1998 AMC 8 Problem 1

Attempt Problem 1 of the 1998 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

For x=7,x = 7, which of the following is the smallest?
6x\dfrac{6}{x}
6x+1\dfrac{6}{x+1}
6x−1\dfrac{6}{x-1}
x6\dfrac{x}{6}
x+16\dfrac{x+1}{6}
Answer: B
Concepts:substitutionfraction
Small hint:
Substitute x=7x = 7 first.
Big hint:
With equal numerators, compare denominators.
Solution:
Substituting x=7x = 7 gives the five values 67,\dfrac{6}{7}, 68,\dfrac{6}{8}, 1,1, 76,\dfrac{7}{6}, and 43.\dfrac{4}{3}.
The first two are the only values less than 1.1. Because they have the same numerator, the fraction with the larger denominator is smaller. Thus, the correct answer is B .
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