Difference between revisions of "1985 AJHSME Problems/Problem 10"
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Therefore, the answer is <math>\boxed{\text{C}}</math> | Therefore, the answer is <math>\boxed{\text{C}}</math> | ||
| + | Edit by mathmagical~ | ||
==See Also== | ==See Also== | ||
Revision as of 20:09, 31 August 2019
Problem
The fraction halfway between 
and 
(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]](http://latex.artofproblemsolving.com/3/2/2/322c90d5df68225b0db4a2f8a9b71d5be2891561.png)
Solution
The fraction halfway between and is simply their average, which is
- For any two fractions that have the property
and 
where a is equal for both fractions;
Set aside the a. Take the average of b and c. For this example, the average of 3 and 5 is 4. Then multiply those numbers. 3*5=15. The number that is exactly in between 
and 
will be 
.
In this example, it would be 
, which simplifies to 
.
Therefore, the answer is
Edit by mathmagical~
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
