Aptitude - Pipes and Cistern - Discussion

pipe ----part filling----lcm of A & B----multiplying factor(cal for how much will fill in 1 unit time)
A ----75/2 ----225 ----225%(75/2)= 6 i.e A can fill 6 ltr in 1 min
B ---- 45 ---- 225 ---- 225/45=5 i.e B can fill 5 ltr in 1 min.

* NOTE : if fraction comes then while calculating LCM take lcm of numerators only.
now given that tank will fill in half-hour means A will fill till half hour ie for 30 min.
in 1 min A is filling 6 ltr so in 30 min =30 *6 =180 ltr fill by pipe A.

Means remeaning will be filled by B. i.e 225-180=45 ltr will be filled by PIPE B.

But pipe B is filling 5 ltr in 1 min so for 45 ltr = 45/5 = 9 min it will take.
so B will open for 9 min.


(here A will continuously open, we only have to stop B. so 1st calculate how much water will A can fill in given time and after that calculate how much time it will take to fill remeaning by closing pipe i.e B )

Thanks @Deepak.

Nice work @Kasi Srinivas.

@Kasi Srinivas.

Nice explanation, anyone can easily understand this, Thanks.

A*(A'work)+B*(B'work)=1,
It means 30(A)+x(B)=1,
Then, x=9.

Best solution:

Parts filled by A in 1 min = 2/75
Parts filled by B in 1 min = 1/45
Now;
In 30 min tank is full so,
Parts filled by A in 30 min = 30*2/75 =60/75
Parts filled by B in X min = X/45
So, 60/75 + X/45 = 1
Therefore ,X = 9 min.

Much simple way is the LCM method.

LCM of (75/2 and 45) = 225, assume this is total unit of work.

Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.

Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.

2/75 + 1/45 = 8/225.

How it can be 11/225?

Can anyone explain how 2/75 comes? please.

For 1min it takes( 2/75).
For 30 min (1/2 hour) it takes 2/75*30=4/5 part.
So remaining only 1 part out of 5 which is going to fill by B.
So (1/5*45) = 9 min.