2010 AMC 10A Problem 11

Attempt Problem 11 of the 2010 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

The length of the interval of solutions of the inequality a≤2x+3≤ba \le 2x + 3 \le b is 10.10. What is b−a?b - a?

66

1010

1515

2020

3030

Answer: D
Concepts:inequalitylinear equation
Difficulty rating: 1280
Small Hint:

Solve the inequality for the two endpoints of xx

Big Hint:

The interval length is b−32−a−32\dfrac{b-3}{2}-\dfrac{a-3}{2}

Solution:

Splitting the inequality into two of them and solving gives us a≤2x+3 a \le 2x + 3 x≥a−32 x \ge \dfrac{a - 3}{2} and 2x+3≤b 2x + 3 \le b x≤b−32. x \le \dfrac{b - 3}{2}.

The range of the solutions is then b−32−a−32=10, \dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10, which then simplifying gives us (b−3)−(a−3)=20 (b - 3) - (a - 3) = 20 b−a=20. b - a = 20.

Thus, D is the correct answer.

Problem 10#10
Full Exam

Problem 11 in Other Years