2011 AMC 12A Problem 14

Attempt Problem 14 of the 2011 AMC 12A 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 2011 AMC 12A solutions, or check the answer key.

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14.

Suppose aa and bb are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b)(a, b) lies above the parabola y=ax2bx?y = ax^2 - bx?

1181\dfrac{11}{81}

1381\dfrac{13}{81}

527\dfrac{5}{27}

1781\dfrac{17}{81}

1981\dfrac{19}{81}

Answer: E
Concepts:parabolabasic probabilitycasework
Difficulty rating: 1690
Solution:

Substituting x=a,x = a, y=b,y = b, the point is above the parabola when b>a3ab,b \gt a^3 - ab, i.e. b(a+1)>a3.b(a + 1) \gt a^3.

For a=1:a = 1: b>12,b \gt \tfrac12, all 99 values work. For a=2:a = 2: b>83,b \gt \tfrac83, so b3,b \ge 3, giving 7.7. For a=3:a = 3: b>274=6.75,b \gt \tfrac{27}{4} = 6.75, so b7,b \ge 7, giving 3.3. For a4,a \ge 4, no b9b \le 9 works.

The count is 9+7+3=199 + 7 + 3 = 19 out of 81,81, so the probability is 1981.\dfrac{19}{81}.

Thus, the correct answer is E.

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