Question

A goods train leaves a station at 6pm, followed by an express train which leaves at $8pm$ and travels $20km/hr$ faster than the goods train. The express train arrives at a station, $1040km$ away, $36min$. before the goods train. Assuming that the speeds of both the trains remain constant between the two stations, calculate the speeds.

• A. $80km/hr$ and $100km/hr$
• B. $20km/hr$ and $40km/hr$
• C. $60km/hr$ and $80km/hr$
• D. $120km/hr$ and $100km/hr$

Answer 1

Answer: (A)

$80km/hr$ and $100km/hr$

Let the speed of the goods train be $xkm/hr$ .

$\therefore$ speed of the express will be $(x+20)km/hr$ .

Let the time taken by goods train to travel $1040km$ be $t_g$

and that taken by express train be $t_e$.

As per the time of departure, $t_e=t_g-2$ and D$=1040km$ .

As per the time of arrival, $t_e=t_g-\frac{36}{60}$

$\therefore$ Total time taken by express train

$t_e=t_g-2-\frac{36}{60}$

$\frac{1040}{x+20}=\frac{1040}{x}-\frac{120+36}{60}$

$\frac{1040}{x+20}-\frac{1040}{x}=-\frac{13}{5}$

On solving the above, we get

$x^2+20x-8000=0$

$\therefore x=-100$ or $x=80$

Since speed cannot be negative, $x=80km/hr$

Speed of goods train is $80km/hr$ and speed of express is $80+20=100km/hr$.