Question

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

Answer 1

Hint: Form an equation using the initial condition i.e. when the speed of the train is usual and another equation using the condition given in the question. And then solve both the equations to find the required value.

Complete step by step answer:
Let the usual speed of the train be $x$ km/hr.
According to the question, the distance covered by the train in the journey is 300 km. And we know that ${\text{speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time taken}}}}$. If the time taken by the train is $t$ hours then we have:
$
   \Rightarrow x = \dfrac{{300}}{t}, \\
   \Rightarrow t = \dfrac{{300}}{x} .....(i) \\
$
After increasing the speed of the train by 5 km/hr, it is taking two hours less. So speed in this case is $\left( {x + 5} \right)$ km/hr. And time taken is $\left( {t - 2} \right)$ hour. So, again using the same formula, ${\text{speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time taken}}}}$we have:
$
   \Rightarrow x + 5 = \dfrac{{300}}{{t - 2}}, \\
   \Rightarrow t - 2 = \dfrac{{300}}{{x + 5}}, \\
   \Rightarrow t = \dfrac{{300}}{{x + 5}} + 2 .....(ii) \\
$
Comparing equations $(i)$ and $(ii)$, we’ll get:
$ \Rightarrow \dfrac{{300}}{x} = \dfrac{{300}}{{x + 5}} + 2$
Taking terms with variables on one side and constants on the other, we’ll get:
$ \Rightarrow \dfrac{{300}}{x} - \dfrac{{300}}{{x + 5}} = 2$
On further simplification, this will give us:
$
   \Rightarrow 300\left( {\dfrac{1}{x} - \dfrac{1}{{x + 5}}} \right) = 2 \\
   \Rightarrow \dfrac{{x + 5 - x}}{{{x^2} + 5x}} = \dfrac{1}{{150}} \\
   \Rightarrow \dfrac{5}{{{x^2} + 5x}} = \dfrac{1}{{150}} \\
$
On cross multiplying the above equation, we’ll get:
$
   \Rightarrow {x^2} + 5x = 750 \\
   \Rightarrow {x^2} + 5x - 750 = 0 \\
$
Now, we can solve the above quadratic equation using factorization:
\[
   \Rightarrow {x^2} + 5x - 750 = 0 \\
   \Rightarrow {x^2} + 30x - 25x - 750 = 0 \\
\]
Taking $x$ common from the first two terms and -25 from the other two terms, we’ll get:
\[ \Rightarrow x\left( {x + 30} \right) - 25\left( {x + 30} \right) = 0\]
Now taking $\left( {x + 30} \right)$ outside from the entire equation, we’ll get:
\[
   \Rightarrow \left( {x - 25} \right)\left( {x + 30} \right) = 0 \\
   \Rightarrow \left( {x - 25} \right) = 0{\text{ or }}\left( {x + 30} \right) = 0 \\
   \Rightarrow x = 25{\text{ or }}x = - 30 \\
\]
Speed cannot be negative. So, ignoring negative values, we have $x = 25$ as a valid solution.
Thus, the usual speed of the train is 25 km/hr.

Note: If we are facing any trouble factorising a quadratic equation, we can also use a direct formula to find out its roots. The roots of the quadratic equation $a{x^2} + bx + c = 0$ can also be determined using the formula:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$