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

Latest revision as of 03:25, 4 December 2020

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?

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

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.

~savannahsolver

~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

@@ Line 17: / Line 17: @@
 <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|before=First Problem|num-a=3}}
+{{AMC12 box|year=2017|ab=B|num-b=1|num-a=3}}
 {{MAA Notice}}
 [[Category:Introductory Algebra Problems]]