### 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 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. |

Swetha Reddy said: (Dec 19, 2019) | |

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

Fathima Sulfikkar said: (Aug 2, 2020) | |

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. |

Mani said: (Feb 18, 2021) | |

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? |

Aruna said: (Jun 8, 2021) | |

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. |

Jamshaid said: (Aug 25, 2022) | |

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

#### Post your comments here:

Name *:

Email : (optional)

» Your comments will be displayed only after manual approval.