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 | + | For positive numbers <math>x</math> and <math>y</math> the operation <math>\spadesuit (x,y)</math> is defined as |
− | <cmath> | + | <cmath>\spadesuit (x,y) = x-\dfrac{1}{y}</cmath> |
− | What is <math>\spadesuit | + | What is <math>\spadesuit (2,\spadesuit (2,2))</math>? |
<math> | <math> | ||
Line 19: | Line 19: | ||
==Solution== | ==Solution== | ||
− | <math>\spadesuit | + | <math>\spadesuit (2,2) =2-\frac{1}{2} =\frac{3}{2}</math>. Then, <math>\spadesuit \left(2,\frac{3}{2}\right)</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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | <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)} = | ||
+ | 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== | ||
+ | https://youtu.be/P7rGLXp_6es | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2010|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 18:58, 31 August 2022
Problem 6
For positive numbers and
the operation
is defined as
What is ?
Solution
. Then,
is
The answer is
Solution 2
Video Solution
~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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