Question

A train has a journey of 320 km to run. After going 1/5th of the distance, the engine breaks and it can only run the remaining part of the journey at 3/4th of the original speed. If it arives 2hrs 40 min late, what is its original speed?​

Answer 1

Answer:

32km/h

Step-by-step explanation:

given,

total distance=320 km

let the original speed of train is x and the time taken by the train to cover the distance at the it's original speed is t

since, time taken=$\frac{distance}{speed}$

  t=$\frac{320}{x}$.........(1)

since, the train run at original speed only for $\frac{1}{5}$ of total distance i.e 32$320\times\frac{1}{5}=64$ and rest distance covered by the train by $\frac{3}{4}$ of original speed i.e 320-64=256 km

since the train reach at it's destination 2hrs 40 min late

$\frac{64}{x} +\frac{256}{\frac{3}{4} \times x }=t+2\frac{40}{60}$

 $\frac{64}{x} +\frac{1024}{3\times x}=\frac{320}{x}+\frac{16}{6}$       (from eq. 1, t=$\frac{320}{x}$ )

x=(128+24-120 )km/hr=32km/hr

hence, the original speed of train is 32 km\hr

Answer 2

Answer:

suppose that train start at 9:00 am and by get it reached at station 11:40 am