Aptitude - Boats and Streams - Discussion
Discussion Forum : Boats and Streams - General Questions (Q.No. 14)
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:
Answer: Option
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.
Discussion:
30 comments Page 2 of 3.
Mohit said:
8 years ago
Downstream distance(x) : upstream distance(y)= 2:1.
Boat speed(u) : stream speed(v)= x+y : x-y.
= 3 : 1 --> answer.
Boat speed(u) : stream speed(v)= x+y : x-y.
= 3 : 1 --> answer.
Einstein said:
8 years ago
How you relate the speed of the man with speed of the boat? Explain it.
Pranay pal said:
9 years ago
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.
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.
Ecoist said:
9 years ago
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.
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.
Rahulj said:
9 years ago
@Niklu Rana, your explanation is very nice.
SOUMIK said:
9 years ago
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.
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.
(1)
Pankaj pateriya said:
9 years ago
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.
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.
(1)
Nithya said:
9 years ago
Thank you all for explaining the solution.
Niklu Rana said:
10 years ago
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.
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.
Omkar said:
1 decade ago
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.
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.
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