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

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==Solution==
 
==Solution==
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Clearly, <math>\text{(I)}</math> must be true (do you see why?)
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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>
  
 
==See Also==
 
==See Also==

Revision as of 17:42, 7 February 2014

Problem

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

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

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

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

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

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

Solution

Clearly, $\text{(I)}$ must be true (do you see why?)

Consider $x=-2013, a=1, y=-2013, b=1$. Clearly, we have $x<a$ and $y<b$. Note that $\text{(II), (III), }$ and $\text{(IV)}$ are false, so our answer is $\boxed{\textbf{(B) 1}}$

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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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