Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Video Solution)
 
(3 intermediate revisions by 3 users not shown)
Line 12: Line 12:
 
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>.
 
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>.
  
−
For <math>a=1</math>, we get <math>b>\frac{1}{2}</math> for our inequality, and thus <math>b</math> can equal any integer from <math>1</math> to <math>9</math>.
+
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>.
  
−
For <math>a=2</math>, we get <math>b>\frac{8}{3}</math> for our inequality, and thus <math>b</math> can equal any integer from <math>3</math> to <math>9</math>.
+
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 equal any integer from <math>7</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>.
  
 
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>
 
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>
 +
 +
==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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_12A_Problems/Problem_14&oldid=173365"