ayeshaoct
Answered

On a 120km track, a train travels the first 30 km at a uniform speed of 30km/hr. How fast must the train travel the next 90 km so as to average 60 km/hr for the entire trip?

Answer :

kvnmurty
distance d1 = 30 km,      v1 = 30 kmph
   => time taken = t1 = d1/v1 = 1 hour

So distance remaining = d2 = 120 - d1 = 90 km
let speed to cover distance d2 = v2  kmph
   => time taken = d2/v2 = 90/v2  hours

average speed = total distance / total duration of time
               60 kmph = 120 km / [1 hour + 90 / v2  hours]
       =>  1 = 2 / [ 1 + 90 / v2 ]
       =>  v2 = 90 kmph
====================================
another way:

   as the total distance = 120 km
   average speed = 60 kmph 
       => total time duration = 120 / 60 = 2 hours
 
   time taken for the first 30 km =  30 km /30 kmph = 1 hours

   Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour
   => speed during 90 km  = 90 kmph

chaitanya6786

Answer:

distance d1 = 30 km,      v1 = 30 kmph

  => time taken = t1 = d1/v1 = 1 hour

So distance remaining = d2 = 120 - d1 = 90 km

let speed to cover distance d2 = v2  kmph

  => time taken = d2/v2 = 90/v2  hours

average speed = total distance / total duration of time

              60 kmph = 120 km / [1 hour + 90 / v2  hours]

      =>  1 = 2 / [ 1 + 90 / v2 ]

      =>  v2 = 90 kmph

====================================

another way:

  as the total distance = 120 km

  average speed = 60 kmph  

      => total time duration = 120 / 60 = 2 hours

 

  time taken for the first 30 km =  30 km /30 kmph = 1 hours

  Hence, time remaining to cover the remaining 90 km = 2 hrs - 1 hrs = 1 hour

  => speed during 90 km  = 90 kmph

Explanation:

Other Questions