2010 AMC 10A Problem 11

Below is the professionally curated solution for Problem 11 of the 2010 AMC 10A, 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.

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Concepts:inequalitylinear equation

Difficulty rating: 1280

11.

The length of the interval of solutions of the inequality a2x+3ba \le 2x + 3 \le b is 10.10. What is ba?b - a?

66

1010

1515

2020

3030

Solution:

Splitting the inequality into two of them and solving gives us a2x+3 a \le 2x + 3 xa32 x \ge \dfrac{a - 3}{2} and 2x+3b 2x + 3 \le b xb32. x \le \dfrac{b - 3}{2}.

The range of the solutions is then b32a32=10, \dfrac{b - 3}{2} - \dfrac{a - 3}{2} = 10, which then simplifying gives us (b3)(a3)=20 (b - 3) - (a - 3) = 20 ba=20. b - a = 20.

Thus, D is the correct answer.

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