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Difference between revisions of "1998 AJHSME Problems/Problem 1"

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==Problem 1==
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==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>
  
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==Solution 1==
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==Solution==
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===Solution 1===
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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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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>
 
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The smaller would go on the numerator, which is <math>6</math>.
 
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The answer choice with <math>\frac{6}{x+1}</math> is <math>\boxed{B}</math>
 
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==Solution 2==
 
  
 
Plugging <math>x</math> in for every answer choice would give
 
Plugging <math>x</math> in for every answer choice would give
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 16:27, 25 November 2022

Problem

For $x=7$, which of the following is the smallest?

$\text{(A)}\ \dfrac{6}{x} \qquad \text{(B)}\ \dfrac{6}{x+1} \qquad \text{(C)}\ \dfrac{6}{x-1} \qquad \text{(D)}\ \dfrac{x}{6} \qquad \text{(E)}\ \dfrac{x+1}{6}$

Solution

Solution 1

Plugging $x$ in for every answer choice would give

$\text{(A)}\ \dfrac{6}{7} \qquad \text{(B)}\ \dfrac{6}{8} \qquad \text{(C)}\ \dfrac{6}{6} \qquad \text{(D)}\ \dfrac{7}{6} \qquad \text{(E)}\ \dfrac{8}{6}$

From here, we can see that the smallest is answer choice $\boxed{B}$

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. AMC logo.png

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