2017 AMC 10B Problems/Problem 3

Problem

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?

$\textbf{(A)}\ y+x^2\qquad\textbf{(B)}\ y+xz\qquad\textbf{(C)}\ y+y^2\qquad\textbf{(D)}\ y+2y^2\qquad\textbf{(E)}\ y+z$

Notice that $y+z$ must be positive because $|z|>|y|$ . Therefore the answer is $\boxed{\textbf{(E) } y+z}$ .

The other choices:

$\textbf{(A)}$ As $x$ grows closer to $0$ , $x^2$ decreases and thus becomes less than $y$ .

$\textbf{(B)}$ $x$ can be as small as possible ( $x>0$ ), so $xz$ grows close to $0$ as $x$ approaches $0$ .

$\textbf{(C)}$ For all $-1<y<0$ , $|y|>|y^2|$ , and thus it is always negative.

$\textbf{(D)}$ The same logic as above, but when $-\frac{1}{2}<y<0$ this time.

~savannahsolver

~IceMatrix

2017 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2017 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions