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

(Solution)
(Solution: 1)
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<math>6(8x+10\pi) = 2 \cdot  6(4x+5\pi) = 2 \cdot 2P = \fbox{4P}</math>
 
<math>6(8x+10\pi) = 2 \cdot  6(4x+5\pi) = 2 \cdot 2P = \fbox{4P}</math>
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== Solution 2 ==
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<math>\fbox{B}</math>
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<math>4(3x+5\pi) = (4 \cdot 3x) + (4 \cdot 5\pi) = 12x + 20\pi = P. </math>
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<math>6(8x+10\pi)= (6 \cdot 8x) + (6 \cdot 10\pi) = 48x + 60\pi = 4P.</math>
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So the answer is <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Revision as of 21:20, 9 April 2019

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

$\fbox{B}$

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

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

Solution 2

$\fbox{B}$

$4(3x+5\pi) = (4 \cdot 3x) + (4 \cdot 5\pi) = 12x + 20\pi = P.$


$6(8x+10\pi)= (6 \cdot 8x) + (6 \cdot 10\pi) = 48x + 60\pi = 4P.$

So the answer is $\fbox{B}$

See also

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