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

Latest revision as of 09:04, 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 $\dfrac{6}{x+1}<\dfrac{6}{x}<\dfrac{6}{x-1}$ (for $x>1$ ) and $\dfrac{x}{6}<\dfrac{x+1}{6}$ . Therefore, we just need to compare $\dfrac{6}{x+1}$ and $\dfrac{x}{6}$ . Plugging in $x=7$ , we get $\dfrac{3}{4}$ and $\dfrac{7}{6}$ , respectively, with $\dfrac{3}{4}<\dfrac{7}{6}$ . Thus, the answer is $\boxed{(B) \dfrac{6}{x+1}}$ .

~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.

@@ Line 1: / Line 1: @@
-==Problem 1==
+==Problem==
 For <math>x=7</math>, which of the following is the smallest?
@@ Line 5: / Line 5: @@
 <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 1==
+==Solutions==
+===Solution 1===
-The smallest fraction would be in the form <math>\frac{a}{b}</math> where <math>b</math> is larger than <math>a</math>.
+Plugging <math>x</math> in for every answer choice would give
-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>
+<math>\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}</math>
-The smaller would go on the numerator, which is <math>6</math>.
+From here, we can see that the smallest is answer choice <math>\boxed{B}</math>
-The answer choice with <math>\frac{6}{x+1}</math> is <math>\boxed{B}</math>
-==Solution 2==
-Plugging <math>x</math> in for every answer choice would give
+===Solution 2===
-<math>\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}</math>
+Note that <math>\dfrac{6}{x+1}<\dfrac{6}{x}<\dfrac{6}{x-1}</math> (for <math>x>1</math>) and <math>\dfrac{x}{6}<\dfrac{x+1}{6}</math>. Therefore, we just need to compare <math>\dfrac{6}{x+1}</math> and <math>\dfrac{x}{6}</math>. Plugging in <math>x=7</math>, we get <math>\dfrac{3}{4}</math> and <math>\dfrac{7}{6}</math>, respectively, with <math>\dfrac{3}{4}<\dfrac{7}{6}</math>. Thus, the answer is <math>\boxed{(B) \dfrac{6}{x+1}}</math>.
-From here, we can see that the smallest is answer choice <math>\boxed{B}</math>
+~By Leon0168
 == See also ==
@@ Line 28: / Line 25: @@
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}

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