Difference between revisions of "2019 AMC 10A Problems/Problem 3"
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Solution by Baolan | Solution by Baolan | ||
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| + | ==Solution 2== | ||
| + | Simple guess and check works. Start with all the square numbers - 1, 4, 9, 16, 25, 36, etc. (probably cap off at like 100 since at that point it wouldn't make sense) If Ana is 9, then Bonita is 3, so their ages were 4x apart the year prior. If Ana is 16, then Bonita is 4, and thus their ages were 5x apart the year prior, which is what we are trying to find. The difference in the ages is 16 - 4 = 12. | ||
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| + | iron | ||
==See Also== | ==See Also== | ||
Revision as of 20:31, 11 February 2019
Contents
Problem
Ana and Bonita were born on the same date in different years, 
years apart. Last year Ana was 
times as old as Bonita. This year Ana's age is the square of Bonita's age. What is 
Solution
Let 
be the age of Ana and 
be the age of Bonita. Then,
![\[A-1 = 5(B-1)\]](http://latex.artofproblemsolving.com/e/e/b/eeb3cc039f87fb845971069da564891c3e1d6dc6.png)
and
![\[A = B^2.\]](http://latex.artofproblemsolving.com/f/4/0/f405bc2666154b29ef09e9275cbcd1340f1b0df9.png)
Substituting the second equation into the first gives us
![\[B^2-1 = 5(B-1).\]](http://latex.artofproblemsolving.com/6/6/c/66cbc568f3f79ad7a644b1e0924952e50d1316be.png)
By using difference of squares and dividing, 
Moreover, 
The answer is 
Solution by Baolan
Solution 2
Simple guess and check works. Start with all the square numbers - 1, 4, 9, 16, 25, 36, etc. (probably cap off at like 100 since at that point it wouldn't make sense) If Ana is 9, then Bonita is 3, so their ages were 4x apart the year prior. If Ana is 16, then Bonita is 4, and thus their ages were 5x apart the year prior, which is what we are trying to find. The difference in the ages is 16 - 4 = 12.
iron
See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
