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2010 AMC 10A Problems/Problem 6: Difference between revisions

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== 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>

Revision as of 15:50, 20 December 2010

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 $\boxed{C}$

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