Difference between revisions of "1998 AJHSME Problems/Problem 1"
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| − | ==Problem | + | ==Problem== |
For <math>x=7</math>, which of the following is the smallest? | For <math>x=7</math>, which of the following is the smallest? | ||
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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> | ||
| − | ==Solution 1== | + | ==Solution== |
| + | ===Solution 1=== | ||
The smallest fraction would be in the form <math>\frac{a}{b}</math> where <math>b</math> is larger than <math>a</math>. | The smallest fraction would be in the form <math>\frac{a}{b}</math> where <math>b</math> is larger than <math>a</math>. | ||
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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> | ||
| − | ==Solution 2== | + | ===Solution 2=== |
Plugging <math>x</math> in for every answer choice would give | Plugging <math>x</math> in for every answer choice would give | ||
Revision as of 11:48, 23 December 2012
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 | ||
