Difference between revisions of "1993 AHSME Problems/Problem 4"

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== Solution ==
 
== Solution ==
  
Note that <math>3 \circ y = 4 \cdot 3 - 3y + 3y = 12 - 3y + 3y = 12</math>, which is satisfied for all values of <math>y</math>. Thus there are more than four solutions, and the answer is <math>\fbox{E}</math>
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Note that <math>3 \circ y = 4 \cdot 3 - 3y + 3y = 12</math>, so <math>3 \circ y = 12</math> is true for all values of <math>y</math>. Thus there are more than four solutions, and the answer is <math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 21:07, 27 May 2021

Problem

Define the operation "$\circ$" by $x\circ y=4x-3y+xy$, for all real numbers $x$ and $y$. For how many real numbers $y$ does $3\circ y=12$?

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) more than 4}$

Solution

Note that $3 \circ y = 4 \cdot 3 - 3y + 3y = 12$, so $3 \circ y = 12$ is true for all values of $y$. Thus there are more than four solutions, and the answer is $\fbox{E}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 26 27 28 29 30
All AHSME Problems and Solutions

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