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

Latest revision as of 16:20, 18 August 2020

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

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

We can see that $8x+10\pi$ is equal to $2(4x+5\pi),$ and we know that $2^2 = 4,$ so the answer is $\boxed{B}\, .$

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 7: / Line 7: @@
 \text{(D) } 8P\quad
 \text{(E) } 18P</math>
 == Solution ==
-<math>\fbox{B}</math>
+We can see that <math>8x+10\pi</math> is equal to <math>2(4x+5\pi),</math> and we know that <math>2^2 = 4,</math> so the answer is <math>\boxed{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 = \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 ==