Difference between revisions of "1995 AHSME Problems/Problem 25"

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Let <math>a</math> be the smallest element, so <math>a+18</math> is the largest element. Since the mode is <math>8</math>, at least two of the five numbers must be <math>8</math>. The last number we denote as <math>b</math>.
 
Let <math>a</math> be the smallest element, so <math>a+18</math> is the largest element. Since the mode is <math>8</math>, at least two of the five numbers must be <math>8</math>. The last number we denote as <math>b</math>.
  
Then their average is <math>\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26</math>. Clearly <math>a \le 8</math>. Also we have <math>b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a</math>. Thus there are a maximum of <math>6</math> values of <math>a</math> which corresponds to <math>6</math> values of <math>b</math>; listing shows that all such values work.  
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Then their average is <math>\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26</math>. Clearly <math>a \le 8</math>. Also we have <math>b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a</math>. Thus there are a maximum of <math>6</math> values of <math>a</math> which corresponds to <math>6</math> values of <math>b</math>; listing shows that all such values work. The answer is <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:59, 10 February 2019

Problem

A list of five positive integers has mean $12$ and range $18$. The mode and median are both $8$. How many different values are possible for the second largest element of the list?

$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$

Solution

Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$, at least two of the five numbers must be $8$. The last number we denote as $b$.

Then their average is $\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26$. Clearly $a \le 8$. Also we have $b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a$. Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$; listing shows that all such values work. The answer is $\boxed{B}$.

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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