Difference between revisions of "2014 AMC 10A Problems/Problem 7"
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==Solution== | ==Solution== | ||
− | + | Let us denote <math>a = x + k</math> where <math>k > 0</math> and <math>b = y + l</math> where <math>l > 0</math>. We can write that <math>x + y < x + y + k + l \implies x + y < a + b</math>. | |
− | + | It is important to note that <math>1</math> counterexample fully disproves a claim. Let's try substituting <math>x=-3,y=-4,a=1,b=4</math>. | |
− | |||
+ | <math>\textbf{(II)}</math> states that <math>x-y<a-b \implies -3 - (-4) < 1 - 4 \implies 1 < -3</math>.Therefore, <math>\textbf{(II)}</math> is false. | ||
− | <math>\textbf{(III)}</math> states that <math>xy<ab \implies (-3)(- | + | <math>\textbf{(III)}</math> states that <math>xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4</math>. Therefore, <math>\textbf{(III)}</math> is false. |
+ | |||
+ | <math>\textbf{(IV)}</math> states that <math>\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-4} < \frac{1}{4} \implies 0.75 < 0.25</math>. Therefore, <math>\textbf{(IV)}</math> is false. | ||
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One of our four inequalities is true, hence, our answer is <math>\boxed{\textbf{(B) 1}}</math> | One of our four inequalities is true, hence, our answer is <math>\boxed{\textbf{(B) 1}}</math> | ||
+ | |||
+ | ~MathFun1000 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 16:35, 8 September 2021
Contents
Problem
Nonzero real numbers ,
,
, and
satisfy
and
. How many of the following inequalities must be true?
Solution
Let us denote where
and
where
. We can write that
.
It is important to note that counterexample fully disproves a claim. Let's try substituting
.
states that
.Therefore,
is false.
states that
. Therefore,
is false.
states that
. Therefore,
is false.
One of our four inequalities is true, hence, our answer is
~MathFun1000
Video Solution
~savannahsolver
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.