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10 Jan 2020, 04:33
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Difficulty:
65%
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Question Stats:
65% (02:55) correct
35%
(02:58)
wrong
based on 314
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A boatman rows to a place 48 km distant and back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. Find the rate of the stream.
A. 1 km/h
B. 2 km/h
C. 3 km/h
D. 4 km/h
E. 5 km/h
A. 1 km/h
B. 2 km/h
C. 3 km/h
D. 4 km/h
E. 5 km/h
25 Sep 2020, 12:58
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Bunuel
You could go through the trouble of setting up equations and solving them. But if you think about it, you can use the fact that the distances with and against the stream are the same to make the problem much easier.
Since the distances are the same, the rates will be inversely proportional to the times. (Note: this wouldn't work if the distances were different each way.) Since the ratio for rates is 4:3, the times will be 3:4. Therefore, the time will be:
> Time, with stream: \(\frac{3}{7} × 14 = 6\) hours
> Time, against stream: \(\frac{4}{7} × 14 = 8\) hours
By the formula \(Rate = \frac{Distance}{Time}\), you get
> Rate, with stream: \(\frac{48}{6} = 8\) km/h
> Rate, against stream: \(\frac{48}{8} = 6\) km/h
Don't be fooled into thinking the answer is 2. The stream adds for the with-stream rate and subtracts for the against-stream rate. So, the no-stream rate must be 7 km/h and the speed of the stream must be (A) 1 km/h.
25 Sep 2020, 07:29
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Boats and streams use the concept of relative speed, where we have one moving object on a surface which is in motion.
In this case, Speeds are added when moving in the same direction, and subtracted when moving in the opposite direction.
In the question above, we are given that for every 4 kms traveled downstream, he moves 3 kms upstream i.e the ratio of speeds \(S_d : S_u = 4 : 3\).
Let \(S_d = 4x\) and \(S_u = 3x\)
\(Total \space Time = \frac{d_u}{S_u} + \frac{d_d}{S_d}\)
\(14 = \frac{48}{4x} + \frac{48}{3x} = \frac{12}{x} + \frac{16}{x} = \frac{28}{x}\)
Therefore x = 2.
Substituting for x, \(S_d = 8\) and \(S_u = 6\)
\(S_s = \frac{S_d \space - \space S_u}{2}= \frac{8 - 6}{2} = 1\)
Option A
It is good to remember to derive the equations when doing questions on Boats and Streams
There are 4 equations based on this.
If \(S_b\) is the Speed of Boat in still water and \(S_s\) is the speed of the stream, then
Speed Downstream, \(S_d = S_b \space + \space S_s\) .... (1)
Speed Upstream, \(S_d = S_b \space - \space S_s\) ..... (2)
Adding Equations (1) and (2), we get \(S_b = \frac{S_d \space + \space S_u}{2}\) ... (3)
Subtracting the Equations, we get \(S_s = \frac{S_d \space - \space S_u}{2}\) ... (4)
Arun Kumar
In this case, Speeds are added when moving in the same direction, and subtracted when moving in the opposite direction.
In the question above, we are given that for every 4 kms traveled downstream, he moves 3 kms upstream i.e the ratio of speeds \(S_d : S_u = 4 : 3\).
Let \(S_d = 4x\) and \(S_u = 3x\)
\(Total \space Time = \frac{d_u}{S_u} + \frac{d_d}{S_d}\)
\(14 = \frac{48}{4x} + \frac{48}{3x} = \frac{12}{x} + \frac{16}{x} = \frac{28}{x}\)
Therefore x = 2.
Substituting for x, \(S_d = 8\) and \(S_u = 6\)
\(S_s = \frac{S_d \space - \space S_u}{2}= \frac{8 - 6}{2} = 1\)
Option A
It is good to remember to derive the equations when doing questions on Boats and Streams
There are 4 equations based on this.
If \(S_b\) is the Speed of Boat in still water and \(S_s\) is the speed of the stream, then
Speed Downstream, \(S_d = S_b \space + \space S_s\) .... (1)
Speed Upstream, \(S_d = S_b \space - \space S_s\) ..... (2)
Adding Equations (1) and (2), we get \(S_b = \frac{S_d \space + \space S_u}{2}\) ... (3)
Subtracting the Equations, we get \(S_s = \frac{S_d \space - \space S_u}{2}\) ... (4)
Arun Kumar
