A train runs at a speed of $90\text{km/h}$ in the first $10\text{min}$, $60\text{km/h}$ in the next $20\text{min}$, and $40\text{km/h}$ in the last $5\text{minutes}$. What is the average speed (in $\text{km/h}$) of the train?

Average speed is defined as the total distance divided by the total time taken.
$$\text{Average Speed}=\frac{\text{Total Distance}}{\text{Total Time}}$$
First, calculate the distance covered in each interval:First interval: Speed = $90\text{km/h}$, Time = $10\text{min}=\frac{10}{60}\text{h}$ $
${\text{Distance}}_1=90\times \frac{10}{60}=15\text{km}$$Secondinterval:Speed=$ 60\,\text{km/h}$,Time=$ 20\,\text{min} = \frac{20}{60}\,\text{h}
$\text{Distance}_2 = 60 \times \frac{20}{60} = 20\,\text{km}
Third interval: Speed = $40\text{km/h}$, Time = $5\text{min}=\frac{5}{60}\text{h}$ $
${\text{Distance}}_3=40\times \frac{5}{60}=\frac{20}{6}=\frac{10}{3}\text{km}$$Now,computeTotalDistanceandTotalTime:$$\text{Total Distance}=15+20+\frac{10}{3}=35+\frac{10}{3}=\frac{105+10}{3}=\frac{115}{3}\text{km}$ $
$$\text{Total Time}=10+20+5=35\text{min}=\frac{35}{60}\text{h}=\frac{7}{12}\text{h}$$
Calculate the average speed:
$$\text{Average Speed}=\frac{\frac{115}{3}}{\frac{7}{12}}=\frac{115}{3}\times \frac{12}{7}=\frac{115\times 4}{7}=\frac{460}{7}\text{km/h}$$
Converting $\frac{460}{7}$ to a mixed fraction:
$$\frac{460}{7}=65\frac{5}{7}\text{km/h}$$