2014 AMC 10A Problem 7

P1-10~3 minPrint

Absolute Value & Inequalities Test small/specific values or special cases to find or verify the answer

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

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

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

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

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

Show solution without answering

Key insight

Adding two inequalities that point the same way is always safe. Subtracting, multiplying and dividing are not, and one example with negative numbers breaks them.

Solution

"Must be true" means true for every allowed choice of numbers. So to reject a statement, one example where it fails is enough. To accept a statement, we need a reason it always works.

Statement I. If $x$ is smaller than $a$ and $y$ is smaller than $b$ , then $x + y$ is smaller than $a + b$ : you are adding a smaller number to a smaller number. This one always holds.
Statement II. Try $x = 1$ , $a = 2$ , $y = 1$ , $b = 10$ . The rules $x < a$ and $y < b$ hold. But $x - y = 0$ and $a - b = -8$ , and $0 < -8$ is false. So II can fail.
Statement III. Negative numbers are the danger with products, so try them. Take $x = -2$ , $a = -1$ , $y = -2$ , $b = -1$ . The rules hold, since $-2 < -1$ . But $xy = 4$ and $ab = 1$ , and $4 < 1$ is false. So III can fail.
Statement IV. Use the same numbers. $\frac{x}{y} = \frac{-2}{-2} = 1$ and $\frac{a}{b} = \frac{-1}{-1} = 1$ . Then $1 < 1$ is false. So IV can fail.

Only statement I must be true.

The answer is $\boxed{\textbf{(B)}\ 1}$ .

Why this works

Adding two inequalities that point the same way is always safe. Subtracting one from another, multiplying or dividing can all go wrong, depending on signs and sizes. When you need an example that breaks a statement, try negative numbers first; a pair like $-2 < -1$ is worth keeping ready.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

(I) Inequalities in the same direction can be added: $x + y < a + b$ . This one is always true.

(II) Subtracting reverses the effect of the second inequality. Take $x = 1, a = 2$ and $y = 1, b = 10$ : then $x - y = 0$ but $a - b = -8$ , so II fails.

(III) Products misbehave with negatives. Take $x = y = -2$ and $a = b = -1$ : both hypotheses hold, but $xy = 4$ and $ab = 1$ , so III fails.

(IV) The same example gives $\frac{x}{y} = 1$ and $\frac{a}{b} = 1$ , so the strict inequality fails.

Exactly one statement must be true.

The answer is $\boxed{\textbf{(B)}\ 1}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Intermediate Algebra, Chapter 12: Inequalities (§12.3, §12.4)