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Difference between revisions of "1988 AJHSME Problems/Problem 1"

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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>
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Each part of the semicircle is one fourth of it, 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==
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{{AJHSME box|year=1988|before=First<br>Problem|num-a=2}}
 
{{AJHSME box|year=1988|before=First<br>Problem|num-a=2}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 15:16, 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]

$\text{(A)}\ 10.05 \qquad \text{(B)}\ 10.15 \qquad \text{(C)}\ 10.25 \qquad \text{(D)}\ 10.3 \qquad \text{(E)}\ 10.6$

Solution

Clearly the arrow marks a value between $10.25$ and $10.5$, so only $\text{C}$ and $\text{D}$ are possible.

Looking, we see that the arrow is closer to $10.3$, so $\boxed{\text{D}}$



Each part of the semicircle is one fourth of it, so we see that the answer is between $10.25$ and $10.5$. THe arrow is pointing a little in front of The middle of $10.25$ and $10.5$.Therefore, we have that $\boxed{\text{D}}$ 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. AMC logo.png

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