Difference between revisions of "1988 AJHSME Problems/Problem 1"
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− | 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>. THe arrow is pointing a little in front of The middle of <math>10.25</math> and <math>10.5</math>.Therefore, we have <math>\boxed{\text{D}}</math> is our answer --- stjwyl | + | 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>. THe arrow is pointing a little in front of The middle of <math>10.25</math> and <math>10.5</math>.Therefore, we have that <math>\boxed{\text{D}}</math> is our answer --- stjwyl |
==See Also== | ==See Also== |
Revision as of 16:16, 28 April 2021
Problem
The diagram shows part of a scale of a measuring device. The arrow indicates an approximate reading of
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
. THe arrow is pointing a little in front of The middle of
and
.Therefore, we have that
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.