GATE2014 EC-2: GA-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

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 2

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>