2017 AMC 10B Problems/Problem 3: Difference between revisions

Latest revision as of 03:25, 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

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 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 1: / Line 1: @@
 ==Problem==
-[QUESTION]
+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>.
+Which of the following numbers is necessarily positive?
-<math>\textbf{(A)}\ [???]\qquad\textbf{(B)}\ [???]\qquad\textbf{(C)}\ [???]\qquad\textbf{(D)}\ [???]\qquad\textbf{(E)}\ [???]</math>
+<math>\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</math>
 ==Solution==
+Notice that <math>y+z</math> must be positive because <math>|z|>|y|</math>. Therefore the answer is <math>\boxed{\textbf{(E) } y+z}</math>.
-{{AMC10 box|year=2017|ab=b|num-b=2|num-a=4}}
+The other choices:
+<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>.
+<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.
+<math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time.
+==Video Solution==
+https://youtu.be/BnkFy36V_WE
+~savannahsolver
+==Video Solution by TheBeautyofMath==
+https://youtu.be/zTGuz6EoBWY?t=525
+~IceMatrix
+==See Also==
+{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
+{{AMC12 box|year=2017|ab=B|num-b=1|num-a=3}}
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

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