Difference between revisions of "2011 AMC 12A Problems/Problem 14"
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== Problem == | == Problem == | ||
| − | 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>? | + | 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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== Solution == | == Solution == | ||
| + | If <math>(a,b)</math> lies above the parabola, then for any <math>a</math>, <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. | ||
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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 equal 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 equal any integer from <math>3</math> to <math>9</math>. | ||
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| + | 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>. | ||
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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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== 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}} | ||
Revision as of 20:38, 10 February 2011
Problem
Suppose 
and 
are single-digit positive integers chosen independently and at random. What is the probability that the point 
lies above the parabola 
?
Solution
If lies above the parabola, then for any , must be greater than 
. We thus get the inequality 
. Solving this for gives us 
. Now note that 
constantly increases when is positive. Then since this expression is greater than 
when 
, we can deduce that must be less than 
in order for the inequality to hold, since otherwise would be greater than and not a single-digit integer.
For 
, we get 
for our inequality, and thus can equal any integer from 
to .
For 
, we get 
for our inequality, and thus can equal any integer from 
to .
For 
, we get 
for our inequality, and thus can equal any integer from 
to .
Finally, if we total up all the possibilities we see there are 
points that satisfy the condition, out of 
total points. The probability of picking a point that lies above the parabola is thus 
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 | |
