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Difference between revisions of "1971 AHSME Problems/Problem 25"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(D) }59}</math>.
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Because the father's age is a two digit number, we know that the father's age must be <math>43</math> so that the second two digits of the original number can be a number in the teens. Let the son's age be <math>x</math>. We know that the original number is <math>4300+x</math>, and the positive difference between their ages is <math>43-x</math>. Thus, we have the equation <math>4300+x-(43-x)=4289</math>, which yields <math>x=16</math>. Thus, the sum of the two ages is <math>43+16=\boxed{\textbf{(D) }59}</math>.
  
 
== See Also ==
 
== See Also ==
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{{AHSME 35p box|year=1971|num-b=25|num-a=27}}
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{{AHSME 35p box|year=1971|num-b=24|num-a=26}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 18:00, 6 August 2024

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 (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
All AHSME Problems and Solutions

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