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

(Solution)
(Solution)
 
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\text{(E) } 18P</math>
 
\text{(E) } 18P</math>
 
== Solution ==
 
== Solution ==
We can see that <math>8x+10p</math>i is equal to <math>2(4x+5pi),</math> and we know that <math>2^2 = 4,</math> so the answer is <math>\boxed{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>
  
 
== See also ==
 
== See also ==

Latest revision as of 17:20, 18 August 2020

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

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 $\boxed{B}\, .$

See also

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