1992 AHSME Problems/Problem 1: Difference between revisions

Latest revision as of 16:20, 18 August 2020

If $3(4x+5\pi)=P$ then $6(8x+10\pi)=$

$\text{(A) } 2P\quad \text{(B) } 4P\quad \text{(C) } 6P\quad \text{(D) } 8P\quad \text{(E) } 18P$

We can see that $8x+10\pi$ is equal to $2(4x+5\pi),$ and we know that $2^2 = 4,$ so the answer is $\boxed{B}\, .$

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 2
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All AHSME Problems and Solutions

@@ Line 7: / Line 7: @@
 \text{(D) } 8P\quad
 \text{(E) } 18P</math>
 == Solution ==
-<math>\fbox{B}</math>
+We can see that <math>8x+10\pi</math> is equal to <math>2(4x+5\pi),</math> and we know that <math>2^2 = 4,</math> so the answer is <math>\boxed{B}\, .</math>
-<math>6(4x+5\pi) = 2 \cdot 3(4x+5\pi) = 2 \cdot P</math>
-<math>6(8x+10\pi) = 2 \cdot  6(4x+5\pi) = 2 \cdot 2P = \fbox{4P}</math>
 == See also ==