Multiple choice

Ten years ago father was $12$ times as old as his son at that time and $10$ years hence he will be twice as old as his son. Find their present ages.

  1. $55$ years, $24$ years
  2. $34$ years, $12$ years
  3. $48$ years, $19$ years
  4. $42$ years, $18$ years
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let F, S be present ages. (F-10) = 12(S-10) => F-10 = 12S - 120 => F = 12S - 110. (F+10) = 2(S+10) => F+10 = 2S + 20 => F = 2S + 10. 12S - 110 = 2S + 10 => 10S = 120 => S = 12. F = 2(12) + 10 = 34.

AI explanation

Let the present age of the father be F and the son be S. Ten years ago, F minus 10 equaled 12 times (S minus 10), simplifying to F equals 12S minus 110. Ten years hence, F plus 10 will equal 2 times (S plus 10), simplifying to F equals 2S plus 10. Equating the two expressions gives 12S minus 110 equals 2S plus 10, meaning 10S equals 120 and S equals 12. Substituting 12 for S means the father is 34 years old.

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