Difference between revisions of "1987 AHSME Problems/Problem 5"
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\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 == | ||
− | {{AHSME box|year=1987|num-b= | + | {{AHSME box|year=1987|num-b=4|num-a=6}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
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 , the total number of measurements. What is the smallest possible value of ?
Solution
Note that , , and . Thus, since the frequencies must be integers, must be divisible by , , and (so that etc. are integers), or in other words, is divisible by . Thus the smallest possible value of is the smallest positive multiple of , which is itself, or .
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 |
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