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.

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.

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 â€˜Tâ€™ hours.
Also, let us consider that the train covers distance â€˜Dâ€™ km and its actual speed is â€˜Sâ€™ 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 â€“ 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 â€“ 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.