A train covered a certain distance at a uniform speed. If the train would have been 10km/hr faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10km/hr, it would have taken 3 hours more than the scheduled time. The distance covered by the train is
A. 300km
B. 400km
C. 600km
D. 260km

Hint: First of all, we will suppose the distance and speed to be variable and then we will use the formula to calculate the time. If we have distance ’d’ and velocity with which the distance is covered is 'v' the time ’t’ to cover the distance is given by,
\[t=\dfrac{\text{distance(d)}}{\text{speed(v)}}\]

Complete step-by-step answer:
We know that, the time 't' is given by,
\[t=\dfrac{d}{v}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (1)}\]
Now, we have been given that, if the train runs 10km/hr faster, then, it would take 2 hours less than the scheduled time.
So, we have,
\[\text{Speed(v) = }\left( v+10 \right)km/hr,\text{ distance = d and time(t) = }\left( t-2 \right)\text{hour}\]
Using the formula, \[\text{time}=\dfrac{\text{distance(d)}}{\text{speed(v)}}\]
\[\Rightarrow \dfrac{d}{v+10}=t-2\]
Substituting the value of t from equation (1), we get:
\[\begin{align}
  & \Rightarrow \dfrac{d}{v+10}=\dfrac{d}{v}-2 \\
 & \Rightarrow \dfrac{d}{v+10}+2=\dfrac{d}{v} \\
 & \Rightarrow \dfrac{d+2v+20}{v+10}=\dfrac{d}{v} \\
\end{align}\]
On cross multiplication, we get:
\[\Rightarrow dv+2{{v}^{2}}+20v=dv+10d\]
On cancelling the similar terms, we get:
\[\Rightarrow 2{{v}^{2}}+20v-10d=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (2)}\]
Again, we have been given that, if the train runs 10km/hr slower then, it would take 3 hours more than the scheduled time.
So, we have,
\[\text{Speed(v }\!\!'\!\!\text{ }\!\!'\!\!\text{ ) = }\left( v-10 \right)km/hr,\text{ distance = d and time(t }\!\!'\!\!\text{ }\!\!'\!\!\text{ ) = }\left( t+3 \right)\text{hour}\]
Using the formula,
\[\begin{align}
  & \text{time}=\dfrac{\text{distance}}{\text{speed}} \\
 & \Rightarrow \dfrac{d}{v-10}=t+3 \\
\end{align}\]
Substituting the value of t from equation (1), we get:
\[\begin{align}
  & \Rightarrow \dfrac{d}{v-10}=\dfrac{d}{v}+3 \\
 & \Rightarrow \dfrac{d}{v-10}-3=\dfrac{d}{v} \\
 & \Rightarrow \dfrac{d-3v+30}{v-10}=\dfrac{d}{v} \\
\end{align}\]
On cross multiplication, we get:
\[\Rightarrow dv-3{{v}^{2}}+30v=dv-10d\]
On cancelling the similar terms, we get:
\[\Rightarrow 3{{v}^{2}}-30v-10d=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (3)}\]
On subtracting equation (2) from equation (3), we get:
\[\begin{align}
  & \Rightarrow {{v}^{2}}-50v=0 \\
 & \Rightarrow v\left( v-50 \right)=0 \\
 & \Rightarrow v=0\text{ and v=50} \\
\end{align}\]
v cannot be equal to 0 as \[\dfrac{d}{v}\] tends to infinity.
So, v = 50 km/hr.
Substituting the value of v = 50 in equation (2), we get:
\[\begin{align}
  & \Rightarrow 2\times {{\left( 50 \right)}^{2}}+20\times 50-10d=0 \\
 & \Rightarrow 2\times 2500+1000-10d=0 \\
 & \Rightarrow 5000+1000-10d=0 \\
 & \Rightarrow 6000-10d=0 \\
 & \Rightarrow 6000=10d \\
 & \Rightarrow d=\dfrac{6000}{10} \\
 & \Rightarrow d=600km \\
\end{align}\]
Hence, the distance is equal to 600km.
Therefore, the correct option is C.

Note: It is very important that you will suppose the speed as well as the distance in km/hr and km respectively. Since, the other conditions are also given in this unit of distance and speed. If the units were different in the question, we would have converted them to a single unit. As soon as students get equation 2, they may try to use the quadratic formula to find the values of v. But, it is not advised to do so since they will end up with terms containing d inside the root, which will simply complicate everything.