Difference between revisions of "1998 AJHSME Problems/Problem 1"
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<math>\text{(A)}\ \dfrac{6}{x} \qquad \text{(B)}\ \dfrac{6}{x+1} \qquad \text{(C)}\ \dfrac{6}{x-1} \qquad \text{(D)}\ \dfrac{x}{6} \qquad \text{(E)}\ \dfrac{x+1}{6}</math> | <math>\text{(A)}\ \dfrac{6}{x} \qquad \text{(B)}\ \dfrac{6}{x+1} \qquad \text{(C)}\ \dfrac{6}{x-1} \qquad \text{(D)}\ \dfrac{x}{6} \qquad \text{(E)}\ \dfrac{x+1}{6}</math> | ||
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
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From here, we can see that the smallest is answer choice <math>\boxed{B}</math> | From here, we can see that the smallest is answer choice <math>\boxed{B}</math> | ||
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+ | ===Solution 2=== | ||
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+ | Note that <math>\dfrac{6}{x+1}<\dfrac{6}{x}<\dfrac{6}{x-1}</math> (for <math>x>1</math>) and <math>\dfrac{x}{6}<\dfrac{x+1}{6}</math>. Therefore, we just need to compare <math>\dfrac{6}{x+1}</math> and <math>\dfrac{x}{6}</math>. Plugging in <math>x=7</math>, we get <math>\dfrac{3}{4}</math> and <math>\dfrac{7}{6}</math>, respectively, with <math>\dfrac{3}{4}<\dfrac{7}{6}</math>. Thus, the answer is <math>\boxed{(B) \dfrac{6}{x+1}}</math>. | ||
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+ | ~By Leon0168 | ||
== See also == | == See also == |
Latest revision as of 09:04, 11 October 2025
Problem
For , which of the following is the smallest?
Solutions
Solution 1
Plugging in for every answer choice would give
From here, we can see that the smallest is answer choice
Solution 2
Note that (for
) and
. Therefore, we just need to compare
and
. Plugging in
, we get
and
, respectively, with
. Thus, the answer is
.
~By Leon0168
See also
1998 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by First question |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.