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Difference between revisions of "1999 AHSME Problems/Problem 3"

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== Problem==
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The number halfway between <math>1/8</math> and <math>1/10</math> is
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<math> \mathrm{(A) \  } \frac 1{80} \qquad \mathrm{(B) \  } \frac 1{40} \qquad \mathrm{(C) \  } \frac 1{18} \qquad \mathrm{(D) \  } \frac 1{9} \qquad \mathrm{(E) \  } \frac 9{80} </math>
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==Solution 1==
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To find the number halfway between <math>\frac{1}{8}</math> and <math>\frac{1}{10}</math>, simply take the arithmetic mean, which is
 
To find the number halfway between <math>\frac{1}{8}</math> and <math>\frac{1}{10}</math>, simply take the arithmetic mean, which is
  
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Thus the answer is choice <math>\boxed{E}.</math>
 
Thus the answer is choice <math>\boxed{E}.</math>
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==Solution 2==
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Note that <math>\frac{1}{10} = 0.1</math> and <math>\frac{1}{8} = 0.125</math>.  Thus, the answer must be greater than <math>\frac{1}{10}</math>. 
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Answers <math>A</math>, <math>B</math>, and <math>C</math> are all less than <math>\frac{1}{10}</math>, so they can be eliminated. 
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Answer <math>D</math> is equivalent to <math>0.\overline{1}</math>, which is <math>0.0\overline{1}</math> away from <math>0.1</math>, and is <math>0.013\overline{8}</math> away from <math>0.125</math>.  These distances are not equal, eliminating <math>D</math>.
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Thus, <math>\boxed{E}</math> must be the answer.  Computing <math>\frac{9}{80} = 0.1125</math> as a check, we see that it is <math>0.1125 - 0.1 = 0.0125</math> away from <math>\frac{1}{10}</math>, and similarly it is <math>0.125 - 0.1125 = 0.0125</math> away from <math>\frac{1}{8}</math>.
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==See Also==
 
==See Also==
  
 
{{AHSME box|year=1999|num-b=2|num-a=4}}
 
{{AHSME box|year=1999|num-b=2|num-a=4}}

Revision as of 12:31, 8 August 2011

Problem

The number halfway between $1/8$ and $1/10$ is

$\mathrm{(A) \ } \frac 1{80} \qquad \mathrm{(B) \ } \frac 1{40} \qquad \mathrm{(C) \ } \frac 1{18} \qquad \mathrm{(D) \ } \frac 1{9} \qquad \mathrm{(E) \ } \frac 9{80}$

Solution 1

To find the number halfway between $\frac{1}{8}$ and $\frac{1}{10}$, simply take the arithmetic mean, which is

$\frac{\frac{1}{8}+\frac{1}{10}}{2}=\frac{\frac{9}{40}}{2}=\frac{9}{80}.$

Thus the answer is choice $\boxed{E}.$

Solution 2

Note that $\frac{1}{10} = 0.1$ and $\frac{1}{8} = 0.125$. Thus, the answer must be greater than $\frac{1}{10}$.

Answers $A$, $B$, and $C$ are all less than $\frac{1}{10}$, so they can be eliminated.

Answer $D$ is equivalent to $0.\overline{1}$, which is $0.0\overline{1}$ away from $0.1$, and is $0.013\overline{8}$ away from $0.125$. These distances are not equal, eliminating $D$.

Thus, $\boxed{E}$ must be the answer. Computing $\frac{9}{80} = 0.1125$ as a check, we see that it is $0.1125 - 0.1 = 0.0125$ away from $\frac{1}{10}$, and similarly it is $0.125 - 0.1125 = 0.0125$ away from $\frac{1}{8}$.


See Also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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