2010 AMC 10A Problems/Problem 11
Problem 11
The length of the interval of solutions of the inequality is . What is ?
Solution 1
Since we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .
Subtract from all of the quantities:
Divide all of the quantities by .
Since we have the range of the solutions, we can make it equal to .
Multiply both sides by 2.
Re-write without using parentheses.
Simplify.
We need to find for the problem, so the answer is
Solution 2
Without loss of generality, let the interval of solutions be (or any real values ). Then, substitute and to . This gives and . So, the answer is . ~ bearjere
Video Solution
~IceMatrix
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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.