2007 AMC 8 Problem 15
Attempt Problem 15 of the 2007 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 2007 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
15.
Let and be numbers with Which of the following is impossible?
Answer: A
Small hint:
Use the fact that and .
Big hint:
For the other choices, one counterexample is enough to show they are possible.
Solution:
We know that and Adding these two inequalities together yields This shows that A is impossible, and therefore the right answer.
To ensure that this is correct, we can show that the other options are possible.
B and C : and
D : and
E : and
Thus, A is the correct answer.
Problem 15 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
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