Question

A passenger train takes 3 h less for a journey of 360 km, if its speed is increased by 10 km/h from its usual speed. Find its usual speed.

Answer 1

Answer:

Hope it will help you.........

Your answer is 30km/hr

Answer 2

$\sf \: Please \: Note : -$

Here, we have to find out the usual speed of the train, so we assume it as x km/h and then translate the word problem into symbolic form leading to a quadratic equation by using the formula :-

$\boxed{ \sf{ Time = \ \frac{Distance}{Speed} }}$

$\sf \: SOLUTION : -$

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

Then, by using the formula,

$\sf \: Time = \frac{Distance}{Speed} \: we \: get$

$\sf \: Time \: taken \: by \: the \: train = \frac{360}{x} h \: \: \: \: \: [given \: distance = 360 \: km]$

If speed is increased by 10 km/h, then the new speed of the train = ( x + 10 ) km/h.

$\therefore \: \sf \: Time \: taken \: by \: the \: train = \bigg( \frac{360}{x + 10} \bigg)h$

According to the question,

$\sf \: \: \: \: \: \: \: \frac{360}{x} = \frac{360}{x + 10} + 3$

$\implies \: \sf \: \frac{360}{x} - \frac{360}{x + 10} = 3$

$\implies \: \sf \: \frac{360(x + 10 - x)}{x(x + 10)} = 3$

$\implies \: \sf \: \frac{360 \times 10}{3} = x(x + 10)$

$\implies \: \sf \: 1200 = {x}^{2} + 10x$

$\implies \: \sf \: {x}^{2} + 10x - 1200 = 0$

$\implies \: \sf \: {x}^{2} + 40x - 30x - 1200 = 0 \: \: \: \: \: [by \: factorisation]$

$\implies \: \sf \: x(x + 40) - 30(x + 40) = 0$

$\implies \: \sf \: (x + 40)(x - 30) = 0$

$\implies \: \sf \: x + 40 = 0 \: or \: x - 30 = 0$

$\implies \: \sf \: x = - 40 \: or \: x = 30$

But speed cannot be negative.

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \therefore \: \sf \: x = 30$

Hence the usual speed of passenger train is 30 km/h.