Difference between revisions of "1994 AJHSME Problems/Problem 13"
Mrdavid445 (talk | contribs) (Created page with "==Problem== The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is <math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(...") |
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<math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math> | <math>\text{(A)}\ \dfrac{1}{10} \qquad \text{(B)}\ \dfrac{1}{5} \qquad \text{(C)}\ \dfrac{5}{24} \qquad \text{(D)}\ \dfrac{7}{24} \qquad \text{(E)}\ \dfrac{5}{12}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | The number halfway between is the average. | ||
| + | |||
| + | <cmath>\frac{\frac16 + \frac14}{2} = \frac{\frac{2}{12} + \frac{3}{12}}{2} = \boxed{\text{(C)}\ \frac{5}{24}}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1994|num-b=12|num-a=14}} | ||
Revision as of 00:43, 23 December 2012
Problem
The number halfway between 
and 
is
Solution
The number halfway between is the average.
![\[\frac{\frac16 + \frac14}{2} = \frac{\frac{2}{12} + \frac{3}{12}}{2} = \boxed{\text{(C)}\ \frac{5}{24}}\]](http://latex.artofproblemsolving.com/3/2/b/32b1376161b36c5c2a79eed29608b8bf71115fc6.png)
See Also
| 1994 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
