Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with '4/3')
 
Line 1: Line 1:
−
4/3
+
== Problem 6 ==
 +
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>
 +
 
 +
What is <math>\spadesuit(2,\spadesuit(2, 2))</math>?
 +
 
 +
<math>
 +
\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
 +
</math>
 +
 
 +
==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>
 +
The answer is <math>\boxed{C}</math>

Revision as of 15:50, 20 December 2010

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 $\boxed{C}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_6&oldid=36196"