Difference between revisions of "1997 AJHSME Problems/Problem 14"

Revision as of 17:51, 16 April 2019

Problem

There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8. What is the difference between the largest and smallest integers in the set?

$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution

JO MAMA

1997 AJHSME (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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==Solution==
-Call the set <math>\{a, b, c, d, e\}</math>, with <math>a \le b \le c \le d \le e</math>.
+JO MAMA
-Since the median, or middle number, is <math>5</math>, we have <math>c=5</math>.
-Since the mode, or most common number, is <math>8</math>, we have <math>d=e=8</math>.
-Thus, the set is now <math>\{a, b, 5, 8, 8\}</math>.  Additionally, since there is only one mode, <math>a</math> and <math>b</math> are both less than <math>5</math> and are distinct from each other.
-Since the mean of the set is <math>5</math>, and there are five numbers in the set, the sum of the numbers in the set is <math>5\times 5 = 25</math>.
-Thus, we have <math>a + b + 5 + 8 + 8 = 25</math>, which leads to <math>a + b = 4</math>
-Since <math>a</math> and <math>b</math> are distinct positive integers, they must equal <math>1</math> and <math>3</math>.
-The smallest number is <math>1</math>, and the largest number is <math>8</math>, giving a difference of <math>7</math>, leading to answer <math>\boxed{D}</math>
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Difference between revisions of "1997 AJHSME Problems/Problem 14"

Revision as of 17:51, 16 April 2019

Problem

Solution

See also