The average speed of the train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 hours, covering a distance of 800 km. find the speed of the train in the onward journey.
Answer
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Hint: We assume the speed of the train in its return journey. From the relation between the return journey and onward journey, we find the speed of the onward journey. We find the total time taken for the journey only and the distance between two places. Using the formula of time relation, we find a linear equation and then solve that to find the solution.
Complete step-by-step answer:
The average speed of the train in the onward journey is 25% more than that in the return journey.
Let’s assume that the speed of the train in its return journey is x km/h.
So, the speed in its onward journey will be 25% more than x.
So, the speed is $x\left( 1+\dfrac{25}{100} \right)=\dfrac{5x}{4}$ km/h.
It’s also given the total time taken for the complete to and fro journey is 17 hours, covering a distance of 800 km.
If the total distance of going and coming back is 800 km. then the distance between two places is $\dfrac{800}{2}=400$ km.
Total time taken is 17 hours but the train halts for one hour on reaching the destination.
So, actual time taken in the journey only is $17-1=16$ hours.
We have a theorem that $time=\dfrac{dis\tan ce}{speed}$.
The time to cover the distance of 400 km in onward journey is ${{t}_{1}}=\dfrac{400}{\dfrac{5x}{4}}=\dfrac{320}{x}$ hour.
The time to cover the distance of 400 km in return journey is ${{t}_{2}}=\dfrac{400}{x}$ hour.
The total time consumed is 16 hours. So, $\dfrac{400}{x}+\dfrac{320}{x}=16$.
We solve the equation to get the value of x.
$\begin{align}
& \dfrac{400}{x}+\dfrac{320}{x}=16 \\
& \Rightarrow x=\dfrac{720}{16}=45 \\
\end{align}$
So, the speed in return journey is 45 km/h.
The speed of the train in the onward journey is $\dfrac{5x}{4}=\dfrac{5\times 45}{4}=56.25$ km/h.
Note: We need to consider only 1 break of 1 hour as after the return journey, there won’t be another break as the journey already ended. The journey took 800 km which means the same distance has been covered twice. So, the actual distance is 400 km.
Complete step-by-step answer:
The average speed of the train in the onward journey is 25% more than that in the return journey.
Let’s assume that the speed of the train in its return journey is x km/h.
So, the speed in its onward journey will be 25% more than x.
So, the speed is $x\left( 1+\dfrac{25}{100} \right)=\dfrac{5x}{4}$ km/h.
It’s also given the total time taken for the complete to and fro journey is 17 hours, covering a distance of 800 km.
If the total distance of going and coming back is 800 km. then the distance between two places is $\dfrac{800}{2}=400$ km.
Total time taken is 17 hours but the train halts for one hour on reaching the destination.
So, actual time taken in the journey only is $17-1=16$ hours.
We have a theorem that $time=\dfrac{dis\tan ce}{speed}$.
The time to cover the distance of 400 km in onward journey is ${{t}_{1}}=\dfrac{400}{\dfrac{5x}{4}}=\dfrac{320}{x}$ hour.
The time to cover the distance of 400 km in return journey is ${{t}_{2}}=\dfrac{400}{x}$ hour.
The total time consumed is 16 hours. So, $\dfrac{400}{x}+\dfrac{320}{x}=16$.
We solve the equation to get the value of x.
$\begin{align}
& \dfrac{400}{x}+\dfrac{320}{x}=16 \\
& \Rightarrow x=\dfrac{720}{16}=45 \\
\end{align}$
So, the speed in return journey is 45 km/h.
The speed of the train in the onward journey is $\dfrac{5x}{4}=\dfrac{5\times 45}{4}=56.25$ km/h.
Note: We need to consider only 1 break of 1 hour as after the return journey, there won’t be another break as the journey already ended. The journey took 800 km which means the same distance has been covered twice. So, the actual distance is 400 km.
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