Mathematics

Solving the first quadratic equation x² + 11x + 30 = 0 gives roots x = -5 and x = -6. Solving the second quadratic y² + 15y + 56 = 0 gives roots y = -7 and y = -8. Since the largest root of x (-5) is greater than the largest root of y (-7), it follows that x > y, making option B correct.

Let the slower pipe fill the tank in x minutes. Then the faster pipe fills it in x/3 minutes. Their combined rate is 1/x + 3/x = 4/x tanks per minute. Given they fill the tank together in 36 minutes, their combined rate is 1/36. So, 4/x = 1/36, which gives x = 144 minutes. However, this contradicts the initial calculation, so let's re-express carefully: slower pipe rate = 1/x, faster pipe rate = 1/(x/3) = 3/x. Together rate = 1/36, so 1/x + 3/x = 4/x = 1/36, thus x = 144 minutes. Therefore, the slower pipe alone takes 144 minutes to fill the tank. The correct answer is C (144 min).

The cistern normally fills in 8 hours, but with the leak, it takes 10 hours. The leak causes the extra 2 hours delay. Using the formula for combined work rates, the leak empties the cistern in 40 hours.

Pipe A fills the tank in 6 hours, so its filling rate is 1/6 tank per hour. With the leak, it takes 9 hours, so the net filling rate is 1/9 tank per hour. The leak's emptying rate is the difference: 1/6 - 1/9 = 1/18 tank per hour, meaning the leak alone empties the tank in 18 hours.

Tap A fills the tank at a rate of 1/12 per hour, B at 1/15 per hour, and C at 1/20 per hour. Tap A is open continuously, while taps B and C are open alternately for one hour each. In every two-hour cycle, the tank is filled by 2*(1/12) + 1/15 + 1/20 = 1/6 + 1/15 + 1/20 = 17/60 of the tank. After three such cycles (6 hours), the tank is 3 * 17/60 = 51/60 full, leaving 9/60 (or 3/20) of the tank to be filled. In the next hour, tap A and tap B are open together, filling 1/12 + 1/15 = 9/60 = 3/20 of the tank, which completes the filling. Therefore, the total time to fill the tank is 6 + 1 = 7 hours, corresponding to option C.

The first quadratic equation x² + 5x + 6 = 0 factors as (x + 2)(x + 3) = 0, giving roots x = -2 and x = -3. The second quadratic equation y² + 9y + 14 = 0 factors as (y + 7)(y + 2) = 0, giving roots y = -7 and y = -2. Both equations share the root -2, but the other roots differ (-3 for x and -7 for y). Since the roots overlap and the other roots do not have a consistent inequality relationship, the relationship between x and y cannot be definitively established. Therefore, the correct answer is D: "If x = y or the relationship between x and y cannot be established."

If three pipes fill the tank in 8 hours, the rate of one pipe is 1/(3*8) = 1/24 of the tank per hour. Two pipes will fill at a rate of 2/24 = 1/12 per hour, so the time taken is 12 hours. However, the calculation shows 12 hours, so option B is correct. Rechecking: One pipe fills in 24 hours, so two pipes fill in 12 hours. Therefore, the correct answer is B, not C. The original correct_answer was B, which is correct. The explanation was missing and is now added. Difficulty is set to easy based on the problem type.

The first pipe fills the tank at a rate of 1/20 tank per minute, and the second at 1/24 tank per minute. Let the tank capacity be x gallons. The waste pipe empties 3 gallons per minute. Working together, the net filling rate is x/15 tanks per minute. Setting up the equation: (x/20 + x/24) - 3 = x/15. Solving for x gives 180 gallons.

Pipe A fills the tanker in 60 minutes, so its rate is 1/60 per minute; pipe B fills it in 40 minutes, so its rate is 1/40 per minute. If B is used for half the time (t/2 minutes), it fills (1/40)*(t/2) = t/80 of the tanker. For the other half (t/2 minutes), both A and B work together, filling (1/60 + 1/40)*(t/2) = (1/24)*(t/2) = t/48. The total filled is t/80 + t/48 = 1 (full tanker). Solving t/80 + t/48 = 1 gives t = 27.5 minutes.

Pipe C fills the tank in 10 minutes, so its rate is 1/10 tank per minute. Pipes A and B fill at rates 1/30 and 1/20 respectively. Combined, the three pipes fill at (1/30 + 1/20 + 1/10) = (2/60 + 3/60 + 6/60) = 11/60 tank per minute. In 3 minutes, total volume filled = 3 * 11/60 = 11/20. Volume contributed by pipe C in 3 minutes = 3 * 1/10 = 3/10 = 6/20. Therefore, proportion of solution R (from pipe C) = (6/20) / (11/20) = 6/11.