Aptitude - Boats and Streams - Discussion

Here, the shortest way is
2(A-B) = A+B.
A/B = 3/1.

Distance is same.

D (up)=D (down); also t (u)= 2t (d)

Speed of boat=b: speed of current = s.
D (u) = D (d).
t (u)(b-s) = t (v)(b+s).
2t (v)(b-s) = t (v)(b+s).
2b-2s = b+s.
2b-b = s+2s.
b = 3s.
b/s = 3/1.

There is a formula.

B/S = (T1 + T2) / (T1 ~ T2).
Here,

* B is the speed of boat in still water.
* S is the speed of stream/current/river.
* T1 is the time taken in upstream/ downstream.
* T2 is the time taken in downstream/ upstream.

Given:

Twice as long to row ' against' the stream (is. Upstream) as to row same distance in 'favor' of the stream(ie. downstream).

Solution:

Upstream : downstream
Time: 2 : 1 ( T1 = 2 *T2 )
So, T1 = 2 and T2 = 1.

Apply in the formula,
B/S = (2+1)/(2-1) = 3/1.
B:S = 3:1.

Hi, but in this question upstream is same as downstream then for both it would be twice then how you could say upstream is x downstream is 2x?

Here we use this formula and simply find the ratio.
Sb/Ss =n+1/n-1 then,
Sb/Ss = 2+1/2-1,
Sb/Ss =3/1 so.

Then we will get,
Sb:Ss = 3:1.

Let speed of boat = b.
Speed of current = c.
Upstream spped = b - c.
Downstream speed= b + c.

Since distance is constant, and time taken in upward is twice the time taken in downstream.
So, 2 * ((d) / (b+c)) = (d) / (b - c).

After solving it.
2b - 2c = b + c.
b/c = 3/1.

If the math is delivered in question as MCQ then it can be solved in a very easy and quick process:.

Let downstream speed = 80.

So, upstream = 40.

So, Boat speed, 60 + current speed 20 = 80.

Boat speed, 60 - current speed 20 = 40,

So ratio = 60 : 20 = 3 : 1, it's informal but by letting this type of numbers we can solve MCQ within a short period.

@Niklu Rana, your explanation is very nice.

How the speed of the boat in still water is (2x+x) /2?

The speed of boat= B, speed of river= R.
upstream=x=B-R -----> (1)
downstream=2x=B+R -----> (2)
So, (1)+(2).

2B=3x
B=x(3/2)
B+R=2x.

=>R=2x-B =>R=2x-(3/2)x =>R=x/2.
B:R=3/2 : 1/2 = 3:1.