Question

A train, travelling at a uniform speed for $360km$, would have taken $48$ minutes less to travel the same distance if its speed were $5km/hr$ more. Find the original speed of the train.

Answer 1

Distance = $360$ km

Let original sped = $Vkm/hr$ and time be $t$ hours.

$\frac{360}{v}=t....\left(1\right)$

However if speed was $5km/hr$ more, then $48$ minutes less is taken

$\frac{360}{V+5}=t-\frac{43}{60}$

$\frac{360}{V+5}=\frac{360}{v}-\frac{4}{5}$

$\frac{360}{V}-\frac{360}{V+5}=\frac{4}{5}$

$360\left(\frac{5}{V(V+5)}\right)=\frac{4}{5}$

$\frac{\left(360\right)\left(5\right)\left(5\right)}{4}=V^2+5V$

$V^2+5V=2250$

$V^2+5V-2250=0$

$V^2+50V-45V-2250=0$

$V(V+50)-45(v+50)=0$

$v=45$ or $V=-50$

$V$ has to be positive

$\therefore$ Original speed = $45km/hr$