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 "1985 AJHSME Problems/Problem 10"
(→‎Solution)
 
Line 27: Line 27:
 
<math>\boxed{\text{C}}</math>
 
<math>\boxed{\text{C}}</math>
  
−
*For any two fractions that have the property <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> where a is equal for both fractions;
+
*For any two fractions that have the property <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> where <math>a</math> is equal for both fractions;
  
 
Set aside the a. Take the average of b and c. For this example, the average of 3 and 5 is 4. Then multiply those numbers. 3*5=15. The number that is exactly in between <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> will be <math>\frac{a*((b+c)/2)}{b*c}</math>.
 
Set aside the a. Take the average of b and c. For this example, the average of 3 and 5 is 4. Then multiply those numbers. 3*5=15. The number that is exactly in between <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> will be <math>\frac{a*((b+c)/2)}{b*c}</math>.

Latest revision as of 07:34, 6 October 2024

Problem

The fraction halfway between $\frac{1}{5}$ and $\frac{1}{3}$ (on the number line) is

[asy] unitsize(12); draw((-1,0)--(20,0),EndArrow); draw((0,-.75)--(0,.75)); draw((10,-.75)--(10,.75)); draw((17,-.75)--(17,.75)); label("$0$",(0,-.5),S); label("$\frac{1}{5}$",(10,-.5),S); label("$\frac{1}{3}$",(17,-.5),S); [/asy]

$\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{2}{15} \qquad \text{(C)}\ \frac{4}{15} \qquad \text{(D)}\ \frac{53}{200} \qquad \text{(E)}\ \frac{8}{15}$

Solution

The fraction halfway between $\frac{1}{5}$ and $\frac{1}{3}$ is simply their average, which is \begin{align*} \frac{\frac{1}{5}+\frac{1}{3}}{2} &= \frac{\frac{3}{15}+\frac{5}{15}}{2} \\ &= \frac{\frac{8}{15}}{2} \\ &= \frac{4}{15} \\ \end{align*}

$\boxed{\text{C}}$

  • For any two fractions that have the property $\frac{a}{b}$ and $\frac{a}{c}$ where $a$ is equal for both fractions;

Set aside the a. Take the average of b and c. For this example, the average of 3 and 5 is 4. Then multiply those numbers. 3*5=15. The number that is exactly in between $\frac{a}{b}$ and $\frac{a}{c}$ will be $\frac{a*((b+c)/2)}{b*c}$.

In this example, it would be $\frac{1((3+5)/2)}{3*5}$, which simplifies to $\frac{4}{15}$.

Therefore, the answer is $\boxed{\text{C}}$

Edit by mathmagical~

Video Solution

https://youtu.be/BcWDI1TvK-Y

~savannahsolver

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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=1985_AJHSME_Problems/Problem_10&oldid=229086"