Question

A family consists of the father, mother, two sons and the youngest daughter. the age of the father is four times the age of the second son. the age of the first son is in the ratio of 3:1 with that of his sister. the mother is 3.5 times older than the second son. the age of the second son is 2/3 times of the first son. the age of the youngest daughter is 5 years. find the sum of all ages.

Answer 1

Daughter's age (D) is 5 years.

Ratio of age of First son: Daughter = 3:1

So age of First son (S1) = 3*5 = 15 Years

Age of Second son (S2) is 2/3 times First son,

S2 = (2/3) * S1

S2 = (2/3) * 15

S2 = 2 * 5

S2 = 10

Age of Second son (S2) = 10 Years.

Age of the Father (F) is 4 times the age of Second son (S2),

F = 4 * S2

F = 4 * 10

F = 40

Age of the Father (F) = 40 Years.

Age of the Mother (M) is 3.5 times older than Second son (S2)

M = 3.5 * S2

M = 3.5 * 10

M = 35

Age of the Mother (M) = 35 Years.

Sum of all ages = D + S1 + S2 + F + M

= 5 + 15 + 10 + 40 + 35

= 105 Years

Total = 105 Years