Difference between revisions of "1985 AJHSME Problems/Problem 10"
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<math>\boxed{\text{C}}</math> | <math>\boxed{\text{C}}</math> | ||
− | *For any two fractions that have the property <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> where a is equal for both fractions; | + | *For any two fractions that have the property <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> where <math>a</math> 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 <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> will be <math>\frac{a*((b+c)/2)}{b*c}</math>. | 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 <math>\frac{a}{b}</math> and <math>\frac{a}{c}</math> will be <math>\frac{a*((b+c)/2)}{b*c}</math>. | ||
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Therefore, the answer is <math>\boxed{\text{C}}</math> | Therefore, the answer is <math>\boxed{\text{C}}</math> | ||
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+ | Edit by mathmagical~ | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/BcWDI1TvK-Y | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 07:34, 6 October 2024
Contents
Problem
The fraction halfway between and (on the number line) is
Solution
The fraction halfway between and is simply their average, which is
- For any two fractions that have the property and where 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.