Aptitude - Pipes and Cistern - Discussion

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.

A - 30 min.
B - 20 min.
C - 10 min.
Now take LCM (30,20,10) = 60.

So,
A - 60/30 = 2 LTR.
B - 3LTR.
C - 6 LTR.

So TOTAL 2+3+6 = 11 LTR.
So proportion of C is 6/11.

So Simple.

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.

@Lalitha

= 1/30+1/20+1/10
So L.C.M of 30,20,10=60
Then 1/30+1/20+1/10=2+3+6/60
=11/60
ok

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

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

Any more short cuts in commercial mathematics.

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

A, B and C can fill a tank in 6 hrs. After 2 hrs C is stopped. Then A and B together fill the tank in 7 hrs.

How much C will take individully take to fill the tank?

@Sumanth Gowda thanks I understand now.