1985 AJHSME Problems/Problem 10: Difference between revisions
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==Video Solution== | |||
https://youtu.be/BcWDI1TvK-Y | |||
~savannahsolver | |||
==See Also== | ==See Also== | ||
Revision as of 08:24, 10 January 2023
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~
Video Solution
~savannahsolver
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
