Discussion :: Boats and Streams - General Questions (Q.No.14)
|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
|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.
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.
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.
|Zameer Basha said: (Aug 20, 2015)|
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)|
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
2B = 3x
B= 1.5 x
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)
=>R=2x-B =>R=2x-(3/2)x =>R=x/2.
B:R=3/2 : 1/2 = 3:1.
|Swetha Reddy said: (Dec 19, 2019)|
|How the speed of the boat in still water is (2x+x) /2?|
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