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

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==Solution==
 
==Solution==
 
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 <math>y+z</math>, the last answer choice, must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.
 
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 <math>y+z</math>, the last answer choice, must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.
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 +
The other choices:
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<math>\textbf{(A)}</math> As <math>x</math> grows closer to <math>0</math>, <math>x^2</math> decreases and thus becomes less than <math>y</math>.
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 +
<math>\textbf{(B)}</math> <math>x</math> can be as small as possible (<math>x>0</math>), so <math>xz</math> grows close to <math>0</math> as <math>x</math> approaches <math>0</math>.
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<math>\textbf{(C)}</math> For all <math>-1<y<0</math>, <math>y>y^2</math>, and thus it is always negative.
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<math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time.
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{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:44, 16 February 2017

Problem

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?

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

Solution

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 (Problems • Answer Key • Resources)
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Problem 4
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All AMC 10 Problems and Solutions

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