Difference between revisions of "2019 AMC 10A Problems/Problem 3"
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==Solution== | ==Solution== | ||
| + | Let A be the age of Ana and B be the age of Bonita. | ||
| + | |||
| + | A-1 = 5(B-1) | ||
| + | A = B^2 | ||
| + | We can solve for B to get B = 4 | ||
| + | Then that means that A = 16. | ||
| + | 16-4 = (D) 12 | ||
| + | |||
| + | Solution by Baolan (someone please LaTeX this) | ||
==See Also== | ==See Also== | ||
Revision as of 17:41, 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 A be the age of Ana and B be the age of Bonita.
A-1 = 5(B-1) A = B^2 We can solve for B to get B = 4 Then that means that A = 16. 16-4 = (D) 12
Solution by Baolan (someone please LaTeX this)
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. 
