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> | + | 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>. | <math>0<x<1</math>, <math>-1<y<0</math>, and <math>1<z<2</math>. | ||
Which of the following numbers is necessarily positive? | Which of the following numbers is necessarily positive? | ||
Line 8: | Line 8: | ||
==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 04:25, 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.