Question

★ Quadratic Equations ★

1: An express train takes 1 hour less then a passenger train to travel 132 km between Mysore & Bangalore ( without taking into consideration the time they stop at intermediate stations ). If the average speed 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

Answer:-

→ Speed of express train = 44 km/h

→ Speed of passenger train = 33 km/h

Explanation:-

Let the speed of the passenger train be v km/h.

Thus, speed of the express train :-

= (v + 11) km/h

Now, let's calculate the time taken by both the trains to travel between Mysore and Bangalore.

Time taken by the passenger train

= Distance/Speed

= [132/v] hrs

Time taken by the express train

= Distance/Speed

= [132/(v + 11)] hrs

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Now according to the question, we get the equation as :-

⇒ [132/v] - [132/(v + 11)] = 1

⇒ [132(v + 11) - 132v]/[v² + 11v] = 1

⇒ [1452 + 132v - 132v] = v² + 11v

⇒ 1452 = v² + 11v

⇒ v² + 11v - 1452 = 0

⇒ v² + 44v - 33v - 1452 = 0

⇒ v(v + 44) - 33(v + 44) = 0

⇒ (v - 33)(v + 44) = 0

⇒ v = 33 km/h ; v = - 44 km/h

But we know that speed cannot be negative.

So, speed of the passenger train :-

= 33 km/h

And, speed of the express train :-

= (33 + 11) km/h

= 44 km/h

Answer 2

Given :-

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

To Find :-

Average speed

Solution :-

Let's assume that average speed is x km

Now

Average speed = x + 11

According to the question

$\sf \dfrac{132}{x+11} = \dfrac{132}{x+1}$

$\sf\dfrac{132}{x+11} = \dfrac{132+x}{x}$

$\sf 132x = (132 + x)(x+11)$

$\sf 132x = x^{2} +132x+11x+1452$

Cancel 132x from both sides

$\sf x^{2} + 11x-1452 = 0$

$\sf x^{2} +(44x - 33x) - 1452 = 0$

$\sf x^{2} + 44x-33x-1452=0$

$\sf x(x+44) - 33(x+44)$

$\sf (x+33)(x-44)$

Either

x = 0 - 33

x = -33 km/h

or

x = 44 + 0

x = 44 km/h

Speed can't be negative so,

Speed of express train = 44 km/h

Speed of passenger train = 44 - 11 = 33 km/h