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

Revision as of 08:54, 2 January 2012

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

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

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 them 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 $\boxed{20\ \textbf{(D)}}$

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

@@ 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==
 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 28: / Line 46: @@
 We need to find <math> b - a </math> for the problem, so the answer is <math> \boxed{20\ \textbf{(D)}} </math>
+== See also ==
+{{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}}