Question

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
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Answer 1

Given,

speed of express train = x

speed of passenger train=(x+11)

distance = 132 km

Time = distance/speed

⇒ Time taken by express train = 132/x

⇒ Time taken by passenger train=  132/(x+11)

According to the question,

⇒ $\frac{132}{x} - \frac{132}{x+11} = 1$

132(x+11)-132x = x(x+11)

132x +1452-132x = x^2 +11x

⇒ x² + 11x-1452 = 0

x² +44x-33x-1452=0

x(x+44) -33(x+44) = 0

(x+44)(x-33) = 0

x = 33 [-44 neglected]

∴ Speed of express train = 33km/hr

∴ speed of passenger train =(x+11) = 44km/hr.

Answer 2

Answer:

Let the average speed of the passenger train be x km/hr.

Then the average speed of express train = (x+11)

Time taken by passenger train=(132/x)hr

Time taken by the express train = {(132)/x+11} hr

ATQ,

(132/(x+11)) hr = ((132/x) −1)hr  (Express train takes 1 hr less)

⇒132x = x²+132x+11x+1452

⇒x²+11x−1452=0

⇒x² + 44x - 33x - 1452 = 0

⇒(x−33)(x+44)=0

⇒(x−33)=0 or (x+44)=0

⇒x=33 or x=−44

Speed can not be negative,

⇒x=33, x+11=44

Hence, the speed of passenger train =33 km/hr and the speed of express train =44 km/hr

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