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 $43+16=\boxed{\textbf{(D) }59}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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 • 31 • 32 • 33 • 34 • 35
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1971 AHSME Problems/Problem 25

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