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Difference between revisions of "2010 AMC 10A Problems/Problem 11"

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==Solution==
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== 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>.  
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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>
  
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Since we have the range of the solutions, we can make them equal to <math> 10 </math>.
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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>
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== Solution 2 ==
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[[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>.
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~ bearjere
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 18:54, 31 August 2022

Problem 11

The length of the interval of solutions of the inequality $a \le 2x + 3 \le b$ is $10$. What is $b - a$?

$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 20 \qquad \mathrm{(E)}\ 30$


Solution 1

Since we are given the range of the solutions, we must re-write the inequalities so that we have $x$ in terms of $a$ and $b$.

$a\le 2x+3\le b$

Subtract $3$ from all of the quantities:

$a-3\le 2x\le b-3$

Divide all of the quantities by $2$.

$\frac{a-3}{2}\le x\le \frac{b-3}{2}$

Since we have the range of the solutions, we can make it equal to $10$.

$\frac{b-3}{2}-\frac{a-3}{2} = 10$

Multiply both sides by 2.

$(b-3) - (a-3) = 20$

Re-write without using parentheses.

$b-3-a+3 = 20$

Simplify.

$b-a = 20$

We need to find $b - a$ for the problem, so the answer is $\boxed{20\ \textbf{(D)}}$

Solution 2

Without loss of generality, let the interval of solutions be $[0, 10]$ (or any real values $[p, 10+p]$). Then, substitute $0$ and $10$ to $x$. This gives $b=23$ and $a=3$. So, the answer is $23-3=\boxed{20\ \textbf{(D)}}$. ~ bearjere

Video Solution

https://youtu.be/kU70k1-ONgM

~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

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