1976 AHSME Problems/Problem 29

Problem 29

Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is $44$ years, then Ann's age is

$\textbf{(A) }22\qquad \textbf{(B) }24\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }28$

Solution

This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$ . Let $y$ equal the difference between their ages, so $y=a-b$ . Know that $y$ is constant because the difference between their ages will always be the same.

Now, let's tackle the equation: $b=$ Ann's age when Barbara was Ann's age when Barbara was $\frac{a}{2}$ . When Barbara was $\frac{a}{2}$ years old, Ann was $\frac{a}{2}+y$ years old. So the equation becomes $b=$ Ann's age when Barbara was $\frac{a}{2}+y$ . Adding on their age difference again, we get $b = \frac{a}{2} + y + y \Rightarrow b = \frac{a}{2} + 2y$ . Substitute $a-b$ back in for $y$ to get $b = \frac{a}{2} + 2(a-b)$ . Simplify: $2b = a + 4(a-b) \Rightarrow 6b = 5a$ . Solving $b$ in terms of $a$ , we have $b = \frac{5a}{6}$ . Substitute that back into the first equation of $a+b=44$ to get $\frac{11a}{6}=44$ . Solve for $a$ , and the answer is $\boxed{\textbf{(B) }24}$ . ~jiang147369

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1976 AHSME Problems/Problem 29

Problem 29

Solution

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