Difference between revisions of "2010 AMC 10A Problems/Problem 6"
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| − | 4/3 | + | == Problem 6 == |
| + | For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit (x,y)</math> is defined as | ||
| + | |||
| + | <cmath>\spadesuit (x,y) = x-\dfrac{1}{y}</cmath> | ||
| + | |||
| + | What is <math>\spadesuit (2,\spadesuit (2,2))</math>? | ||
| + | |||
| + | <math> | ||
| + | \mathrm{(A)}\ \dfrac{2}{3} | ||
| + | \qquad | ||
| + | \mathrm{(B)}\ 1 | ||
| + | \qquad | ||
| + | \mathrm{(C)}\ \dfrac{4}{3} | ||
| + | \qquad | ||
| + | \mathrm{(D)}\ \dfrac{5}{3} | ||
| + | \qquad | ||
| + | \mathrm{(E)}\ 2 | ||
| + | </math> | ||
| + | |||
| + | ==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> | ||
| + | The answer is <math>\boxed{C}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/P7rGLXp_6es | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2010|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 06:39, 26 May 2020
Contents
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 
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. 
