Aptitude Shortcuts and Mind Tricks for Boats and Streams Problems Type-II – Download in PDF:
Dear Readers,Here we have given the Aptitude Shortcuts and Mind Tricks for Boats and Streams Problems Type-II. Candidates those who are preparing for the upcoming examination can also download this in PDF.
EXAMPLE QUESTION:
A man can row (28/3) km/hr in still water and finds that it takes him thrice as much time to row upstream than as to row downstream the same distance in the stream. What is the speed of the current?
GIVEN
Speed of man in still water, a = (28/3) km/hr
Time taken to travel upstream is thrice as much as the time taken to travel downstream
SOLUTION
NORMAL METHOD
Let the speed upstream = u km/hr
Then the speed downstream = 3u km/hr
Now the Speed of man in still water = (1/2) × (speed downstream + speed upstream)
Therefore, Speed in still water = (1/2) × (3u + u) = (1/2) × (4u)
                                                = 2u km/hr
Now speed of man in still water is given as (28/3) km/hr in the question
Therefore, 2u = (28/3)
                  u = (14/3)
So, Speed upstream = u = (14/3) km/hr
Speed downstream = 3u = 3 × (14/3) km/hr = 14 km/hr
Now, Speed of Current = (1/2) × (speed downstream – speed upstream)
                                    = (1/2) × (14 – [14/3])
                                    = (1/2) × ([42 – 14]/3)
                                    = (1/2) × (28/3)
Therefore, the speed of current = (14/3) km/hr
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ALTERNATE METHOD
Speed of current = Speed of man in still water × ([T1-T2] / [T1+T2])
T1Â = time taken to go upstream
T2Â = time taken to go downstream
As per the question, T1 = 3T2
Where, speed of man in still water = (28/3) km/hr
Speed of current = (28/3) × ([3T2-T2] / [3T2+T2])
                          = (28/3) × (2/4)
                          = (28/3) × (1/2)
Therefore, Speed of current = (14/3) km/hr
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