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

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
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

@@ Line 10: / Line 10: @@
 == 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 ==