Question

A passenger train takes 2 hours less for a journey of 300 km, if its speed is increased by 5 km/ hr from its usual speed. Find its usual speed.

Class:- X


Chapter - 3

Linear Equations in Two Variables

Answer 1

Solution:-

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

Distance = 300 km.

Now, Time = Distance / Speed

=> Time = 300 / x

New Speed = x + 5 km/hr

=> Time = 300 / x+5

$= > \frac{300}{x} - \frac{300}{x + 5} = 2 \\ \\ = > \frac{300(x + 5) - 300x}{x(x + 5)} = 2 \\ \\ = > 300x + 1500 - 300x = 2( {x}^{2} + 5x) \\ \\ = > 1500 = 2( {x}^{2} + 5x) \\ \\ = > 2(750) = 2( {x}^{2} + 5x) \\ \\ = > {x}^{2} + 5x - 750 = 0 \\ \\ = > {x}^{2} + (30 - 25)x - 750 = 0 \\ \\ = > {x}^{2} + 30x - 25x - 750 = 0 \\ \\ = > x(x + 30) - 25(x + 30) = 0 \\ \\ = > (x - 25)(x + 30) = 0 \\ \\ = > x = 25 \: \: and \: \: x = - 30(neglect)$

Answer 2

Answer:

Let the usual speed of the train be y km / hr .

A.T.Q.

$\displaystyle{\frac{300}{y} -\frac{300}{y+5} =2 }$

300 ( y + 5 ) - 300 y = 2 y ( y + 5 )

y² + 5 y - 750 = 0

y² + 30 y - 25 y - 750 = 0

( y + 30 ( y -25 ) = 0

y = - 30 or y = 25

Since , speed of train can't be negative .

Therefore , the usual speed of the train is 25 km / hr .