1989 AMC 8 Problem 8 — Solution | LIVE by Po-Shen Loh

LIVE

English

1989 AMC 8 Problem 8

Attempt Problem 8 of the 1989 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 1989 AMC 8 solutions, or check the answer key.

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

8.

What is the value of

$(2 \times 3 \times 4)\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)?$

$1$

$3$

$9$

$24$

$26$

Answer: E

Concepts:distributive property fraction

Difficulty rating: 820

Solution:

Since $2 \times 3 \times 4 = 24,$ distribute it over the sum: $24 \cdot \frac{1}{2} + 24 \cdot \frac{1}{3} + 24 \cdot \frac{1}{4}.$

This equals $12 + 8 + 6 = 26.$

Thus, the correct answer is E .

← Problem 7#7

Problem 9#9 →

Problem 8 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 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 · 1998 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