Aptitude - Boats and Streams - Discussion

14. 

A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

[A]. 2 : 1
[B]. 3 : 1
[C]. 3 : 2
[D]. 4 : 3

Answer: Option B

Explanation:

Let man's rate upstream be x kmph.

Then, his rate downstream = 2x kmph.

(Speed in still water) : (Speed of stream) = 2x + x : 2x - x
2 2

   = 3x : x
2 2

   = 3 : 1.


Vikas said: (Oct 6, 2010)  
How speed of boat in still water is (2x+x) /2 ?

Amit 7677 said: (Nov 17, 2010)  
Speed should be greater than x but less than 2x.hence mean is taken of 2x and x i.e. 2x+x/2

Ani said: (Oct 10, 2011)  
Why should it be greater than x but lessthan 2x?

Rajendra said: (Feb 8, 2012)  
Can u explain how the below expression frames
2x+x/2

Suhas said: (Jul 12, 2012)  
Let speed of boat in downstream be 'x' kmph then speed of boat in upstream is 'x/2' kmph because time taken is twice hence.

A be the speed of boat and b be the speed of stream.
A+B = B.
A-B = X/2.
Hence A=3/4 and B=1/4.

Ratio is 3:1.

Ravitheja said: (Jan 6, 2015)  
Speed of still water is A = X.

Speed upstream is x/2.

Speed downstream B = 2xX = 2X+X = 3X.

B/A = 3/1 = 3:1.

Try said: (Mar 6, 2015)  
Let the speed of man in still water = M.

Let the speed of stream = S.

Let speed of man in direction of stream = X.

Therefore,

M + S = X (Man in direction of stream).

M - S = X/2 (Man takes twice as much time it means his speed is 1/2*X).

-------------.

2M = 1.5*X.
=> M = 0.75*X.
=> S = 0.25*X.
=> M/S = 0.75/0.25 = 3/1.

R V said: (May 21, 2015)  
Suppose speed upstream is x kmph.

Speed of current y kmph.

Time suppose is = t.

So (x+y)*t = (x-y)*2t because twice time taken from here we get.

y = x/3.

Now we need.

(x+y)/2:(x-y)/2.

Put value of why we get 2:1.

I think its right answer.

Srikanth said: (Jun 25, 2015)  
Let time taken to travel against the stream is 2,

Time taken to travel in favour of stream is 1,

As we know D=S*T,

As Distance is same (constant) so we can neglect it,

Speed downstream (Sd) = D/t =>1/2-----take it as a,

Speed upstream (Su) = D/t =>1-----take it as b,

We need to find ratio of the speed of the boat (in still water) to the stream is, i.e.

1/2(a+b):1/2(a-b).

==>1/2(1+1/2):1/2(1-1/2).

==>3:1.

Zameer Basha said: (Aug 20, 2015)  
Dear @Srikanth.

In favour of stream means 'downstream' ==> (sd) = 1/1---taking it as a,

Against the stream means 'upstream' ==> (Su) = 1/2---taking it as b, now speed of stream will be positive value ((Sd)-(Su))/2.

Am I correct here reply.

Omkar said: (Sep 18, 2015)  
Let t1 = 2t2.

T1/T2 = 2/1.

But time is inversely proportional to speed. Speed = Distance/Time.

So s1/s2 = 1/2.

Speed of boat in still water = 1/2(2+1) = 3/2.

Speed of stream = 1/2(2-1) = 1/2.

3/2/1/2 = 3:1.

Niklu Rana said: (Nov 1, 2015)  
Let the speed of still water = X and the speed of stream = Y;

As given here, Speed against the stream = 1/2*(Speed in favor of the stream);

Therefor 2X-2Y = X+ Y;

So that, X/Y = 3/1.

Nithya said: (Jun 29, 2016)  
Thank you all for explaining the solution.

Pankaj Pateriya said: (Jul 14, 2016)  
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.

Soumik said: (Jul 29, 2016)  
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.

Rahulj said: (Aug 16, 2016)  
@Niklu Rana, your explanation is very nice.

Ecoist said: (Sep 27, 2016)  
Let's call Speed of Boat as B and Speed of Stream as S.
Distance = Time x Velocity.
Let's call Time with T.

Distance with Favour is (B + S) x T.
Distance against the stream is (B - S) x T.


Since the distance is twice if the stream is in favour
So, (B + S) x T = (B - S) x T x 2,
B + S = 2B - 2S,
B = 3S.
So, the ratio is 3 : 1.

Pranay Pal said: (Mar 24, 2017)  
Let,

Boat speed=b
Speed of water= w

Upstream speed =b"w= x kmph---> eqn a.
Downstream speed= b+w= 2x kmph
b+w= 2x kmph
b+w= 2(b"w) kmph
3w= b kmph

Putting value of w in equation a.
M/x = 3/2
Answer would be 3:2.

Einstein said: (Aug 23, 2017)  
How you relate the speed of the man with speed of the boat? Explain it.

Mohit said: (Sep 29, 2017)  
Downstream distance(x) : upstream distance(y)= 2:1.

Boat speed(u) : stream speed(v)= x+y : x-y.
= 3 : 1 --> answer.

Vishi said: (Dec 9, 2017)  
Boat speed(B) +stream speed (S).

B+S = x
B-S = 2x
BY adding
2B = 3x
B= 1.5 x
By substracting
2S = -x
S= -.5 on neglecting - sign .5x
Ration becomes .5 : 1.5
Which is 1:3 also
Stream speed : boat speed.

Sajith said: (Mar 31, 2018)  
Let d= distance and x = time.
Speed in still water = 1/2(speed in down stream + speed in up stream).
= 1/2 (d/x + d/2x) => eqn 1.
Rate of stream = 1/2( speed in down stream - speed in up stream).
= 1/2(d/x - d/2x) => eqn 2;
We need eqn 1 : eqn 2.

On simplification, we will get 3:1.

Manoj said: (Jun 7, 2018)  
The distance is same so,
D/t = x.
where x is the speed of the boat in downstream.
D/t = 2x in upstream case.

Shalinee said: (Jun 10, 2018)  
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.

Ragi K R said: (Jun 30, 2019)  
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.

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