1995 AMC 8 Problem 11
Below is the professionally curated solution for Problem 11 of the 1995 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 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).
Difficulty rating: 1030
11.
Jane can walk any distance in half the time it takes Hector to walk the same distance. They set off in opposite directions around the outside of the -block area as shown. When they meet for the first time, they will be closest to
Solution:
The perimeter of the region is blocks, so when Jane and Hector meet they have together walked blocks. Since Jane walks twice as fast, she covers blocks and Hector covers
Starting from the middle of the bottom edge, Hector walks blocks (to then up to ), and Jane walks blocks (to up to then across the top to ). They meet at
Thus, the correct answer is D .
Problem 11 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 · 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