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

Revision as of 11:39, 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 $\boxed{\textbf{(E) } y+z}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 ==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>.
 {{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}

Page

Toolbox

Search

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

Revision as of 11:39, 16 February 2017

Problem

Solution