Difference between revisions of "1992 AHSME Problems/Problem 1"

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== Solution ==
 
== Solution ==
 
<math>\fbox{B}</math>
 
<math>\fbox{B}</math>
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<math>6(4x+5\pi) = 2 \cdot 3(4x+5\pi) = 2 \cdot P</math>
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<math>6(8x+10\pi) = 2 \cdot  6(4x+5\pi) = 2 \cdot 2P</math>
  
 
== See also ==
 
== See also ==

Revision as of 20:37, 2 March 2017

Problem

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

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

Solution

$\fbox{B}$

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

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

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
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