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

Latest revision as of 17:58, 31 August 2022

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

Solution 2

$\spadesuit (x, y) \text{is defined as } x - \frac{1}{y} \text{. Hence } \spadesuit (2,\spadesuit(2, 2)) =2 - \frac{1}{\spadesuit (2, 2)} = 2 - \frac{1}{2 - \frac{1}{2}}=2-\frac{1}{\frac{3}{2}}=2-\frac{2}{3}=\frac{4}{3}\Longrightarrow \boxed{\textbf{(C) } \frac{4}{3}}$

Video Solution

https://youtu.be/P7rGLXp_6es

~IceMatrix

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 ==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>

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