Subject
Quantitative Aptitude
Topic
Speed & Distance
A train has to travel the distance between Mumbai and Chennai, equal to 1000 km, at a constant speed. It travelled quarter of the way with the specified speed and stopped for 1 hour, to arrive at Chennai on time, it had to increase its speed by 20 km/h for the rest of the way. Next time the train stopped half-way for 30 minutes. By what value must it increase its speed for the remaining half of the distance to arrive at Chennai as per the schedule?
A train has to travel the distance between Mumbai and Chennai, equal to 1000 km, at a constant speed. It travelled quarter of the way with the specified speed and stopped for 1 hour, to arrive at Chennai on time, it had to increase its speed by 20 km/h for the rest of the way. Next time the train stopped half-way for 30 minutes. By what value must it increase its speed for the remaining half of the distance to arrive at Chennai as per the schedule?
ক)
10.12 kmphখ)
14.4 kmphগ)
16.87 kmphঘ)
16 kmphঙ)
Can’t be determinedExplanation
Let the regular speed of the train = x km/hr. Now, time = distance/speed ∴ regular time taken by the train to travel the full distance = (1000/x) hrs Case-1: Distance travelled at regular speed = 1000/4 = 250km ∴ time taken to travel this distance = (250/x) hrs ---(1) Waiting time = 1 hour ⇒ for the remaining 750 kms, the train increases its speed by 20 km/h (i.e. x + 20) ∴ time taken to travel last 750 km = 750/(x + 20) ---(2) From the given data, we can clearly deduce that: Regular time = (time taken to travel first 250 km) + (waiting time) + (time taken to travel last 750 km) ⇒ x 2 + 20x – 15000 = 0 ⇒ D = b 2 – 4ac = 400 - 4 × (-15000) = 60400 ⇒ x = (-b ± √D)/2a Omitting the negative value of speed, X = 112.85 km/h Now, in the second case: The train travels first 500 kms at its regular speed and waits for 30 minutes at the midway stop. In this case: waiting time = ½ hrs Let y = the speed with which it travels the remaining distance. Again, it reaches the destination in regular time. Substituting x = 112.85, ⇒ y = 127.22 km/h Increase in usual speed = y – x = 127.22 – 112.85 ~ 14.4 km/hrRelated questions
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