Question

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

Answer 1

Answer

Usual speed of train = x = 25 km/hr.

Explanation

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

A train takes 2 hours less for a journey of 300 km.

Time = 2 hrs.

Distance = 300 km

Time = Distance/Speed

Time = 300/x

If its speed is increased by 5 km/h from its usual speed.

Speed = (x + 5) km

Distance = 300 km

Time = 300/(x + 5)

As per given condition,

→ 300/x - 300/(x + 5) = 2

→ 300 [(x + 5) - (x)]/[x(x + 5)] = 2

→ 300 (5)/(x² + 5x) = 2

→ 1500 = 2x² + 10x

→ 2x² + 10x - 1500 = 0

Taking 2 as common

→ x² + 5x - 750 = 0

Split the middle term

→ x² + 30x - 25x - 750 = 0

→ x(x + 30) - 25(x + 30) = 0

→ (x - 25) (x + 30) = 0

→ x = 25, - 30(neglected)

Therefore,

The usual speed of the train is 25 km/hr.

Answer 2

QUESTION

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

SOLUTION

Case ( i )

Train has to cover

distance = 300 km

Let,

usual speed of train = x  km/h

and,

time taken = y hrs

so,

time taken( y hrs) by train to cover 300 km with speed x km/h

(using formula

time = distance/speed )

$y=\frac{300}{x} -------eqn (1)$

and ,

Case ( ii )

distance to be covered (will be same) = 300 km

speed (is increased by 5 km/h )= ( x + 5) km/h

time taken (is decreased) = ( y - 2 )  hrs

taking,

time taken by train ( y-2) hrs , distance to be covered 300 km and speed (x+5) km/h

$y-2 = \frac{300}{x+5} ------eqn(2)$

(putting eqn(1) , y = 300/x  in eqn (2))

$\frac{300}{x} -2=\frac{300}{x+5} \\\\(taking LCM)\\\\\frac{300-2x}{x} =\frac{300}{x+5} \\\\ (cross \\ multiplying)\\\\ 300x+1500 - 2x^{2} - 10x = 300 x\\\\(300 x will be eliminated being on both sides )\\\\-2x^{2} -10x+1500 = 0\\(multiplying by -1 both sides )\\\\2x^{2} +10x-1500=0\\\\2x^{2} +60x - 50x - 1500 = 0\\\\2x ( x + 30) - 50 ( x + 30) = 0\\\\(2x-50)(x+30) = 0$

From here we will get

x = -30

and

x = 50 / 2 = 25

since,

speed cannot be negative

therefore neglecting negative value

we will take

x = 50/2 = 25  km/h

HENCE USUAL SPEED OF TRAIN WILL BE  25 KM/HR.