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 | + | 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> | ||
Revision as of 09:53, 31 July 2011
Contents
Problem 1
For 
, which of the following is the smallest?
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 | ||
