Question

A train travels 360 km at a uniform speed. If the speed has been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the Train

Answer 1

Ace

Let the original speed of the train be x km/h.

Time taken to cover a distance of 360 km =  360/x hours.

New speed of the train = (x+5) km/h.

Time taken to cover a distance of 360 km at new speed = 360/x+5 hours.

Since, the train takes 1 hour less time,

∴ 360/x - 360/ x+5 = 1

⇒360 (x+5-x)/x(x+5) = 1

⇒360 (5) = x² + 5x

⇒1800 = x² + 5x

⇒x² + 5x - 1800 = 0

⇒x² + 45x - 40x - 1800 = 0

⇒x (x+45) - 40( x +45) = 0

⇒(x+45) (x-40) = 0

⇒x = (-45), 40

But since speed cannot be in negative.

∴ x = 40 km/hr.

Hence, the original speed of the train is 40 km/h.

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Answer 2

Answer :

Let the initial speed of the train be 'x' km/hr.

Hence, New speed is (x + 5) km / hr.

Time to cover 360 km = Distance / Speed = 360/x hours.

New time after increasing the speed = 360/(x + 5) hours

From the given condition,

( 60 min = 60/60 hours )

$\tt{ \frac{360}{x + 5} = \frac{360}{x} - \frac{60}{60}}$

$\tt{\therefore \frac{360}{x} - \frac{360}{x + 5} = \frac{60}{60}}$

Now,

Dividing both sides by 360,

$\tt{\therefore \frac{1}{x} - \frac{1}{x + 5} = \frac{60}{60 \times 360}}$

$\tt{\therefore \frac{x + 5 - x}{x(x + 5)} = \frac{1}{360}}$

$\tt {\frac{5}{ {x}^{2} + 5x } = \frac{1}{360}}$

$\tt{x {}^{2} + 5x =18 00}$

$\tt{x {}^{2} + 5x - 1800 = 0}$

Splitting the middle term,

$\tt{x {}^{2} + 45x - 40x - 1800 = 0}$

$\tt{x(x + 45) - 40(x + 45) = 0}$

$\tt{(x - 40)(x + 45) = 0}$

$\tt{x - 40 = 0} \: \: \: \: or \: \: \: \: \tt{ x + 45 = 0}$

$\tt{x = 40 \: \: \: \: or \: \: \: \: x = - 45}$

Here,

We got 2 values of x.

But , we know that, Speed can never be negative.

So, x = 40 km/hr.

Thus,

The speed of the train is 40 km/hr.