lets see who can solve it
amar caught anearlier train to home today than which he usually takes. his wife normally drives down in a car to pick him up from the station. today he set out on foot from the station to meet his wife on part of the way. he reached home 12 min earlier than he would have done had he waited for his wife. the car can travel at a uniform speed that is 5 times amar speed on foot. amar reached home at striking 6'o clock.
what time wouls amar have reached home, if his wife, forewarned of his change of plans, had met him at the station??
n/a
guy wud have reachd at 5:12 had he told hos wify about his early coming
Let x be the distance from station to home."y" be the speed of guy on foot."x1" be the distance that the guy travelled on foot before he meets his wife.Now lets think from wifes side.Earlier she used to travel 2x distance from home to station to home,time taken would be 2x/5y hr.Today she travelled (x-x1) distance only before she meets her husband.Time taken to reach back home will be 2(x-x1)/5y hr.Difference between these times is 12 mts i.e 1/5 hr
Therefore the eqn becomes 2x/5y - 2(x-x1)/5y =1/5 hr
2x1=y
Time taken by the guy = x1/y = 1/2 hr =30 mts
These 30 mts would have been saved had he informed his wife about his plan.So,he would have reached at 6:12 "o clock - 30 mts = 5:42 " o clock.
what are these types of questions called??