1985 AJHSME Problems/Problem 10

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~

See Also

1985 AJHSME (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS