2465.21 – Stephanie and Bonnie's Ages

Stephanie and Bonnie compare ages. They find that Bonnie is as old as Stephanie was when Bonnie was as old as Stephanie had been when Bonnie was half as old as Stephanie is now. If the sum of their present ages is 44 years, what is Stephanie’s age?

Solution

Let $s$ and $b$ be Stephanie and Bonnie’s current ages. Let “was” be $x$ years ago and “had been” be $y$ years ago.

Here is what we know:

(1) “Bonnie is as old as Stephanie was”: $b = s – x$ ,

(2) "when Bonnie was as old as Stephanie had been”: $b – x = s – y$ ,

(3) "when Bonnie was half as old as Stephanie is now”: $b – y = s/2$ .

We also know

(4) that $s + b = 44$ .

Here we go:

by (4) $s = 44 – b,$

by (1)

\begin{aligned} b = (44 - b) – x \\ 2b = 44 – x \\ b = (44 – x)/2 \end{aligned}

by (2)

\begin{aligned} (44 – x)/2 – x = 44 – b – y \\ 44 – x – 2x = 88 – 2b – 2y \\ 44 – 3x = 88 - 2(44 – x)/2 – 2y \\ 44 – 3x = 88 – 44 + x – 2y \\ 44 – 4x = 44 – 2y \\ 4x = 2y \\ 2x = y \end{aligned}

by (3)

\begin{aligned} (44 – x)/2 – 2x = (44 – b)/2 \\ 44 – x – 4x = 44 – b \\ -5x = -(44 – x)/2 \\ 10x = 44 – x \\ 11x = 44 \end{aligned}

We now know that $x = 4; y = 8.$ Finally, $b = s – 4$ (1) and $s + (s – 4) = 44$ (4), so $2s = 48$ . Thus, $s = 24$ (Stephanie) and $b = 24 – 4 = 20$ (Bonnie).

Check:

(1) 20 = 24 - 4,

(2) 20 – 4 = 24 – 8,

(3) 20 – 8 = 24/2,

(4) 24 + 20 = 44.