Aptitude - Pipes and Cistern - Discussion

Here pipes A and B are used for filling the tanks,
A and B takes 5 and 6 hours respectively to fill the tank .
for 1 hour ,A can fill 1/5th of the tank and B can fill 1/6th of the tank.
now tank C is used foe emptying the tank(water is running off from the tank)..it can empty the tank in 12 hours..means for 1 hour it empties 1/12th of water from the tank

NOTE:
A and B are filling the tank whereas C is used for emptying the tank.
therefore add A and B and subtract it from c
now consider how much water gets filled in the tank in 1 hour,
1/5+1/6-1/12=17/60
for 1 hour----17/60
? hours----1
(note: for 1 hour-- 17 parts got filled up out of 60 parts
for "?"hours-- 60 parts get filled up out of 60 (60/60=1)

Here 1 indicates tank gets filled)
just by cross multiplying it
(1*1)=(?*17/60)
?=(17/60)/(1*1)
60/17 hours it can fill the tank.

Dear Sir,

I think the pipes that filling the tank operates under constant pressure. So the flow rate is constant. But the pipe that is draining is operating under varying head, ie. Under varying pressure. So the draining of tank is not uniform, but proportional to the square root of the head. Hence the problem can't be solved using simple mathematics. For centuries we put it in wrong way.

(1/5 + 1/6 - 1/12) = (6+5/30 - 1/12)
= (11/30 -1/12)
= (132-30/360)
= (102/360)dividing numerator &denominator by 2
= (51/180)dividing numerator &denominator by 2
= (17/60)

Hi guys use this concepts to find the solution in easily
Concept 1:
A fill tank in x hrs and B fill in Y hrs and C in z hrs means together fill=?
A&B=X*y/x+y
let that A&B ans as p
then A,B&C=p*z/p+z
Concept 2:
A fill tank in x hrs and B empty in Y hrs then both together work how much time is to empty=?
=x*y/y-x

@Kaveri.

Pipe A =1/5 hrs & B=1/6 hrs, & C=/12 hrs.

LCM of 5, 6 , 12 = 60.
Units or liter (whatever) of :
A = 5 (hours)/60(units) = 12 units.
B = 6 (hours)/60(units) = 10 units.
C = 12 (hours)/60(units) = 5 units.

Now, add A + B(inlet pipe) - C (outlet pipe)/ 60.
--> 22 - 5/60.
---> 17/60.

Thanks manasa but I can't understand one point ie;c is used for emptying of tank & when a&b are used to filling of water so when we use the 3 pipes simultaneously then at any time or at any point the will not be FILL so how can we conclude with a simple answer.

First take a, b, c = 5, 6, 12.
Take a lcm of a, b, c = 60.
And divide by 5/60, 6/60, 12/60.

We get the answer of 12, 10,-5 because c can empty the tank it takes it as a negative term.
12 + 10 - 5 = 17.
Then divide by 60/17.
We get 3.58.

So, the answer is3(9/17).

A PIPE - 5HRS.
B PIPE- 6HRS.
C PIPE - 12HRS TO EMPTY IN TANK, SO THAT,

A-5 B-6 C-12 (TOTAL L.C.M 60 UNIT TANK CAPACITY).
A-12UNIT, B-10UNIT ,C-5UNIT.

TOGETHER(A+B)-EMPTY(C) = 12+10-5 = 17.
TOTAL UNIT 60/17 = 3.52

(1/5 + 1/6 - 1/12) = (6+5/30 - 1/12).
= (11/30 -1/12),
= (132-30/360),
= (102/360)dividing numerator &denominator by 2,
= (51/180)dividing numerator &denominator by 3,
= (17/60).

@Neha

Just forget that LCM part if 17tank is filled in 60hours the how many tank will be filled in 1 hour.

17tank = 60hours
1tank = x,
17 * x = 60 * 1tank = x = 60/17.