Mathematics

If 6 men take 9 days to complete the work, the total work is 6 × 9 = 54 man-days. To finish in 3 days, the number of men required is 54 ÷ 3 = 18 men.

Let the original speed be x kmph and distance be d km. From the problem, (d/(x+3)) = (d/x) - 2/3 hours and (d/(x-2)) = (d/x) + 2/3 hours. Solving these equations yields d = 36 km. Therefore, option A is correct.

Let the total work be W. Given A and B together take 12 days, so their combined rate is W/12. Similarly, B and C take 15 days (rate W/15), and C and A take 20 days (rate W/20). Adding all three rates: (A+B) + (B+C) + (C+A) = W/12 + W/15 + W/20 = 2(A+B+C). Calculating the sum: (1/12 + 1/15 + 1/20)W = (5/60 + 4/60 + 3/60)W = (12/60)W = (1/5)W. So, 2(A+B+C) = W/5, thus A+B+C = W/10. Since A+B = W/12, then C = (A+B+C) - (A+B) = W/10 - W/12 = (6/60 - 5/60)W = W/60. Similarly, from C+A = W/20, A = W/20 - C = W/20 - W/60 = (3/60 - 1/60)W = W/30. Therefore, A alone can do the work in 30 days. However, the calculation shows A alone takes 30 days, but the options given are 20, 30, 40, and none of these, and the correct answer is marked as B (30 days). The explanation above confirms 30 days is correct, so the original correct_answer is correct. The initial calculation was incorrect in the explanation; correcting it shows A alone takes 30 days.

P can do the work in 24 days, so P's 1 day work = 1/24. Q can do the work in 9 days, so Q's 1 day work = 1/9. R can do the work in 12 days, so R's 1 day work = 1/12. Combined, Q and R's 1 day work = 1/9 + 1/12 = 7/36. In 3 days, Q and R complete (7/36) * 3 = 7/12 of the work. The remaining work is 1 - 7/12 = 5/12. P will finish the remaining work in (5/12) / (1/24) = 10 days. Therefore, the correct answer is A (10).

Given the combined work rates of pairs, we can set up equations to find individual rates. Solving these shows X takes 30 days, Y takes 40 days, and Z takes 60 days to complete the work alone.

P and Q's work rates are 1/15 and 1/10 of the work per day respectively. Together, their combined rate is 1/15 + 1/10 = 1/6 of the work per day. They work together for 2 days, completing 2 × 1/6 = 1/3 of the work. The remaining work is 2/3, which P completes alone at a rate of 1/15 per day. Time taken by P alone to finish the remaining work is (2/3) ÷ (1/15) = 10 days. Total time to finish the work is 2 + 10 = 12 days. Therefore, the correct answer is A (12 days).

P can complete the work in 20 days, so P's work rate is 1/20 per day. In 4 days, P alone completes 4/20 = 1/5 of the work. The remaining work is 1 - 1/5 = 4/5. Q can complete the work in 12 days, so Q's work rate is 1/12 per day. Together, P and Q's combined work rate is 1/20 + 1/12 = (3/60 + 5/60) = 8/60 = 2/15 per day. To complete the remaining 4/5 of the work at a rate of 2/15 per day, the time required is (4/5) ÷ (2/15) = (4/5) × (15/2) = 6 days. Total time to complete the work is 4 days (P alone) + 6 days (P and Q together) = 10 days. Therefore, the correct answer is B (10 days).

Q’s work rate = 1/12 per day. Since P works twice as fast as Q, P’s rate = 2/12 = 1/6 per day. Combined rate = 1/6 + 1/12 = 1/4 per day, so together they complete the work in 4 days.

Let P take x days to complete the work. Then Q takes x/2 days and R takes x/3 days. Their daily work rates are 1/x, 2/x, and 3/x respectively. Together, they complete 1/2 of the work per day, so (1/x + 2/x + 3/x) = 6/x = 1/2. Solving gives x = 12, so Q takes 12/2 = 6 days to finish the work alone.

The leak empties the tank in 6 hours, so leak rate = 1/6 tank per hour. The tank empties in 8 hours with inlet open, so net emptying rate = 1/8 tank per hour. The inlet fills at 4 liters per minute = 240 liters per hour. The net emptying rate is leak rate minus inlet rate: 1/6 - inlet rate = 1/8, so inlet rate = 1/6 - 1/8 = 1/24 tank per hour. Since inlet fills 240 liters per hour, 1/24 tank = 240 liters, so tank capacity = 240 * 24 = 5760 liters.