Question

A train after travelling 50 km meet with an accident and then proceeds at $ {{\left( \dfrac{3}{4} \right)}^{th}} $ of its former speed and arrives at its destination 35 minutes late. Had an accident occurred 24 kms further, it would have reached its destination only 25 minutes late. The speed of the train is ? \[\]
A.36 kmph\[\]
B.48 kmph\[\]
C.54 kmph\[\]
D.58 kmph\[\]

Answer 1

Hint: We denote the distance to be covered after the accident as $ y $ km and the speed before the accident as $ x $ km/hr. We use the formula for a time from speed and distance $ t=\dfrac{d}{v} $ to make linear equations in $ x,y $ for the actual condition reaching 35 minutes late. We make another linear equation for the hypothetical condition of reaching 25 minutes. We solve the linear equations to get $ x $. \[\]

Complete step by step answer:
We know that if an object with speed $ v $ covers a distance $ d $ , then the time taken by the object to cover that distance is given by
\[t=\dfrac{d}{v}\]
We are given in the question that A train after travelling 50 km meet with an accident and then proceeds at $ {{\left( \dfrac{3}{4} \right)}^{th}} $ of its former speed and arrives at its destination 35 minutes late. Let us assume the distance the train has to cover after the accident as $ y $ km and the speed of the train before the accident is $ x $ km/hr. So the speed of the train after the accident is $ \dfrac{3}{4}x $ km/hr. The delay in arrival is the time difference between the time the train takes to cover the rest distance $ y $ after the accident and time it would have taken without the accident. So we have;
\[\begin{align}
  & \dfrac{y}{\dfrac{3}{4}x}-\dfrac{y}{x}=\dfrac{35}{60} \\
 & \Rightarrow \dfrac{4y}{3x}-\dfrac{y}{x}=\dfrac{7}{12} \\
 & \Rightarrow \dfrac{y}{x}\left( \dfrac{4}{3}-1 \right)=\dfrac{7}{12} \\
 & \Rightarrow \dfrac{y}{x}\times \dfrac{1}{3}=\dfrac{7}{12} \\
 & \Rightarrow 4y-7x=0......\left( 1 \right) \\
\end{align}\]
We are again given the hypothesis that if an accident would have occurred 24 kms further, it would have reached its destination only 25 minutes late. Now the rest distance the train has to cover is $ y-24 $ km. Similarly we have
\[\begin{align}
  & \dfrac{y-24}{\dfrac{3}{4}x}-\dfrac{y-24}{x}=\dfrac{25}{60} \\
 & \Rightarrow \dfrac{4\left( y-24 \right)}{3x}-\dfrac{\left( y-24 \right)}{x}=\dfrac{5}{12} \\
 & \Rightarrow \dfrac{y-24}{x}\left( \dfrac{4}{3}-1 \right)=\dfrac{5}{12} \\
 & \Rightarrow \dfrac{y-24}{x}\times \dfrac{1}{3}=\dfrac{5}{12} \\
\end{align}\]
We simplify and cross multiply to have
\[\begin{align}
  & \Rightarrow 12\left( y-24 \right)=3x\times 5 \\
 & \Rightarrow 4\left( y-24 \right)=5x \\
 & \Rightarrow 4y-5x=76.....\left( 2 \right) \\
\end{align}\]
We subtract equation (1) from equation (2) to have;
\[\begin{align}
  & \Rightarrow 2x=96 \\
 & \Rightarrow x=\dfrac{96}{2}=48 \\
\end{align}\]
So speed of the train is 48 km/hr and correct option is B.\[\]

Note: We note that be careful that the speed is given in the unit km/hr and the speed is given in minutes. So we have converted the equations by dividing by 60. We can alternatively solve the time difference between the original and hypothetical situation that is $ 35-25=10 $ minutes to cover the extra 24 km. In actual situation the train takes $ \dfrac{24}{\dfrac{3}{4}x} $ minutes time and in original situation without the accident the train takes $ \dfrac{24}{x} $ minutes. We get the equation $ \dfrac{24}{\dfrac{3}{4}x}-\dfrac{24}{x}=\dfrac{10}{60} $ from where we can get $ x=48 $.