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

Revision as of 12:21, 26 December 2015

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 $x=-3,y=-2,a=1,b=1$

$\textbf{(II)}$ states that $x-y<a-b \implies -3-(-2)<1-1 \implies 1<0$ Since this is false, $\textbf{(II)}$ must also be false.

$\textbf{(III)}$ 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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 One of our four inequalities is true, hence, our answer is <math>\boxed{\textbf{(B) 1}}</math>
-==Solution 2==
-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.
 ==See Also==