QUESTIONS ON PIPES & CISTERN (PART-I)
QUESTIONS ON PIPES & CISTERN (PART-I)
Two pipe p and q can fill a cistern in 12 and 15 minutes respectively. If both are opened together and at the end of 3 minutes the first one is closed; how much longer will cistern take to fill ?
A) 35/4 minutes
B) 33/4 minutes
C) 30 minutes
D) 9 minutes
In such a question find the work of a person or thing to be finished in 1 unit of time
Now p’s 1 minute work = 1/12
and q’s 1 minute work = 1/15
They work together for 3 minutes
So the work to be finished by both in 3 minutes = p’s 3 minute work + q’s 3 minute work
= (1/12)*3 + (1/15)*3
= (1/4) + (1/5) = 9/20
Therefore the remaining work (to be finished by ‘q’ alone) = 1 – (9/20) = 11/20
Now q’s 1/15 work takes 1 minute
So its 11/20 work will take = 15*(11/20) = 33/4 minutes (option ‘B’)
A and B can fill a cistern in 75/2 min and 45 min respectively. Both the pipes are opened together. The cistern will be filled just in half an hour if the pipe B is turned off after?
A) 9 minutes
B) 5 minutes
C) 10 minutes
D) 15 minutes
The part of cistern to be filled by A in 1 minute = 2/75
Therefore in half an hour i.e. 30 minutes it will fill the part of cistern = 30*(2/75) = 60/75 = 4/5
So the empty part of the cistern still, which B will fill = 1 – 4/5 = 1/5
Now the part of the cistern to be filled by 1 minute = 1/45
Hence the time taken by it to fill 1/5 of the whole = 45/5 = 9 minutes.
Obviously after 9 minutes it must be turned off. (option ‘A’)
A cylindrical overhead tank is filled by two pumps P1 and P2. P1 can fill the tank in 8 hr while P2 can fill the tank in 12 hr. There is another pipe P3 which can empty the tank in 8 hr. Both the pumps are opened simultaneously. The supervisor of the tank, before going out on a work, sets a timer to open P3 when the tank is half filled so that the tank is exactly filled up by the time he is back. Due to technical fault P3 opens when the tank is one-third filled. If the supervisor comes back as per the plan what percent of the tank is still empty?
To know this we need to know the time in which the supervisor had to come back.
Pump P3 was to open when the tank was filled half according to the supervisor’s plan. Therefore calculate the time to be taken by the pumps 1 and 2 working together in which half of the tank will be filled.
Now part of the tank to be filled by P1 + P2 in one hour = 1/8 + 1/12 = 5/24
Therefore the time in which half of the tank was to filled by both of them working together = (24/5)/2 = 12/5 hours
Now the pump P3 was to open
Obviously, part of the tank to be filled by all of them in one hour = 1/8 + 1/12 – 1/8 = 1/12
Therefore time to be taken by all of them to fill the remaining half of the tank = 12/2 = 6 hours
Thus, total time in which the supervisor had to return = 12/5 + 6 = 42/5 hours
Now we just have to calculate the part of the tank filled by all the tanks in 42/5 hours
1/3 of the tank filled by the pumps P1 + P2 in = (24/5)/3 = 8/5 hours (from above)
Remaining time in which all the pumps have to work together = 42/5 – 8/5 = 34/5 hours
Part of the tank to be filled by all the pumps working together in 34/5 hours = (1/12)*(34/5) = 17/30
So the part of the tank filled in total = 1/3 + 17/30 = 27/30
Therefore empty part of the tank on return of the supervisor = 1 – 27/30 = 1/10
In percentage = (1/10)*100 = 10% (option ‘D’)
A and B can fill a cistern in 20 and 30 minutes respectively, while C can empty it in 15 minutes. If A, B, C are kept open successively for 1 minute each, how soon will the cistern is filled?
A) 165 minutes
B) 170 minutes
C) 167 minutes
D) 175 minutes
As all the cisterns are kept open successively for 1 minute each for 3 minutes, we should find the part of the cistern filled in 3 minutes i.e. one interval of time; so it is 1/20 + 1/30 – 1/15 = 1/60
It’s obvious that the pipe C which is emptying the cistern will not have to be opened in the last interval, otherwise the cistern will never fill; so we need to know the combined work of A and B in the last interval i.e. last 3 minutes.
So, A’s 1 minute work + B’s 1 minute work i.e. 2 minute work = 1/20 + 1/30 = 5/60 = 1/12
It does mean all the pipes A, B and C work simultaneously to fill = 1 – 1/12 = 11/12 of the cistern
Time taken to fill 1/60 of the cistern by all the pipes working together = 3 minutes
Hence, time taken in all to fill 11/12 of the cistern = 165 minutes
So, total time to fill the full cistern = Time taken for 11/12 of the part + time taken for 1/12 of the part
= 165 + 2 = 167 minutes (option ‘C’)
One pipe can fill a tank three times as fast as another pipe. If the two pipes together can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in?
