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Difference between revisions of "2014 AMC 10A Problems/Problem 7"

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\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}</math>
 
\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}</math>
  
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4</math>
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<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4</math>
  
 
==Solution==
 
==Solution==
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First, we note that <math>\textbf{(I)}</math> must be true by adding our two original inequalities. <cmath>x<a, y<b</cmath> <cmath>\implies x+y<a+b</cmath>
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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>.
  
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Though one may be inclined to think that <math>\textbf{(II)}</math> must also be true, it is not, for we cannot subtract inequalities.  
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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>.
  
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In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting <cmath>x=-3,y=-2,a=1,b=1</cmath>
 
  
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<math>\textbf{(II)}</math> states that <math>x-y<a-b \implies -3-(-2)<1-1 \implies 1<0</math> Since this is false, <math>\textbf{(II)}</math> must also be false.
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<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.
  
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<math>\textbf{(III)}</math> states that <math>xy<ab \implies (-3)(-2)<1\cdot 1 \implies 6<1</math>. This is also false, thus <math>\textbf{(III)}</math> is false.
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<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.
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<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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<math>\textbf{(IV)}</math> states that <math>\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-2}<\frac{1}{1}\implies 1.5<1</math>. This is false, so <math>\textbf{(IV)}</math> is false.
 
  
 
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>
  
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==Solution 2==
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~MathFun1000
  
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Also, with some intuition, we could have plugged <math>0=X</math>, <math>1=A</math>, <math>-3=Y</math>, and <math>-2=B</math> and then plugged these values into the equations to see which ones held.
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==Video Solution==
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https://youtu.be/0QeqCeojr6Q
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2014|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2014|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Algebra Problems]]

Latest revision as of 15:35, 8 September 2021

Problem

Nonzero real numbers $x$, $y$, $a$, and $b$ satisfy $x < a$ and $y < b$. How many of the following inequalities must be true?

$\textbf{(I)}\ x+y < a+b\qquad$

$\textbf{(II)}\ x-y < a-b\qquad$

$\textbf{(III)}\ xy < ab\qquad$

$\textbf{(IV)}\ \frac{x}{y} < \frac{a}{b}$

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

Let us denote $a = x + k$ where $k > 0$ and $b = y + l$ where $l > 0$. We can write that $x + y < x + y + k + l \implies x + y < a + b$.

It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$.


$\textbf{(II)}$ states that $x-y<a-b \implies -3 - (-4) < 1 - 4 \implies 1 < -3$.Therefore, $\textbf{(II)}$ is false.

$\textbf{(III)}$ states that $xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4$. Therefore, $\textbf{(III)}$ is false.

$\textbf{(IV)}$ states that $\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-4} < \frac{1}{4} \implies 0.75 < 0.25$. Therefore, $\textbf{(IV)}$ is false.


One of our four inequalities is true, hence, our answer is $\boxed{\textbf{(B) 1}}$

~MathFun1000

Video Solution

https://youtu.be/0QeqCeojr6Q

~savannahsolver

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
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Problem 6 		Followed by
Problem 8
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All AMC 10 Problems and Solutions

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