AMC 10 · 2022 · #3
Grade 6 arithmeticProblem
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?
Pick an answer.
AMC 10 2022 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: 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 $|\text{first} - \text{second}|$.
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) $5$
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
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.
💡 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.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!
💡 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.
5.OA.A.1 Step 3 Read off the first and second numbers using $z = 7$: $\text{first} = 6 \cdot 7 = 42$ and $\text{second} = 7 + 40 = 47$.
💡 Working back from $z = 7$ through the two simple expressions gives the actual numbers — Grade 5 "evaluate a numerical expression".
6.NS.C.7 Step 4 - Compute the absolute difference.
- The second number is larger, so $|\text{first} - \text{second}| = |42 - 47| = 5$.
💡 Absolute value is just the distance between two numbers — Grade 6 "ordering and absolute value".
6.EE.B.6 Use the third number as the dial. From the clues, $\text{first} = 6z$ and $\text 6.EE.B.7 Guess. Try $z = 5$: total $= 30 + 45 + 5 = 80$. Too small. Try $z = 10$: total $ 5.OA.A.1 Read off the first and second numbers using $z = 7$: $\text{first} = 6 \cdot 7 = 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.1Use 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.6Use 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.7Solve 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.7Understand 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 $5$.
⭐ 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 $5$.