Skip to main content

The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km

The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is:
[A] 45 km/hr
[B] 47.5 km/hr
[C] 52 km/hr
[D] 56.25 km/hr


ANS: D
Explanation:
Let the speed in return journey be x km/hr.
Then, speed in onward journey =(125/100)x = (5/4)x km/hr
So. Speed in onward journey = [(5/4)*45] km/hr = 56.25 km/hr

Comments

  1. Let the speed in the return journey be x km/hr
    Thrn speed in onward journey=(125/100)x=(5/4)x km/hr
    Here time taken =17-1=16 hrs
    Then 400/x + 400×4/5x =16
    x=45
    SO, speed in onward journey={5/4*45}km/hr
    =56.25 Km/hr

    ReplyDelete
  2. This comment has been removed by the author.

    ReplyDelete

Post a Comment

Popular posts from this blog

Find the number of perfect squares in the given series 2013, 2020, 2027,.

Find the number of perfect squares in the given series 2013, 2020, 2027,................, 2300   (Hint 44^2=1936) a. 1   b. 2    c. 3   d. Can’t be determined Answer: a Explanation: The given series is an AP with common difference of 7. So the terms in the above series are in the form of 2013 + 7k.  We have to find the perfect squares in this format in the given series. Given that 44^2 = 1936. Shortcut: To find the next perfect square, add 45th odd number to 44^2. So 45^2 = 1936 + (2 x 45 -1) = 2025 46^2 = 2025 + (2 x 46 - 1) = 2116 47^2 = 2116 + (2 x 47 - 1) = 2209 Now subtract 2013 from the above numbers and divide by 7. Only 2209 is in the format of 2013 + 7k.  One number satisfies.