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

Latest revision as of 20:07, 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 $\fbox{E}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == Solution ==
-Note that <math>3 \circ y = 4(3) - 3(y) + (3)(y) = 12 - 3y + 3y = 12</math> for all values of <math>y</math>. This implies that the answer is <math>\fbox{E}</math>, more than four solutions.
+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>
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Difference between revisions of "1993 AHSME Problems/Problem 4"

Latest revision as of 20:07, 27 May 2021

Problem

Solution

See also