Difference between revisions of "2002 AMC 12A Problems/Problem 4"

m (Solution)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
<math>4(90-x)=(180-x)</math>
+
We can create an equation for the question, <math>4(90-x)=(180-x)</math>
  
 
<math>360-4x=180-x</math>
 
<math>360-4x=180-x</math>
Line 14: Line 14:
 
<math>3x=180</math>
 
<math>3x=180</math>
  
<math>x=60 \Rightarrow \mathrm {(B)}</math>
+
After simplifying, we get <math>x=60 \Rightarrow \mathrm {(B)}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2002|ab=A|num-b=3|num-a=5}}
 
{{AMC12 box|year=2002|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:43, 7 April 2014

Problem

Find the degree measure of an angle whose complement is 25% of its supplement.

$\mathrm{(A) \ 48 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 75 } \qquad \mathrm{(D) \ 120 } \qquad \mathrm{(E) \ 150 }$


Solution

We can create an equation for the question, $4(90-x)=(180-x)$

$360-4x=180-x$

$3x=180$

After simplifying, we get $x=60 \Rightarrow \mathrm {(B)}$

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
All AMC 12 Problems and Solutions

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

Invalid username
Login to AoPS