AMC 10 · 2022 · #3

Grade 6 arithmetic

systems-of-equationslinear-equations-two-varabsolute-value convert-to-algebrawork-backwards ↑ Prerequisites: linear-equations-one-var

📏 Short solution 💡 1 insight

Problem

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Three numbers (call them first, second, third) add up to $96$. The first is $6$ times the third, and the third is $40$ less than the second. Find $|\text{first} - \text{second}|$.

Givens: $\text{first} + \text{second} + \text{third} = 96$; $\text{first} = 6 \times \text{third}$; $\text{third} = \text{second} - 40$ (equivalently $\text{second} = \text{third} + 40$); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: $|\text{first} - \text{second}|$

Understand

Restated: Three numbers (call them first, second, third) add up to $96$. The first is $6$ times the third, and the third is $40$ less than the second. Find $|\text{first} - \text{second}|$.

Plan

Primary tool: #6 Guess and Check

Secondary: #11 Work Backwards, #7 Identify Subproblems

All three numbers can be described from the third: $\text{first} = 6z$, $\text{second} = z + 40$. That makes the third number a single dial we can turn. Tool #6 (Guess and Check) lets us test a small whole number for $z$ and see if the total is $96$ — much friendlier than naming three variables. Tool #11 (Work Backwards) and #7 (Subproblems) take over once $z$ is found: plug it back to recover the first and second numbers, then compute their absolute difference.

Execute — Answer: E

#7 Identify Subproblems 6.EE.B.6 Step 1

Use the third number as the dial.
From the clues, $\text{first} = 6z$ and $\text{second} = z + 40$.
So the three numbers are $6z, \; z+40, \; z$ — every guess for $z$ gives one complete trio.

$$\text{first} = 6z, \quad \text{second} = z + 40, \quad \text{third} = z$$

💡 Naming the smallest piece $z$ and writing the others in terms of it is the Grade 6 "use a variable to stand for the unknown" idea — one number controls all three.

#6 Guess and Check 6.EE.B.7 Step 2

Guess.
Try $z = 5$: total $= 30 + 45 + 5 = 80$.
Too small.
Try $z = 10$: total $= 60 + 50 + 10 = 120$.
Too big.
Try $z = 7$: total $= 42 + 47 + 7 = 96$.
Match!

$$z=5: 30+45+5 = 80, \; z=10: 60+50+10 = 120, \; z=7: 42+47+7 = 96 \;\checkmark$$

💡 Two directed guesses sandwich $z$ between $5$ and $10$; one more lands on $7$ — Grade 6 "solve $px+q=r$" without needing pencil-and-paper algebra.

#11 Work Backwards 5.OA.A.1 Step 3

Read off the first and second numbers using $z = 7$: $\text{first} = 6 \cdot 7 = 42$ and $\text{second} = 7 + 40 = 47$.

$$\text{first} = 42, \quad \text{second} = 47$$

💡 Working back from $z = 7$ through the two simple expressions gives the actual numbers — Grade 5 "evaluate a numerical expression".

#11 Work Backwards 6.NS.C.7 Step 4

Compute the absolute difference.
The second number is larger, so $|\text{first} - \text{second}| = |42 - 47| = 5$.

$$|42 - 47| = 5 \;\Rightarrow\; \textbf{(E)}$$

💡 Absolute value is just the distance between two numbers — Grade 6 "ordering and absolute value".

[1] #7 6.EE.B.6 Use the third number as the dial. From the clues, $\text{first} = 6z$ and $\text

[2] #6 6.EE.B.7 Guess. Try $z = 5$: total $= 30 + 45 + 5 = 80$. Too small. Try $z = 10$: total $

[3] #11 5.OA.A.1 Read off the first and second numbers using $z = 7$: $\text{first} = 6 \cdot 7 =

[4] #11 6.NS.C.7 Compute the absolute difference. The second number is larger, so $|\text{first}

Review

Reasonableness: Verify the trio: $42 + 47 + 7 = 96 \checkmark$. First is six times third: $42 = 6 \cdot 7 \checkmark$. Third is $40$ less than second: $7 = 47 - 40 \checkmark$. All clues hold, so $|42 - 47| = 5$ is solid, matching choice (E).

Alternative: Tool #13 (Convert to Algebra) for a one-line proof. Substituting $\text{first} = 6z$ and $\text{second} = z + 40$ into the sum: $6z + (z+40) + z = 96 \Rightarrow 8z = 56 \Rightarrow z = 7$. Then $|6z - (z+40)| = |5z - 40| = |35 - 40| = 5$. Same answer.

CCSS standards used (min grade 6)

5.OA.A.1 Use parentheses, brackets, or braces in numerical expressions and evaluate (Evaluating $6 \cdot 7$ and $7 + 40$ to recover the first and second numbers once $z = 7$ is known.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Letting $z$ stand for the third number and writing the first as $6z$, the second as $z + 40$, so one dial controls all three.)
6.EE.B.7 Solve real-world problems by writing and solving equations (Treating "sum equals $96$" as an equation in $z$ and finding the value ($z = 7$) by directed guessing.)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Reading $|42 - 47|$ as the distance between the two numbers, which is $5$.)

⭐ This AMC 10 problem only needs Grade 6 "name the unknown and try small numbers" — once the third number turns out to be $7$, the gap between $42$ and $47$ is just $5$.