A train 180 meters long running at 54 km/h passes a stationary point tracking observer. Find the time parameter required to cross. MCQ with Answer and Explanation

A train 180 meters long running at 54 km/h passes a stationary point tracking observer. Find the time parameter required to cross.
A. 14 seconds
B. 12 seconds
C. 10 seconds
D. 16 seconds
Answer: Option B
Solution (By JKSSB Mock Tests)
Speed of train = 54 * (5/18) = 15 m/s. Time taken = 180 / 15 = 12 seconds.

Discuss this Question (0)

No comments yet. Be the first to start the discussion!

Practice More Time Speed and Distance Questions

Question #1
A merchant van covers 120 km at 40 km/h, 180 km at 60 km/h and 100 km at 50 km/h. Find the net average speed.
A. 52 km/h
B. 50 km/h
C. 48 km/h
D. 54 km/h

Correct Answer: Option B


Explanation:
Total distance = 120 + 180 + 100 = 400 km. Total time = (120/40) + (180/60) + (100/50) = 3 + 3 + 2 = 8 hours. Average speed = 400 / 8 = 50 km/h.

This question belongs to: Maths Time Speed and Distance
Question #2
A person row a boat 16 km upstream in 4 hours. If the speed of the current is 1 km/h, find the speed of the boat in still water.
A. 5 km/h
B. 4 km/h
C. 7 km/h
D. 6 km/h

Correct Answer: Option A


Explanation:
Upstream speed = 16 / 4 = 4 km/h. Upstream Speed = Speed in still water - Current speed => 4 = Speed in still water - 1 => Speed in still water = 5 km/h.

This question belongs to: Maths Time Speed and Distance
Question #3
A car covers a distance of 360 km at a uniform speed. If the speed had been 15 km/h more, it would have taken 2 hours less for the journey. Find the original speed of the car.
A. 60 km/h
B. 45 km/h
C. 55 km/h
D. 50 km/h

Correct Answer: Option B


Explanation:
Let original speed be s. 360/s - 360/(s+15) = 2 => 180/s - 180/(s+15) = 1 => 180(15) = s(s+15) => 2700 = s(s+15). Solving s^2 + 15s - 2700 = 0 gives (s - 45)(s + 60) = 0. Since speed is positive, s = 45 km/h.

This question belongs to: Maths Time Speed and Distance