Difference between revisions of "1971 AHSME Problems/Problem 25"
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== Solution == | == Solution == | ||
| − | <math>\boxed{\textbf{(D) }59}</math>. | + | 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 == | ||
Revision as of 18:55, 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 
. The sum of their ages was
Solution
Because the father's age is a two digit number, we know that the father's age must be 
so that the second two digits of the original number can be a number in the teens. Let the son's age be 
. We know that the original number is 
, and the positive difference between their ages is 
. Thus, we have the equation 
, which yields 
. Thus, the sum of the two ages is 
.
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 25 |
Followed by Problem 27 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
