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

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The answer is <math>\boxed{C}</math>
The answer is <math>\boxed{C}</math>


==Video Solution==
https://youtu.be/P7rGLXp_6es
~IceMatrix


== See Also ==
== See Also ==

Revision as of 05:39, 26 May 2020

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 $\boxed{C}$

Video Solution

https://youtu.be/P7rGLXp_6es

~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

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