Aptitude - Pipes and Cistern - Discussion

Nice work @Kasi Srinivas.

Thanks @Deepak.

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 )

I didn't get this sum please give me some tricks to understand the problem.

Well said @Kasi Srinivas.

@Sarika.


Just take LCM of 37.5 and 45=450(total units).
You need 30mins to fully fill so(450/30)=15mins u have in total if Both worked non stop
now A works for a full period so 450/75 (convert 37.5 into the whole num as used for LCM)=6 mins.

So, B has worked 15-6=9mins.

Thank you @Pintu Francis.

I Can't understand this. Please, someone explain to me. clearly.

For this type of question, we can approach like;

Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y').

At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is,
y(1-(t/x))= 45(1-((30*2)/75)) = 9min.

Thanks for explaining the answer @Kasi.