1987 AJHSME Problems/Problem 4: Difference between revisions

Revision as of 12:04, 25 May 2009

Martians measure angles in clerts. There are $500$ clerts in a full circle. How many clerts are there in a right angle?

$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250$

The right angle is $1/4$ of the circle, hence it contains $500/4=125\rightarrow \boxed{\text{C}}$ clerts.

1987 AJHSME (Problems • Answer Key • Resources)
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 AJHSME/AMC 8 Problems and Solutions

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 ==Solution==
-The right angle is <math>1/4</math> of the circle, hence it contains <math>500/4=\boxed{125}</math> clerts.
+The right angle is <math>1/4</math> of the circle, hence it contains <math>500/4=125\rightarrow \boxed{\text{C}}</math> clerts.
 ==See Also==
-[[1987 AJHSME Problems]]
+{{AJHSME box|year=1987|num-b=3|num-a=5}}
+[[Category:Introductory Geometry Problems]]