2017 AMC 12B Problem 2

Attempt Problem 2 of the 2017 AMC 12B 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 2017 AMC 12B solutions, or check the answer key.

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

Real numbers $x,$ $y,$ and $z$ satisfy the inequalities

$0 \lt x \lt 1, \quad -1 \lt y \lt 0,$ and $1 \lt z \lt 2.$

Which of the following numbers is necessarily positive?

$y + x^2$

$y + xz$

$y + y^2$

$y + 2y^2$

$y + z$

Answer: E

Concepts:inequality counterexample

Difficulty rating: 1020

Solution:

Adding $y \gt -1$ and $z \gt 1$ gives $y + z \gt 0,$ so $y + z$ is always positive. Each of the other four choices can be made negative: with $x = \tfrac18,$ $y = -\tfrac14,$ $z = \tfrac32,$ every one of $y + x^2,$ $y + xz,$ $y + y^2,$ and $y + 2y^2$ is negative.

Thus, the correct answer is E.

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