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 "1987 AHSME Problems/Problem 5"

m (→‎See also)
m (Fixed formatting)
 
(2 intermediate revisions by the same user not shown)
Line 11: Line 11:
 
\textbf{(D)}\ 25 \qquad
 
\textbf{(D)}\ 25 \qquad
 
\textbf{(E)}\ 50</math>
 
\textbf{(E)}\ 50</math>
 +
 +
== Solution ==
 +
Note that <math>12.5\% = \frac{1}{8}</math>, <math>25\% = \frac{1}{4}</math>, and <math>50\% = \frac{1}{2}</math>. Thus, since the frequencies must be integers, <math>N</math> must be divisible by <math>2</math>, <math>4</math>, and <math>8</math> (so that <math>\frac{N}{8}</math> etc. are integers), or in other words, <math>N</math> is divisible by <math>8</math>. Thus the smallest possible value of <math>N</math> is the smallest positive multiple of <math>8</math>, which is <math>8</math> itself, or <math>\boxed{\text{B}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 11:25, 1 March 2018

Problem

A student recorded the exact percentage frequency distribution for a set of measurements, as shown below. However, the student neglected to indicate $N$, the total number of measurements. What is the smallest possible value of $N$?

\[\begin{tabular}{c c}\text{measured value}&\text{percent frequency}\\ \hline 0 & 12.5\\ 1 & 0\\ 2 & 50\\ 3 & 25\\ 4 & 12.5\\ \hline\ & 100\\ \end{tabular}\]

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 50$

Solution

Note that $12.5\% = \frac{1}{8}$, $25\% = \frac{1}{4}$, and $50\% = \frac{1}{2}$. Thus, since the frequencies must be integers, $N$ must be divisible by $2$, $4$, and $8$ (so that $\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$. Thus the smallest possible value of $N$ is the smallest positive multiple of $8$, which is $8$ itself, or $\boxed{\text{B}}$.

See also

1987 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1987_AHSME_Problems/Problem_5&oldid=92539"