Difference between revisions of "1999 AHSME Problems/Problem 3"
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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> | <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> | ||
| − | ==Solution 1== | + | ==Solutions== |
| + | ===Solution 1=== | ||
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> | ||
| − | ==Solution 2== | + | ===Solution 2=== |
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>. | 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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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>. | 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 13:36, 13 February 2019
Problem
The number halfway between 
and 
is
Solutions
Solution 1
To find the number halfway between 
and 
, simply take the arithmetic mean, which is
Thus the answer is choice 
Solution 2
Note that 
and 
. Thus, the answer must be greater than .
Answers 
, 
, and 
are all less than , so they can be eliminated.
Answer 
is equivalent to 
, which is 
away from 
, and is 
away from 
. These distances are not equal, eliminating .
Thus, 
must be the answer. Computing 
as a check, we see that it is 
away from , and similarly it is 
away from .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
