A man can row at a speed of 5 km/hr in still water. If he rows a certa

Let the distance traveled upstream be D
We know, Speed= Distance / Time

Time for traveling upstream = Distance/Speed = D/ 5-1.5 = D /3.5

Distance traveled upstream be D
Time for traveling upstream = Distance/Speed = D/ 5+1.5 = D /6.5

Total Time = D /3.5 + D /6.5

Average Speed = Total Distance/ Total Time = D + D / (D /3.5 + D /6.5 ) = (2D * 7 * 5 * 13) / (200 D) 4.55

Given: A man can row at a speed of 5 km/hr in still water.
Asked: If he rows a certain distance upstream and back to the starting point in a river which flows at 1.5 km/hr, what is his average speed for the double journey?

Let the average speed for the double journey be x kmh and distance for one side be D km

D/(5-1.5) + D/(5+1.5) = 2D/x
D/3.5 + D/6.5 = 2D/x
2/7 + 2/13 = 2/x
20/7*13 = 1/x
x = 7*13/20 = 91/20 = 4.55 kmh

IMO B

When the distance is same for "to and fro" Journey we can use the formula
Average Speed= 2*X*Y/(X+Y)
Where X and Y are the speeds for to and fro journey respectively
Average Speed= 2*6.5*3.5/6.5+3.5
=91/20
=4.55 km/h
B:)

Solution:

We see that his upstream speed is 5 - 1.5 = 3.5 km/hr and his downstream speed is 5 + 1.5 = 6.5 km/hr. We can let the one-way distance be 45.5 km. So, the time traveling upstream is 45.5/3.5 = 13 hours and the time traveling downstream is 45.5/6.5 = 7 hours. Therefore, the average speed for the entire round-trip journey is (45.5 + 45.5)/(13 + 7) = 91/20 = 4.55 km/h.

Answer: B

Such a great question to really understand Weighted Average and the fact that Average Speed is just a “Time-Weighted” Average.

The effective speeds are as follows:

(5 + 1.5) = 6.5 ——— downstream

5 ——— if there were no current, in still water

(5 - 1.5) = 3.5——- upstream

Since the distance is the same for both legs of the round trip, the Time completely determines near which effective speed the average speed will lie.

Since he is rowing much faster with the current down stream (almost twice as fast), he will spend less time rowing at the speed of 6.5 and more time rowing at the speed of 3.5.

Therefore, the speed of 3.5 will be “weighted” more heavily than the speed of 6.5, and the average speed will lie closer to 3.5.

In other words, the average speed will lie much closer to the 3.5 speed-data point, since more time will spent rowing at that speed. D and E can be eliminated.

The average speed is not going to be his upstream speed of 3.5, because he does spend an entire leg of the journey traveling at 6.5. So A can not be the answer.

And C is the classic average speed trap-answer. The Average Speed is NOT the arithmetic mean of the two effective speeds: 5. So C must be wrong.

The only answer it could be is B) 4.55

You can get there completely by looking at the answer choices and using logic.

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