Difference between revisions of "1985 AJHSME Problem 10"

(Created page with "== 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); d...")
 
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unitsize(12);
 
unitsize(12);
 
draw((-1,0)--(20,0),EndArrow);
 
draw((-1,0)--(20,0),EndArrow);
draw((0,-.75)--(0,.75));
+
draw((0,-.25)--(0,.25));
draw((10,-.75)--(10,.75));
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draw((10,-.25)--(10,.25));
draw((17,-.75)--(17,.75));
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draw((17,-.25)--(17,.25));
 
label("$0$",(0,-.5),S);
 
label("$0$",(0,-.5),S);
 
label("$\frac{1}{5}$",(10,-.5),S);
 
label("$\frac{1}{5}$",(10,-.5),S);

Latest revision as of 23:07, 15 June 2025

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,-.25)--(0,.25)); draw((10,-.25)--(10,.25)); draw((17,-.25)--(17,.25)); 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

$\frac{1}{5} = \frac{3}{15}$ and $\frac{1}{3} = \frac{5}{15}.$ The fraction in between is $\boxed{\text{(C)} \quad \frac{4}{15}}.$