Question

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer 1

Hint: First of all consider the original time taken, speed of the train and distance as T, S, and D respectively. Then use the formula, \[\text{Speed = }\dfrac{\text{Distance}}{\text{Time}}\]. Use the same formula and substitute in it the decreased time and increased speed. Also, make another equation substituting increased time and decreased speed. Finally, solve these 3 equations to find D, T, and S.

Complete step-by-step answer:
We are given a train that covers a certain distance at a uniform speed. If the train would have been 10 km/hr faster, it would take 2 hours less than the scheduled time. Also, if the train were 10 km/hr slower, it would take 3 hours more. We have to find the distance covered by train.
Let us consider that the scheduled time of the train is ‘T’ hours.
Also, let us consider that the train covers distance ‘D’ km and its actual speed is ‘S’ km/hr.
We know that \[\text{Speed = }\dfrac{\text{Distance travelled}}{\text{Time taken}}.....\left( i \right)\]
By applying this for the given train, we get,
\[S=\dfrac{D}{T}.....\left( ii \right)\]
We are given that if the speed of the train is increased by 10 km/hr, then the time taken would decrease by 2 hours to travel the same distance.
So, we get,
Net speed of the train \[=\left( S+10 \right)km/hr\]
Net time is taken by train \[=\left( T-2 \right)\text{hours}\]
Distance = D km
By substituting these in equation (ii), we get,
\[\left( S+10 \right)=\dfrac{D}{\left( T-2 \right)}....\left( iii \right)\]
Also, we are given that if the speed of the train is decreased by 10 km/hr then the time taken would increase by 3 hours.
So we get,
New speed of train = (S – 10) km/hr
New time taken by train = (T + 3) hours
Distance = D km
 By substituting these in equation (ii), we get,
\[\left( S-10 \right)=\dfrac{D}{\left( T+3 \right)}....\left( iv \right)\]
By adding (iii) and (iv), we get,
\[\left( S+10 \right)+\left( S-10 \right)=\dfrac{D}{\left( T-2 \right)}+\dfrac{D}{\left( T+3 \right)}\]
By simplifying the above equation, we get,
\[\Rightarrow 2S=D\left[ \dfrac{\left( T+3 \right)+\left( T-2 \right)}{\left( T-2 \right)\left( T+3 \right)} \right]\]
\[\Rightarrow 2S=D\left( \dfrac{2T+1}{\left( T-2 \right)\left( T+3 \right)} \right)\]
By substituting \[S=\dfrac{D}{T}\] from equation (ii), we get,
\[\Rightarrow 2\dfrac{D}{T}=\dfrac{D\left( 2T+1 \right)}{\left( T-2 \right)\left( T+3 \right)}\]
By canceling like terms and cross multiplying the above equation, we get,
\[2\left( T-2 \right)\left( T+3 \right)=T\left( 2T+1 \right)\]
By simplifying the above equation, we get
\[\begin{align}
  & \Rightarrow 2\left( {{T}^{2}}+3T-2T-6 \right)=2{{T}^{2}}+T \\
 & \Rightarrow 2{{T}^{2}}+2T-12=2{{T}^{2}}+T \\
\end{align}\]
By canceling the like terms, we get,
\[\Rightarrow 2T-12=T\]
Or, \[2T-T=12\]
T = 12 hours
By substituting T = 12 in equation (ii), we get,
\[S=\dfrac{D}{12}\]
By multiplying 12 on both sides, we get
\[\Rightarrow 12S=D....\left( v \right)\]
By substituting T = 12 hours in equation (iii), we get,
\[\left( S+10 \right)=\dfrac{D}{12-2}=\dfrac{D}{10}\]
By multiplying 10 on both sides, we get
\[\Rightarrow 10\left( S+10 \right)=D....\left( vi \right)\]
By equating D from equation (v) and (vi), we get,
\[\begin{align}
  & 12S=10\left( S+10 \right) \\
 & \Rightarrow 12S=10S+100 \\
\end{align}\]
Or, \[12S-10S=100\]
\[\Rightarrow 2S=100\]
By dividing 2 on both sides, we get,
S = 50 km/hr
By substituting S = 50 km/hr in equation (v), we get
D = 12 x 50 = 600 km
Therefore, we get the distance covered by the train is equal to 600 km.

Note: Students must keep in mind that if the distance is constant, time and speed are inversely proportional to each other that is, if speed increases, time decreases and vice – versa. Students can also check the values of D, T and S by substituting them in various equations. For example, if we substitute D = 600 km, T = 12 hours and S = 50 km/hr in the following equation,
\[\left( S-10 \right)=\dfrac{D}{T+3}\]
We get, \[\left( 50-10 \right)=\dfrac{600}{12+3}\]
\[\Rightarrow 40=\dfrac{600}{15}\]
\[\Rightarrow 40=40\]
Since, LHS = RHS, therefore our answer is correct.