1995 AHSME Problems/Problem 25

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 $\boxed{B}$ .

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1995 AHSME Problems/Problem 25

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