Let's denote the total capacity of the tank by $C$ liters.

1. On the first day, half of the petrol is used, so the amount of petrol used is: $\frac{1}{2}C$ Therefore, the remaining petrol after the first day is: $C - \frac{1}{2}C = \frac{1}{2}C$

2. On the second day, one-third of the remaining petrol is used, so the amount of petrol used on the second day is: $\frac{1}{3} \left( \frac{1}{2}C \right) = \frac{1}{6}C$ Therefore, the remaining petrol after the second day is: $\frac{1}{2}C - \frac{1}{6}C = \frac{3}{6}C - \frac{1}{6}C = \frac{2}{6}C = \frac{1}{3}C$

We are given that after the second day, there is 20 liters of petrol left in the tank. So, $\frac{1}{3}C = 20$

Solving for $C$: $C = 20 \times 3 = 60$

Therefore, the capacity of the tank of the car is 60 liters.

Would you like further details or have any questions about this solution?

Here are 5 questions you might want to ask next:

1. How do you solve problems involving fractions and remaining quantities?
2. Can you explain another example involving fractions and remaining quantities?
3. What are some common pitfalls in solving problems with multiple steps and fractions?
4. How do you approach solving equations with fractions in general?
5. What are the real-world applications of solving problems involving fuel consumption and capacity?

Tip: When dealing with fractions, it's often helpful to simplify the problem step by step and keep track of the remaining quantity after each step to avoid confusion.