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IBPS PO Quantitative Aptitude Boats And Streams Test 1

2440.The current of a stream runs at the rate of 2 km per hr. A motor boat goes 10 km upstream and back again to the starting point in 55 min. Find the speed of the motor boat in still water?
22 km/hr
12 km/hr
20 km/hr
16 km/hr
Explanation:

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

2445.A boat covers a certain distance downstream in 4 hours but takes 6 hours to return upstream to the starting point. If the speed of the stream be 3 km/hr, find the speed of the boat in still water
15 km/hr
12 km/hr
13 km/hr
14 km/hr
Explanation:

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

$\left(x + 3\right)4 = \left(x - 3\right)6$

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

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

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

2447.A man can row three-quarters of a kilometre against the stream in 11$ \dfrac{1}{4} $ minutes and down the stream in 7$ \dfrac{1}{2} $ minutes. The speed in km/hr of the man in still water is:
2
3
4
5
Explanation:

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.

2450.The speed of a boat in still water is 8 kmph. If it can travel 1 km upstream in 1 hr, what time it would take to travel the same distance downstream?
1 minute
2 minutes
3 minutes
4 minutes
Explanation:

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 $=\dfrac{1}{15}\text{ hr = }\dfrac{1 \times 60}{15}\text{= 4 minutes}$

2454.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?
2 : 1
3 : 2
8 : 3
Cannot be determined
Explanation:

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.

2455.In a river flowing at 2 km/hr, a boat travels 32 km upstream and then returns downstream to the starting point. If its speed in still water be 6 km/hr, find the total journey time.
10 hours
12 hours
14 hours
16 hours
Explanation:

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

2460.A man can row 7.5 kmph in still water and he finds that it takes him twice as long to row up as to row down the river. Find the rate of stream.
10 km/hr
2.5 km/hr
5 km/hr
7.5 km/hr
Explanation:

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 $=\dfrac{1}{2}(2x-x) = \dfrac{x}{2}= \dfrac{5}{2} = 2.5\text{ km/hr}$

2461.A man can row at 5 kmph in still water. If the velocity of current is 1 kmph and it takes him 1 hour to row to a place and come back, how far is the place?
2.4 km
2.5 km
3 km
3.6 km
Explanation:

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.

2462.Tap A can fill the tank completely in 6 hrs while tap B can empty it by 12 hrs. By mistake, the person forgot to close the tap B, As a result, both the taps, remained open. After 4 hrs, the person realized the mistake and immediately closed the tap B. In how much time now onwards, would the tank be full?
2 hours
4 hours
5 hours
1 hour
Explanation:

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

2463.A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is:
1 km/hr
1.5 km/hr
2 km/hr
2.5 km/hr
Explanation:

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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