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

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

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

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

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

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

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

First, we note that $\textbf{(I)}$ must be true by adding our two original inequalities. $x<a, y<b$ $\implies x+y<a+b$

Though one may be inclined to think that $\textbf{(II)}$ must also be true, it is not, for we cannot subtract inequalities.

In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting $x=-3,y=-2,a=1,b=1$ .

$\textbf{(III)}$ states that $xy<ab \implies (-3)(-2)<1\cdot 1 \implies 6<1$ . This is also false, thus $\textbf{(III)}$ is false.

$\textbf{(IV)}$ states that $\frac{x}{y}<\frac{a}{b} \implies \frac{-3}{-2}<\frac{1}{1}\implies 1.5<1$ . This is false, so $\textbf{(IV)}$ is false.

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

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

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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.		Though one may be inclined to think that <math>\textbf{(II)}</math> must also be true, it is not, for we cannot subtract inequalities.

−	In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting <cmath>x=-2,y=-2,a=1,b=1</cmath> ~~that gives <cmath>-2-(-2)<1-1</cmath>, which is not the case, thus <math>\textbf{(II)}</math> is false~~.	+	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>.