Difference between revisions of "1998 AJHSME Problems/Problem 1"
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In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is <math>x+1</math> or <math>8</math> | In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is <math>x+1</math> or <math>8</math> | ||
− | The smaller would go on the numerator, which is <math> | + | The smaller would go on the numerator, which is <math>6</math>. |
The answer choice with <math>\frac{6}{x+1}</math> is <math>\boxed{B}</math> | The answer choice with <math>\frac{6}{x+1}</math> is <math>\boxed{B}</math> | ||
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== See also == | == See also == | ||
− | {{AJHSME box|year=1998|before= | + | {{AJHSME box|year=1998|before=First question|num-a=2}} |
* [[AJHSME]] | * [[AJHSME]] | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:51, 31 October 2016
Problem
For , which of the following is the smallest?
Solution
Solution 1
The smallest fraction would be in the form where is larger than .
In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is or
The smaller would go on the numerator, which is .
The answer choice with is
Solution 2
Plugging in for every answer choice would give
From here, we can see that the smallest is answer choice
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.