Multiple choice

The ages of two persons differ by $16$ years. If $6$ years ago, the elder one be $3$ times as old as the younger one, find their present ages.

  1. 15 years and 31 years

  2. 14 years and 30 years

  3. 12 years and 28 years

  4. 10 years and 26 years
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let ages be x and x+16. Six years ago, (x+16-6) = 3(x-6), which simplifies to x+10 = 3x-18. Solving gives 2x = 28, so x = 14. The ages are 14 and 30.

AI explanation

Let the present age of the younger person be x years. Since their ages differ by 16 years, the elder person is x plus 16 years old. Six years ago, the elder one was 3 times as old as the younger one, giving the equation x plus 10 equals 3 times x minus 6. Solving this linear equation yields x equals 14, meaning the younger person is 14 years old. The elder person is 14 plus 16, equaling 30 years old.

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