Difference between revisions of "2010 AMC 10A Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | <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> | + | <math>\spadesuit (2,2) =2-\frac{1}{2} =\frac{3}{2}</math>. Then, <math>\spadesuit \left(2,\frac{3}{2}\right)</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> | ||
| + | |||
| + | ==Solution 2== | ||
| + | <cmath>\spadesuit (x, y) \text{is defined as } x - \frac{1}{y} \text{. Hence } \spadesuit (2,\spadesuit(2, 2)) =2 - \frac{1}{\spadesuit (2, 2)} = | ||
| + | 2 - \frac{1}{2 - \frac{1}{2}}=2-\frac{1}{\frac{3}{2}}=2-\frac{2}{3}=\frac{4}{3}\Longrightarrow \boxed{\textbf{(C) } \frac{4}{3}}</cmath> | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 17:58, 31 August 2022
Problem 6
For positive numbers 
and 
the operation 
is defined as
![\[\spadesuit (x,y) = x-\dfrac{1}{y}\]](http://latex.artofproblemsolving.com/2/9/c/29c8e1b1ec789c2d4943123570f9dccf63d5c69d.png)
What is 
?
Solution
. Then, 
is 
The answer is 
Solution 2
![\[\spadesuit (x, y) \text{is defined as } x - \frac{1}{y} \text{. Hence } \spadesuit (2,\spadesuit(2, 2)) =2 - \frac{1}{\spadesuit (2, 2)} = 2 - \frac{1}{2 - \frac{1}{2}}=2-\frac{1}{\frac{3}{2}}=2-\frac{2}{3}=\frac{4}{3}\Longrightarrow \boxed{\textbf{(C) } \frac{4}{3}}\]](http://latex.artofproblemsolving.com/6/4/0/640f9a729e2bc1e9b42281e136ec705d57892f30.png)
Video Solution
~IceMatrix
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. 
