1. **Problem statement:** We have a group of people who participated to get a vehicle licence. - $\frac{1}{9}$ fail the written test. - $\frac{1}{18}$ fail the medical test. - From the remaining, $\frac{4}{5}$ pass the practical test. - 20 people fail the practical test. We need to find: (i) Fraction who fail both written and medical tests. (ii) Fraction who pass the practical test as a fraction of total. (iii) Total number of participants. (iv) Number of people unable to get the licence. 2. **(i) Fraction who fail both written and medical tests:** - Those who fail written test: $\frac{1}{9}$ - Those who fail medical test: $\frac{1}{18}$ Assuming these are independent groups, the fraction who fail both is the sum of these fractions (since failing either test means failing): $$\frac{1}{9} + \frac{1}{18} = \frac{2}{18} + \frac{1}{18} = \frac{3}{18} = \frac{1}{6}$$ So, $\frac{1}{6}$ fail either the written or medical test. 3. **(ii) Fraction who pass the practical test as a fraction of total:** - Remaining after written and medical failures: $1 - \frac{1}{6} = \frac{5}{6}$ - Of these, $\frac{4}{5}$ pass the practical test. Fraction passing practical test of total: $$\frac{5}{6} \times \frac{4}{5} = \frac{4}{6} = \frac{2}{3}$$ 4. **(iii) Total number of participants:** - Those who fail practical test are the remaining $\frac{1}{5}$ of the $\frac{5}{6}$ who passed written and medical tests. Fraction failing practical test: $$\frac{5}{6} \times \frac{1}{5} = \frac{1}{6}$$ Given 20 people fail practical test, so: $$\frac{1}{6} \times \text{Total} = 20 \implies \text{Total} = 20 \times 6 = 120$$ 5. **(iv) Number of people unable to get the licence:** - People failing written or medical test: $\frac{1}{6} \times 120 = 20$ - People failing practical test: 20 (given) Total unable to get licence: $$20 + 20 = 40$$ **Final answers:** (i) $\frac{1}{6}$ (ii) $\frac{2}{3}$ (iii) 120 (iv) 40