Today Rebecca, who is 34 years old, and her daughter, who is 8 years old, celebrate their birthdays. How many years will pass before Rebecca’s age is twice her daughters age?

(A) 10

(B) 14

(C) 18

(D) 22

(E) 26

This is a classic “age problem.” These problems almost always lead to a system of linear equations that is easily solved by substitution. This problem is a rare exception to that rule, and we’ll only need one variable. I’ll provide that solution in a moment, but first we’re going to solve the problem by using the answers – always a good idea in an age problem.

When checking your answers, it’s optimal to start with (B) or (D). I’ll start with (D).

(D) Rebecca = 34 + 22 = 56. Daughter = 8 + 22 = 30.

So, after 22 years, Rebecca is less than twice her daughter’s age. Too much time has passed… 22 is too big as is 26. We can eliminate (D) and (E). Now we check (B).

(B) Rebecca = 34 + 14 = 48. DaughterÂ = 8 + 14 = 22.

So, after 14 years, Rebecca is more than twice her daughter’s age. Not enough time has passes…14 is too small.

**The correct answer is (C).**

Now for the algebraic solution. Let \(t\) = the amount of time that needs to pass before Rebecca is exactly twice her daughter’s age.

\(34 + t = 2(8 + t)\)

\(34 + t = 16 + 2t\)

\(34 – 16 = 2t – t\)

\(18 = t\)