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

(Solution: 1)
(Solution 2)
Line 14: Line 14:
  
 
<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>
 
== Solution 2 ==
 
<math>\fbox{B}</math>
 
 
<math>4(3x+5\pi) = (4 \cdot 3x) + (4 \cdot 5\pi) = 12x + 20\pi = P. </math>
 
 
 
<math>6(8x+10\pi)= (6 \cdot 8x) + (6 \cdot 10\pi) = 48x + 60\pi = 4P.</math>
 
 
So the answer is <math>\fbox{B}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 20:25, 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

$\fbox{B}$

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

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

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png