GATE ECE 2014 Set 2 | GA | Question: 10

Question

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?

• A. $4$ times the draining rate
• B. $3$ times the draining rate
• C. $2.5$ times the draining rate
• D. $2$ times the draining rate

Comments

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!

Answer 1

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

Comments

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

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

Answer 2

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.

Answer 3

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

Comments

better share ur approach then posting answer , we all know the answer

Answer 4

4 A. 4 times the draining rate

Answer 5

A. Option is true because at the same rate four pump required or one with four times filling. one to maintain level and 3 to fill in 10 minutes

Answer 6

60 min to empty full tank, negative

Relative speed/rate = rate to fill – rate to empty

speed/rate = distance/time, consider distance 1 litre

To fill 1/x, x is time to fill, since distance constant.

Relative rate/ speed = 1/x – 1/60

Given time to half fill = 10 min, to completely fill = 10*2 = 20 min

=> 1/20 = 1/x – 1/60

=> 3/60 + 1/60 = 1/x

=> 4/60 = 1/x

which is 4 * 1/60 <draining rate>