The problem involves finding the average weight of six individuals: A, B, C, D, E, and F. We are provided with the following information:

1. The average weight of A, B, and C is 40 kg: $\frac{(A+B+C)}{3} = 40 \Rightarrow A+B+C = 120$ (Equation 1)

2. The average weight of B, D, and E is 42 kg: $\frac{(B+D+E)}{3} = 42 \Rightarrow B+D+E = 126$ (Equation 2)

3. The weight of F is equal to that of B: $F = B$ (Equation 3)

Now, we need to find the average weight of A, B, C, D, E, and F.

By adding Equation 1 and Equation 2:

$(A+B+C) + (B+D+E) = 120+126$ $A+2B+C+D+E = 246$

Since F = B, we replace B with F:

$A + B + C + D + E + F = 246$

Now, the average weight of A, B, C, D, E, and F is:

$\frac{(A+B+C+D+E+F)}{6} = \frac{246}{6} = 41$

Thus, the average weight is 41 kg.