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Boats & Streams
1.A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(13+4\right)\phantom{\rule{thinmathspace}{0ex}}km/hr\\ & =17\phantom{\rule{thinmathspace}{0ex}}km/hr\\ & \text{Time}\phantom{\rule{thinmathspace}{0ex}}\text{taken}\phantom{\rule{thinmathspace}{0ex}}\text{to}\phantom{\rule{thinmathspace}{0ex}}\text{travel}\phantom{\rule{thinmathspace}{0ex}}68\phantom{\rule{thinmathspace}{0ex}}km\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\frac{68}{17}\phantom{\rule{thinmathspace}{0ex}}hrs\\ & =4\phantom{\rule{thinmathspace}{0ex}}hrs\end{array}$

2.A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. The man's speed against the current is:

Explanation:

Solution:
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.

3.A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

Explanation:

Solution:
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.
$\begin{array}{rl}& ⇒\left(x×8\frac{4}{5}\right)=\left(y×4\right)\\ & ⇒\frac{44}{5}x=4y\\ & ⇒y=\frac{11}{5}x\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{ratio}\\ & =\left(\frac{y+x}{2}\right):\left(\frac{y-x}{2}\right)\\ & =\left(\frac{16x}{5}×\frac{1}{2}\right):\left(\frac{6x}{5}×\frac{1}{2}\right)\\ & =\frac{8}{5}:\frac{3}{5}\\ & =8:3\end{array}$

4.In one hour, a boat goes 11 km/hr along the stream and 5 km/hr against the stream. The speed of the boat in still water (in km/hr) is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{stillwater}\\ & =\frac{1}{2}\left(11+5\right)\phantom{\rule{thinmathspace}{0ex}}kmph\\ & =8\phantom{\rule{thinmathspace}{0ex}}kmph\end{array}$

5.A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water?

Explanation:

Solution:
$\begin{array}{rl}& \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(\frac{16}{2}\right)kmph=8\phantom{\rule{thinmathspace}{0ex}}kmph.\\ & \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{upstream}\\ & =\left(\frac{16}{4}\right)kmph=4\phantom{\rule{thinmathspace}{0ex}}kmph.\\ & \therefore \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{still}\phantom{\rule{thinmathspace}{0ex}}\text{water}\\ & =\frac{1}{2}\left(8+4\right)kmph=6\phantom{\rule{thinmathspace}{0ex}}kmph\end{array}$

6.The speed of a boat in still water in 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(15+3\right)kmph=18\phantom{\rule{thinmathspace}{0ex}}kmph.\\ & \text{Distance}\phantom{\rule{thinmathspace}{0ex}}\text{travelled}\\ & =\left(18×\frac{12}{60}\right)km=3.6\phantom{\rule{thinmathspace}{0ex}}km\end{array}$

7.A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{stream}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}mph.\phantom{\rule{thinmathspace}{0ex}}\text{Then},\\ & \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}=\left(10+x\right)\phantom{\rule{thinmathspace}{0ex}}mph,\\ & \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{upstream}=\left(10-x\right)\phantom{\rule{thinmathspace}{0ex}}mph.\\ & \therefore \frac{36}{\left(10-x\right)}-\frac{36}{\left(10+x\right)}=\frac{90}{60}\\ & ⇒72x×60=90\left(100-{x}^{2}\right)\\ & ⇒{x}^{2}+48x-100=0\\ & ⇒\left(x+50\right)\left(x-2\right)=0\\ & ⇒x=2\phantom{\rule{thinmathspace}{0ex}}mph\end{array}$

8.A boat covers a certain distance downstream in 1 hour, while it comes back in 11/2 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{speed}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{boat}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{still}\phantom{\rule{thinmathspace}{0ex}}\text{water}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}kmph\text{,}\phantom{\rule{thinmathspace}{0ex}}\text{then}\\ & \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(x+3\right)\phantom{\rule{thinmathspace}{0ex}}kmph\\ & \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{upstream}\\ & =\left(x-3\right)\phantom{\rule{thinmathspace}{0ex}}kmph\\ & \therefore \left(x+3\right)×1=\left(x-3\right)×\frac{3}{2}\\ & ⇒2x+6=3x-9\\ & ⇒x=15\phantom{\rule{thinmathspace}{0ex}}kmph\end{array}$

9.A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?

Explanation:

Solution:
$\begin{array}{rl}& \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(\frac{1}{10}×60\right)km/hr=6\phantom{\rule{thinmathspace}{0ex}}km/hr\\ & \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{upstream}=2\phantom{\rule{thinmathspace}{0ex}}km/hr\\ & \text{Speed}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{still}\phantom{\rule{thinmathspace}{0ex}}\text{water}\\ & =\frac{1}{2}\left(6+2\right)km/hr=4\phantom{\rule{thinmathspace}{0ex}}km/hr\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{time}\\ & =\left(\frac{5}{4}\right)hrs\\ & =1\frac{1}{4}hrs\\ & =1\phantom{\rule{thinmathspace}{0ex}}hr\phantom{\rule{thinmathspace}{0ex}}15\phantom{\rule{thinmathspace}{0ex}}min.\end{array}$

10.A man can row three-quarters of a kilometre against the stream in 111/4 minutes and down the stream in 71/2 minutes. The speed (in km/hr) of the man in still water is:

$\begin{array}{rl}& \text{We}\phantom{\rule{thinmathspace}{0ex}}\text{can}\phantom{\rule{thinmathspace}{0ex}}\text{write}\phantom{\rule{thinmathspace}{0ex}}\text{three - quarters}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{akilometre}\phantom{\rule{thinmathspace}{0ex}}\text{as750}\phantom{\rule{thinmathspace}{0ex}}\text{metres}.\\ & \text{and}\phantom{\rule{thinmathspace}{0ex}}11\frac{1}{4}\phantom{\rule{thinmathspace}{0ex}}\text{minutes}\phantom{\rule{thinmathspace}{0ex}}\text{as}\phantom{\rule{thinmathspace}{0ex}}\text{675}\phantom{\rule{thinmathspace}{0ex}}\text{seconds}\text{.}\\ & \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{upstream}\\ & =\left(\frac{750}{675}\right)m/\mathrm{sec}=\frac{10}{9}m/\mathrm{sec}\\ & \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{downstream}\\ & =\left(\frac{750}{450}\right)m/\mathrm{sec}=\frac{5}{3}m/\mathrm{sec}\\ & \therefore \text{Rate}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{still}\phantom{\rule{thinmathspace}{0ex}}\text{water}\\ & =\frac{1}{2}\left(\frac{10}{9}+\frac{5}{3}\right)m/\mathrm{sec}\\ & =\frac{25}{18}\phantom{\rule{thinmathspace}{0ex}}m/\mathrm{sec}\\ & =\left(\frac{25}{18}×\frac{18}{5}\right)km/hr\\ & =5\phantom{\rule{thinmathspace}{0ex}}km/hr\end{array}$