Aptitude - Pipes and Cistern - Discussion

How 1/3 got over there? please explain.

How is 1/2-3/7 possible? Please explain indetail.

It takes 2 hours to fill the tank.

It takes 2 1/3 (7/3 hrs)hours to fill the tank including the leak.
Now,
Tank fill in 1 hour=1/2,
Tank fill in 1 hour including leak = 3/7,
Tank leak for 1 hour= tank fill in 1 hour- tank fill including tank leak in 1 hour = 1/2 - 3/7=1/14,
So, 1/14 of the tank will be empty in 1 hour.
Now, 1/14*14 of the tank will be empty in 1*14 hour. (Multiply 14 on both sides),
The Complete tank will be empty by leak in 14 hours.

Let consider 12-litre tank and pump fill in 2 hours(120 mins) w/o leakage.
with leakage, it takes 2 hours 20 mins,
means its take 20 mins extra,
so pump fill 2 litres extra in leakage condition,
total 14 litres fill by the pump but in tank 12 litres left,
means in 2hrs 20mins leakage leaks only 2 litres,
the total is 12 litres,
2 litres x6=12 litres.
7/3 hours(2 hrs 20 mins)x6 = 14 hours.

Good explanation. Thanks @Paushali Mukhoty.

@Jamsheed.

Always, we can consider total work be 1.
So
Hour ---- Work completed.
x ---- 1
3 ---- ?
By, cross multiplication,
3/X of the work.
Hope you understand.

It can also solved by;

Let the Filling pump be 'A' and leakage be 'B'.
A takes 2 hrs to fill, In 1 hr -it will complete 1/2 of the work, ie half of the tank.

In addition of B,
A-B=2 1/3 hrs, In 1 hr they complete 1/21/3 which is;
A-B=3/7.

We can substitute A,
1/2 -B=3/7
B=1/14.

So, In hr, B drains 1/14 part of the tank.
Total work-1.
Hr -- work
1 -- 1/14
? -- 1.

1/1/14= 14 hours.

Thanks @Ash.

Without leakage a takes 2 hours with,
leakage 7/3 hour take lcm of 7/3 *3=7and 2*3=6.
lcm of 6 and 7=42 it takes more 1/3 hour so divide 42 by 3 =42/3 =14.

Through LCM method:

Pump A fills in = 2hrs.
Leaking pump = 7/3 hrs.

Take LCM of 2 and 3 = 6 units,
Therefore,
Pump A = 3 u/h,
Leaking pump = 14 u/h,

Hence, the leaking pump can drain all the water in = 14 hours.