A train completes a non-stop journey in 8 , t | vedclass

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(B) Let the original speed of the train be $s \, 	ext{km/h}$.The time taken at this speed is $T_1 = 8 \, 	ext{hours}$.The distance $D$ covered is $D

Vedclass · Accepted Answer

$25$

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(B) Let the original speed of the train be $s \, 	ext{km/h}$.The time taken at this speed is $T_1 = 8 \, 	ext{hours}$.The distance $D$ covered is $D = 	ext{speed} 	imes 	ext{time} = 8s$.If the speed is increased by $5 \, 	ext{km/h}$, the new speed is $(s + 5) \, 	ext{km/h}$.The new time taken is $6 \, 	ext{hours} \, 40 \, 	ext{minutes} = 6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3} \, 	ext{hours}$.Since the distance remains the same, we equate the two expressions for distance:$8s = (s + 5) 	imes \frac{20}{3}$Multiply both sides by $3$:$24s = 20(s + 5)$$24s = 20s + 100$$4s = 100$$s = 25 \, 	ext{km/h}$.Thus, the lowest speed of the train is $25 \, 	ext{km/h}$.

$A$ train completes a non-stop journey in $8 \, \text{hours}$. If it had travelled $5 \, \text{km/h}$ faster, it would have completed the journey in $6 \, \text{hours} \, 40 \, \text{minutes}$. What is its lowest speed in $\text{km/h}$?

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