Difference between revisions of "2010 AMC 10A Problems/Problem 6"
Atmchallenge (talk | contribs) |
|||
Line 1: | Line 1: | ||
== Problem 6 == | == Problem 6 == | ||
− | For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit | + | For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit (x,y)</math> is defined as |
− | <cmath> | + | <cmath>\spadesuit (x,y) = x-\dfrac{1}{y}</cmath> |
− | What is <math>\spadesuit | + | What is <math>\spadesuit (2,\spadesuit (2,2))</math>? |
<math> | <math> | ||
Line 19: | Line 19: | ||
==Solution== | ==Solution== | ||
− | <math>\spadesuit | + | <math>\spadesuit (2,2) =2-\frac{1}{2} =\frac{3}{2}</math>. Then, <math>\spadesuit (2,\frac{3}{2})</math> is <math>2-\frac{1}{\frac{3}{2}} = 2- \frac{2}{3} = \frac{4}{3}</math> |
The answer is <math>\boxed{C}</math> | The answer is <math>\boxed{C}</math> | ||
Revision as of 21:31, 1 November 2015
Problem 6
For positive numbers and the operation is defined as
What is ?
Solution
. Then, is The answer is
See Also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.