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2017 AMC 10B Problem 3

2017 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 2 min · position P1-10

Real numbers $x$ , $y$ , and $z$ satisfy the inequalities $0<x<1$ , $-1<y<0$ , and $1<z<2$ . Which of the following numbers is necessarily positive?

A)
$y+x^2$
B)
$y+xz$
C)
$y+y^2$
D)
$y+2y^2$
E)
$y+z$

Answer

Key insight

Since y > -1 and z > 1, the sum y + z exceeds 0; every other choice can be pushed negative by taking y close to -1.

Solution

Choice (E): from $y > -1$ and $z > 1$ , adding gives $y + z > 0$ . So (E) is always positive.

To be sure the others fail, produce one counterexample for each. Take $y$ near $-1$ where it hurts most:

(A) $x = 0.1$ , $y = -0.9$ : $y + x^2 = -0.89 < 0$ .
(B) $x = 0.1$ , $z = 1.5$ , $y = -0.9$ : $y + xz = -0.75 < 0$ .
(C) $y = -0.5$ : $y + y^2 = -0.25 < 0$ .
(D) $y = -0.25$ : $y + 2y^2 = -0.25 + 0.125 < 0$ .

The answer is $\boxed{\textbf{(E)}\ y+z}$ .

Why this works

"Necessarily positive" means positive for every allowed triple, so only a proof (adding two inequalities) confirms a choice, while a single bad example rules one out. For numbers between $-1$ and $0$ , the square is smaller in absolute value than the number, so $y + y^2$ and similar expressions are negative when $|y|$ is small enough.

The trap

Assuming y + y^2 or y + 2y^2 is positive because squares are positive, when y = -1/2 makes y + y^2 negative.

Common mistakes

Assuming y + y^2 or y + 2y^2 is positive because squares are positive, when y = -1/2 makes y + y^2 negative.
Testing only one convenient triple such as $x = 0.5$ , $y = -0.5$ , $z = 1.5$ , which happens to make several choices positive.

Techniques

Bound the quantity above/below or estimate to pin it down · Test small/specific values or special cases to find or verify the answer