A couple got married 9 years ago when the age of wife was 20% less than her husband. 6 years from now the age of wife will be only 12.5% less than her husband. Now they have six children including single, twins and triplets and the ratio of their ages is 2 : 3 : 4 respectively. What can be the maximum possible value for the present age of this family? 

Let us assume that, Present age of wife is W years and present age of husband is H years .
9 years ago :-

→ Wife age (W) = 80% of Husband age(H)

→ (W - 9) = (80/100)(H - 9)

→ (W - 9)/(H - 9) = 4/5

→ 5W - 45 = 4H - 36

→ 5W - 4H = 9 -------------- Eqn.(1)

6 years from now :-

→ Wife age (W) = 87.5% of Husband age(H)

→ (W + 6) = (875/1000)(H + 6)

→ (W + 6)/(H + 6) = 7/8

→ 8W + 48 = 7H + 42

→ 8W - 7H = (-6) -------------- Eqn.(2)

Multiply Eqn.(1) by 8 and Eqn.(2) by 5 and subtracting the result we get,

→ 8(5W - 4H) - 5(8W - 7H) = 8*9 - 5*(-6)

→ 40W - 40W - 32H + 35H = 72 + 30

→ 3H = 102

→ H = 34 years .

putting value of H in Eqn.(1),

→ 5W - 4 * 34 = 9

→ 5W = 9 + 136

→ 5W = 145

→ W = 29 years .

Now, Since couple married 9 years ago , the maximum age of any child must be less than 9 years.

given that, the ratio of their ages is 2:3:4 .

therefore, Their ages will be,

• 2 years, 3 years and 4 years .
• 4 years, 6 years and 8 years . (we need maximum so, will take this.)

hence,

→ The maximum possible age of family = H + W + Single(4) + twins(6) + triplets(8)

→ The maximum possible age of family = 34 + 29 + 1*4 + 2*6 + 3*8

→ The maximum possible age of family = 34 + 29 + 4 + 12 + 24

→ The maximum possible age of family = 103 years (Ans.)