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    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.
    A A: 48 years, B: 36 years, C: 24 years, D: 12 years Correct Answer Incorrect Answer
    B A: 27 years, B: 21 years, C: 15 years, D: 9 years Correct Answer Incorrect Answer
    C A: 29 years, B: 23 years, C: 17 years, D: 11 years Correct Answer Incorrect Answer
    D A: 25 years, B: 27 years, C: 18 years, D: 5 years Correct Answer Incorrect Answer
    E A: 30 years, B: 24 years, C: 18 years, D: 12 years Correct Answer Incorrect Answer

    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 × 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 – 6 = (9 × 6) - 6 = 54 – 6 = 48 years B = 7x – 6 = (7× 6) – 6 = 42- 6 = 36 years C = 5x – 6 = (5 × 6) - 6 = 30 – 6 = 24 years D = 3x – 6 = (3 × 6) - 6 = 18 – 6 = 12 years

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