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

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==Solution==
 
==Solution==
===Solution 1===
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=== Solution 1 ===  
 
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Thisone
The smallest fraction would be in the form <math>\frac{a}{b}</math> where <math>b</math> is larger than <math>a</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>6</math>.
 
 
 
The answer choice with <math>\frac{6}{x+1}</math> is <math>\boxed{B}</math>
 
  
 
===Solution 2===
 
===Solution 2===

Revision as of 18:09, 24 December 2020

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

Thisone

Solution 2

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 (ProblemsAnswer KeyResources)
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

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