Difference between revisions of "2010 AMC 10A Problems/Problem 11"
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Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. | Since we are given the range of the solutions, we must re-write the inequalities so that we have <math> x </math> in terms of <math> a </math> and <math> b </math>. | ||
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<math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math> | <math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math> | ||
| − | Since we have the range of the solutions, we can make | + | Since we have the range of the solutions, we can make it equal to <math> 10 </math>. |
<math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math> | <math> \frac{b-3}{2}-\frac{a-3}{2} = 10 </math> | ||
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We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math> | We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math> | ||
| + | |||
| + | ~ I don't know who made this solution but it's actually really clear, hope the person that wrote this knows that :) - fowlertip | ||
| + | |||
| + | == Solution 2 == | ||
| + | [[Without loss of generality]], let the interval of solutions be <math>[0, 10]</math> (or any real values <math>[p, 10+p]</math>). Then, substitute <math>0</math> and <math>10</math> to <math>x</math>. This gives <math>b=23</math> and <math>a=3</math>. So, the answer is <math>23-3=\boxed{20\ \textbf{(D)}}</math>. | ||
| + | ~ bearjere | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 22:30, 13 January 2025
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 
~ I don't know who made this solution but it's actually really clear, hope the person that wrote this knows that :) - fowlertip
Solution 2
Without loss of generality, let the interval of solutions be ![$[0, 10]$](http://latex.artofproblemsolving.com/2/8/7/2872eb0dc1de0c6367b541403e6997b3e04d53ce.png)
(or any real values ![$[p, 10+p]$](http://latex.artofproblemsolving.com/f/9/2/f927920459b87899f56bc84f479c2082146d3600.png)
). 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
