AMC 8 · 2007 · #15

Grade 6 logic

Archetype Small-Domain Constrained Search

if-then-reasoninglogical-deductioninterval-arithmetic caseworkcontradiction-elementary ↑ Prerequisites: interval-arithmeticif-then-reasoning

📏 Medium solution 💡 1 insight

Problem

Let $a, b$ and $c$ be numbers with $0 < a < b < c$ . Which of the following is
impossible?

$\mathrm{(A)} \ a + c < b \qquad \mathrm{(B)} \ a \cdot b < c \qquad \mathrm{(C)} \ a + b < c \qquad \mathrm{(D)} \ a \cdot c < b \qquad \mathrm{(E)}\frac{b}{c} = a$

Pick an answer.

(A)

$a + c < b$

(B)

$$a \cdot b < c$$

(C)

$a + b < c$

(D)

$$a \cdot c < b$$

(E)

$$\dfrac{b}{c} = a$$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Three positive numbers satisfy $0 < a < b < c$. Of the five inequalities listed, which one can NEVER be true?

Givens: $a$, $b$, $c$ are numbers with $0 < a < b < c$; Answer choices: (A) $a + c < b$, (B) $a \cdot b < c$, (C) $a + b < c$, (D) $a \cdot c < b$, (E) $\dfrac{b}{c} = a$

Unknowns: Which one of the five statements is impossible under the condition $0 < a < b < c$

Understand

Restated: Three positive numbers satisfy $0 < a < b < c$. Of the five inequalities listed, which one can NEVER be true?

Givens: $a$, $b$, $c$ are numbers with $0 < a < b < c$; Answer choices: (A) $a + c < b$, (B) $a \cdot b < c$, (C) $a + b < c$, (D) $a \cdot c < b$, (E) $\dfrac{b}{c} = a$

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check

The question asks which choice is impossible, so Tool #3 (Eliminate Possibilities) is the direct fit: rule out every choice that can happen, and the lone survivor is the answer. To rule a choice OUT of the suspect list, we use Tool #6 (Guess and Check) — pick concrete numbers that satisfy $0 < a < b < c$ AND the choice's inequality. To rule a choice IN as impossible, we use the basic rules of inequalities: adding a positive number makes things larger. We are deliberately avoiding Tool #13 (Algebra) because no equation needs solving — small numerical experiments and one transitivity step are enough.

Execute — Answer: A

#6 Guess and Check 6.EE.B.8 Step 1

Test (B) $a \cdot b < c$ with whole numbers.
Try $a = 1$, $b = 2$, $c = 5$.
The order $0 < 1 < 2 < 5$ holds, and $a \cdot b = 1 \cdot 2 = 2 < 5 = c$.
So (B) CAN happen — eliminate it.

$a = 1,\ b = 2,\ c = 5 \;\Rightarrow\; a \cdot b = 2 < 5 = c$ ✓

💡 When $a = 1$, the product $a \cdot b$ equals $b$, which is automatically less than $c$. An easy hit.

#6 Guess and Check 6.EE.B.8 Step 2

Test (C) $a + b < c$ with whole numbers.
Try $a = 1$, $b = 2$, $c = 10$.
The order $0 < 1 < 2 < 10$ holds, and $a + b = 3 < 10 = c$.
So (C) CAN happen — eliminate it.

$a = 1,\ b = 2,\ c = 10 \;\Rightarrow\; a + b = 3 < 10 = c$ ✓

💡 Make $c$ much bigger than $a$ and $b$ and the sum $a + b$ stays small by comparison.

#6 Guess and Check 5.NF.B.5 Step 3

Test (D) $a \cdot c < b$.
Whole numbers will not work here: if $a \geq 1$ then $a \cdot c \geq c > b$.
So let $a$ be a small fraction.
Try $a = \dfrac{1}{2}$, $b = 1$, $c = \dfrac{3}{2}$.
The order $0 < \tfrac{1}{2} < 1 < \tfrac{3}{2}$ holds, and $a \cdot c = \tfrac{1}{2} \cdot \tfrac{3}{2} = \tfrac{3}{4} < 1 = b$.
So (D) CAN happen — eliminate it.

$a = \tfrac{1}{2},\ b = 1,\ c = \tfrac{3}{2} \;\Rightarrow\; a \cdot c = \tfrac{3}{4} < 1 = b$ ✓

💡 Multiplying by a fraction less than $1$ makes things smaller. With $a < 1$, the product $a \cdot c$ can drop below $b$.

