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

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

Revision as of 11:19, 2 January 2020

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.

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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