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.​

Answer 1

${\large{\sf{\underline{\underline{\purple{ Solution:-}}}}}}$

Let us say, the average speed of passenger train = x km/h.

Average speed of express train = (x + 11) km/h

Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

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

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

$⇒ \frac{132 \times 11}{x(x + 11)} = 1$

$⇒ 132 × 11 = x(x + 11)$

$⇒ x2² + 11x – 1452 = 0$

$⇒ x2² + 44x -33x -1452 = 0$

$⇒ x(x + 44) -33(x + 44) = 0$

$⇒ (x + 44)(x – 33) = 0$

$⇒ x = – 44, 33$

As we know, Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be $→$ $\small\boxed{\red{33 + 11 = 44\:km/h}}$

Answer 2

Answer:

Let us say, the average speed of passenger train = x km/h.

Average speed of express train = (x + 11) km/h

Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

(\frac{132}{x} ) - ( \frac{132}{x + 11} ) = 1(

x

132

)−(

x+11

132

)=1

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

x(x+11)

132(x+11−x)

=1

⇒ \frac{132 \times 11}{x(x + 11)} = 1⇒

x(x+11)

132×11

=1

⇒ 132 × 11 = x(x + 11)⇒132×11=x(x+11)

⇒ x2² + 11x – 1452 = 0⇒x2²+11x–1452=0

⇒ x2² + 44x -33x -1452 = 0⇒x2²+44x−33x−1452=0

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

⇒ (x + 44)(x – 33) = 0⇒(x+44)(x–33)=0

⇒ x = – 44, 33⇒x=–44,33

As we know, Speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be →→