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Difference between revisions of "2014 AMC 10B Problems/Problem 18"

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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>
 
<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 1==
 
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.  
 
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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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>
 
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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==Solution 2==
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Note that <math>x_1 + \ldots + x_{11} = 110</math> let <math>x_6 = 9</math> so <math>x_1 + \ldots + x_5 + x_7 + \ldots + x_{11} = 101</math>. To maximize the value of <math>x_i</math> where <math>i</math> ranges from <math>1</math> to <math>11</math>, we let any <math>7</math> elements be <math>1,2,\ldots,7</math> so <math>x_1 + x_2 + x_3 = 57</math>. Now we have to let one of above <math>3</math> values = <math>8</math> hence <math>x_1 + x_2 = 49</math> now let <math>x_1 = 35</math>, <math>x_2 = 14</math> hence <math>\boxed{\textbf{(E) }35}</math> is the answer.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}
 
{{AMC10 box|year=2014|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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Solution 2:
 
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Note that x1 + ... + x11 = 110
 
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let x6 = 9
 
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so x1 + ... + x5 + x7+ ... + x11 = 101
 
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now to maximise the value of xi where i ranges from 1 to 11, we let any 7 elements be 1,2,...,7
 
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so x1 + x2 + x3 = 57
 
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now we have to let one of above 3 values = 8 hence x1 + x2 = 49
 
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now let x1 = 35 , x2 =14 hence 35 is the answer
 

Latest revision as of 16:44, 30 January 2021

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 1

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 $\boxed{\textbf{(E) }35}$


Solution 2

Note that $x_1 + \ldots + x_{11} = 110$ let $x_6 = 9$ so $x_1 + \ldots + x_5 + x_7 + \ldots + x_{11} = 101$. To maximize the value of $x_i$ where $i$ ranges from $1$ to $11$, we let any $7$ elements be $1,2,\ldots,7$ so $x_1 + x_2 + x_3 = 57$. Now we have to let one of above $3$ values = $8$ hence $x_1 + x_2 = 49$ now let $x_1 = 35$, $x_2 = 14$ hence $\boxed{\textbf{(E) }35}$ is the answer.

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 10 Problems and Solutions

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