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 it's usual speed.

Answer 1

$\boxed{ \bf \: Question:-}$

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

$\red{ \bullet} \: \rm \: Note : - \: Here, \: we \: have \: to \: find \: out \: the \: usual \\ \rm speed \: of \: the \: train, \: so \: we \: assume \: it \: as \: \red{x} \: kmh \: and \\ \rm then \: translate \: the \: word \: problem \: into \: symbolic \: form \\ \rm leading \: to \: a \: quadratic \: equation \: by \: using \: the \: formula : - \\ \\ \implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf \: Time = \frac{Distance}{Speed}}$

$\boxed{ \bf \: Solution:-}$

$\rm \: Let \: the \: usual \: speed \: of \: the \: train \: = x \: km/h.$

$\rm \: Then, \: by \: using \: the \: formula,$

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

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

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

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

$\rm \: According \: to \: the \: question,$

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

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

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

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

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

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

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

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

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

$\implies \: \: \: \: \: \: \: \: \: \rm \: x + 40 = 0$

$\rm \: \: \: \: or \: \: \: \: \: \: \: \: \: \: \: \: \: \: x - 30 = 0$

$\rm \: But \: speed \: cannot \: be \: negative.$

$\therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: x = 30$

$\rm \: Hence, \: the \: usual \: speed \: of \: passenger \: train \: is \: \red{30} \: km/h.$