Difference between revisions of "2017 AMC 10B Problems/Problem 3"

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==Problem==
 
==Problem==
  
Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satify the inequalities
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Real numbers <math>x</math>, <math>y</math>, and <math>z</math> satisfy the inequalities
 
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.
 
<math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>.
 
Which of the following numbers is necessarily positive?
 
Which of the following numbers is necessarily positive?

Revision as of 12:53, 16 February 2017

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?

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

Solution

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.

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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