Difference between revisions of "2019 AMC 10A Problems/Problem 3"
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<math>\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15</math> | <math>\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15</math> | ||
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==Solution 1== | ==Solution 1== | ||
Revision as of 12:15, 28 July 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 1
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 2 (Guess and Check)
Simple guess and check works. Start with all the square numbers - 
, 
, 
, 
, 
, 
, etc. (probably stop at around 
since at that point it wouldn't make sense). If Ana is , then Bonita is 
, so in the previous year, Ana's age was times greater than Bonita's. If Ana is , then Bonita is , and Ana's age was times greater than Bonita's in the previous year, as required. The difference in the ages is 
.
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. 
