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

m (Solution)
Line 8: Line 8:
  
 
==Solution==
 
==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}}
 
{{AMC10 box|year=2017|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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 (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

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

Invalid username
Login to AoPS