2017 AMC 10B Problems/Problem 3

Revision as of 12:44, 16 February 2017 by E power pi times i (talk | contribs) (Solution)


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

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


We start from the last answer choice because the answer for these type of questions are usually in the last few answer choices. Notice that $y+z$, the last answer choice, 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.

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 10 Problems and Solutions

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