Mathematics

Let the rate of pipe A be x (tank per hour). Then pipe B's rate is 2x, and pipe C's rate is 4x (since C is twice as fast as B, and B is twice as fast as A). The combined rate of all three pipes is x + 2x + 4x = 7x. They fill the tank together in 5 hours, so their combined rate is 1/5 tank per hour. Thus, 7x = 1/5, giving x = 1/35. Therefore, pipe A alone takes 35 hours to fill the tank. This matches option C.

When tossing four coins, the total number of outcomes is 2^4 = 16. The number of ways to get exactly two heads and two tails is the number of combinations of 4 coins taken 2 at a time, which is C(4,2) = 6. Therefore, the probability is 6/16 = 3/8.

The cistern normally fills in 9 hours, but due to the leak, it takes 10 hours. The leak causes the filling to slow down by 1 hour. Using the formula for combined work rates, the leak empties the cistern in 90 hours, which corresponds to option D.

Pipe P fills the cistern in 18 minutes, so its rate is 1/18 per minute. Pipe Q fills it in 24 minutes, so its rate is 1/24 per minute. Let pipe P be turned off after t minutes. Then, the total work done in 12 minutes is (1/18)*t + (1/24)*12 = 1 (full cistern). Solving for t gives t = 8.5 minutes. Thus, pipe P must be turned off after 8.5 minutes.

Given the curved surface area (CSA) of the cylinder is 264 m² and volume is 924 m³, we use the formulas: CSA = 2πrh and Volume = πr²h. From CSA, 2πrh = 264, so rh = 264/(2π). From volume, πr²h = 924, so r²h = 924/π. Dividing volume by CSA: (πr²h) / (2πrh) = r/2 = 924/264 = 3.5, so r = 7. Using rh = 264/(2π) and substituting r=7, h = 6. Diameter = 2r = 14. Therefore, the ratio of diameter to height = 14:6 = 7:3. This matches option B. The initial explanation incorrectly matched the ratio to option D, but the correct ratio is 7:3, option B.

The first tap fills the cistern in 10 minutes, so its rate is 1/10 per minute. The second tap fills it in 15 minutes, so its rate is 1/15 per minute. Together, without the waste pipe, their combined rate is 1/10 + 1/15 = 1/6 per minute. When the waste pipe is open, they fill the cistern in 18 minutes, so the net filling rate is 1/18 per minute. The waste pipe's emptying rate is the difference between the combined filling rate and the net rate: 1/6 - 1/18 = 1/9 per minute. This means the waste pipe empties the full cistern in 9 minutes. Therefore, the correct answer is D (9 min).

Pipe A can fill the tank in 2 minutes, so its rate is 1/2 tank per minute. Pipe B can fill the tank in 15 minutes, so its rate is 1/15 tank per minute. When both pipes are open together, their combined rate is 1/2 + 1/15 = 15/30 + 2/30 = 17/30 tank per minute. For the first 4 minutes, both pipes work together, filling (17/30) × 4 = 68/30 = 2.2667 tanks, which is more than the tank capacity, meaning the tank would be filled before 4 minutes. This indicates an inconsistency in the problem as stated (pipe A fills the tank too quickly). However, if the problem intended pipe A to fill the tank in 12 minutes (a common variant), then the calculation proceeds as follows: Pipe A's rate = 1/12 tank per minute, pipe B's rate = 1/15 tank per minute, combined rate = 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20 tank per minute. In 4 minutes, they fill (3/20) × 4 = 12/20 = 3/5 of the tank. Remaining tank = 1 - 3/5 = 2/5. Pipe B alone fills at 1/15 tank per minute, so time to fill remaining = (2/5) ÷ (1/15) = (2/5) × 15 = 6 minutes. Total time = 4 + 6 = 10 minutes. This matches option B. Therefore, the original problem likely had a typo in pipe A's filling time. The corrected answer is 10 minutes (option B).

Let the time taken by pipe A to fill the cistern be x hours. Then pipe B takes x + 6 hours. Together, their combined rate is 1/4 cistern per hour. So, 1/x + 1/(x+6) = 1/4. Solving this equation gives x = 8 hours, which is the time taken by pipe A alone.

The total number of bulbs is 9, with 4 defective and 5 good bulbs. The probability that at least one bulb is good is 1 minus the probability that all chosen bulbs are defective. The number of ways to choose 4 defective bulbs out of 4 is 1, and total ways to choose any 4 bulbs out of 9 is C(9,4)=126. So, probability all are defective = 1/126. Therefore, probability at least one is good = 1 - 1/126 = 125/126.

Pipe A fills the cistern in 4 hours, so its rate is 1/4 per hour; pipe B fills in 5 hours, rate 1/5 per hour; pipe C empties in 2 hours, rate -1/2 per hour. Pipes open sequentially at 1 AM (A), 2 AM (B), and 3 AM (C). Calculating the net volume over time shows the cistern will be empty at 7 PM.