Difference between revisions of "1987 AJHSME Problems/Problem 4"

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==Problems==
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==Problem==
  
 
Martians measure angles in clerts.  There are <math>500</math> clerts in a full circle.  How many clerts are there in a right angle?
 
Martians measure angles in clerts.  There are <math>500</math> clerts in a full circle.  How many clerts are there in a right angle?
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<math>\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250</math>
 
<math>\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 125 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 250</math>
  
==Solution==
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==Solution 1 (Ratio)==
  
 
The right angle is <math>1/4</math> of the circle, hence it contains <math>500/4=125\rightarrow \boxed{\text{C}}</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.
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{{AJHSME box|year=1987|num-b=3|num-a=5}}
 
{{AJHSME box|year=1987|num-b=3|num-a=5}}
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:17, 16 July 2024

Problem

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

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

Solution 1 (Ratio)

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

See Also

1987 AJHSME (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 AJHSME/AMC 8 Problems and Solutions

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