Question

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

Answer 1

Step 1

Let the original speed of train$=skm/h$

We know that

$Time=\frac{dis\tan ce}{speed}$

According to the question, we have,

Time is taken by train $=\frac{360}{s}hour$

And, Time taken by train its speed increase $5km/h=360/(s+5)$

Step 2

Now, It is given that,

Time taken by train in first – time taken by train in 2nd case =$48min=48/60hour$

$$\begin{gathered} 360/s–360/(s+5)=48/60=4/5 \\ 360(1/s–1/(s+5\left)\right)=4/5 \\ 360\times 5/4(5/(s²+5s\left)\right)=1 \\ 450\times 5=s²+5s \\ s²+5s-2250=0 \\ s=(-5\pm \surd (25+9000\left)\right)/2 \\ =(-5\pm \surd (9025))/2 \\ =(-5\pm 95)/2 \\ =-50,45 \\ Buts\ne -50becausespeedcannotbenegative \\ So,s=45km/h \end{gathered}$$

Hence, original speed of train $=45km/h$