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 the usual speed of the train.

Answer 1

$\huge{\bigstar{\bigstar{\underline{\underline\blue{\mathcal{\red{solution!!...}}}}}}}$

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

A.T.Q.

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...

❗hope!!...it helps uhh...❗

Answer 2

Let the usual speed of the train = x km/hr.

Distance = 300 km [ Given ]

Time taken = Distance/ Time = 300/x hr ......( i )

Increased speed of the train = y km/hr.

y = x + 5 ............( ii )

Time taken = 300/y hr ..........( iii )

According to the question,

(Scheduled time) - (time by increasing the speed) = 2 hr.

$: \implies \frac{300}{x} - \frac{300}{y} = 2$

Putting the value of y from ( ii ) , we get

$: \implies \frac{ 300}{x} - \frac{300}{x + 5} = 2$

$: \implies \frac{300x + 1500 - 300x}{x(x + 5)} = 2$

$: \implies \frac{1500}{x(x + 5)} = 2$

$: \implies \: 2x(x + 5) = 1500$

$: \implies \: {x}^{2} + 5x = 750$

$: \implies \: {x}^{2} + 5x - 750 = 0$

$: \implies \: {x}^{2} + 30x - 25x - 750 = 0$

$: \implies \: x(x + 30) - 25(x + 30) = 0$

$: \implies \: (x + 30)(x - 25) = 0$

$\sf{ \bigstar \: x = - 30 \: or \: x = 25}$

But Speed cannot be negative.

Hence, the usual speed = 25 km/hr.