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1985 AJHSME Problems/Problem 10: Difference between revisions

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==See Also==
==See Also==


[[1985 AJHSME Problems]]
{{AJHSME box|year=1985|num-b=9|num-a=11}}
[[Category:Introductory Algebra Problems]]

Revision as of 18:54, 14 May 2009

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*}

$\boxed{\text{C}}$

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
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