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

Revision as of 23:06, 9 June 2011

Problem 1

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 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 nominator, 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 $\boxed{B}$

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

@@ Line 24: / Line 24: @@
 == See also ==
-{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+{{AJHSME box|year=1998|before=First question|num-a=2}}
 * [[AJHSME]]
 * [[AJHSME Problems and Solutions]]
 * [[Mathematics competition resources]]

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

Revision as of 23:06, 9 June 2011

Contents

Problem 1

Solution 1

Solution 2

See also