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Difference between revisions of "2014 AMC 10A Problems/Problem 7"
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Nonzero real numbers <math>x</math>, <math>y</math>, <math>a</math>, and <math>b</math> satisfy <math>x < a</math> and <math>y < b</math>. How many of the following inequalities must be true?
 
Nonzero real numbers <math>x</math>, <math>y</math>, <math>a</math>, and <math>b</math> satisfy <math>x < a</math> and <math>y < b</math>. How many of the following inequalities must be true?
  
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<math>\textbf{(I)} x+y < a+b\qquad</math>
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<math>\textbf{(I)}\ x+y < a+b\qquad</math>
  
 
<math>
 
<math>
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\textbf{(II)} x-y < a-b\qquad</math>
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\textbf{(II)}\ x-y < a-b\qquad</math>
  
 
<math>
 
<math>
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\textbf{(III)} xy < ab\qquad</math>
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\textbf{(III)}\ xy < ab\qquad</math>
  
 
<math>
 
<math>
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\textbf{(IV)} \frac{x}{y} < \frac{a}{b}</math>
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\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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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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 +
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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 +
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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) \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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One of our four inequalities is true, hence, our answer is <math>\boxed{\textbf{(B) 1}}</math>
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 +
~MathFun1000
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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 16:35, 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)
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

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