Difference between revisions of "1999 AHSME Problems/Problem 3"
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+ | == Problem== | ||
+ | The number halfway between <math>1/8</math> and <math>1/10</math> is | ||
+ | |||
+ | <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> | ||
+ | |||
+ | ==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 | ||
Line 4: | Line 12: | ||
Thus the answer is choice <math>\boxed{E}.</math> | Thus the answer is choice <math>\boxed{E}.</math> | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
==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 14:36, 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 |
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