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}

6x1\dfrac{6}{x-1}

x6\dfrac{x}{6}

x+16\dfrac{x+1}{6}

Answer: B
Concepts:substitutionfraction
Difficulty rating: 450
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 .

Full Exam

Problem 1 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8