Difference between revisions of "1994 AJHSME Problems/Problem 13"
(→Uh Oh Unluky 13!) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | |||
| + | |||
The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is | The number halfway between <math>\dfrac{1}{6}</math> and <math>\dfrac{1}{4}</math> is | ||
Latest revision as of 12:24, 30 October 2016
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
