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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 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.

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 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.

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