Question

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

Answer 1

Hint- In order to find the speed of both the trains, first we will assume the speed of the faster train as variable, then by using the relation between time distance and speed we can make equations and solve them to find the speeds.

Complete step-by-step answer:

Let the speed of a fast train be x km/hr.
Speed of slower train $= x - 10km/hr$

Given that: a fast train takes 1 hour less than a slower one covering 200 km.
As we know $\left[ {time = \dfrac{{distance}}{{speed}}} \right]$
Time taken by faster train = time taken by slower train – 1
$\Rightarrow \dfrac{{200}}{x} = \dfrac{{200}}{{x - 10}} - 1 \\\ \Rightarrow \dfrac{{200}}{x} = \dfrac{{200 - x + 10}}{{x - 10}} \\\ \Rightarrow 200 \times \left( {x - 10} \right) = \left( {210 - x} \right) \times \left( x \right) \\\ \Rightarrow 200x - 2000 = 210x - {x^2} \\\ \Rightarrow {x^2} - 10x - 2000 = 0 \\\ \Rightarrow {x^2} - 50x + 40x - 2000 = 0 \\\ \Rightarrow \left( {x - 50} \right)\left( {x + 40} \right) = 0 \\\ x = 50{\text{ and }}x = - 40 \\\$
Since, speed cannot be negative, therefore the speed of faster train is $50km/hr$
And speed of the slower train is $40km/hr$

Note- In order to solve these types of problems, first of all remember the basic relation between speed, time and distance. Second, read the statements carefully and gather the information given in the question. At last start solving the question according to the conditions. The basic requirement is you must know how to solve algebraic and quadratic equations.