Difference between revisions of "2019 AMC 10A Problems/Problem 3"
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Solving this quadratic yields <math>B=4.</math> Moreover, <math>A=B^2=16.</math> | Solving this quadratic yields <math>B=4.</math> Moreover, <math>A=B^2=16.</math> | ||
| − | The answer is <math>16-4 = 12 \implies \boxed{D}.</math> | + | The answer is <math>16-4 = 12 \implies \boxed{\textbf{(D)}}.</math> |
Solution by Baolan | Solution by Baolan | ||
Revision as of 18:52, 9 February 2019
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)
Solving this quadratic yields 
Moreover, 
The answer is 
Solution by Baolan
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. 
