In a family, there are two parents and four children. The father is twice as old as the mother was when the father was the age the mother is now. The sum of the ages of all six family members is 140 years. If the youngest child is 4 years old and the age difference between each child is 2 years, how old are the parents?
The father is 40 years old, and the mother is 32 years old.

Here's the reasoning:

1. Let F represent the father's age, M represent the mother's age, and C1, C2, C3, and C4 represent the ages of the four children.
2. We know that the sum of their ages is 140 years: F + M + C1 + C2 + C3 + C4 = 140.
3. The youngest child (C1) is 4 years old, and there is a 2-year age difference between each child. So, C2 = 6, C3 = 8, and C4 = 10.
4. Substituting these values into the equation from step 2, we get: F + M + 4 + 6 + 8 + 10 = 140.
5. Simplifying this equation, we obtain: F + M = 112.
6. Now let's consider the information about the father's age being twice as old as the mother was when the father was the age the mother is now. This can be represented as: F = 2(M - (F - M)).
7. Expanding this equation, we get: F = 2M - 2F + 2M.
8. Rearranging terms, we find: 3F = 4M.
9. Since we already have an equation for F + M (from step 5), we can solve this system of equations using substitution or elimination methods.
10. Using substitution, we can express F in terms of M: F = (4/3)M.
11. Substituting this expression into the equation F + M = 112, we get: (4/3)M + M = 112.
12. Simplifying this equation, we find that M = 32.
13. Now, using the expression F = (4/3)M, we can calculate the father's age: F = (4/3)(32) = 40.
14. Therefore, the father is 40 years old, and the mother is 32 years old.