Time and Work

Mr. A's work rate is 1/6 work per hour, and Mr. B's work rate is 1/12 work per hour. Together, their combined rate is 1/6 + 1/12 = 1/4 work per hour, so they will complete the work in 4 hours.

Sam's work rate is 1/4 job per day, Lisa's is 1/6, and Tom's is 1/2. Combined, their rate is 1/4 + 1/6 + 1/2 = 0.25 + 0.1667 + 0.5 = 0.9167 jobs per day. The time to complete one job together is 1 / 0.9167 ≈ 1.09 days.

The total man-hours required to paint the house is calculated by multiplying the number of workers by the hours they work: 325 workers × 8 hours = 2600 man-hours. To complete the same job in one day with only 8 workers, the total hours needed per worker is 2600 man-hours ÷ 8 workers = 325 hours. Since the question asks how many hours 8 workers should work in one day to finish the job, the mathematically correct answer is 325 hours, which is not among the options provided. However, given the options, none are correct for the problem as stated. The original answer choice 'D' (13 hours) is incorrect based on the calculation. The question or options appear flawed or ambiguous. The correct calculation is 325 hours, but since this is not an option, the question should be revised for clarity or options corrected. For now, the correct answer based on the problem is 325 hours, which is not listed.

The total work is calculated as 12 men × 6 hours = 72 man-hours. After working for 1 hour, the 12 men complete 12 man-hours of work, leaving 60 man-hours remaining. When 8 more men join, making 20 men in total, the remaining work of 60 man-hours will be completed in 60 ÷ 20 = 3 hours. Therefore, the correct answer is A (3 hours).

The first person completes 1/10 of the work per day, and the second person completes 1/15 per day. Together, they complete 1/10 + 1/15 = 1/6 of the work per day, so they finish the work in 6 days.

If 10 men take 12 days to complete the work, the total work is 10 × 12 = 120 man-days. With 12 men working, the time taken is 120 ÷ 12 = 10 days. However, the calculation shows 10 days, which matches option B, so let's recheck carefully. Actually, 10 men × 12 days = 120 man-days. For 12 men, days = 120 ÷ 12 = 10 days. So the correct answer is B, not C. The original correct_answer was 'X', which is invalid. The correct answer is B (10 days).

A and B together complete the work in 15 days, so their combined work rate is 1/15 per day. B alone completes the work in 20 days, so B's work rate is 1/20 per day. Therefore, A's work rate is (1/15 - 1/20) = (4/60 - 3/60) = 1/60 per day, meaning A alone can complete the work in 60 days. However, since 60 is option A, the correct answer should be A, not D. Rechecking the calculation: A's rate = 1/15 - 1/20 = (4/60 - 3/60) = 1/60, so A alone takes 60 days. Therefore, correct answer is A.

Twelve men take 6 hours to finish the work, so total work = 12 × 6 = 72 man-hours. After working for 1 hour, 12 men complete 12 man-hours of work, leaving 60 man-hours remaining. Now, 20 men work to complete the remaining 60 man-hours. Time required = 60 ÷ 20 = 3 hours. Therefore, 20 men will take 3 more hours to finish the remaining work, making the correct answer option A (3 hours).

Pipe A fills 1/20 of the cistern per minute, pipe B fills 1/30 per minute, and pipe C empties 1/40 per minute. Net fill rate = 1/20 + 1/30 - 1/40 = (6/120) + (4/120) - (3/120) = 7/120 per minute. Time to fill = 1 / (7/120) = 120/7 = 17 1/7 minutes. However, the calculation shows 17 1/7 minutes, which matches option C, so the original answer is correct. Rechecking the calculation: 1/20=0.05, 1/30=0.0333, 1/40=0.025; net=0.05+0.0333-0.025=0.0583; time=1/0.0583=17.14 minutes, which is 17 1/7 minutes. Therefore, the original correct answer C is correct. The initial confusion was in the explanation. The correct answer is C.

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.

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.

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).

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 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.

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.

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.

The first pipe fills the tank in 18 minutes, so its rate is 1/18 tank per minute. The second pipe fills in 15 minutes, rate 1/15. The outlet empties in 45 minutes, rate -1/45. Combined rate = 1/18 + 1/15 - 1/45 = (5/90 + 6/90 - 2/90) = 9/90 = 1/10, so the tank fills in 10 minutes. However, recalculating carefully: 1/18 + 1/15 = (5/90 + 6/90) = 11/90; subtract 1/45 (2/90) gives 9/90 = 1/10. So the correct time is 10 minutes, which corresponds to option D. Therefore, the original correct_answer 'D' is correct. The explanation was missing and is now provided. Difficulty is medium due to rate calculations.

The cistern normally fills in 8 hours, so its filling rate is 1/8 per hour. With the leak, it takes 10 hours, so the net filling rate is 1/10 per hour. The leak's emptying rate is the difference: 1/8 - 1/10 = 1/40 per hour, meaning the leak alone empties the cistern in 40 hours.

