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Difference between revisions of "2014 AMC 12B Problems/Problem 11"

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== Problem ==
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==Problem==
  
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Let <math>\mathcal P</math> be the parabola with equation <math>y = x^2</math> and let <math>Q = (20, 14)</math>. There are real numbers <math>r</math> and <math>s</math> such that the line through <math>Q</math> with slope <math>m</math> does not intersect <math>\mathcal P</math> if and only if <math>r < m < s</math>. What is <math>r + s</math>?
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A list of <math>11</math> positive integers has a mean of <math>10</math>, a median of <math>9</math>, and a unique mode of <math>8</math>. What is the largest possible value of an integer in the list?
  
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<cmath> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}}\ 52\qquad\textbf{(E)}\ 80 </cmath>
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<math> \textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35 </math>
  
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== Solution ==
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==Solution==
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The line through <math>Q</math> only meets the parabola if the system of equations <math>y = x^2</math> and <math>y = m(x - 20) + 14</math> has one or more solution(s) (the second equation comes from point-slope form). We can equate <math>y</math> and get the equation <math>x^2 = mx - 20m + 14</math>. Now if we move all the terms to the left, we get <math>x^2 - mx - (20m - 14) = 0</math>. We want this equation to have no solutions, and are trying to find the bounds of <math>r</math> and <math>s</math> for <math>m</math> such that this is unsolvable. A quadratic is unsolvable across the reals if its discriminant is less than 0. So, we set the discriminant to be less than 0:
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<cmath>\sqrt{m^2 - 80m - 56} \le 0</cmath>
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We start off with the fact that the median is <math>9</math>, so we must have <math>a, b, c, d, e, 9, f, g, h, i, j</math>, listed in ascending order. Note that the integers do not have to be distinct.
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We can square each side, keeping track of extraneous solutions: the resulting equation is not just <math>m^2 - 80m - 14 = 0</math>, but <math>\pm(m^2 - 80m - 66) = 0</math>. The solutions to this equation are <math>r</math> and <math>s</math>: to get their sum we use Vieta's Formulas and get <math>r + s = \pm80</math>. Because <math>-80</math> is not a valid answer choice, we can know for certain that the correct answer is <math>\boxed{(\text{E}) 80}</math>.
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Since the mode is <math>8</math>, we have to have at least <math>2</math> occurrences of <math>8</math> in the list. If there are <math>2</math> occurrences of <math>8</math> in the list, we will have <math>a, b, c, 8, 8, 9, f, g, h, i, j</math>. In this case, since <math>8</math> is the unique mode, the rest of the integers have to be distinct. So we minimize <math>a,b,c,f,g,h,i</math> in order to maximize <math>j</math>. If we let the list be <math>1,2,3,8,8,9,10,11,12,13,j</math>, then <math>j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33</math>.  
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Next, consider the case where there are <math>3</math> occurrences of <math>8</math> in the list. Now, we can have two occurrences of another integer in the list. We try <math>1,1,8,8,8,9,9,10,10,11,j</math>. Following the same process as above, we get <math>j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35</math>. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is <math>\boxed{\textbf{(E) }35}</math>
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== See also ==
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{{AMC12 box|year=2014|ab=B|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 09:13, 3 March 2015

Problem

A list of $11$ positive integers has a mean of $10$, a median of $9$, and a unique mode of $8$. What is the largest possible value of an integer in the list?

$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$

Solution

We start off with the fact that the median is $9$, so we must have $a, b, c, d, e, 9, f, g, h, i, j$, listed in ascending order. Note that the integers do not have to be distinct.

Since the mode is $8$, we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$. In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$. If we let the list be $1,2,3,8,8,9,10,11,12,13,j$, then $j = 11 \times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$.

Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$. Following the same process as above, we get $j = 11 \times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{\textbf{(E) }35}$

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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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