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1998 AJHSME Problems/Problem 1

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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} ...

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} \qqu...

From here, we can see that the smallest is answer choice \boxed{B}

See also

1998 AJHSME (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
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