1992 AHSME Problems/Problem 1: Difference between revisions

Revision as of 19:37, 2 March 2017

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

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

$\fbox{B}$

$6(4x+5\pi) = 2 \cdot 3(4x+5\pi) = 2 \cdot P$

$6(8x+10\pi) = 2 \cdot 6(4x+5\pi) = 2 \cdot 2P$

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

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 == Solution ==
 <math>\fbox{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</math>
 == See also ==