AMC 10 · 2014 · #7
Grade 7 arithmeticNonzero real numbers x, y, a, and b satisfy x<a and y<b. How many of the following inequalities must be true?
(I) x+y<a+b
(II) x−y<a−b
(III) xy<ab
(IV) yx<ba
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $x$, $y$, $a$, $b$ are all nonzero real numbers; $x < a$; $y < b$; Four candidate inequalities: (I) $x+y<a+b$, (II) $x-y<a-b$, (III) $xy<ab$, (IV) $\frac{x}{y}<\frac{a}{b}$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: How many of the four inequalities must be true for every allowed set of values
Understand
Restated: 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.
Givens: $x$, $y$, $a$, $b$ are all nonzero real numbers; $x < a$; $y < b$; Four candidate inequalities: (I) $x+y<a+b$, (II) $x-y<a-b$, (III) $xy<ab$, (IV) $\frac{x}{y}<\frac{a}{b}$; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #6 Guess and Check
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
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 $1$.
Execute — Answer: B
7.EE.B.4 Step 1 Prove (I) always holds
- Inequalities pointing the same way can be added.
- Since $x<a$ and $y<b$, add them left-to-left and right-to-right.
- The direction is preserved, so the sum inequality is true no matter what the numbers are.
💡 If each part on the left is smaller than its partner on the right, the two left parts together stay smaller than the two right parts together.
6.NS.C.7 Step 2 Pick one counterexample
- For the other three, one failing example is enough.
- Choose negatives that satisfy both givens: let $x=-3$, $a=-1$ (so $-3<-1$), and $y=-4$, $b=-2$ (so $-4<-2$).
- All four are nonzero, and both conditions $x<a$ and $y<b$ hold.
💡 On the number line $-3$ sits left of $-1$ and $-4$ sits left of $-2$, so this really is a legal case to test.
7.NS.A.1 Step 3 Test (II): subtraction
- Compute both sides of $x-y<a-b$ with the chosen values.
- Subtracting a negative adds.
- Both sides come out equal, so the strict inequality fails.
💡 Flipping $y<b$ around to subtract it reverses that inequality, so the differences no longer line up the safe way.
7.NS.A.2 Step 4 Test (III): multiplication
- Compute both sides of $xy<ab$.
- A negative times a negative is positive, and the more-negative factors give the larger product, so the left side ends up bigger, not smaller.
💡 Two smaller (more negative) numbers multiply to a bigger positive, so "smaller inputs" does not mean "smaller product."
7.NS.A.2 Step 5 Test (IV): division
- Compute both sides of $\frac{x}{y}<\frac{a}{b}$.
- A negative divided by a negative is positive.
- The left quotient is larger, so this inequality fails too.
💡 Dividing two negatives cancels the signs, and the size ratio can land the "smaller" pair on top.
7.EE.B.4 Step 6 Count the survivors
- Only (I) is forced to hold; the single test case breaks (II), (III), and (IV) at once.
- So exactly one inequality must be true.
💡 Adding same-direction inequalities is the one operation guaranteed to be safe; subtracting, multiplying, and dividing can all flip with negatives.
7.EE.B.4 Inequalities pointing the same way can be added. Since $x<a$ and $y<b$, add them 6.NS.C.7 For the other three, one failing example is enough. Choose negatives that satisf 7.NS.A.1 Compute both sides of $x-y<a-b$ with the chosen values. Subtracting a negative a 7.NS.A.2 Compute both sides of $xy<ab$. A negative times a negative is positive, and the 7.NS.A.2 Compute both sides of $\frac{x}{y}<\frac{a}{b}$. A negative divided by a negativ 7.EE.B.4 Only (I) is forced to hold; the single test case breaks (II), (III), and (IV) at Review
Reasonableness: 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).
Alternative: Instead of a single shared counterexample, attack each of (II)-(IV) structurally: (II) rewrites as needing $-y<-b$, but $y<b$ gives $-y>-b$, so it can fail; (III) and (IV) can fail whenever the values are negative because sign changes flip the comparison. Only (I), the sum of two same-direction inequalities, has no way to fail. This again yields one always-true inequality, confirming (B).
CCSS standards used (min grade 7)
7.EE.B.4Use variables to represent quantities and construct simple equations and inequalities (Adding the two given inequalities in general symbols to prove (I) always holds, and reasoning about which inequalities are forced.)6.NS.C.7Understand ordering and absolute value of rational numbers (Choosing negative values and verifying $-3<-1$ and $-4<-2$ so the counterexample legally satisfies $x<a$ and $y<b$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Evaluating the differences $x-y$ and $a-b$ with negative numbers to test inequality (II).)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Evaluating the products $xy$, $ab$ and the quotients $\frac{x}{y}$, $\frac{a}{b}$ with negatives to test inequalities (III) and (IV).)
⭐ 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.
⭐ 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.
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