Question

A bus takes 2.5 hrs more to cover a distance of 50 km when its speed reduced by 10 kmph .find its usual speed

Answer 1

let the usual speed of bus be x km/ h.
time taken by it to cover a distance of
50km= 50/x h.
when speed of bus is reduced by 10 km/h
the new speed=(x-10) km/h
time taken by it to cover 50km with new speed
=50/(x-10)
now, according to question
$\frac{50}{x - 10} = \frac{50}{x} + 2.5$
$\frac{50}{x - 10} = \frac{50 + 2.5x}{x}$
$50x = (x - 10)(50 + 2.5x)$
$50x = 50x + 2.5 {x}^{2} - 500 - 25x$
$2.5 {x}^{2} - 25x - 500 = 0$
$dividing \: both \: sides \: by \: 2.5$
${x}^{2} - 10x - 200 = 0$
$splitting \: the \: middle \: term \: we \: get$
${x}^{2} - 20x + 10x - 200$
$x(x - 20) + 10(x - 20) = 0$
$(x - 20)(x + 10) = 0$
$hence \: either \: (x - 20) = 0 \: or \: (x + 10) = 0$
$x = 20 \: or \: x = - 10$
$rejecting \: x = - 10 \: because \: speed \: can \: not \: be \: negative \: we \: get \: x = = 20$
$hence \: the \: usual \: speed \: of \: bus \: is \: 20kmperh$