1990 AMC 8 Problem 14

Below is the professionally curated solution for Problem 14 of the 1990 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 8 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:basic probability ratio and proportion

Difficulty rating: 860

14.

A bag contains only blue balls and green balls. There are $6$ blue balls. If the probability of drawing a blue ball at random from this bag is $\dfrac14,$ then the number of green balls in the bag is

$12$

$18$

$24$

$30$

$36$

Solution:

Since blue balls make up $\dfrac14$ of the bag and there are $6$ of them, the total is $6 \cdot 4 = 24$ balls.

So the number of green balls is $24 - 6 = 18$ .

Thus, the correct answer is B .

← Problem 13 Full Exam Problem 15 →

Problem 14 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 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