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

Revision as of 17:09, 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 $\boxed{B}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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

@@ Line 6: / Line 6: @@
 ==Solution==
-===Solution 1===
+=== Solution 1 ===
+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===

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