2014 AMC 10A Problems/Problem 7

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. $\newline$ $\textbf{(III)}$ states that $xy<ab \implies (-3) \cdot (-4) < 1 \cdot 4 \implies 12 < 4$ . Therefore, $\textbf{(III)}$ is false. $\newline$ $\textbf{(IV)}$ 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. $\newline$ One of our four inequalities is true, hence, our answer is $\boxed{\textbf{(B) 1}}$

~MathFun1000 (Majority of Solution)

Video Solution

https://youtu.be/0QeqCeojr6Q

~savannahsolver

2014 AMC 10A (Problems • Answer Key • Resources)
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2014 AMC 10A Problems/Problem 7

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