A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of her mother. The wife is six years younger to her husband and the sister in three years older than her brother. The age of the mother is:

The correct answer is (b), 48 years.

Let $f$ be the father’s age, $s$ be the son’s age, and $d$ be the daughter’s age. We know that $f = 4s$ and $d = \frac{1}{3}m$, where $m$ is the mother’s age. We also know that the wife is six years younger than her husband, so $m = f – 6$. Finally, the sister is three years older than her brother, so $d = s + 3$.

Substituting the first equation into the second equation, we get $d = \frac{1}{3}(f – 6)$. Substituting the third equation into the second equation, we get $d = s + 3$.

Equating these two equations, we get $\frac{1}{3}(f – 6) = s + 3$. Multiplying both sides by 3, we get $f – 6 = 3s + 9$. Adding 6 to both sides, we get $f = 3s + 15$.

Substituting this into the first equation, we get $3s + 15 = 4s$. Subtracting 3s from both sides, we get $15 = s$. Therefore, the son’s age is 15 years.

Substituting this into the third equation, we get $d = 15 + 3 = 18$. Therefore, the daughter’s age is 18 years.

Finally, substituting the first equation into the fourth equation, we get $m = (4)(15) – 6 = 48$. Therefore, the mother’s age is 48 years.