Question
Five years ago, the average age of a family of
four members was 25 years. Six years hence, the ratio of their ages will be 9:7:5:3. Find the approximate present ages of each family member.Solution
Let the ages of the four family members be A, B, C, and D, respectively. Five years ago, the average age of a family of four members was 25 years: The total age of the family five years ago = 4 Γ 25 = 100 The sum of their ages five years ago = A + B + C + D = 100 The total present age of the family = (A + 5) + (B + 5) + (C + 5) + (D + 5) = 100+20 =120 Six years hence, The total age of the family =(A+5+6) + (B+5+6) + (C+5+6) + (D+5+6) = 120+24 =144 Six years hence, the ratio of their ages will be 9:7:5:3 (9x) + (7x) + (5x) + (3x) = 144 x = 144 / 24 x = 6 A = 9x β 6 = (9 Γ 6) - 6 = 54 β 6 = 48 years B = 7x β 6 = (7Γ 6) β 6 = 42- 6 = 36 years C = 5x β 6 = (5 Γ 6) - 6 = 30 β 6 = 24 years D = 3x β 6 = (3 Γ 6) - 6 = 18 β 6 = 12 years
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