Walking at 7/9 of his baseline typical speed, a courier agent reaches his office terminal 12 minutes late. Find his typical baseline time parameter. MCQ with Answer and Explanation

Walking at 7/9 of his baseline typical speed, a courier agent reaches his office terminal 12 minutes late. Find his typical baseline time parameter.
A. 46 minutes
B. 50 minutes
C. 42 minutes
D. 38 minutes
Answer: Option C
Solution (By JKSSB Mock Tests)
New speed = 7/9 of usual speed => New time = 9/7 of usual time. Difference = 2/7 of usual time = 12 minutes. 1/7 of usual time = 6 minutes => Usual time = 42 minutes.

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Practice More Time Speed and Distance Questions

Question #1
Excluding regular stops, an express cargo van tracks at 84 km/h and including stops it tracks at 70 km/h. How many minutes does it stop per hour?
A. 10 minutes
B. 15 minutes
C. 12 minutes
D. 8 minutes

Correct Answer: Option A


Explanation:
Stoppage time per hour = (84 - 70) / 84 = 14 / 84 = 1/6 hour. In minutes = (1/6) * 60 = 10 minutes.

This question belongs to: Maths Time Speed and Distance
Question #2
A high-speed train clears a structural platform watchpost signal pillar in 6 seconds and passes across the full platform itself, which measures 200 meters long, in 16 seconds. Find the speed index of the train in km/h.
A. 72 km/h
B. 80 km/h
C. 84 km/h
D. 64 km/h

Correct Answer: Option A


Explanation:
Time taken to cross the platform span alone = 16 - 6 = 10 seconds. Speed of train = 200 / 10 = 20 m/s. In km/h index = 20 * (18/5) = 72 km/h.

This question belongs to: Maths Time Speed and Distance
Question #3
A person walks from his house to his office at a speed of 4 km/h and reaches 10 minutes late. If he walks at 5 km/h, he reaches 5 minutes early. Find the distance between his house and office.
A. 8 km
B. 5 km
C. 4 km
D. 6 km

Correct Answer: Option B


Explanation:
Time difference = 10 - (-5) = 15 minutes = 1/4 hour. Let distance be d. d/4 - d/5 = 1/4 => d/20 = 1/4 => d = 5 km.

This question belongs to: Maths Time Speed and Distance