Ex. A train travells at average speed of 100 km / hr, it stops for 3 mins after travelling 75 kms of diatance. How long it takes to reach 600 kms from the starting point.
Solution : Time taken to travel 600 kms= 600 / 100 = 6 hrs
But it stop after travelling 75 kms , so number of stoping point in 600 kms will be= 600 / 75 = 8, but the last stoping point is actual end stoping so
Number of stoping point will be 7, and time taken= 7*3= 21 minutes
So total time = 6 hrs 21 mins
Ex. A is faster than B . A and B each walk 24 km. The sum of their speeds is 7 km / hr and the sum of their time taken is 14 hrs. Then A's speed is equal to :
Solution : Let A's speed = x km / hr and B's speed is = 7 - x km / hr
(x-3) (x-4) = 0
x=3, x=4, So A's speed is 4 km / hr, B's speed is 3 km / hr
Ex. A man on tour travels first 180 km at 60 km / hr and next 180 km at speed of 80 km / hr . The average speed of first 360 km of the tour is :
Solution : Total time taken = ( 180 / 60 ) + ( 180 / 80 ) = 21 / 4 hrs
Ex. A train running at 3 / 7 of its own speed reached the destination in 14 hours, how much time could be saved if the train would have run at its own speed ?
Solution : New speed = 3 / 7 of normal speed
So, New Time will be = 7 / 3 of normal time. (Invers relation )
As 7 / 3 of normal time is = 14 hours
So, normal time = (14 * 3 / 7 ) = 12 hrs
So time saved = 14 - 12 = 2 hours
Ex. Speed ratio of two school buses A and B in covering a certain distance is 4 : 5, If A takes 30 minutes more than B covering the distance, then time taken by B to reach the destination is :
Solution : Since speed ratio is 4 : 5
Time ratio will be 5 : 4 , let A takes 5x hrs and B takes 4x hrs to reach the destination then ,
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