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The average speed of a 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 from journey is 17 hours, covering a distance of 800 km. What is the speed of the train in the onward journey?

I know how to solve this question by long method which is very time consuming. Is there any other short way to solve this question?

I solved this question in the following way:

Let x be the distance. total distance is 800 km, so x = 400 km (assuming the train follows the same path in both directions)

Speed outbound V1 = 400/t1, where t1 is time taken

Speed back V2 = 400/t2, where t2 is time taken

V1 = (1.25)V2 t1 + t2 + 1 = 17

V1 = (1.25)V2 400/t1 = (1.25)(400/t2) 1/t1 = (1.25/t2) 1.25 t1 = t2

t1 + t2 + 1 = 17 t1 + (1.25 t1) + 1 = 17 2.25 t1 = 16 t1 = 7.111 hours t2 = 8.888 hours total time = t1 + t2 + 1 = 16.999, check

V1 = 400/7.111 = 56.25 km/hour V2 = 400/8.888 = 45 km/hour

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  • $\begingroup$ Mind showing what your "time-consuming" method is? I have a solution, but I'm sure it's the one you've used. $\endgroup$ – Kevin Zakka May 14 '15 at 13:54
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The train halts for one hour at the station, so the total time travelled is 16 hours. Therefore, $v_{avg} = \frac{800}{16}= 50km/h$
You can also find that: $v_{avg}= \frac{v_1 + v_2}{2}$

Then its easy to find $v_2$:
$v_1 = 1.25v_2$ $100 = 2.25v_2$
$v_2 = 44.4 km/h$
$v_1 = 44.4 * 1.25 = 55.6km/h$

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