1971 AHSME Problems/Problem 25

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

Because the father's age is a two digit number, we know that the father's age must be $43$ so that the second two digits of the original number can be a number in the teens. Let the son's age be $x$. We know that the original number is $4300+x$, and the positive difference between their ages is $43-x$. Thus, we have the equation $4300+x-(43-x)=4289$, which yields $x=16$. Thus, the sum of the two ages is $43+16=\boxed{\textbf{(D) }59}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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