Difference between revisions of "1988 AJHSME Problems/Problem 1"
(→Solution) |
(→Solution) |
||
| Line 23: | Line 23: | ||
Looking, we see that the arrow is closer to <math>10.3</math>, so <math>\boxed{\text{D}}</math> | Looking, we see that the arrow is closer to <math>10.3</math>, so <math>\boxed{\text{D}}</math> | ||
| − | Each part of the semicircle is one | + | |
| + | |||
| + | |||
| + | Each part of the semicircle is one fourth, so we see that the answer is between <math>10.25</math> and <math>10.5</math>. So, we have <math>\boxed{\text{D}}</math> is our answer --- stjwyl | ||
==See Also== | ==See Also== | ||
Revision as of 15:13, 28 April 2021
Problem
The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of
![[asy] draw((-3,0)..(0,3)..(3,0)); draw((-3.5,0)--(-2.5,0)); draw((0,2.5)--(0,3.5)); draw((2.5,0)--(3.5,0)); draw((1.8,1.8)--(2.5,2.5)); draw((-1.8,1.8)--(-2.5,2.5)); draw((0,0)--3*dir(120),EndArrow); label("$10$",(-2.6,0),E); label("$11$",(2.6,0),W); [/asy]](http://latex.artofproblemsolving.com/e/e/a/eea86891575ef6a6c7c90ff41c59a2a847689a3b.png)
Solution
Clearly the arrow marks a value between 
and 
, so only 
and 
are possible.
Looking, we see that the arrow is closer to 
, so 
Each part of the semicircle is one fourth, so we see that the answer is between and . So, we have is our answer --- stjwyl
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
