Aptitude - Pipes and Cistern
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- Pipes and Cistern - Formulas
- Pipes and Cistern - General Questions
- Pipes and Cistern - Data Sufficiency 1
- Pipes and Cistern - Data Sufficiency 2
| Part filled by (A + B + C) in 3 minutes = 3 | ![]() |
1 | + | 1 | + | 1 | ![]() |
= | ![]() |
3 x | 11 | ![]() |
= | 11 | . |
| 30 | 20 | 10 | 60 | 20 |
| Part filled by C in 3 minutes = | 3 | . |
| 10 |
Required ratio = |
![]() |
3 | x | 20 | ![]() |
= | 6 | . |
| 10 | 11 | 11 |
| Net part filled in 1 hour | ![]() |
1 | + | 1 | - | 1 | ![]() |
= | 17 | . |
| 5 | 6 | 12 | 60 |
The tank will be full in |
60 | hours i.e., 3 | 9 | hours. |
| 17 | 17 |
hours to fill the tank. The leak can drain all the water of the tank in:
| Work done by the leak in 1 hour = | ![]() |
1 | - | 3 | ![]() |
= | 1 | . |
| 2 | 7 | 14 |
Leak will empty the tank in 14 hrs.
minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
x |
![]() |
2 | + | 1 | ![]() |
+ (30 - x). | 2 | = 1 |
| 75 | 45 | 75 |
|
11x | + | (60 -2x) | = 1 |
| 225 | 75 |
11x + 180 - 6x = 225.
x = 9.
Suppose, first pipe alone takes x hours to fill the tank .
Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.
|
1 | + | 1 | = | 1 |
| x | (x - 5) | (x - 9) |
|
x - 5 + x | = | 1 |
| x(x - 5) | (x - 9) |
(2x - 5)(x - 9) = x(x - 5)
x2 - 18x + 45 = 0
(x - 15)(x - 3) = 0
x = 15. [neglecting x = 3]

