Aptitude - Pipes and Cistern - Discussion

Don't make this type of questions complicated.

I am giving you simple solution which I gives in my class for this type of questions.

Just take LCM of all given values.

30, 20, 10 = 60 (LCM).

Assume that capacity of tank is 60 liters.

Now find the capacity of each pipe of filling the tank by.

* Total capacity of tank in liters / time taken by specific pipe.

1) Pipe A capacity 60 / 30 = 2 liters every min(solution P).
2) Pipe B capacity 60 / 20 = 3 liters every min(solution Q).
3) Pipe c capacity 60 / 10 = 6 liters every min(solution R).

Clearly if all works together than the capacity is 2+3+6 = 11.

Liters per minute. Now out of 11 pipe c (solution R) have 6 liters so in all solution in 1 minute solution are share should 6/11 (don't need to multiply by 3 minutes because the ratio will remain same for 1, 2, 3, 4. Any number of minutes) its very easy to solve any problem of pipe and cistern using this method, it looks lengthy in reading but if you practice I assure you, everyone of you will solve this type of questions in few seconds.

Just to make it easier for you guys let me show you my trick to solve this particular question.

So, first we have to find out the total capacity of the tank so for that we have to find out the lcm of the above 3 no's.

i.e 30/20/10 so lcm would be 60 and this 60 is the capacity of the tank.
Now let us find out the speed at which the tank will be filled so just divide the total capacity with the given time.
i.e : 60/30, 60/20, 60/10

Now we have the the speed at which this tank will be filled i.e 2lt/min, 3lt/min and 6lt/min.
If a pipe can fill 2 lt of chemical in 1 min so how much will be the amount of chemical after 3 min. i.e 6lt similarly for b it would be 9lt and for c it will be 18lt and we have the calculate the ratio of chemical l after 3 min.

So the ratio can be calculated as shown below:

18(amount of chemical after 3min)/18 + 9 ++6 = 18/33ie 6/11 (solution).

Pipe A fills the tank in 30 min
in one min pipe A fills 1/30 of the tank
pipe B fills the tank in 20 min
in one min pipe B fills 1/20 of the tank
pipe C fills the tank in 10 min
in one min pipe C fills 1/10 of the tank
part of the tank filled when all the three pipes are opened for one min is {(1/30)+(1/20)+(1/10)}=11/60
the part of the tank filled when all three pipes are opened for
three min is =3*(11/60)=33/60=11/20
now the chemical solution R present after 3 min=3*(part of the tank filled by C for one min)
=3*1/10=3/10
propotion of sol R in the tank after 3min=(chemical soln r in the tank after 3 min)/(total chemical soln in the tank)
=(3/10)/(11/20)=(3*20)/(11*10)=60/110=6/11
hence the ans is 6/11

Part filled by A in one min is given by=1/30 min.
Part filled by B in one min is given by=1/20 min.
Part filled by C in one min is given by=1/10 min.

Now, total (A+B+C)filled in 1 min is given as (1/30+1/20+1/10) by taking lcm and we get 11/20
then in the qn they said after 3 min proportion of R ....clearly we are knowing that (A=p),(B=Q),(C=R).

So we should consider the value of c.
To calculate the proportion of R after 3 min,
Now part filled by C in 3 min is 3x(1/10)=3/10,
the proportion of R=total filled /c filled,
= (3/10)x(20/11),
= 6/11.

Hi,

Please any one check whether my solution is right?

'A' fills tank =30 mins.
'B' fills tank =20 mins.
'C'fills tank =10 mins.

So if we find the LCM for this means we'll assume it as our tanks capacity,

Here its LCM is 60
The tanks capacity is 60 litres

A' fills d tank= 2 lit/min
B' fills d tank=3 lit / min
C' fills d tank=6 lit / min

(A+B+C)'s fill d tank = 11 lit/min

A , B , C respectively discharge P,Q,R respectively

So C it self discharge 6/11 lit

Hi Sashank Singh

Let me explain in detail.

(A,B and C)Three pipes work together 2 hours from that completed work is
(1/6)*2=1/3
The remaining work is 2/3.
This 2/3 work is done by A and B alone 7hrs.
That is
2/3 =7hrs
1 =? =>7*3/2=21/2hrs
A,B&c=6hrs
A&B=21/2hrs
lets take A,B&C total work time is X and A,B work time is Yhrs.
To find A alone use the formula as (X*Y)/(Y-X)
(6*21/2)/(21/2-6)=14

@Mohsin & @Prabhas.


The part filled by C in 3 mins is 3/10 and part filled by all three combined together is 11/20. To find the proportion of solution are we have to divide 3/10 by 11/20, thus multiplied by 20/11*.

If we divide a number say 25 by 5 we have 2 options.

I) 25/5 i.e- 5 or.
II) 25 x INVERSE OF 5 (I. E- 1/5) = 25x 1/5.
Both are one and the same. Hope this clears.

Hey guys listen properly.

A part filled by 3 min
for a=1/30,b=1/20,c=1/20
first u add this tree and multiply by 3 min k then u'l get 11/20
discharge chemical solutions P,Q and R respectively its equal to A,Band C
observe carefully P=A,Q=B,R=C;
proportion of the solution R in the liquid in the tank after 3 minutes where 3min s nothing but 3(C).....3(1/10) ie 3/10

Now u got ans.

First take the LCM of all the time taken by A, B, and C which we will give us total work = 60, and then take out the efficiency by the formula (total work/time) A-> 2 ,B-> 3 and C-> 6 and then take out the Ratio of C's R liquid proportion (6/11) by multiplying both the numerator and denominator by 3.
And the result we will get is 6/11.

At any time proportion of solution R is same with respect to other two solution,

In 1 minute A will drop =1/30.
" " "" " " B will drop =1/20.
" " " " " C will drop =1/10.

Multiplying 60 to each one,
A:B:C = 2:3:6.
Since solution R corresponds to C we have = 6/(6+3+2) = 6/11.