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

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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>
 
<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>
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==Solution 2==
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<cmath>\spadesuit (x, y) \text{is defined as } x - \frac{1}{y} \text{. Hence } \spadesuit (2,\spadesuit(2, 2)) =2 - \frac{1}{\spadesuit (2, 2)} =
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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}}</cmath>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 06:18, 29 June 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, $\spadesuit (2,\frac{3}{2})$ 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

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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