1992 AHSME Problems/Problem 1: Difference between revisions

Revision as of 20:20, 9 April 2019

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

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

$\fbox{B}$

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

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

$\fbox{B}$

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

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

So the answer is $\fbox{B}$

1992 AHSME (Problems • Answer Key • Resources)
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All AHSME Problems and Solutions

@@ Line 14: / Line 14: @@
 <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 ==