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

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== Problem 6 ==
 
== Problem 6 ==
For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit(x, y)</math> is defined as  
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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>
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<cmath>\spadesuit (x,y) = x-\dfrac{1}{y}</cmath>
  
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?
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What is <math>\spadesuit (2,\spadesuit (2,2))</math>?
  
 
<math>
 
<math>
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==Solution==
 
==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>
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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>
 
The answer is <math>\boxed{C}</math>
 
The answer is <math>\boxed{C}</math>
  

Revision as of 21:31, 1 November 2015

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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