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Kritisood
A man rows at the rate of 5 mph in still water. If the river runs at the rate of 1 mph, it takes him 1 hour to row to a place and come back. What is the total, upstream and downstream, distance covered by him in miles?

A) 2
B) 2.4
C) 2.8
D) 4
E) 4.8

Searched the forum for a similar question before posting


Logical
The speed upstream will be 5-1 and downstream will be 5+1.
If the entire route was traveled at the lesser speed, the distance would be 4 miles in 1 hour. But the speed for half the distance is 6mph, so surely the distance is more than 4.
Only 4.8 possible

Average speed
Since the distance is traveled in exactly in 1 hour, we are looking at the average speed for entire distance.
When distance traveled is same, the average speed =\(\frac{2*6*4}{6+4}=4.8\)

Algebraic
Let the distance traveled one way be d, then time taken = \(\frac{d}{4}+\frac{d}{6}=1……d=2.4\)
Total distance =2d=2*2.4=4.8 miles


E
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still water speed = 5mph
upstream speed = 4mph
downstream = 6mph
we need to find avg speed
2 *6*4 / (6+4) = 48/10 ; 4.8
option E

Kritisood
A man rows at the rate of 5 mph in still water. If the river runs at the rate of 1 mph, it takes him 1 hour to row to a place and come back. What is the total, upstream and downstream, distance covered by him in miles?

A) 2
B) 2.4
C) 2.8
D) 4
E) 4.8

Searched the forum for a similar question before posting
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Solution:

To be able to solve this question. we need to be aware of one fact/property:

If the speed of the boat in still water = x and the speed of the stream is y.

Then speed of the boat upstream \(= x-y\) and speed of the boat downstream \(= x+y\).

We are given speed of boat in still water \(= 5mph\) and speed of river/stream \(= 1mph\).

Thus speed of the boat upstream \(= 5-1=4mph\) and speed of the boat downstream \(= 5+1=6mph\)

Let the distance travelled be \(D\). So, distance travelled upstream and downstream \(=\frac{D}{2}\) respectively.

We are given the time of total journey \(= 1hour\)

So. we can write \(\frac{D}{2\times 4} + \frac{D}{2\times 6} = 1\)

\(⇒ \frac{D}{8} + \frac{D}{12} = 1\)

\(⇒ \frac{3D+2D}{24} = 1\)

\(⇒ D=\frac{24}{5}=4.8\)

Hence the right answer is Option E.
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The speed traveling against the current is = (5 - 1) = 4 mph

The speed traveling WITH the current is = (5 + 1) = 6 mph


In both cases we have a constant distance for each leg of the journey.

Over a constant distance, the Ratio of Speeds is Inversely Proportional to the Ratio of Time traveled

Speed UP : Speed DOWN = 4 : 6 = 2 : 3

Time spent traveling UP : Time spent traveling DOWN = 3 : 2

(3/5) of the Total Time is spent traveling at 4 mph

(2/5) of the Total Time is spent traveling at 6 mph

Total time is 1 hour

Leg 1: (4 mph) * (3/5 hr) = 12/5 miles

Leg 2: (6 mph) * (2/5 hr) = 12/5 miles

Total distance = 24/5 =


4.8 miles

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Hey guys,

Start with the standard formula for distance, rate, and time:

Disance = Rate * Time

\(Time = \frac{Distance}{Rate}\)

Total Time = Time to go upstream + Time to go downstream = 1 hour

Time to go upstream = \(\frac{D}{4mph}\)

Time to go downstream = \(\frac{D}{6mph}\)

Total Time = 1 = \(\frac{D}{4mph}\) + \(\frac{D}{6mph}\)

1 = \(\frac{3D}{12}\) + \(\frac{2D}{12}\)

1 = \(\frac{5D}{12}\)

5D = 12

D = \(\frac{12}{5}\)

A little long division gives you the answer:

The distance was 2.4 miles

Since he went upstream and downstream, the total distance is 2D

2.4 * 2 = 4.8

The answer is E
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Speed upstream ( against current ) = 5-1 = 4m/h
Speed downstream ( with current ) = 5+1 = 6m/h
Distance travelled is same x in both scenarios.
total distance = 2x
Let time be t when upstream and 1-t when downstream so that total time is 1 hour.

Upstream : x = 4t
Downstream : x = 6(1-t)

Solving both gives, x = 12/5 and total distance is 24/5 =4.8m
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