Question

solve the attachment​

Answer 1

Answer:

Given :-

A train travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if it's speed were 5 km/h more.

To Find :-

What is the original speed of the train.

Formula Used :-

$\longmapsto \sf\boxed{\bold{\pink{Time =\: \dfrac{Distance}{Time}}}}$

Solution :-

Let, the original speed of the train be x km/hr.

Given :

Distance = 360 km

Then,

Time taken by the train initially is $\sf \dfrac{360}{x}$

And, the speed was increased by 5 km/hr

Then, the time taken by the train is $\sf \dfrac{360}{x + 5}$

According to the question,

↦ $\sf \dfrac{360}{x} - \dfrac{360}{x + 5} =\: \dfrac{48}{60}$

↦ $\sf \dfrac{1}{x} - \dfrac{1}{x + 5} =\: \dfrac{48}{60} \times \dfrac{1}{360}$

↦ $\sf \dfrac{5}{{x}^{2} + 5x} =\: \dfrac{1}{450}$

↦ $\sf {x}^{2} + 5x =\: 5 \times 450$

↦ $\sf {x}^{2} + 5x =\: 2250$

↦ $\sf {x}^{2} + 5x - 2250 =\: 0$

↦ $\sf {x}^{2} - 45x + 50x - 2250 =\: 0$

↦ $\sf x(x - 45) + 50(x - 45) =\: 0$

↦ $\sf (x - 45) (x + 50) =\: 0$

↦ $\sf x - 45 =\: 0$

➠ $\sf\bold{\red{x =\: 45}}$

Either,

↦ $\sf x + 50 =\: 0$

➠ $\sf\bold{\green{x = - 50}}$

As we can't take speed as negative (- ve).

So , x = 45

$\therefore$ The original speed of the train is 45 km/hr .