{3}{20}x=3$$ Now divide both sides by $\frac{3}{20}$: $$x=3\div \frac{3}{20}$$ $$x=3\cdot \frac{20}{3}$$ $$x=20$$ 5. **Check the answer.** Food: $\frac{7}{10}\cdot 20=14$ Remaining after food: $$20-14=6$$ Stationery: half of $6$ is $3$ Left over: $3$ 6. **Final answer.** Alan’s allowance was **20**.