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?
Solution
Correct Answer: Option B
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)
\(\begin{array}{l} \Rightarrow \frac{{1000}}{x} = \frac{{250}}{x} + 1 + \frac{{750}}{{x + 20}}\\ \Rightarrow \frac{{750}}{x} - \frac{{750}}{{x + 20}} = 1\\ \Rightarrow \frac{{20}}{{{x^2} + 20x}} = \frac{1}{{750}} \end{array}\)
⇒ x2 + 20x – 15000 = 0
⇒ D = b2 – 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.
\(\begin{array}{l} \Rightarrow \frac{{1000}}{x} = \frac{{500}}{x} + \frac{1}{2} + \frac{{500}}{y}\\ \Rightarrow \frac{{500}}{x} = \frac{1}{2} + \frac{{500}}{y} \end{array}\)
Substituting x = 112.85,
⇒ y = 127.22 km/h
Increase in usual speed = y – x = 127.22 – 112.85 ~ 14.4 km/hr
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