Difference between revisions of "2017 AMC 10B Problems/Problem 3"
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==Problem== | ==Problem== | ||
| − | + | 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)}\ | + | <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== | ==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>. | ||
| + | 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}} | {{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}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 03:25, 4 December 2020
Problem
Real numbers 
, 
, and 
satisfy the inequalities
, 
, and 
.
Which of the following numbers is necessarily positive?
Solution
Notice that 
must be positive because 
. Therefore the answer is 
.
The other choices:
As grows closer to 
, 
decreases and thus becomes less than .
can be as small as possible (
), so 
grows close to as approaches .
For all , 
, and thus it is always negative.
The same logic as above, but when 
this time.
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/zTGuz6EoBWY?t=525
~IceMatrix
See Also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
