2010 AMC 10A Problems/Problem 6

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
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

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

Page

Toolbox

Search

2010 AMC 10A Problems/Problem 6

Contents

Problem 6

Solution

Solution 2

Video Solution

See Also