Let the speed of the boat in still water $=x$ km/hr
Speed of the current = 2 km/hr
Then, speed downstream $=\left(x+2\right)$ km/hr
speed upstream $=\left(x-2\right)$ km/hr
Total time taken to travel 10 km upstream and back = 55 minutes $=\dfrac{55}{60}$ hour = $\dfrac{11}{12}$ hour
$\Rightarrow \dfrac{10}{x-2} + \dfrac{10}{x+2} = \dfrac{11}{12}$
$120\left(x+2\right) + 120\left(x-2\right) = 11\left(x^2-4\right)$
$240x = 11x^2 - 44$
$11x^2 - 240x - 44 = 0$
$11x^2 - 242x +2x - 44 = 0$
$11x\left(x-22\right)+2\left(x-22\right)=0 $
$ \left(x-22\right)(11x+2)=0$
$x=22\text{ or }\dfrac{-2}{11}$
Since $x$ cannot be negative, $x$ = 22
i.e., speed of the boat in still water = 22 km/hr
Let the speed of the water in still water = $x$
Given that speed of the stream = 3 kmph
Speed downstream $=\left(x+3\right)$ kmph
Speed upstream $=\left(x-3\right)$ kmph
He travels a certain distance downstream in 4 hour and come back in 6 hour.
ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour
since distance = speed × time, we have
$\left(x + 3\right)4 = \left(x - 3\right)6$=>$\left(x + 3\right)2 = \left(x - 3\right)3$
=>$2x + 6 = 3x - 9$
=>$x = 6+9 = 15\text{ kmph}$
We can write three-quarters of a kilometre as 750 metres,
and 11$\dfrac{1}{4}$ minutes as 675 seconds.
| Rate upstream =$ \left(\dfrac{750}{675} \right) $m/sec=$ \dfrac{10}{9} $m/sec. |
| Rate downstream =$ \left(\dfrac{750}{450} \right) $m/sec=$ \dfrac{5}{3} $m/sec. |
| $\therefore$ Rate in still water =$ \dfrac{1}{2} \left(\dfrac{10}{9} +\dfrac{5}{3} \right) $m/sec |
| =$ \dfrac{25}{18} $m/sec |
| =$ \left(\dfrac{25}{18} \times\dfrac{18}{5} \right) $km/hr |
= 5 km/hr.
Speed of the boat in still water = 8 km/hr
Speed upstream $=\dfrac{1}{1}$ = 1 km/hr
Speed of the stream = 8-1 = 7 km/hr
Speed downstream = (8+7) = 15 km/hr
Time taken to travel 1 km downstream $=\dfrac{1}{15}\text{ hr = }\dfrac{1 \times 60}{15}\text{= 4 minutes}$
Let the mans 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.
| $\Rightarrow \left(x \times 8\dfrac{4}{5} \right)$ = $y$ x 4 |
| $\Rightarrow \dfrac{44}{5} x $ =4$ y $ |
| $\Rightarrow y $ =$ \dfrac{11}{5} x $. |
| $\therefore$ Required ratio =$ \left(\dfrac{y + x}{2} \right) $:$ \left(\dfrac{y - x}{2} \right) $ |
| =$ \left(\dfrac{16x}{5} \times\dfrac{1}{2} \right) $:$ \left(\dfrac{6x}{5} \times\dfrac{1}{2} \right) $ |
| =$ \dfrac{8}{5} $:$ \dfrac{3}{5} $ |
= 8 : 3.
speed of the boat = 6 km/hr
Speed downstream = (6+2) = 8 km/hr
Speed upstream = (6-2) = 4 km/hr
Distance travelled downstream = Distance travelled upstream = 32 km
Total time taken = Time taken downstream + Time taken upstream
$= \dfrac{32}{8} + \dfrac{32}{4} = \dfrac{32}{8} + \dfrac{64}{8} = \dfrac{96}{8}$ = 12 hr
Given that, time taken to travel upstream = 2 × time taken to travel downstream
When distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream
Let speed upstream $=x$
Then speed downstream $=2x$
we have, $\dfrac{1}{2}(x + 2x)$ = speed in still water
$\Rightarrow \dfrac{1}{2}(3x) = 7.5$
$\Rightarrow 3x = 15$
$\Rightarrow x = 5$
i.e., speed upstream = 5 km/hr
Rate of stream $=\dfrac{1}{2}(2x-x) = \dfrac{x}{2}= \dfrac{5}{2} = 2.5\text{ km/hr}$
Speed downstream = (5 + 1) kmph = 6 kmph.
Speed upstream = (5 - 1) kmph = 4 kmph.
Let the required distance be $ x $ km.
| Then,$ \dfrac{x}{6} $+$ \dfrac{x}{4} $= 1 |
$\Rightarrow$ 2$ x $ + 3$ x $ = 12
$\Rightarrow$ 5$ x $ = 12
$\Rightarrow x $ = 2.4 km.
Tap A can fill the tank completely in 6 hours
=> In 1 hour, Tap A can fill $\dfrac{1}{6}$ of the tank
Tap B can empty the tank completely in 12 hours
=> In 1 hour, Tap B can empty $\dfrac{1}{12}$ of the tank
i.e., In one hour, Tank A and B together can effectively fill $\left(\dfrac{1}{6}-\dfrac{1}{12}\right)=\dfrac{1}{12}$ of the tank
=> In 4 hours, Tank A and B can effectively fill $\dfrac{1}{12}\times4=\dfrac{1}{3}$ of the tank.
Time taken to fill the remaining $\left(1-\dfrac{1}{3}\right)=\dfrac{2}{3}$ of the tank $=\dfrac{\left(\dfrac{2}{3}\right)}{\left(\dfrac{1}{6}\right)}$ = 4 hours
Suppose he move 4 km downstream in $ x $ hours. Then,
| Speed downstream =$ \left(\dfrac{4}{x} \right) $km/hr. |
| Speed upstream =$ \left(\dfrac{3}{x} \right) $km/hr. |
| $\therefore \dfrac{48}{(4/x)} $+$ \dfrac{48}{(3/x)} $= 14 or $ x $ =$ \dfrac{1}{2} $. |
So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.
| Rate of the stream =$ \dfrac{1}{2} $(8 - 6) km/hr = 1 km/hr. |
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