Word Problems and Applications

Correct answer: A. 40 km/h

Use the relation time = distance/speed. Let the original speed be x km/h. The original time is 360/x hours, while the increased-speed time is 360/(x + 5) hours. The statement gives 360/x − 360/(x + 5) = 1. On simplifying, 1800/[x(x + 5)] = 1, so x² + 5x − 1800 = 0. Factorisation gives (x − 40)(x + 45) = 0. Since speed must be positive, x = 40 km/h. Thus option A is correct. Checking, the original time is 9 hours and the new time is 360/45 = 8 hours, exactly one hour less. The other options do not produce this one-hour difference.

Correct answer: B. 60 km/h

Let the actual speed be x km/h. The actual travel time is 300/x hours, whereas at the reduced speed it is 300/(x − 10) hours. Since the reduced speed causes one extra hour, 300/(x − 10) − 300/x = 1, with x > 10. Combining the fractions gives 3000/[x(x − 10)] = 1, hence x² − 10x − 3000 = 0. This factors as (x − 60)(x + 50) = 0. The negative root is not a physical speed, so x = 60 km/h. Option B is correct. Verification: at 60 km/h the time is 5 hours; at 50 km/h it is 6 hours, exactly one hour more. The remaining options fail this time comparison.