Let the time taken by the first pipe be x hours. Then the second pipe takes (x - 5) hours, and the third pipe takes (x - 9) hours (since the second pipe is 4 hours slower than the third). The first two pipes together fill the tank in the same time as the third pipe alone. Using the rates, 1/x + 1/(x-5) = 1/(x-9). Solving this equation gives x = 15 hours for the first pipe.

The filling rate is 1/4 per hour and the emptying rate is 1/9 per hour. Net filling rate = 1/4 - 1/9 = (9 - 4)/36 = 5/36 per hour. Time to fill = 1 / (5/36) = 36/5 = 7.2 hours. However, the question asks for the time after which the cistern gets filled, so the correct answer is 7.2 hours, which corresponds to option D. Therefore, the original correct answer D is correct, but the explanation was missing and the options need clarification.

Pipe A fills the tank in 15 minutes, so its rate is 1/15 tank per minute. Pipe B fills the tank in 20 minutes, so its rate is 1/20 tank per minute. Both pipes open for 4 minutes fill (1/15 + 1/20) * 4 = (4/60 + 4/80) = (1/15 + 1/20) * 4 = (4/60 + 3/60) = 7/60 of the tank. Remaining tank = 1 - 7/60 = 53/60. Pipe B alone fills the rest at 1/20 tank per minute, so time = (53/60) / (1/20) = (53/60) * 20 = 17.67 minutes total. Total time = 4 + 13.67 = 17.67 minutes, but this contradicts the options, so rechecking: Actually, the calculation is: Both pipes fill in 4 minutes: (1/15 + 1/20)*4 = (4/60 + 3/60) = 7/60 tank. Remaining = 1 - 7/60 = 53/60. Pipe B alone fills at 1/20 per minute, so time = (53/60) * 20 = 17.67 minutes. Total time = 4 + 17.67 = 21.67 minutes, which is not an option. Re-examining: The initial calculation was incorrect. Let's calculate carefully: Rate A = 1/15, Rate B = 1/20. Combined rate = 1/15 + 1/20 = (4/60 + 3/60) = 7/60 tank per minute. In 4 minutes, they fill 7/60 * 4 = 28/60 = 14/30 = 7/15 tank. Remaining tank = 1 - 7/15 = 8/15. Pipe B alone fills at 1/20 tank per minute, so time to fill remaining = (8/15) / (1/20) = (8/15) * 20 = 160/15 = 10.67 minutes. Total time = 4 + 10.67 = 14.67 minutes = 14 minutes 40 seconds. So the correct answer is D (14 min 40 sec). Explanation corrected accordingly.

The pump fills the tank in 2 hours, so its rate is 1/2 tank per hour. With the leak, it takes 7/3 hours (2 1/3 hours) to fill the tank, so the effective filling rate is 3/7 tank per hour. The leak's draining rate is the difference between the pump's rate and the effective rate: 1/2 - 3/7 = 1/14 tank per hour, meaning the leak alone can drain the tank in 14 hours. However, since the question asks 'The leak can drain all the water of the tank in?', the correct answer corresponds to 14 hours, which is option D. Therefore, the original correct_answer 'D' is correct, but the explanation was missing and has been added. The difficulty is medium as it involves fractional time and rate calculations.

Pipe A fills the tank at a rate of 1/5 tank per hour, Pipe B fills at 1/6 tank per hour, and Pipe C empties at 1/12 tank per hour. The combined filling rate when all three pipes are open is (1/5 + 1/6 - 1/12) = (12/60 + 10/60 - 5/60) = 17/60 tank per hour. Therefore, the time taken to fill the tank is the reciprocal of the combined rate, which is 60/17 hours, approximately 3.529 hours or 3 hours 31 minutes. This matches option C (3 9/17 hours). The initially given correct answer B is incorrect based on this calculation.

The combined filling rate of the two pipes is (1/20 + 1/30) = 1/12 tank per hour. Due to the leak, only two-thirds of this water remains, so the effective filling rate is (2/3) × (1/12) = 1/18 tank per hour. Therefore, the total time to fill the tank is 18 hours.

Pipe P fills 1/12 of the cistern per minute, and pipe Q fills 1/15 per minute. Together, they fill (1/12 + 1/15) = 9/60 = 3/20 of the cistern per minute. After 3 minutes, they fill 3 * 3/20 = 9/20 of the cistern. Remaining is 11/20. Pipe Q alone fills 1/15 per minute, so time to fill remaining is (11/20) / (1/15) = 11/20 * 15 = 8.25 minutes or 8 1/4 minutes. Adding the initial 3 minutes gives total time 11 1/4 minutes, so the cistern takes 8 1/4 minutes longer after pipe P is turned off.