1995 AMC 8 Problem 11
Attempt Problem 11 of the 1995 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 1995 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
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
Answer: D
Small hint:
When they first meet, together they have covered the full -block loop
Big hint:
Jane walks twice as fast, so she covers blocks while Hector covers
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
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- 2007 AMC 8
- 2008 AMC 8
- 2009 AMC 8
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