A train covers a distance of 10 text{ km} in | vedclass

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Question

(D) Initial speed of the train $= \frac{	ext{Distance}}{	ext{Time}} = \frac{10 	ext{ km}}{12 	ext{ min}} = \frac{10 	ext{ km}}{12/60 	ext{ hr}}

Vedclass · Accepted Answer

$13 	ext{ min } 20 	ext{ sec}$

Vedclass · Answer

(D) Initial speed of the train $= \frac{	ext{Distance}}{	ext{Time}} = \frac{10 	ext{ km}}{12 	ext{ min}} = \frac{10 	ext{ km}}{12/60 	ext{ hr}} = \frac{10 	imes 60}{12} 	ext{ km/hr} = 50 	ext{ km/hr}$.If the speed is decreased by $5 	ext{ km/hr}$,the new speed $= 50 - 5 = 45 	ext{ km/hr}$.Time taken to cover the same distance of $10 	ext{ km}$ at the new speed:Time $= \frac{	ext{Distance}}{	ext{New Speed}} = \frac{10}{45} 	ext{ hours}$.Converting into minutes: $\frac{10}{45} 	imes 60 	ext{ minutes} = \frac{10 	imes 4}{3} 	ext{ minutes} = \frac{40}{3} 	ext{ minutes}$.$\frac{40}{3} 	ext{ minutes} = 13 \frac{1}{3} 	ext{ minutes} = 13 	ext{ minutes } + (\frac{1}{3} 	imes 60) 	ext{ seconds} = 13 	ext{ minutes } 20 	ext{ seconds}$.

$A$ train covers a distance of $10 \text{ km}$ in $12 \text{ minutes}$. If its speed is decreased by $5 \text{ km/hr}$,the time taken by it to cover the same distance will be:

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