2015 AMC 10A Problems/Problem 8: Difference between revisions
| Line 14: | Line 14: | ||
To solve the system of equations: | To solve the system of equations: | ||
p | <math>p=3c-4</math> | ||
p | <math>p=4c-12</math> | ||
3c-4 | <math>3c-4=4c-12</math> | ||
c | <math>c=8</math> | ||
p | <math>p=20.</math> | ||
Let <math>x</math> be the number of years until Pete is twice as old as Claire. | Let <math>x</math> be the number of years until Pete is twice as old as Claire. | ||
Revision as of 16:20, 19 December 2018
- The following problem is from both the 2015 AMC 12A #6 and 2015 AMC 10A #8, so both problems redirect to this page.
Problem
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be 
: 
?
Solution
This problem can be converted to a system of equations. Let 
be Pete's current age and 
be Claire's current age.
The first statement can be written as 
. The second statement can be written as 
To solve the system of equations:
Let 
be the number of years until Pete is twice as old as Claire.
The answer is 
.
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
