The average age of a family of five members is 26. If the youngest member of the family is now ten years, what was the average of the family one year before the youngest member was born?

The present average age of the five family members is given as $$26 \text{ years}$$.

Therefore, the present total age of all five members is
$$5 \times 26 = 130 \text{ years}$$

The youngest member is now $$10 \text{ years}$$ old.

Hence, the combined present age of the other four members is
$$130 - 10 = 120 \text{ years}$$

We need the family’s average age one year before the youngest member was born.
The youngest was born $$10$$ years ago, so “one year before” that point is $$10 + 1 = 11$$ years ago.

At that moment:

• The youngest member had not yet been born, so the family had only $$4$$ members.
• Each of those four members was $$11$$ years younger than they are now.

Thus, the total age of the four members at that time was
$$120 - 4 \times 11 = 120 - 44 = 76 \text{ years}$$

The required average age is then
$$\frac{76}{4} = 19 \text{ years}$$

Option C which is: 19 years