Question & Answer
QUESTION

A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of two trains.

ANSWER Verified Verified
Hint: We are going to solve the given problem with the help of relation between Time, Speed and Distance to form an equation with the given information and by simplifying the equation the speeds of trains will be computed.

Complete step-by-step answer:
It is given that the two trains are covering a distance of 200 km.
$\therefore $Let Distance be D = 200 km
Let time taken for a fast train to reach destination be ${T_1}hr$ and time taken for the slow train to reach destination be ${T_2}hr$.
Given that the fast train reaches the destination one hour before the slow train.
$ \Rightarrow {T_2} - {T_1} = 1hr$-- (1)
Let the speed of the fast train be $S$ km/hr. Then the speed of the slow train will be $(S - 10)$ km/hr.
We know that $D = S \times T$
Equation (1) can be written as
$\dfrac{D}{{S - 10}} - \dfrac{D}{S} = 1$
$ \Rightarrow \dfrac{{200}}{{S - 10}} - \dfrac{{200}}{S} = 1$
On simplifying the above equation we get
\[
   \Rightarrow S(200) - (S - 10)200 = S(S - 10) \\
   \Rightarrow 200S - 200S + 2000 = {S^2} - 10S \\
   \Rightarrow {S^2} - 10S - 2000 = 0 \\
 \]
Now we got the equation in form of quadratic equation, by factorization we get
$
   \Rightarrow {S^2} - 50S + 40S - 2000 = 0 \\
   \Rightarrow S(S - 50) + 40(S - 50) = 0 \\
   \Rightarrow (S - 50)(S + 40) = 0 \\
   \Rightarrow (S - 50) = 0{\text{ }}\& {\text{ }} (S + 40) = 0 \\
   \Rightarrow S = 50{\text{ }}\& {\text{ }}S = - 40 \\
$

Speed is never negative so we take speed S = 50 kmph
$\therefore $ The speed of fast train is S = 50 kmph and speed of slow train is S-10 = 40 kmph

Note: Distance = Speed x Time, Speed is directly proportional to distance but inversely proportional to time. That means if speed is more we can cover more distance and at the same time with more speed the time taken to cover the distance will be reduced.