Difference between revisions of "2015 AMC 10A Problems/Problem 8"
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==Solution== | ==Solution== | ||
− | This problem can be converted to a system of equations. Let <math>p</math> be | + | This problem can be converted to a system of equations. Let <math>p</math> be Trahee's current age and <math>c</math> be her brother's current age. |
The first statement can be written as <math>p-2=3(c-2)</math>. The second statement can be written as <math>p-4=4(c-4)</math> | The first statement can be written as <math>p-2=3(c-2)</math>. The second statement can be written as <math>p-4=4(c-4)</math> | ||
Line 25: | Line 25: | ||
<math>p=20.</math> | <math>p=20.</math> | ||
− | Let <math>x</math> be the number of years until | + | Let <math>x</math> be the number of years until Trahee is twice as old as her brother. |
<math>20+x=2(8+x)</math> | <math>20+x=2(8+x)</math> | ||
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<math>x=4</math> | <math>x=4</math> | ||
− | The answer is <math>\boxed{\textbf{ | + | The answer is <math>\boxed{\textbf{ }4}</math>. |
==Video Solution== | ==Video Solution== |
Latest revision as of 10:47, 24 February 2021
- The following problem is from both the 2015 AMC 12A #6 and 2015 AMC 10A #8, so both problems redirect to this page.
Contents
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 Trahee's current age and be her brother'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 Trahee is twice as old as her brother.
The answer is .
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.