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The average speed of a train in the onward journey is $25\, \%$ less than that of 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 $800 \,km$. The speed of the train in the onward journey is (in $km/h$)
$45$
$47.06$
$43.75$
$56.25$
Solution
Let the average speed of train for return journey be $'x'$ $km/hr$
Average speed in the onward journey $=\frac{3}{4} x \,km / hr$
i.e. Average speed for the whole journey $=\frac{800}{16}=50 \,km / hr$
$50=\frac{2(x)\left(\frac{3}{4} x\right)}{x+\frac{3}{4} x}=\frac{\frac{3}{2} x^{2}}{\frac{7}{4} x}=\frac{3}{2} \times \frac{4}{7} x=\frac{6}{7} x$
$x=\frac{7 \times 50}{6}\, km / hr$
Speed of train in onward journey $=\frac{3}{4} \times \frac{7 \times 50}{6}=\frac{175}{4}=43.75 km / hr$
