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

(Solution)
m (Solution)
 
(3 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
<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>
 
== Solution ==
 
== 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 8/-16 = -1/2. Therefore the solution is E.
+
<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 ==

Latest revision as of 15:07, 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 (ProblemsAnswer KeyResources)
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

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