The average speed of a train in the onward jo | vedclass

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(C) Let the speed of the return journey be $v_r = x\, 	ext{km/h}$.Then,the speed of the onward journey is $v_o = x - 0.25x = 0.75x = \frac{3}{4}x\, \

Vedclass · Accepted Answer

$43.75$

Vedclass · Answer

(C) Let the speed of the return journey be $v_r = x\, 	ext{km/h}$.Then,the speed of the onward journey is $v_o = x - 0.25x = 0.75x = \frac{3}{4}x\, 	ext{km/h}$.The total distance for the to and fro journey is $800 + 800 = 1600\, 	ext{km}$.The total time spent is $17\, 	ext{hours}$,including a $1\, 	ext{hour}$ halt. Thus,the actual travel time is $17 - 1 = 16\, 	ext{hours}$.Time taken for onward journey $t_o = \frac{800}{v_o} = \frac{800}{3x/4} = \frac{3200}{3x}$.Time taken for return journey $t_r = \frac{800}{v_r} = \frac{800}{x}$.Total travel time $t_o + t_r = 16\, 	ext{hours}$.$\frac{3200}{3x} + \frac{800}{x} = 16$.Multiply by $3x$: $3200 + 2400 = 48x$.$5600 = 48x \implies x = \frac{5600}{48} = \frac{350}{3} \approx 116.67\, 	ext{km/h}$.Speed of the onward journey $v_o = \frac{3}{4}x = \frac{3}{4} 	imes \frac{350}{3} = \frac{350}{4} = 87.5\, 	ext{km/h}$.Wait,re-evaluating the problem statement: If the total distance covered is $800\, 	ext{km}$ for the whole trip (i.e.,$400\, 	ext{km}$ one way):$t_o + t_r = 16 \implies \frac{400}{3x/4} + \frac{400}{x} = 16 \implies \frac{1600}{3x} + \frac{1200}{3x} = 16 \implies \frac{2800}{3x} = 16 \implies 48x = 2800 \implies x = 58.33$.$v_o = \frac{3}{4} 	imes 58.33 = 43.75\, 	ext{km/h}$.Thus,the speed of the onward journey is $43.75\, 	ext{km/h}$.

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