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>.
  
<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.
  
 
<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.
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==Video Solution==
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https://youtu.be/BnkFy36V_WE
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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https://youtu.be/zTGuz6EoBWY?t=525
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~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
{{AMC12 box|year=2017|ab=B|before=First Problem|num-a=3}}
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{{AMC12 box|year=2017|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 04:25, 4 December 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.

Video Solution

https://youtu.be/BnkFy36V_WE

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/zTGuz6EoBWY?t=525

~IceMatrix

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
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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