Difference between revisions of "2010 AMC 10A Problems/Problem 6"

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<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(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>
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== See Also ==
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{{AMC10 box|year=2010|ab=A|num-b=5|num-a=7}}

Revision as of 19:38, 1 January 2012

Problem 6

For positive numbers $x$ and $y$ the operation $\spadesuit?(x, y)$ is defined as

\[?\spadesuit(x, y) = x -\dfrac{1}{y}\]

What is $\spadesuit?(2,\spadesuit??(2, 2))$?

$\mathrm{(A)}\ \dfrac{2}{3} \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ \dfrac{4}{3} \qquad \mathrm{(D)}\ \dfrac{5}{3} \qquad \mathrm{(E)}\ 2$

Solution

$\spadesuit??(2, 2) = 2 - \frac{1}{2} = \frac{3}{2}$. Then, $\spadesuit??(2, \frac{3}{2})$ is $2-\frac{1}{\frac{3}{2}} = 2- \frac{2}{3} = \frac{4}{3}$ The answer is $\boxed{C}$


See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
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
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