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

Revision as of 12:06, 9 February 2014

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

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

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

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

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

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}}\ 3\qquad\textbf{(E)}\ 4$ (Error compiling LaTeX. Unknown error_msg)

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 $\boxed{\textbf{(B) 1}}$

(Solution by bestwillcui1)

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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 ==Solution==
-Clearly, <math>\text{(I)}</math> must be true (do you see why?)
+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>
-Consider <math>x=-2013, a=1, y=-2013, b=1</math>. Clearly, we have <math>x<a</math> and <math>y<b</math>. Note that <math>\text{(II), (III), }</math> and <math>\text{(IV)}</math> are false, so our answer is <math>\boxed{\textbf{(B) 1}}</math>
+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=-3,y=-2,a=1,b=1</cmath>
+<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.
+<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.
+<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>
+(Solution by bestwillcui1)
 ==See Also==