The speed index tracking profile of an express locomotive is exactly 20% higher than a commercial car. Both take off from terminal M together and reach terminal N, 120 km away, at the same instant. On the corridor route, the train logs a total stall of 10 minutes at intermediate stations. Find the car speed. MCQ with Answer and Explanation
The speed index tracking profile of an express locomotive is exactly 20% higher than a commercial car. Both take off from terminal M together and reach terminal N, 120 km away, at the same instant. On the corridor route, the train logs a total stall of 10 minutes at intermediate stations. Find the car speed.
A. 140 km/h
B. 100 km/h
C. 150 km/h
D. 120 km/h
Answer: Option D
Solution (By JKSSB Mock Tests)
Let car speed be c, then train tracking speed is 1.2c. Time difference parameter = 120/c - 120/(1.2c) = 10/60 hours = 1/6 hours. 120/c * (1 - 1/1.2) = 1/6 => 120/c * (0.2/1.2) = 1/6 => 120/c * (1/6) = 1/6 => c = 120 km/h.
Explanation:
Time difference = 30 minutes = 0.5 hours. Let distance be d. d / 45 - d / 60 = 0.5 => (4d - 3d) / 180 = 0.5 => d / 180 = 0.5 => d = 90 km.
A student walks from his quarter block to the academy center at 4 km/h and arrives 8 minutes late. If he manages a pace of 5 km/h, he arrives 4 minutes early. Find the track distance.
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