#6 Guess and Check 6.RP.A.1 Step 4

Test (E) $\dfrac{b}{c} = a$.
Since $b < c$, the ratio $\dfrac{b}{c}$ is some number less than $1$, so $a$ must be a fraction.
Try $a = \dfrac{1}{2}$, $c = 4$, and set $b = a \cdot c = 2$.
The order $0 < \tfrac{1}{2} < 2 < 4$ holds, and $\dfrac{b}{c} = \dfrac{2}{4} = \dfrac{1}{2} = a$.
So (E) CAN happen — eliminate it.

$a = \tfrac{1}{2},\ b = 2,\ c = 4 \;\Rightarrow\; \tfrac{b}{c} = \tfrac{2}{4} = \tfrac{1}{2} = a$ ✓

💡 Pick $a$ first, then pick any $c$ you like; the equation $b = a \cdot c$ tells you exactly what $b$ has to be.

#3 Eliminate Possibilities 6.EE.B.8 Step 5

Only (A) is left.
Show it is impossible without examples — use the rules of inequalities.
We have $b < c$ (given).
Since $a > 0$, adding $a$ to the right side of $b < c$ keeps the inequality, giving $b < c + a$, i.e., $b < a + c$.
That is the exact opposite of (A) $a + c < b$, so (A) cannot hold for any allowed $a, b, c$.

$$\underbrace{b < c}_{\text{given}}\ \text{and}\ \underbrace{a > 0}_{\text{given}}\ \Rightarrow\ b < c < c + a = a + c \;\Rightarrow\; b < a + c \;\Rightarrow\; \textbf{(A)}$$

💡 Adding a positive number to the bigger side keeps it bigger. So $a + c$ is even larger than $c$, which is already larger than $b$. There is no room for $a + c$ to drop below $b$.

[1] #6 6.EE.B.8 Test (B) $a \cdot b < c$ with whole numbers. Try $a = 1$, $b = 2$, $c = 5$. The

[2] #6 6.EE.B.8 Test (C) $a + b < c$ with whole numbers. Try $a = 1$, $b = 2$, $c = 10$. The ord

[3] #6 5.NF.B.5 Test (D) $a \cdot c < b$. Whole numbers will not work here: if $a \geq 1$ then $

[4] #6 6.RP.A.1 Test (E) $\dfrac{b}{c} = a$. Since $b < c$, the ratio $\dfrac{b}{c}$ is some num

[5] #3 6.EE.B.8 Only (A) is left. Show it is impossible without examples — use the rules of ineq

Review

Reasonableness: Double-check (A) with a numerical try. Pick $a = 1$, $b = 2$, $c = 3$: then $a + c = 4$, which is greater than $b = 2$. Pick $a = \tfrac{1}{10}$, $b = \tfrac{1}{2}$, $c = 1$: then $a + c = 1.1$, again greater than $b = 0.5$. No matter how small $a$ is, the inequality $a + c < b$ keeps failing because $c$ alone is already larger than $b$ and we are only adding to it. The four eliminated choices each came with a concrete witness, so the lone survivor is indeed (A).

Alternative: Tool #5 (Look for a Pattern): notice that (B), (C), (D), (E) all let one side stay small (either by multiplying by a fraction or by setting up a ratio), while (A) tries to make a SUM smaller than one of its parts. For positive numbers, $a + c$ must be larger than $c$ — and $c$ is already larger than $b$. The pattern "sum of positives $>$ each part" instantly flags (A) as the only structural violation.

CCSS standards used (min grade 6)

6.EE.B.8 Write an inequality of the form $x > c$ or $x < c$ to represent a constraint or condition in a real-world or mathematical problem (Reading and reasoning with the inequality $0 < a < b < c$, and using the rule that adding a positive number to one side of an inequality preserves the direction, to prove $b < a + c$.)
5.NF.B.5 Interpret multiplication as scaling (resizing) (Recognizing that multiplying by a fraction less than $1$ shrinks a number, which is the key to constructing the example for choice (D).)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities (Interpreting $\dfrac{b}{c} = a$ as a ratio condition and building the example $a = \tfrac{1}{2}, b = 2, c = 4$ from it.)

⭐ Adding a positive number always makes things bigger — so $a + c$ can never sink below $b$. Spot that one rule and this AMC 8 inequality puzzle is decided.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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