A cylindrical water tower with radius 5 m and height 8 m is

Question

A cylindrical water tower with radius 5 m and height 8 m is 3/4 full at noon. Every minute, .08π m3 is drawn from tank, while .03π m3 is added. Additionally, starting at 1pm and continuing each hour on the hour, there is a periodic drain of 4π m3. From noon, how many hours will it take to drain the entire tank?

A. 20 2/7
B. 20 6/7
C. 21
D. 21 3/7
E. 22

A. 20 2/7
B. 20 6/7
C. 21
D. 21 3/7
E. 22

maddyboiler · Accepted Answer

This one got me, it is so hard to pay attention to every detail and think so much. I initially went with D. So D is wrong why why?

Alright so 
initial volume = (3/4)×∏×5²×8 = 150∏
Relative Drain/ min = .08∏ - .03∏ = .05∏ m³/min drain
Relative drain / hour = .05∏×60 = 3∏ m³/ hr

Every one hour starting from 1pm, 4∏ m³ of water is drained. It means that only at the hour the water is drained and NOT “in that 1 hour“

So after 1 hr the relative drain would be 3∏ + 4∏ = 7∏m³ of water drain

What i did initially was formed an equation 150∏ = 7∏×n (n is the number of hrs) so ended up with 21 3/7. This wrong

Look at this way
after 21 hrs the amount of water drain will be 21×7∏ = 147∏ m³
Left over water in the tank after 21 hrs = 3∏ m³
From above we know that it take 1 more hour to drain that 3∏ m³.

So ans is 22hrs

KarishmaB · Answer

Every minute, .05 m^3 is drawn from the tank. In one hour, water drawn from the tank will be 60*.05 = 3 m^3. Since at the end of 21st hr, 3 m^3 water is left in the tank, at the end of the 22nd hour, the entire 3 m^3 water will be drawn and there will be nothing left to drain out in periodic draining.

Further, in 3/7 hrs, 
(3/7)*(60)*(.05) = 9/7 m^3 water is drawn

I do not know why they write 3(0.5) = 1.5

pavanpuneet · Answer

at 12: 3/4 * 200 = 150 pie m3
at 12 - 1: -.08 + .03 = -.05 * 60 = 3pie m3
at 1 - 147 pie m3 ===> 3 pie m3 + 4 pie m3 = 7 pie m3 ==> so in hour effectively 7 so it takes 21 hours for 147.

thus 21 (1- thereon)+ 1 (12-1) = 22 hours.

Marcab · Answer

Another easy approach to this question is:

Volume to be drained= 150pi
Every minute, the volume that drains is : (0.08pi-0.03pi) metre cube or 0.05pi metre cube.
Every hour 4pi metre cube also drains out, but remember this starts 1 hour later.

We can set up an equation here:
 0.08*number  of  minutes  in  y  hours + 4(y-1)=150 
or
 0.08*60*y + 4y=154 
or
 7y=154 
Therefore y= 22 hours

I found this approach rather easier.

fra · Answer

12 noon: 150π m3

For every minute there is a net drain of 0.08π-0.03π=0.05π m3. So for every hour=0.05*60=3π m3/hour
This means at 1pm there is 150π-3π=147π m3

And from 1pm there is an additional drain of 4π m3 so net drain= 3π+4π=7π m3/hour
So the time taken to drain 147π m3 is 147π/7π=21 hours. and lets not forget the 1 hour between 12 and 1pm.

Therefore, total time taken is 21+1=22 hours

luckyme17187 · Answer

3/4 of 200pi = 150pi.

0.08pi removed & 0.03pi added for each minute. for hours 4.8pi removed & 1.8pi added. consolidated will be 3.0pi removed for each hour. Note:  This starts at noon.

147pi.

starting from 1pm for each hour 4 pi is removed from the total. (along with the regular 3.0pi) , it will be 7pi removed from 1 pm .

147pi/7pi = 21 hours.

Adding the hour between 12 to 1 pm - 21+1 = 22 hours.

ssriva2 · Answer

hi,
I have not understood why we have used .08 instead of .05(net drainage)
please help

EgmatQuantExpert · Answer

Hi  ssriva2 ,

You are right, it has to be 0.05, which is net drainage and not 0.08. The equation writes 0.08 but solves using 0.05. Assuming 0.08 in the equation will not give the right answer.

Regards
Harsh

mathivanan · Answer

Volume of the cylinder is 150pi
At 1 pm the volume is 147pi (the net drain is 3pi)
Then onwards, the net drain is 7pi per hour.
Setting AP we get 21 hours.
Therefore, total time taken is 22 hours.

LogicGuru1 · Answer

AT 12 O' CLOCK ) Water at the tank = π*5^2*8*\frac{3}{4} = 150π

Water Removed = \frac{0.08π}{min} ;

Water Added= \frac{0.03π}{min}

Net change in water volume =  \frac{0.05π}{min}

AT 1 PM ) Water at the tank = 150π -0.05π*60 ==>150π-3π=147π (  TIME TAKEN =1 hour)

Now starting at 1 PM , in addition to 3π another 4π of water is removed every hour 
so Net Loss of water ever hour = 3π+4π =7π
Let t be the time taken by tank to completely empty

147π-7π*t=0

7π*t=147π

t= \frac{147π}{7π} = 21 
TOTAL TIME = 21  hours +  1  HOUR (FROM 12 O CLOCK TO 1 O CLOCK) =   22

ANSWER IS E

Divyadisha · Answer

Volume of water present in the tank= 3/4 π r^2h= 3/4π *5*5*8= 150π

Every minute after noon till 1:00 pm the net water that is drained out was= .08π m3- .03π m3 = .05 π cubic meter 
In one hr from noon to 1:00 pm, total water loss was = .05 π *60= 3 π cubic meter

At 1:00 pm 150π-3π= 147π water is remaining that has to be drained out.

In addition to 3π, 4π water is also draining out of the tank, making total 7π water draining out after 1:00 pm

147π water will be drained out in 147π/7π= 21 hrs after 1 pm

Add 1 hr to this as water was drained out from noon to 1:00 pm, making total 21+1= 22 hrs

E is the answer

debtanuB · Answer

Let, the time taken was 't' hours   . Lets assume its 'n'.

Now, pie x 5^2 x (8) x 3/4 + (-0.08pie + 0.03pie) x t x 60 - n x 4 x pie = 0
=>(150 -3t)/4 = n

So, substituting the values in the options, we find only 22 (option E) gives us n=21 (integer) rest all give us decimal no.s .

Hence, Answer is option E.

gmat800live · Answer

"Starting at 1pm and continuing each hour on the hour, there is a periodic drain..."

Guys, is this GMAT wording? It sounds super ambiguous to me.

If 150pi is volume at noon... if at 1pm you start having an extra drain of 4pi... then at 1pm the volume should be 150-3-4 ? But it seems the question assumes that at 1pm the extra 4pi drain has not started... and only kicks in at 2pm.

Why is this? "Starting at 1pm..." seems to imply that at 1pm the extra drain already is taking effect?

I will be grateful if you could help me understand this. Thank you.

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