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

Revision as of 21:03, 10 May 2020

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

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

@@ Line 49: / Line 49: @@
 == 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}}