Difference between revisions of "1987 AJHSME Problems/Problem 4"
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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== | + | ==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. |
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1987|num-b=3|num-a=5}} |
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:17, 16 July 2024
Problem
Martians measure angles in clerts. There are 
clerts in a full circle. How many clerts are there in a right angle?
Solution 1 (Ratio)
The right angle is 
of the circle, hence it contains 
clerts.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
