Any solution to this from any of the Geniuses here? Here is a problem :- In the morning a car travelling at uniform speed drops a man to his office from his home and returns home. This is repeated in the evening to pick him up. Everyday this happens by the clock. There is no traffic on the road and no time is wasted in boarding or alighting. One day, the office closes one hour earlier. Unable to contact his home or the driver, he starts walking towards home at a uniform speed of 4km/hr as soon as his office closes. The driver starts from home at his usual time and picks up the man somewhere enroute, and returns home. The man is happy to find that he arrived 10 minutes earlier than usual. What is the constant speed of the car ?
Answers
Given : One day, the office closes one hour earlier. Unable to contact his home or the driver, he starts walking towards home at a uniform speed of 4km/hr as soon as his office closes. The driver starts from home at his usual time and picks up the man somewhere enroute, and returns home
To find : if Man arrived 10 minutes earlier than usual. What is the constant speed of the car
Solution:
Man Saved 10 minutes
This Saving of timing is of car going From that meeting point to office & Coming Back to that point
so Car meet at that point 5 min before the Usual Time
But Office Closed 1 hr before the usual time hence
person has walked for 55 mins
Distance Covered by person = 4 * (55/60) = 55/15 = 11/3 km
Distance to be covered by Car in 5 mins = 11/3 km
Hence Speed of Car = (11/3)/(5/60) = 44 km/hr
the constant speed of the car = 44 km/hr
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