Difference between revisions of "2010 AMC 10A Problems/Problem 11"
(Added solution 2 WLOG) |
|||
| (9 intermediate revisions by 9 users not shown) | |||
| Line 1: | Line 1: | ||
| + | == Problem 11 == | ||
| + | The length of the interval of solutions of the inequality <math>a \le 2x + 3 \le b</math> is <math>10</math>. What is <math>b - a</math>? | ||
| + | |||
| + | <math> | ||
| + | \mathrm{(A)}\ 6 | ||
| + | \qquad | ||
| + | \mathrm{(B)}\ 10 | ||
| + | \qquad | ||
| + | \mathrm{(C)}\ 15 | ||
| + | \qquad | ||
| + | \mathrm{(D)}\ 20 | ||
| + | \qquad | ||
| + | \mathrm{(E)}\ 30 | ||
| + | </math> | ||
| + | |||
| + | |||
| + | == Solution 1 == | ||
| + | |||
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>. | ||
| Line 11: | Line 29: | ||
<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> | ||
| Line 28: | Line 46: | ||
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> | ||
| + | |||
| + | == 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== | ||
| + | https://youtu.be/kU70k1-ONgM | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | == See also == | ||
| + | {{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}} | ||
| + | {{AMC12 box|year=2010|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:54, 31 August 2022
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 ![$[0, 10]$](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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
