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Difference between revisions of "1985 AJHSME Problems/Problem 10"

(New page: ==Problem== The fraction halfway between <math>\frac{1}{5}</math> and <math>\frac{1}{3}</math> (on the number line) is <asy> unitsize(12); draw((-1,0)--(20,0),EndArrow); draw((0,-.75)--(...)
 
Line 18: Line 18:
 
==Solution==
 
==Solution==
  
−
{{Solution}}
+
The fraction halfway between <math>\frac{1}{5}</math> and <math>\frac{1}{3}</math> is simply their [[Arithmetic mean|average]], which is
 +
<cmath>\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*}</cmath>
 +
 
 +
<math>\boxed{\text{C}}</math>
  
 
==See Also==
 
==See Also==
  
 
[[1985 AJHSME Problems]]
 
[[1985 AJHSME Problems]]

Revision as of 22:29, 13 January 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

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