1. Stating problem 19: Price of chicken is 3 times price of bread. Total spent on 1 bread + 1 chicken = RM5 - RM1 = RM4. 2. Let price of bread = $x$. Then price of chicken = $3x$. 3. Equation from total cost: $x + 3x = 4$ 4. Simplify: $4x = 4$ 5. Solve for $x$: $x = 1$ 6. So price of bread = RM1, price of chicken = $3 \times 1 = RM3$. 7. Cost for 3 breads and 2 chickens: $3 \times 1 + 2 \times 3 = 3 + 6 = RM9$. 1. Stating problem 20: Car uses 50 liters petrol for 510 km. Need petrol for two-way trip of $2 \times 754.8 = 1509.6$ km. 2. Petrol consumption rate: $\frac{50}{510}$ liters per km. 3. Total petrol needed: $\frac{50}{510} \times 1509.6 = \frac{50 \times 1509.6}{510}$ liters. 4. Calculate: $\frac{50 \times 1509.6}{510} = \frac{75480}{510} = 148$ liters. Final answers: - Question 19: RM9 - Question 20: 148 liters