Aptitude Boats And Streams Practice QA

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

=>$\left(x + 3\right)2 = \left(x - 3\right)3$

=>$2x + 6 = 3x - 9$

=>$x = 6+9 = 15\text{ kmph}$

We can write three-quarters of a kilometre as 750 metres,

and 11$\dfrac{1}{4}$ minutes as 675 seconds.

   = 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.

   = 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.

$\Rightarrow$ 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,

So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.