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