In a class of 5 students, average weight of the 4 lightest students is

Let a, b, c, d and e be the weights of the 5 students such that a ≤ b ≤ c ≤ d ≤ e. Since the total weight of the 4 lightest students is 4 x 40 = 160 kg, we have a + b + c + d = 160. Similarly, since the total weight of the 4 heaviest students is 4 x 45 = 180 kg, we have b + c + d + e = 180. We need to determine the maximum and minimum value of (a + b + c + d + e)/5.

The maximum average weight of all 5 students occurs when a is as large as possible. However, since the average weight of the 4 lightest students is 40 kg, a is at most 40, so the maximum average weight is:

(a + b + c + d + e)/5 = (a + 180)/5 = a/5 + 180/5 = 40/5 + 36 = 8 + 36 = 44 kg

The minimum average weight of all 5 students occurs when e is as small as possible. However, since the average weight of the 4 heaviest students is 45 kg, e is at least 45 kg, so the minimum average weight is:

(a + b + c + d + e)/5 = (160 + e)/5 = 160/5 + e/5 = 32 + 45/5 = 32 + 9 = 41 kg

Therefore, the difference between the maximum and minimum average weight is 44 - 41 = 3 kg.

Answer: C