Quantitative Aptitude
BOATS AND STREAMS MCQs
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to travel 68 km downstream = \(\left(\frac{68}{17}\right)hrs = 4hrs.\)
Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr.
Man's rate against the current = (12.5 - 2.5) km/hr = 10 km/hr.
Let the man's rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
\(\left(x\times8\frac{4}{5}\right) = (y\times4)\)
\(\frac{44}{5}x = 4y\)
\(y= \frac{11}{5}x\)
So, Required ratio = \(\left(\frac{x+y}{2}\right):\left(\frac{x-y}{2}\right)\)
\(\left(\frac{16x}{5}\times\frac{1}{2}\right) :\left(\frac{16x}{5}\times\frac{1}{2}\right)\)
\(\frac{8}{5}:\frac{3}{5}\)
= 8:3
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 - x) km/hr.
So, \(\frac{30}{(15+x)}+\frac{30}{(15-x)}=4\frac{1}{2}\)
\(\frac{900}{(225-x^{2})}= \frac{9}{2}\)
9x2 = 225
x2 = 25
x = 5 km/hr.