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. 50 minutes
B. 42 minutes
C. 46 minutes
D. 38 minutes
Answer: Option B
Solution (By JKExamLibrary)
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.

This question belongs to: Maths Time Speed and Distance

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Practice More Time Speed and Distance Questions

Question #1 Report Error
A train 150 meters long is running at 72 km/h. How long will it take to completely pass a platform of length 250 meters?
A. 25 seconds
B. 15 seconds
C. 30 seconds
D. 20 seconds

Correct Answer: Option D


Explanation:
Speed of train = 72 * (5/18) = 20 m/s. Total distance to cross platform = 150 + 250 = 400 meters. Time taken = 400 / 20 = 20 seconds.

This question belongs to: Maths Time Speed and Distance
Question #2 Report Error
A racing car covers a track lap of 5000 meters in 1 minute 40 seconds. Find its speed in km/h.
A. 180 km/h
B. 150 km/h
C. 200 km/h
D. 220 km/h

Correct Answer: Option A


Explanation:
Time = 1 minute 40 seconds = 100 seconds. Speed = 5000 / 100 = 50 m/s. In km/h = 50 * (18/5) = 180 km/h.

This question belongs to: Maths Time Speed and Distance
Question #3 Report Error
An athlete loops a regular hexagonal field of side 50 meters. If his speed is 15 km/h, find the time taken to complete one full lap.
A. 80 seconds
B. 64 seconds
C. 72 seconds
D. 90 seconds

Correct Answer: Option C


Explanation:
Perimeter of the hexagonal field = 6 * 50 = 300 meters. Speed = 15 km/h = 15 * (5/18) = 25/6 m/s. Time = Distance / Speed = 300 / (25/6) = 300 * 6 / 25 = 12 * 6 = 72 seconds.

This question belongs to: Maths Time Speed and Distance