Aptitude - Pipes and Cistern - Discussion

Can you explain about the Remaining part?

(A+B+C) - (A+B) can give you the answer.

(A+B+C) =1/6 and (A+B+C) in 2hrs=2/6 and remaining part 1-2/6=2/3.

So (A+B) in 7 hrs is 2/ (3*7) =2/21.

1/6-2/21=1/14 so answer is 14.

@All.

Here, I've taken tank capacity as 42 litres.
A + B + C = 6 hours.
Remaining part by A + B in 7 hours.
So, LCM of 6 and 7 is 42.
If we take LCM's, cancellations will be easy in further steps.

How 1/6 - 2/21 = 1 /14? Explain please.

Assume total work = 6.
So, efficiency of a + b + c = 1.
We know w = e * t.
w = 1 * 2 {given}
= 2.

Balance = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.

Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.

Let C alone can fill the tank in x hours.

2 hours work of A,B & C + 7 hours work of A & B = 1,
According to the condition,
2/6 + 7(1/6 - 1/x) = 1,
2/6 + 7 (x - 6/6x) = 1,
2/6 + 7x - 42/6x = 1,
2x + 7x - 42/6x = 1,
9x - 42 = 6x,
9x - 6x = 42,
3x = 42,
x = 42/3,
x = 14.

Not getting this, please explain it.

A .+ B + C = 6 hrs
2 hrs = 1/3 work,
remaining = 2/3 work.
A+B, 2/3 work= 7 hrs,
Total work 3/3 i.e 1 = 7 hrs + 7/2 hrs = 21/2 hrs
Now consider LCM of A + B + C & A+B.
i.e 6 & 21/2 is 42
Taking efficiency: A + B + C = 7units
A+B = 4 units.

So C's efficiency= A+ B+C -( A+ B).
= 7-4 = 3.
Total work is LCM i.e 42 & efficiency of C is 3.
So total time= 42/3 = 14.

first understand the question properly.

A, B, C all opened can fill in 6 hours.
ie 1/A+1/B+1/C=1/6________(x)
ie 1/6th of tank is filled if all the three operate for an hour.
So if all 3 operate for 2 hours 2/6th of the tank is filled.
So remaining is 2/3rd.(1-2/6).

After 2 hours, C is closed.
to fill the 2/3rd by A & B it takes 7 hours.
[to find the time by c alone we need to find 1/A+1/B to substitute in (x)..]
So A and B in 1 hour can complete 2/(3*7).
ie 1/A+1/B=2/21. sub in x.
1/C=1/6-2/21.
We get 1/C =1/14.
So, C takes 14 hours to fill the tank.

Please tell me, how to solve this question using LCM method?