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

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

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.

@@ 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>.
@@ Line 11: / Line 29: @@
 <math> \frac{a-3}{2}\le x\le \frac{b-3}{2} </math>
-Since we have the range of the solutions, we can make them equal to <math> 99 </math>.
+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>
@@ Line 19: / Line 37: @@
 <math> (b-3) - (a-3) = 20 </math>
-Simplify
+Re-write without using parentheses.
+<math> b-3-a+3 = 20 </math>
+Simplify.
-<math> b-3-a+3 = 20 \RightArrow b-a = 20 </math>
+<math> b-a = 20 </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}}

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