answer:Let's denote Carla's current age as C and Louis's current age as L. According to the information given, in 6 years, Carla will be 30 years old. Therefore, we can write the following equation for Carla's current age: C + 6 = 30 Now, let's solve for C: C = 30 - 6 C = 24 So, Carla is currently 24 years old. We are also told that the sum of the current ages of Carla and Louis is 55. We can write this as an equation: C + L = 55 Now, we can substitute the value of C (which is 24) into this equation to find L: 24 + L = 55 Now, let's solve for L: L = 55 - 24 L = 31 Therefore, Louis is currently boxed{31} years old.

answer:Let's denote the rate at which pipe A fills the cistern as A litres per minute, and the rate at which pipe B fills the cistern as B litres per minute. We know that pipe B fills the cistern in 5 minutes, so B = 39.99999999999999 litres / 5 minutes = 8 litres per minute. When all three pipes are open, the net rate at which the cistern is emptied is the sum of the rates of filling and emptying. Since the cistern is emptied in 1 hour (60 minutes) when all pipes are open, we can write the following equation: A + B - C = - (Total Volume / Time to Empty) A + 8 - 14 = - (39.99999999999999 / 60) Now we solve for A: A - 6 = - (39.99999999999999 / 60) A = - (39.99999999999999 / 60) + 6 Let's calculate the value of A: A = - (39.99999999999999 / 60) + 6 A = - (0.6666666666666665) + 6 A = 6 - 0.6666666666666665 A = 5.3333333333333335 litres per minute Now that we have the rate at which pipe A fills the cistern, we can calculate the time it takes for pipe A to fill the cistern: Time for A to fill the cistern = Total Volume / Rate of A Time for A = 39.99999999999999 litres / 5.3333333333333335 litres per minute Let's calculate the time: Time for A = 39.99999999999999 / 5.3333333333333335 Time for A ≈ 7.5 minutes Therefore, it takes pipe A approximately boxed{7.5} minutes to fill the cistern.

answer:Let's assume all 12 boats rented were big boats. Then, the number of small boats would be: (6 times 12 - 58) div (6 - 4) = 14 div 2 = 7 (boats) Therefore, the number of big boats would be: 12 - 7 = 5 (boats) Answer: There were 5 big boats and 7 small boats. Thus, the answer is: boxed{5, 7}. Assuming all 12 boats were big boats, they could seat a total of 12 times 6 = 72 people, which is 14 more than the known 58 people. Since one big boat can seat 2 more people than one small boat (6 - 4 = 2), there must be a total of 14 div 2 = 7 small boats. Therefore, there are 12 - 7 = 5 big boats in total. This problem can be solved directly by making an assumption, similar to the "chickens and rabbits in the same cage" problem. It can also be considered a problem with two unknowns, which is easier to solve with equations. The key is to find the correct relationship between quantities, set one unknown as x, and express the other unknown in terms of x, then solve the equation accordingly.

answer:Let n be the number of members in the chess club. 1. When arranged in 10 rows, n equiv 6 pmod{10} (because n+4 must be divisible by 10). 2. When arranged in 11 rows, n equiv 6 pmod{11} (because n+5 must be divisible by 11). Using the Chinese Remainder Theorem: - We are looking for n such that n equiv 6 pmod{10} and n equiv 6 pmod{11}. - The least common multiple of 10 and 11 is 110. - Thus, n = 110k + 6 for some integer k. To find k such that 300 leq 110k + 6 leq 400: - 294 leq 110k leq 394 - 2.67 leq k leq 3.58 - Since k must be an integer, k = 3. Substituting k = 3 into n = 110k + 6 gives: - n = 110 times 3 + 6 = 330 + 6 = 336. Thus, there are boxed{336} members in the chess club.