Aptitude - Pipes and Cistern - Discussion

For this kind of question, Use formula (1 - full tank time / 2nd pipe time)* 1st pipe time.
Depends on the question what they are asking the 1st and 2nd times will be swapped. if 1st pipe time is asked the 1st pipe time should be outside like the above formula, or else 2nd one time should be outside.

Here full time is half an hr should be converted to min 1/2*60 = 30min.

Given data:
Full time = 30min.
1st pipe time= 37 * 1/2min => 37*2=74, 74+1= 75/2min.
2nd pipe time= 45min.

Apply formula
(1 - 30/75/2) * 45,
30 and 75 cancelled (1 - 6/15/2) * 45.

1 * 15/2 = 15/2 => 15/2 - 6 = 15-12/2 => 3/2.

(3/2 / 15/2)*45 remainders gets cancelled and remaining 3/15*45= 15 and 45 gets cancelled by 3 so 3 * 3 = 9min.

So, if u follow the above formula u can get the answer. You must be sure which is asked for if 1st one then outside the bracket if 2nd one then it should be outside.

Here b's time is asked so b's time is outside in the formula.

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 )

@ Gopal , Priya , Vishnupriya :

Both pipes A and B are opened initially.
Now these both pipes (A and B) fill the cistern together for some time...let us say they are filling the cistern together for x mins.
Which implies pipe B is turned off after x min (according to the question).

If u understood this u can follow the problem..

Pipes A & B are filling the tank for x mins => x*(A+B) --- (1)

As after x mins pipe B closed, pipe A alone filling the tank.

He said cistern will be filled in half an hour ie 30 mins.

=> pipe A alone is filling the remaining tank in (30-x) mins
=> A*(30-x)---(2)

So, part filled by (A + B) in x minutes + Part filled by A in (30 -x) minutes = 1.

here 1 indicates the tank is full.

by substituting we get,


x*(2/75 + 1/45) + (30-x)* 2/75 = 1

on solving we get x = 9.

- K @ $ ! (kasi srinivas)

Simply put A can fill in 75/2 minutes.

Therefore in 1 min it fills 2/75 part.

Therefore in 30 min it will fill 2*30/75 = 4/5th part.

Now after 30 min amount remaining to be filled will be 1-4/5 = 1/5.

(In above subtraction 1 means whole cistern), so now B has to fill 1/5 i.e. remaining part. Question is how many minute does B takes to fill 1/5 part of the cistern.

We know from question that B takes 45 min to fill the cistern.

Therefore in 1 minute fill 1/45th part.

Therefore x *1/45 = 1/5 (where x= minute it actually takes to fill the cistern).

Solving X = 9 minutes.

It is easy. It is given that the cistern will be filled in just 30 minutes. And we don't know that who between A and B takes how much time from 30 minutes to fill the tank.

Suppose A takes x minutes. Then B will take (30-x) minutes.

But here is given that B is turned off in some time. We don't know when. So take it as Y minute. After Y minute only A will fill remaining part in 30-Y minute.

So first A+B will fill for Y minute and then B is turned off. So only A will fill in 30-Y minute.

So the equation is (A+B)Y + A(30-Y) = 30 minute.

If pipe B is turned off after x min, which means it is opened for x mins along with pipe A.
therefore for x mins both pipes A & B is filling the tank ie : x*(A+B)----(1)
As after x mins pipe B closed, pipe A alone filling the tank.
In question it asked in total 30 min tank should be filled.
therefore pipe A alone fills the remaining tank in (30-x) mins ie : A*(30-x)-----(2)
Hence the total work is obtained as equ (1) + (2) = 1

Let us understand the problem first, we can easily analyze that Here Pipe A is run for HALF an hour.

Total part filled by part A alone in half an hour is {2/75}*30 = (4/5).

Now (1- (4/5) ) , that is (1/5) part is still remaining. And it is filled by Pipe B and after that it is made off.

So (1/5) part filled by Pipe B in,

= 45*(1/5) = 9 minute.

Hence after 9 minute pipe B will made off.

If we reverse the answer to the common formula:

9x11/225+21x2/75.

We will get answer = 1.

But we actually want to get answer = 30 minutes, right?

This explanation is clearly not logical and misleading.

We can't assume minutes unit is same with part unit.

You can't multiply them with minutes and suddenly make the equation "=1" because it's clearly not same unit.

@Sandip: That's what aptitude is meant for, i.e., How much knowledge we've dragged with us from our primary schooling, which indicated how consistent the candidate was throughout his academic period. Aptitude marks tells this apparently. And that's what recruiter is looking for, a consistency. Plus it tells your mental ability to grasp the things and to sort it out

In 1 minute, both together will fill 2/75 + 1/45 = 11/225 of the cistern.

Cistern should be thus filled in 225/11 minutes.

But cistern filled in 30 minutes.

Therefore, B turned after 30 - 225/11 = (330 - 225)/11.

= 105/11.

= 9(6/11).

Hence B turned off after 9 minutes.