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 that of her mother. The wife is 6 years younger to her husband and the sister is 3 years older then her brother. The mother's age is = ?

142 years
248 years
354 years
463 years

Answer:- 1

Explanation:-

Solution:

\begin{aligned} M \to Mother, F \to Father, S \to Son and D \to Daughter \\ F = 4 S, D = \frac{1}{3} M, M = F - 6 a n d S = D - 3 \\ ∴ M = 3 D = 3 (S + 3) \\ = 3 S + 9 = \frac{3}{4} F + 9 = \frac{3}{4} (M + 6) + 9 \\ = \frac{3}{4} M + \frac{3}{4} \times 6 + 9 \\ \Rightarrow (M - \frac{3}{4} M) = (\frac{9}{2} + 9) \\ \Rightarrow \frac{1}{4} M = \frac{27}{2} \\ \Rightarrow M = (\frac{27}{2} \times 4) \\ = 54 years \end{aligned}

$\eqalign{ & M \to {\text{Mother, }}F \to {\text{Father, }}S \to {\text{Son and }}D \to {\text{Daughter}} \cr & F = {\text{ }}4S,{\text{ }}D{\text{ }} = \frac{1}{3}M,{\text{ }}M = {\text{ }}F - 6{\text{ }}and{\text{ }}S = {\text{ }}D - 3 \cr & \therefore M = 3D = 3\left( {S + 3} \right) \cr & = 3S + 9 = \frac{3}{4}F + 9 = \frac{3}{4}\left( {M + 6} \right) + 9 \cr & = \frac{3}{4}M + \frac{3}{4} \times 6 + 9 \cr & \Rightarrow \left( {M - \frac{3}{4}M} \right) = \left( {\frac{9}{2} + 9} \right) \cr & \Rightarrow \frac{1}{4}M = \frac{{27}}{2} \cr & \Rightarrow M = \left( {\frac{{27}}{2} \times 4} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 54\,{\text{years}} \cr}$
∴ The mother is 54 years old.