Aptitude - Pipes and Cistern - Discussion

@Vinay Kumar Gupta is absolutely Right.

The Answer is 10 Minutes And 40 Seconds.

Sol:
A=15minutes
B=20 minutes.

Taking LCM of Both Which comes 60.
Therefore Total capacity of the Tank is 60 litres,

For A: 15 * 4 = 60.
For B: 20 * 3 = 60.

It Means Pipe A Fills 4 Litres In 1 Minute.
And Pipe B Fills 3 Litres In 1 Minute,
Both ate filling 3+4=7 litres In 1 minute.
Both are Working For 4 minutes together.
Therefore, 7*4=28 litres.
Now total capacity - 28litres
60-28=32.

Now Pipe B Have To Fill These Remaining 32 litres,
As pipe B Fills 3 Litres in 1 minute.
So 3*10 = 30.
30 litres in 10 minutes.
As Pipe B Fills 3 litres in one minute means 1 litre in 20 seconds.
So, the remaining 2 litres Will take 40 seconds.
Therefore The Answer will be 10 minute's and 40 seconds.

@Sriganesh.

B can fill the tank in 1 min = 1/20 parts or work done of the tank.
After 4 min, B can alone to fill the remaining part of the tank?
min = 8/15 remaining parts or work done of the tank.
By cross multiply
1/20 parts = 1 min.
8/15 parts = ? min.
(8/15 * 1)/1/20 = ?
8/15 * 1 * 20 = ?
8/3*4 = ?
?=32/3=10 (2/3) min =10 min 40 sec (2/3 min convert to seconds 2/3 * 60=40 sec).

Thus, Total Time Required to fill the tank = 4 min + 10 min 40 sec.
= 14 min 40 sec.

@All.

Better don't confuse with the complex solutions! as, the main part of the quantitative exam is to minimize time, all should look for a shortest solution, (as given by @Kalpana) but not merely a solution (with complex steps).

For any question, a diagrammatic representation should be drawn roughly, to get a quick solutions. Also, 'learn by solving' is the key to prepare any exam.

All the best!

@Leela.

To fulfil the condition they equate to 1. A part of work + remaining part of work = 1. To be precise, you and your friend are doing a work by sharing. You completed half the work and your friend completed half the remaining work. Ie 1/2 + 1/2 = 1.

This logic is being used in this kind of problems.

Another method could be : (1/15 + 1/20)*4 + (1/20)*x-4 = 1

7/15 + (X-4)/20 = 1

Hence x=44/3.

= 14(2/3) mins

= 14 min 40 seconds ( 2/3 x 60 = 40 seconds)

Here 7/15 is the part filled by (A + B) in 4 mins.

and (x-4) is the total time taken to fill the tank by B alone.

Here 1 means show full tank.

Its so simple:
After 4 minute the a was turned off means till 4 minutes a and b work together.
So,
4(1/15+i/20) = 7/15.

Let total time take = x.

Than 4 minutes after a is turned off so,

7/15+(x-4)*i/20 = 1.

x = 14 min. 40sec.

@Sudan.

Change fraction into mixed fraction.

Eg: 7/2 can be written in form on 3(1/2).

Divide numerator with denominator and whatever remainder you got write it in bracket.

Eg: a/b can be written as c(remainder/b).

a=15;
b=20;
l.c.m of 15,20=60;
a's 1Min=60/15 => 4;
b's 1Min=60/20 => 3;
(a+b)'s 4Min= (4+3)*4=28;
Now 60-28 = 32;
b remaining time = 32/3 => 10.66.
Now add 10.66+4 =>14.66.

LCM of 15 & 20 is 60.
(4 + 3) * 4 = 28.
60 - 28 = 32.
4 + 32/3 =14min = 40sec.

Another method could be : 4/15+x/20=1

x=44/3 i.e 14 min 40 sec.

4/15 is the part filled by A in 4 mins

and x is the total time taken to fill the tank by B.