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It takes $30$ minutes to empty a half-full tank by draining it at a constant rate. It is decided to simultaneously pump water into the half-full tank while draining it. What is the rate at which water has to be pumped in so that it gets fully filled in $10$ minutes? 

  1. $4$ times the draining rate 
  2. $3$ times the draining rate
  3. $2.5$ times the draining rate
  4. $2$ times the draining rate
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Let V be the total volume, $S_{D}$ is the speed of draining and  $S_{P}$ is the speed of pumping.

It takes 30 minutes to empty a half-full tank by draining it at a constant rate

Half full tank = V / 2            

Speed of draining ( $S_{D}$ ) = (V/2) / 30

                            $S_{D}$ = V / 60

What is the rate at which water has to be pumped in so that it gets fully filled in 10 minutes

we have to fill in 10 minutes.

So, water drained in 10 minutes = (V/60) * 10

                                                    = V/6

water remained in tank = (V/2 – V/6) = V/3

water we have to pump = V – V/3 = 2V/3

Speed of pumping water ( $S_{P}$ ) = (2V/3) / 10

                                           $S_{P}$  = V / 15

  So, you can clearly see-

$S_{P}$  =   4 * $S_{D}$

Answer – A

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@ASNR1010

thanks bro!

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6 Answers

19 votes
19 votes
Best answer

Let the capacity of tank be $1$ litre.

Draining rate $=\dfrac{0.5\text{ litre}}{30\text{ minutes}}=\dfrac{1}{60} \text{ litre/min}$

Let filling rate be $x \text{ litre/min}$

In $1$ min tank gets $x- \left(\dfrac{1}{60}\right) \text{litre}$ filled.

To fill the remaining half part we need $10 \text{ minutes}$

$x-\dfrac{1}{60} \text{ litre}\to 1 \text{ min}$

$0.5 \text{ litre}\to 10 \text{ mins}$

$\frac{0.5}{\left(x-\frac{1}{60}\right)}=10$

Solving, we get $x= \dfrac{4}{60}$ which is $4$ times more than draining rate.
So, option A

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1 comment

Please correct the typo from $\dfrac{0.5}{\left(\dfrac{x-1}{60}\right)} = 10$

to $\dfrac{0.5}{\left(x-\dfrac{1}{60}\right)} = 10$

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3 votes
3 votes
Here is my approach:

A empty in the 30 minutes, so rate is 1 litre/minute (negative rate)

Now let B be a pump filling tank.

so now A+B work together and fill tank in 10 minutes , in that period B will drain some water i.e 10 minutes*1=10L

so net work , A-10=30

A=40 so ,A should work 4 times more as to fill the tank in 10 minutes.
1 vote
1 vote

A 444 times the draining rate 44444   444 times the draining rate444  times the draining rate erfgvguhn

1 vote
1 vote

4 A. 4 times the draining rate

Answer:

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