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Difference between revisions of "2007 AMC 10A Problems/Problem 2"

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<math>\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}</math>
 
<math>\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}</math>
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== Solution ==
 
== Solution ==
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<cmath>\frac{6 @ 2}{6 \# 2} = \frac{(6)\times (2) - (2)^2}{(6) + (2) - (6) \cdot (2)^2} = \frac{8}{-16} = \frac{-1}{2} \Rightarrow \mathrm{(A)}</cmath>
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<math>6@2</math> must be equal to <math>6*2-2^2</math> which is 8. <math>6\# 2</math> is equal to <math>6+2-6*2^2</math> which is <math>8-24 = -16</math>. Therefore <math>\frac{6@2}{6\# 2}</math> must be equal to <math>\frac{8}{-16} = -\frac{1}{2}</math>. Therefore the solution is <math>A</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 14:07, 27 January 2021

Problem

Define $a@b = ab - b^{2}$ and $a\#b = a + b - ab^{2}$. What is $\frac {6@2}{6\#2}$?

$\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}$

Solution

$6@2$ must be equal to $6*2-2^2$ which is 8. $6\# 2$ is equal to $6+2-6*2^2$ which is $8-24 = -16$. Therefore $\frac{6@2}{6\# 2}$ must be equal to $\frac{8}{-16} = -\frac{1}{2}$. Therefore the solution is $A$.

See also

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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