AMC 10 · 2014 · #7

Grade 7 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

logical-deductionsign-analysis guess-and-checkcasework ↑ Prerequisites: logical-deduction

📏 Long solution 💡 2 insights

Problem

Four nonzero real numbers x, y, a, b satisfy x ＜ a and y ＜ b. Of the four inequalities x+y ＜ a+b, x-y ＜ a-b, xy ＜ ab, and

\frac{x}{y}

＜

\frac{a}{b}

, decide how many are forced to hold for every choice of numbers meeting the two conditions.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

The question asks which inequalities are forced. There are two jobs. To confirm one is always true you reason in general symbols (Tool #4): add the two given inequalities and the sum-inequality falls out for every choice. To knock an inequality down you only need one counterexample, so Tool #6 (Guess and Check) is the workhorse — pick one concrete set of numbers that obeys x ＜ a and y ＜ b and watch which candidates fail. Choosing negative values is the smart guess, because negatives are exactly what flip subtraction, products, and quotients. One well-chosen test case disproves three of the four at once. Tool #3 then tallies the survivors: one always-true statement means the count is 1.

1STEP 1

Prove (I) always holds

Same-direction inequalities can be added, so adding x ＜ a to y ＜ b keeps the direction and forces x+y ＜ a+b.

x ＜ a and y ＜ b → x+y ＜ a+b

2STEP 2

Pick one counterexample

One failing example kills the other three: take x=-3, a=-1, y=-4, b=-2, all nonzero with x ＜ a and y ＜ b.

x=-3, a=-1, y=-4, b=-2 → x ＜ a, y ＜ b ✓

3STEP 3

Test (II): subtraction

Subtracting a negative adds: x-y = 1 and a-b = 1 come out equal, so the strict inequality (II) fails.

x-y = -3-(-4) = 1, a-b = -1-(-2) = 1, 1 ＜ 1 is false

4STEP 4

Test (III): multiplication

Two negatives multiply to a positive, and the more-negative pair wins: xy = 12 but ab = 2, so (III) fails.

xy = (-3)(-4) = 12, ab = (-1)(-2) = 2, 12 ＜ 2 is false

5STEP 5

Test (IV): division

A negative over a negative is positive: $\frac{x}{y}$ = $\frac{3}{4}$ beats $\frac{a}{b}$ = $\frac{1}{2}$ , so (IV) fails as well.

x/y = -3/-4 = 3/4, a/b = -1/-2 = 1/2, 3/4 ＜ 1/2 is false

6STEP 6

Count the survivors

Only (I) survives — one test case breaks (II), (III), and (IV) at once, so exactly 1 inequality must be true.

True always: (I) only → 1 → (B)

Answer

Statement (I) is proved in general, not just for one example, so it genuinely must be true. The other three each died to a single legal test case (x=-3,a=-1,y=-4,b=-2), which is all it takes to defeat a "must be true" claim. It also matches intuition: adding same-direction inequalities is always valid, while subtraction reverses one inequality and multiplication/division by negatives can flip magnitudes. Exactly one is forced, so the count is 1, choice (B).

💡Key takeaway

You can always add two inequalities that point the same way, but subtracting, multiplying, or dividing them can flip the result — one set of negative numbers is enough to prove it.

Prove (I) always holds
Pick one counterexample
Test (II): subtraction
Test (III): multiplication
Test (IV): division
Count the survivors