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Bunuel
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Pipe A fills a tank of capacity 700 liters at the rate of 40 liters a minute. Another pipe B fills the same tank at the rate of 30 liters a minute. A pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute. If pipe A is kept open for a minute and then closed and pipe B is open for a minute and then closed and then pipe C is open for a minute and then closed and the cycle is repeated, when will the tank be full?

In three minutes net gain is 40+30-20=50 liter;

After 13 cycles (13*3=39 minutes) they fill 13*50=650 liters.

Then in 1 minute pipe A will add 40 liters, 10 liters to be filled;

Then to fill 10 liters pipe B will need 10/30=1/3 min.

So total time \(39+1+\frac{1}{3}=40\frac{1}{3}\) min.

Yes, right....:) Thanks
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How do you get that cycle = 13???
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@ bibha

divide 700/50... u get 14 cycles.
but u know that before 20
then take away the last cycle. so u have 13 cycles (39 mins), where the tank is 650 ls full.
let's take the 14th cycle. the first minute, the tank gets to 650 + 40 = 690ls.
there's only 10 ls left to fill the tank.
the second step of the cycle is pipe B that fills 30 ls in 1 minute. since we only need 10 ls, we only need 1/3rd of a minute.

so we have 13 cycles plus 1 min of pipe A and 1/3rd minute of pipe B.
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I still dont get it.... i mean I understand everything but the logic as to why we should NOT consider 14 cycles as we have already done the filling in and filling out calculations.
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I still dont get it.... i mean I understand everything but the logic as to why we should NOT consider 14 cycles as we have already done the filling in and filling out calculations.

The sequence of the events is:
Pipe A adds \(40\) liters/m
Pipe B adds \(30\) liters/m
Pipe C subtracts \(20\) liters/m
=NET \(+50\)

When you have done 13 circles in 40 minutes, there are 650 liters and 50 to go.
Now the squence repeats
Pipe A adds \(40\) liters/m----50-40=10 liters to go and 1 minute more added
Pipe B adds \(30\) liters/m----so 10 liters will be filled in \(\frac{1}{3}\) of a minute
STOP
Pipe C subtracts 20 liters/m

Tot time \(40\) minutes and \(\frac{1}{3}\)
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aiming4mba
Given: Pipe A fills a tank of capacity 700 liters at the rate of 40 liters a minute. Another pipe B fills the same tank at the rate of 30 liters a minute. A pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute.

Asked: If pipe A is kept open for a minute and then closed and pipe B is open for a minute and then closed and then pipe C is open for a minute and then closed and the cycle is repeated, when will the tank be full?
Tank filling progress = (40 + 30 - 20) + (40 + 30 - 20) + ...... 
= 50 + 50 + 50 + ...... 13 times + 40 + 30/3

Since 700 = 50*14

Tank will be full in minutes = (3*13 + 1) minutes 20 seconds = 40 minutes 20 seconds

IMO D
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TECHNICALLY, the tank will be full at both 42 minutes and 40 minutes 20 seconds - HENCE THE QUESTION STEM IS AMBIGUOUS and CHOICE A and CHOICE D BOTH ARE CORRECT.

QS should lead us to the right answer by saying something like "when will the tank be FIRST full" OR "what is the EARLIEST TIME the tank will be full"
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