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

Revision as of 08:36, 11 October 2025

Problem

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

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

Solutions

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 $\boxed{B}$

Solution 2

Note that $\frac{6}{x+1}<\frac{6}{x}<\frac{6}{x-1}$ (for $x>1$ ) and $\frac{x}{6}<\frac{x+1}{6}$ . Therefore, we just need to compare $\frac{6}{x+1}$ and $\frac{x}{6}$ . Plugging in $x=7$ , we get $\frac{3}{4}$ and $\frac{7}{6}$ , respectively, with $\frac{3}{4}<\frac{7}{6}$ . Thus, the answer is $\boxed{(B) \frac{6}{x+1}}$ .

~By Leon0168

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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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>
-==Solution==
+==Solutions==
 ===Solution 1===

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