Question

A train was delayed by a semaphore for $16$ minutes and made up for the delay on a section of $80$ km travelling with the speed $10$ km per hour higher than that which accorded the schedule. Find the speed of the train which accorded the schedule.

Answer 1

Let the speed accorded be $x$ km/h

let the time taken be $t$ hours

$\therefore xt=80$ km

It it left $16$ min late, therefore to cover it should increase the speed$\therefore$ $(t-\frac{16}{60})(x+10)=80$

$xt+10t-\frac{16}{60}x-\frac{16}{6}=80$

$10\times \frac{80}{x}-\frac{16}{60}x-\frac{16}{6}=0$

$\frac{48000-{16x}^2-160x}{60x}=0$

$48000-16x^2-160x=0$

$x^2+10x-3000=0$

$x^2+60x-50x-3000=0$

$x(x+60)-50(x+60)=0$

$(x-50)(x+60)=0$

$x=50$km/h

$t=\frac{8}{5}\times 60=96$ min