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. 120 km/h
B. 150 km/h
C. 100 km/h
D. 140 km/h
Answer: Option A
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.
Two points A and B are 260 km apart. Train P leaves A for B at 50 km/h at 9:00 AM. Train Q leaves B for A at 60 km/h at 10:00 AM. At what time will they cross each other?
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