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

## 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

The smallest fraction would be in the form $\frac{a}{b}$ where $b$ is larger than $a$.

In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is $x+1$ or $8$

The smaller would go on the numerator, which is $6$.

The answer choice with $\frac{6}{x+1}$ is $\boxed{B}$

### 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 $\boxed{B}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 