2011 AMC 12A Problems/Problem 14

Problem

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

$\textbf{(A)}\ \frac{11}{81} \qquad \textbf{(B)}\ \frac{13}{81} \qquad \textbf{(C)}\ \frac{5}{27} \qquad \textbf{(D)}\ \frac{17}{81} \qquad \textbf{(E)}\ \frac{19}{81}$

Solution

If $(a,b)$ lies above the parabola, then for any $a$ , $b$ must be greater than $y(a)$ . We thus get the inequality $b>a^3-ba$ . Solving this for $b$ gives us $b>\frac{a^3}{a+1}$ . Now note that $\frac{a^3}{a+1}$ constantly increases when $a$ is positive. Then since this expression is greater than $9$ when $a=4$ , we can deduce that $a$ must be less than $4$ in order for the inequality to hold, since otherwise $b$ would be greater than $9$ and not a single-digit integer.

For $a=1$ , we get $b>\frac{1}{2}$ for our inequality, and thus $b$ can equal any integer from $1$ to $9$ .

For $a=2$ , we get $b>\frac{8}{3}$ for our inequality, and thus $b$ can equal any integer from $3$ to $9$ .

For $a=3$ , we get $b>\frac{27}{4}$ for our inequality, and thus $b$ can equal any integer from $7$ to $9$ .

Finally, if we total up all the possibilities we see there are $19$ points that satisfy the condition, out of $9 \times 9 = 81$ total points. The probability of picking a point that lies above the parabola is thus $\frac{19}{81} \rightarrow \boxed{\textbf{E}}$

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2011 AMC 12A Problems/Problem 14

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