# The ratio of speed of man in still water to the speed of the current is:

#### Ganesh Ujwal

P,Q,R are three towns on a river which flows uniformly. Q is equidistant from P and R. A man rows P to Q and back in 10 hr. He can row from P to R in 4 hr. The ratio of speed of man in still water to the
speed of the current is:

I can able to draw the figure, can't proceed anymore.

#### skipjack

Forum Staff
As the man can row from P to Q in 2 hrs and back from Q to P in 8 hrs,
8/2 = (s + c)/(s - c) = (s/c + 1)/(s/c - 1), and so s/c = 5/3.

#### Ganesh Ujwal

8/2 = (s + c)/(s - c) = (s/c + 1)/(s/c - 1), and so s/c = 5/3.
what is s & c ?

In question, they didn't provide values for downstream & upstream. So, how can you equate (s + c)/(s - c) directly to 8/2?

#### skipjack

Forum Staff
I used s and c for the speeds of the man (relative to the water) and of the current respectively, so s + c and s - c are the man's downstream and upstream speeds in relation to the ground.

For a fixed distance, such as the distance between P and Q, covered at constant speed in relation to the ground, that speed is inversely proportional to the time taken to travel the distance (speed = distance/time), which implies that the ratio of the man's times for his journeys from Q to P and from P to Q is the inverse of the ratio of his speeds for those journeys.

#### Ganesh Ujwal

In question, they didn't provide values for downstream & upstream. So, how can you equate (s + c)/(s - c) directly to 8/2?

So Q to P can be downstream or upstream. Looks like ambiguous to me.

#### skipjack

Forum Staff
For a given distance, rowing downstream takes less time than rowing upstream, so he goes downstream for 2 hours from P to Q and goes upstream for 8 hours from Q to P.

Ganesh Ujwal

#### DarnItJimImAnEngineer

So Q to P can be downstream or upstream. Looks like ambiguous to me.
That's the nice thing about algebra. You can actually assume the current is flowing from R to P and still solve the problem. You will just find that the current speed is negative.

#### Ganesh Ujwal

That's the nice thing about algebra.
My question is related to time & distance. Not about variables like we see in algebra.

#### Ganesh Ujwal

For a given distance, rowing downstream takes less time than rowing upstream, so he goes downstream for 2 hours from P to Q and goes upstream for 8 hours from Q to P.
From this statements I can find ratios of Downstream speed & Upstream speed. Question is about ratio of speed of man in still water to the speed of the current. So help me out here.

#### skipjack

Forum Staff
That's a matter of solving 8/2 = (s + c)/(s - c) = (s/c + 1)/(s/c - 1).

Subtracting 1 from both sides gives 3 = 2/(s/c - 1), so 2/3 = s/c - 1.
Hence s/c = 2/3 + 1 = 5/3.