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

Revision as of 10:57, 4 July 2013

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 $\boxed{C}$

2010 AMC 10A (Problems • Answer Key • Resources)
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2010 AMC 10A Problems/Problem 6: Difference between revisions

Revision as of 10:57, 4 July 2013

Problem 6

Solution

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