Aptitude - Pipes and Cistern - Discussion

Total 60 unit. (LCM of 10 20 30).

A in 1 min = 2 units.
B in 1 min = 3 units.
C in 1 min = 6 units.

A + B + C = (6 + 3 + 2) = 11 units.
So, C's contribution is 6/11 units.

First take the LCM of all the time taken by A, B, and C which we will give us total work = 60, and then take out the efficiency by the formula (total work/time) A-> 2 ,B-> 3 and C-> 6 and then take out the Ratio of C's R liquid proportion (6/11) by multiplying both the numerator and denominator by 3.
And the result we will get is 6/11.

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.

Part filled by A in one min is given by=1/30 min.
Part filled by B in one min is given by=1/20 min.
Part filled by C in one min is given by=1/10 min.

Now, total (A+B+C)filled in 1 min is given as (1/30+1/20+1/10) by taking lcm and we get 11/20
then in the qn they said after 3 min proportion of R ....clearly we are knowing that (A=p),(B=Q),(C=R).

So we should consider the value of c.
To calculate the proportion of R after 3 min,
Now part filled by C in 3 min is 3x(1/10)=3/10,
the proportion of R=total filled /c filled,
= (3/10)x(20/11),
= 6/11.

A in 1 min = 1/30.
B in 1 min = 1/20.
C in 1 min = 1/10.
1/30 + 1/20 + 1/10 => (2 + 3 + 6)/60 => 11/60.
Now, three pipes in 3 min = 3 * 11/60 => 11/20.
C share in 3 min = 3 * 1/10 => 3/10.
Required ratio = (3/10 * 20/11) => 6/11.

Why made reciprocal of 11/20? Please explain it.

Good, Thanks for explaining.

A pipe can fill 10% in three minutes that is 100÷3 * 100=10% b can fill 15% in three minutes 100/20=5.

5*3minutes =15%.
Similarly, C can fill in three minutes 30% so in three minutes entire work is 55% so C proportion is30÷55 = 6÷11.

A = 30min fill 100%.
B = 20min fill 100%.
C = 10min fill 100%.

if 3 min filling done all together then,
A =3min fill 10%,
B =3min fill 15%,
C =3min fill 30%.

P =10%,
Q = 15%,
R = 30%.

Total = 55% total tank filled.

Proportion of soln R in total liquid,
=R/Total,
=30/55,
=6/11.

If pipe c is empty the tank. Put negative sign Why use + sign?