Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==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 smal...")
 
(→‎Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
Call the set <math>\{a, b, c, d, e\}</math>, with <math>a \le b \le c \le d \le e</math>.
+
When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. <math>\boxed{D}</math>
−
 
 
−
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>
 
  
 
== See also ==
 
== See also ==
Line 28: Line 14:
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 22:29, 17 December 2020

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

When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. $\boxed{D}$

See also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AJHSME_Problems/Problem_14&oldid=139910"