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Difference between revisions of "2011 AMC 12A Problems/Problem 14"

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== Problem ==
 
== Problem ==
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Suppose <math>a</math> and <math>b</math> are single-digit positive integers chosen independently and at random. What is the probability that the point (a,b) lies above the parabola <math>y=ax^2-bx</math>?
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Suppose <math>a</math> and <math>b</math> are single-digit positive integers chosen independently and at random. What is the probability that the point <math>(a,b)</math> lies above the parabola <math>y=ax^2-bx</math>?
  
 
<math>
 
<math>
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\textbf{(C)}\ \frac{5}{27} \qquad
 
\textbf{(C)}\ \frac{5}{27} \qquad
 
\textbf{(D)}\ \frac{17}{81} \qquad
 
\textbf{(D)}\ \frac{17}{81} \qquad
−
\textbf{(E)}\ \frac{10}{81} </math>
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\textbf{(E)}\ \frac{19}{81} </math>
  
 
== Solution ==
 
== Solution ==
 +
If <math>(a,b)</math> lies above the parabola, then <math>b</math> must be greater than <math>y(a)</math>. We thus get the inequality <math>b>a^3-ba</math>. Solving this for <math>b</math> gives us <math>b>\frac{a^3}{a+1}</math>. Now note that <math>\frac{a^3}{a+1}</math> constantly increases when <math>a</math> is positive. Then since this expression is greater than <math>9</math> when <math>a=4</math>, we can deduce that <math>a</math> must be less than <math>4</math> in order for the inequality to hold, since otherwise <math>b</math> would be greater than <math>9</math> and not a single-digit integer. The only possibilities for <math>a</math> are thus <math>1</math>, <math>2</math>, and <math>3</math>.
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 +
For <math>a=1</math>, we get <math>b>\frac{1}{2}</math> for our inequality, and thus <math>b</math> can be any integer from <math>1</math> to <math>9</math>.
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 +
For <math>a=2</math>, we get <math>b>\frac{8}{3}</math> for our inequality, and thus <math>b</math> can be any integer from <math>3</math> to <math>9</math>.
 +
 +
For <math>a=3</math>, we get <math>b>\frac{27}{4}</math> for our inequality, and thus <math>b</math> can be any integer from <math>7</math> to <math>9</math>.
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 +
Finally, if we total up all the possibilities we see there are <math>19</math> points that satisfy the condition, out of <math>9 \times 9 = 81</math> total points. The probability of picking a point that lies above the parabola is thus <math>\frac{19}{81} \rightarrow \boxed{\textbf{E}}</math>
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 +
==Video Solution==
 +
 +
https://www.youtube.com/watch?v=u23iWcqbJlE
 +
~Shreyas S
 +
 +
this links to problem 11...
 +
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=13|num-a=15|ab=A}}
 
{{AMC12 box|year=2011|num-b=13|num-a=15|ab=A}}
 +
{{MAA Notice}}

Latest revision as of 22:06, 14 April 2022

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 $b$ 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. The only possibilities for $a$ are thus $1$, $2$, and $3$.

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

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

For $a=3$, we get $b>\frac{27}{4}$ for our inequality, and thus $b$ can be 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 $\frac{19}{81} \rightarrow \boxed{\textbf{E}}$

Video Solution

https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S

this links to problem 11...

See also

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

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