A fast train takes 3 hours less than a slow train for a journey of 600km. If the speed of the slow train is 10km/hr less than that of the first train, find the speed of the two trains.

Hint: Take the speed of slow trains as ‘S’. Find the speed of the fast train and use the formula of speed to find the expression connecting distance and speed to the time taken 3hrs. Solve the quadratic equation formed and find the speed of the train.

Complete step-by-step answer:
Given that there are 2 trains, a slow train and a fast train. Now let us assume the speed of the slow train as S.
It is said that the speed of a fast train is 10 km/h more than that of the slow train. Hence the speed of the fast train can be taken as (S + 10).
We are given the total distance of the journey = 600km.
The time difference of the journey is 3 hours.
We know the basic formula of speed is distance by time.
\[\therefore \]Speed = Distance / Time.
\[\therefore \]Time = Distance / Speed.
Now let us formulate a condition on time for both slow and fast trains. Both their distances are the same.
(distance of slow train/speed of slow train) – (distance of fast train/speed of fast train) = 3.
\[\dfrac{600}{S}-\dfrac{600}{S+10}=3\]
Now let us simplify the above expression.
\[600\left[ \dfrac{1}{S}-\dfrac{1}{S+10} \right]=3\]
Cross multiply and simplify.
\[\dfrac{s+10-s}{s\left( S+10 \right)}=\dfrac{3}{600}\]
\[\dfrac{10}{s\left( s+10 \right)}=\dfrac{3}{600}\]
Cross multiply the expression.
\[\begin{align}
  & \dfrac{10\times 600}{3}=s\left( s+10 \right) \\
 & {{s}^{2}}+10s=2000 \\
 & \therefore {{s}^{2}}+10s-2000=0 \\
\end{align}\]
The above equation is similar to a quadratic equation, the quadratic equation general form is \[a{{x}^{2}}+bx+c=0\], comparing both the equations.
We can say that, a = 1, b = 10, c = -2000.
Now let us substitute these values in the quadratic formula.
\[\begin{align}
  & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-10\pm \sqrt{{{10}^{2}}-4\times 1\times \left( -2000 \right)}}{2\times 1} \\
 & =\dfrac{-10\pm \sqrt{100+8000}}{2}=\dfrac{-10\pm \sqrt{8100}}{2}=\dfrac{-10\pm 90}{2} \\
\end{align}\]
Thus we got the roots are \[\left( \dfrac{-10+90}{2} \right)\]and \[\left( \dfrac{-10-90}{2} \right)\]= 40 and -50.
Neglect S = -50, as speed cannot be negative.
Thus we got the speed of slow trains as, S = 40 km/hr.
Similarly, the speed of a fast train, S + 10 = 40 + 10 = 50 km/hr.
\[\therefore \]Speed of slow train = 40 km/hr.
Speed of fast train = 50 km/hr.

Note: Remember that both speeds of the trains are different. We have used the formula of speed here. Remember the basic formulae like these as they are used frequently in problems like these. Here time and speed varies only the distance is the same, that’s why we take time = distance/speed.