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Question

(D) Step $1$: Calculate the distance between Bettiah and Motihari.Distance = $	ext{Speed} 	imes 	ext{Time}$.Given, $	ext{Speed} = 50 \, 	ext{km/h

Vedclass · Accepted Answer

$40$

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(D) Step $1$: Calculate the distance between Bettiah and Motihari.Distance = $	ext{Speed} 	imes 	ext{Time}$.Given, $	ext{Speed} = 50 \, 	ext{km/h}$ and $	ext{Time} = 44 \, 	ext{min} = \frac{44}{60} \, 	ext{h}$.Distance = $50 	imes \frac{44}{60} = \frac{50 	imes 44}{60} = \frac{220}{6} \, 	ext{km} = \frac{110}{3} \, 	ext{km}$.Step $2$: Calculate the new speed.New Speed = $50 + 5 = 55 \, 	ext{km/h}$.Step $3$: Calculate the new time required to cover the same distance.$	ext{New Time} = \frac{	ext{Distance}}{	ext{New Speed}} = \frac{110/3}{55} \, 	ext{h}$.$	ext{New Time} = \frac{110}{3 	imes 55} = \frac{2}{3} \, 	ext{h}$.Step $4$: Convert the time into minutes.$	ext{New Time in minutes} = \frac{2}{3} 	imes 60 = 40 \, 	ext{min}$.

$A$ bus started its journey from Bettiah and reached Motihari in $44 \, \text{min}$ with an average speed of $50 \, \text{km/h}$. If the average speed of the bus is increased by $5 \, \text{km/h}$, how much time will it take to cover the same distance? (in $\text{min}$)

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