1971 AHSME Problems/Problem 25

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Problem

A teen age boy wrote his own age after his father's. From this new four place number, he subtracted the absolute value of the difference of their ages to get $4,289$. The sum of their ages was

$\textbf{(A) }48\qquad \textbf{(B) }52\qquad \textbf{(C) }56\qquad \textbf{(D) }59\qquad  \textbf{(E) }64$

Solution

$\boxed{\textbf{(D) }59}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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