A man wearing a hat stands on a bridge across a river. At an instant his hat falls into the river and starts flowing along with the stream at the same speed of the stream.At the same instant the man jumps into the river and starts swimming in the directionopposite to the stream, like that he swims for 10 minute , then he turns back and starts swimming in the direction of the stream to retrieve the hat. He retrieves the hatunder another bridge which is 1000 yards away down the stream from the first bridge.What is the speed of the river?
Let the swimming speed of man be = a yds/min.
speed of stream = b yds/min.
Then the man swims at (a-b) yds/min. against the stream.
and (a+b) yds/min in direction with the stream.
Lets assume the man covered a distance of x in 10 min swim up the stream at speed (a-b).
The time taken by him to cover the same distance down the stream at speed (a+b) is...
=> 10(a-b)/(a+b)
The total time of travel(being same for both), hat and the man be 't'.
Time taken by hat to travel from first bridge to the bridge down stream 1000 yards
away at speed b is 't' minutes.
Time taken by man to travel from first bridge to the bridge down stream 1000 yards
away at speed (a+b) will be [t-{10+[(10(a-b)/(a+b))]}] minutes.
There for we have for hat b*t=1000 yds. ------> (1)
for man .... (a+b)[t-(10(a-b)/(a+b))] = 1000 yds. ------> (2)
solving equations (1) and (2) we get
t=20 minutes.
Answer : so, speed of the stream is 1000/20= 50 yds/min.
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A man wearing a hat stands on a bridge across a river. At an instant his hat falls into the river and starts flowing along with the stream at the same speed of the stream.At the same instant the man jumps into the river and starts swimming in the directionopposite to the stream, like that he swims for 10 minute , then he turns back and starts swimming in the direction of the stream to retrieve the hat. He retrieves the hatunder another bridge which is 1000 yards away down the stream from the first bridge.What is the speed of the river?
speed of stream = b yds/min.
Then the man swims at (a-b) yds/min. against the stream.
and (a+b) yds/min in direction with the stream.
Lets assume the man covered a distance of x in 10 min swim up the stream at speed (a-b).
The time taken by him to cover the same distance down the stream at speed (a+b) is...
=> 10(a-b)/(a+b)
The total time of travel(being same for both), hat and the man be 't'.
Time taken by hat to travel from first bridge to the bridge down stream 1000 yards
away at speed b is 't' minutes.
Time taken by man to travel from first bridge to the bridge down stream 1000 yards
away at speed (a+b) will be [t-{10+[(10(a-b)/(a+b))]}] minutes.
There for we have for hat b*t=1000 yds. ------> (1)
for man .... (a+b)[t-(10(a-b)/(a+b))] = 1000 yds. ------> (2)
solving equations (1) and (2) we get
t=20 minutes.
Answer : so, speed of the stream is 1000/20= 50 yds/min.
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