Multiple choice

Ten years ago a father was six times as old as his daughter. After $10$ years, he will be twice as old as his daughter. Determine their present age.

  1. Present age of father $=10$ years
    Present age of daughter $=5$ years
  2. Present age of father $=13$ years
    Present age of daughter $=26$ years
  3. Present age of father $=20$ years
    Present age of daughter $=18$ years
  4. present age of father $= 40$ years
    Present age of daughter $=15$ years
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let F and D be present ages. Ten years ago: (F-10) = 6(D-10). In 10 years: (F+10) = 2(D+10). Solving the system: F-10 = 6D-60 => F = 6D-50. Substituting: 6D-50+10 = 2D+20 => 4D = 60 => D = 15. Then F = 6(15)-50 = 40.

AI explanation

Let the present ages of the father and daughter be F and D. Ten years ago, F - 10 = 6(D - 10), so F = 6D - 50. Ten years hence, F + 10 = 2(D + 10), meaning 6D - 50 + 10 = 2D + 20, which gives 4D = 60 and D = 15. Substituting D into the first equation gives F = 6(15) - 50 = 40. The present age of the father is 40 years and the present age of the daughter is 15 years.

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