Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
|Part filled by (A + B + C) in 3 minutes = 3||\(\Bigg(\)||1||+||1||+||1||\(\Bigg)\)||=||\(\Bigg(\)||3 x||11||\(\Bigg)\)||=||11||.|
|Part filled by C in 3 minutes =||3||.|
|Required ratio =||\(\Bigg(\)||3||x||20||\(\Bigg)\)||=||6||.|
Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
|Net part filled in 1 hour||\(\Bigg(\)||1||+||1||-||1||\(\Bigg)\)||=||17||.|
|The tank will be full in||60||hours i.e., 3||9||hours.|
A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
|Work done by the leak in 1 hour =||\(\Bigg(\)||1||-||3||\(\Bigg)\)||=||1||.|
Leak will empty the tank in 14 hrs.
Two pipes A and B can fill a cistern in 37 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||\(\Bigg(\)||2||+||1||\(\Bigg)\)||+ (30 - x).||2||= 1|
|⇒||11x||+||(60 -2x)||= 1|
⇒ 11x + 180 - 6x = 225.
⇒ x = 9.
A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
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.
|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]