Multiple choice

Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

  1. Salims age $= 68$ years, Daughters age $= 24$ years.
  2. Salims age $= 58$ years, Daughters age $= 74$ years.
  3. Salims age $= 23$ years, Daughters age $= 14$ years.
  4. Salims age $= 38$ years, Daughters age $= 14$ years.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let S and D be current ages. (S-2) = 3(D-2) => S-2 = 3D-6 => S = 3D-4. (S+6) = 2(D+6) + 4 => S+6 = 2D+12+4 => S = 2D+10. 3D-4 = 2D+10 => D = 14. S = 3(14)-4 = 38.

AI explanation

Let Salim's present age be $S$ and his daughter's present age be $D$. Two years ago, their ages were $S-2$ and $D-2$, giving the equation $S-2 = 3(D-2)$, which simplifies to $S = 3D - 4$. Six years later, their ages will be $S+6$ and $D+6$, so $S+6 = 2(D+6) + 4$, which simplifies to $S = 2D + 10$. Equating the two expressions for $S$ gives $3D - 4 = 2D + 10$, meaning $D = 14$. Substituting $D$ into either equation yields $S = 38$, making Salim 38 years old and his daughter 14 years old.

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