A) 81 minutes
B) 108 minutes
C) 144 minutes
D) 192 minutes
Let the slower pipe alone fills the tank in x minutes
As the faster pipe is three times efficient as the slower one is; time taken by it to fill the tank = x/3 minutes
Therefore 1 minute work of the slower pipe + 1 minute work of the faster pipe = 1 minute work of both the pipes
=> 1/x + 3/x = 1/36
solving x = 144
So the slower pipe will fill the tank in = 144 minutes (option ‘C’)
RULE: If A is x times efficient than B, and working together they finish a work in y days; then time taken by A = [y(x+1)]/x; and time taken by B = y(x+1)
As the faster pipe i.e. A is faster 3 times than B; so x = 3 and y = 1
Here the slower pipe is B; so time taken by it = 36*(3+1) = 144 minutes (option ‘C’)
If two pipes function together, the tank will be filled in 12 hours. One of them fills the tank in 10 hours faster than the other. How many hours does the faster pipe take to fill the tank?
A) 30 hours
B) 60 hours
C) 20 hours
D) 25 hours
Let time taken by the slower pipe = x hours
So time taken by the faster pipe = (x – 10) hours
Now slower pipe’s 1 hour work = 1/x
And faster pipe’s 1 hour work = 1/(x – 10)
And their given total work in 1 hour = 1/12
Therefore 1/x + 1/(x – 10) = 1/12
Now instead of solving the equation algebraically try hit and trial method in this equation as solving the equation that way will be time consuming; of course x = 30
Hence time to be taken by faster pipe alone = x – 10 = 30 – 10 = 20 hours (option ‘C’)
8 taps are fitted to a water tank. Some of them are water taps to fill the tank and the remaining taps are outlets taps used to empty the tank. Each water tap can fill the tank in 12 hours and each outlet tap can empty in 36 hours. On opening all the taps together, the tank is filled in 3 hours. The number of water taps is?
Let the number of water taps = x
Thus the number of outlet taps = 8 – x
As each water tap can fill the tank in 12 hrs
Thus, part of the tank filled by x taps in 1 hour = x/12
As each outlet tap can empty the tank in 36 hrs
Thus, part of the tank emptied by (8 – x) outlet taps in 1 hour = (8 – x)/36
Therefore part of the tank to be filled in 1 hour = x/12 – (8 -x)/36 = (x – 2)/9
Total time to fill the tank therefore = 9/(x – 2) hrs
But the tank is filled in 3 hrs
Thus, 9/(x – 2) = 3
=> x = 5 (option ‘B’)
A tank is fitted with two taps. The first tap can fill the tank completely in 45 minutes and the second tap can empty the full tank in one hour. If both the taps are opened alternatively for one minute, then in how many hours the empty tank will be filled completely?
A) 2 hours 55 minutes
B) 3 hours 40 minutes
C) 4 hours 48 minutes
D) 5 hours 53 minutes
Both the taps are kept open alternatively for 1 minute each, we’ll find the part of the tank filled in 2 minutes i.e. one interval of time; so it is 1/45 – 1/60 = 1/180
It’s obvious that the second tap which is emptying the tank will not have to be opened in the last interval, otherwise the tank will never fill; so we need to know the work of the first tap in the last 1 minute.
So, work of tap first in 1 minute = 1/45
It does mean both the taps work simultaneously to fill = 1 – 1/45 = 44/45 of the tank.
Time taken to fill 1/180 of the tank by both the taps working together = 2 minutes
Hence, time taken in all to fill 44/45 of the tank = 352 minutes
So, total time to fill the full tank = time taken for 44/45 of the part + time taken for 1/45 of the part
= 352 + 1 = 353 minutes = 5 hours 53 minutes (option ‘D’)
Two taps can fill a tank in 12 and 15 minutes respectively. First tap is replaced with a tap which is double of its radius while second is replaced with half of its radius. If they work together, how much time they will take to fill the tank?
A) 15/8 minutes
B) 20/7 minutes
C) 3 minutes
D) 4 minutes
Remember that the capacity of a pipe increases/decreases in the ratio of the area of the cross-section it has.
Now capacities of the pipe-A
Earlier : πR²
Later as the radius is doubled its capacity = π(2R)²
Therefore capacity increased = [π(2R)²]/[πR²] = 4 times
Capacities of the pipe-B
Earlier = πr²
Later as the radius is now half = π(r/2)²
Therefore capacity decreased = [π(r/2)²]/[πr²)]/ = 1/4 times
Now 1 minute work of pipe-A earlier = 1/12; therefore its 4 times work in 1 minute now = 4/12
And 1 minute work of pipe-B earlier = 1/15; therefore 1/4 times work now = 1/60
There combined work now = 4/12 + 1/60 = 21/60 = 7/20
Hence total time to be taken = 20/7 minutes (option ‘B’)
In what time a cistern can be filled by 3 pipes running together, whose diameters are 1 cm, 4/3 cm, 2 cm, when the largest pipe alone fills it in 61 minutes? The amount of water flowing in by each pipe being proportional to square of its diameter.
A) 30 minutes
B) 64 minutes
C) 36 minutes
D) 45 minutes
The rates at which water is flowing from each pipe:
1² = 1; (4/3)² = 16/9; 2² = 4
The rate of the largest pipe = 4
The rate at which now the cistern is to be filled = 1 + 16/9 + 4 = 61/9
Time taken if the water flows at the rate of 4 = 61 minutes
Time taken if it flows at the rate of 1 = 61*4 minutes
Therefore time to be taken if water flows at the rate of 61/9 = (61*4)/(61/9) = (61*4)*(9/61) = 36 minutes (option ‘C’)