# A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km per hour, it takes 2 hours less in the journey. Find the original speed of the train.

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Hint: Convert the problem statement into equations based on the speed formula and the conditions given and then solve the equations.

Given, distance to be covered by the train = 300 km

Let us suppose the original speed of the train be $x$ km per hour and $t$ hours be the original time taken by the train to cover 300 km distance with a speed of x km per hour.

As we know that, ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$

$ \Rightarrow x = \dfrac{{300}}{t}{\text{ }} \to {\text{(1)}}$

According to the problem, now the speed of the train is increased from $x$ km per hour to $\left( {x + 5} \right)$ km per hour and the time taken for the journey is reduced from $t$ hours to $\left( {t - 2} \right)$ hours.

For this case, $

\left( {x + 5} \right) = \dfrac{{300}}{{\left( {t - 2} \right)}} \Rightarrow x = \dfrac{{300}}{{\left( {t - 2} \right)}} - 5 = \dfrac{{300 - 5\left( {t - 2} \right)}}{{\left( {t - 2} \right)}} = \dfrac{{300 - 5t + 10}}{{\left( {t - 2} \right)}} \\

\Rightarrow x = \dfrac{{310 - 5t}}{{t - 2}}{\text{ }} \to {\text{(2)}} \\

$

Now, since the LHS of equations (1) and (2) are equal, their RHS will also be equal.

i.e., $

\dfrac{{300}}{t}{\text{ }} = \dfrac{{310 - 5t}}{{t - 2}} \Rightarrow 300\left( {t - 2} \right) = t\left( {310 - 5t} \right) \Rightarrow 300t - 600 = 310t - 5{t^2} \\

\Rightarrow 5{t^2} - 10t - 600 = 0 \\

\Rightarrow {t^2} - 2t - 120 = 0 \\

$

Now, by factorising the above quadratic equation we have

$

\Rightarrow {t^2} - 2t - 120 = 0 \Rightarrow {t^2} + 10t - 12t - 120 = 0 \Rightarrow t\left( {t + 10} \right) - 12\left( {t + 10} \right) = 0 \\

\Rightarrow \left( {t + 10} \right)\left( {t - 12} \right) = 0 \\

$

i.e., Either $t + 10 = 0$ or $t - 12 = 0$

$ \Rightarrow t = - 10$ or $t = 12$

Since, time is always positive and never negative.

Therefore, $t = - 10$ is neglected and only possible value is $t = 12$.

Now put $t = 12$ in equation (1), we get

$x = \dfrac{{300}}{{12}} = 25$

Therefore, the original speed of the train is 25 km per hour.

Note- In these types of problems, the problem statement is reduced into important equations which are used further to deduce the unknowns. Here, in this problem the distance covered by train is fixed (i.e., even when the speed of the train and the time taken by the train is changed, the distance covered by the train remains fixed).

Given, distance to be covered by the train = 300 km

Let us suppose the original speed of the train be $x$ km per hour and $t$ hours be the original time taken by the train to cover 300 km distance with a speed of x km per hour.

As we know that, ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}$

$ \Rightarrow x = \dfrac{{300}}{t}{\text{ }} \to {\text{(1)}}$

According to the problem, now the speed of the train is increased from $x$ km per hour to $\left( {x + 5} \right)$ km per hour and the time taken for the journey is reduced from $t$ hours to $\left( {t - 2} \right)$ hours.

For this case, $

\left( {x + 5} \right) = \dfrac{{300}}{{\left( {t - 2} \right)}} \Rightarrow x = \dfrac{{300}}{{\left( {t - 2} \right)}} - 5 = \dfrac{{300 - 5\left( {t - 2} \right)}}{{\left( {t - 2} \right)}} = \dfrac{{300 - 5t + 10}}{{\left( {t - 2} \right)}} \\

\Rightarrow x = \dfrac{{310 - 5t}}{{t - 2}}{\text{ }} \to {\text{(2)}} \\

$

Now, since the LHS of equations (1) and (2) are equal, their RHS will also be equal.

i.e., $

\dfrac{{300}}{t}{\text{ }} = \dfrac{{310 - 5t}}{{t - 2}} \Rightarrow 300\left( {t - 2} \right) = t\left( {310 - 5t} \right) \Rightarrow 300t - 600 = 310t - 5{t^2} \\

\Rightarrow 5{t^2} - 10t - 600 = 0 \\

\Rightarrow {t^2} - 2t - 120 = 0 \\

$

Now, by factorising the above quadratic equation we have

$

\Rightarrow {t^2} - 2t - 120 = 0 \Rightarrow {t^2} + 10t - 12t - 120 = 0 \Rightarrow t\left( {t + 10} \right) - 12\left( {t + 10} \right) = 0 \\

\Rightarrow \left( {t + 10} \right)\left( {t - 12} \right) = 0 \\

$

i.e., Either $t + 10 = 0$ or $t - 12 = 0$

$ \Rightarrow t = - 10$ or $t = 12$

Since, time is always positive and never negative.

Therefore, $t = - 10$ is neglected and only possible value is $t = 12$.

Now put $t = 12$ in equation (1), we get

$x = \dfrac{{300}}{{12}} = 25$

Therefore, the original speed of the train is 25 km per hour.

Note- In these types of problems, the problem statement is reduced into important equations which are used further to deduce the unknowns. Here, in this problem the distance covered by train is fixed (i.e., even when the speed of the train and the time taken by the train is changed, the distance covered by the train remains fixed).

Last updated date: 25th Sep 2023

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