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AMC 8 · 2008 · #13

Grade 6 arithmetic
Archetype Pair-Sum Invariant
systems-of-equationslinear-equations-two-varmulti-digit-arithmetic identify-subproblems ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic
📏 Short solution 💡 3 insights

Problem

Mr. Harman needs to know the combined weight in pounds of three boxes he wants to mail. However, the only available scale is not accurate for weights less than 100100 pounds or more than 150150 pounds. So the boxes are weighed in pairs in every possible way. The results are 122122, 125125 and 127127 pounds. What is the combined weight in pounds of the three boxes?

Pick an answer.

(A)
160
(B)
170
(C)
187
(D)
195
(E)
354
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Toolkit + CCSS Solution

Understand

Restated: Three boxes are weighed two at a time, giving pair totals of $122$, $125$, and $127$ pounds. Find the combined weight of all three boxes.

Givens: There are three boxes with unknown individual weights; Each pair of boxes is weighed once, so there are three pair weights; The pair weights are $122$, $125$, and $127$ pounds; Answer choices: (A) $160$, (B) $170$, (C) $187$, (D) $195$, (E) $354$

Unknowns: The combined weight $x + y + z$ of all three boxes

Understand

Restated: Three boxes are weighed two at a time, giving pair totals of $122$, $125$, and $127$ pounds. Find the combined weight of all three boxes.

Givens: There are three boxes with unknown individual weights; Each pair of boxes is weighed once, so there are three pair weights; The pair weights are $122$, $125$, and $127$ pounds; Answer choices: (A) $160$, (B) $170$, (C) $187$, (D) $195$, (E) $354$

Plan

Primary tool: #11 Find an Invariant

Secondary: #4 Introduce a Variable

We are not asked for each box's weight — only the total. Tool #11 (Find an Invariant) says to look for a quantity that stays the same no matter which pair you pick. The key invariant here is the count: when you add all three pair weights, every box gets counted exactly twice. Tool #4 (Introduce a Variable) lets us name the box weights $x$, $y$, $z$ so we can write that observation as a clean equation. Then the total $x + y + z$ falls out by dividing by $2$ — no need to solve for $x$, $y$, $z$ separately.

Execute — Answer: C

#4 Introduce a Variable 6.EE.B.6 Step 1
  • Name the box weights.
  • Let $x$, $y$, and $z$ be the weights of the three boxes in pounds.
  • The three pair weighings give one equation each.
$$x + y = 122,\ \ x + z = 125,\ \ y + z = 127$$

💡 Using letters for unknown weights is the Grade 6 move: "use variables to represent numbers and write expressions when solving a real-world problem."

#11 Find an Invariant 6.EE.A.3 Step 2
  • Add all three equations.
  • Look at how often each variable appears on the left side: $x$ shows up in the first two equations, $y$ in the first and third, $z$ in the second and third.
  • Every variable appears exactly twice.
$$(x + y) + (x + z) + (y + z) = 122 + 125 + 127$$

💡 The "each box counted twice" pattern is the invariant. It does not depend on which pair weights you got — only on the fact that every pair was weighed.

#11 Find an Invariant 6.EE.A.3 Step 3
  • Combine like terms on the left and add on the right.
  • The left side becomes $2x + 2y + 2z$, which factors as $2(x + y + z)$.
$$2x + 2y + 2z = 374 \;\Rightarrow\; 2(x + y + z) = 374$$

💡 Factoring out the $2$ makes the combined weight $x + y + z$ visible as a single block.

#4 Introduce a Variable 6.EE.B.7 Step 4

Divide both sides by $2$ to isolate the combined weight.

$$x + y + z = \dfrac{374}{2} = 187 \;\Rightarrow\; \textbf{(C)}$$

💡 One-step equation: divide both sides by the same nonzero number. The answer is the combined weight in pounds.

[1] #4 6.EE.B.6 Name the box weights. Let $x$, $y$, and $z$ be the weights of the three boxes in
[2] #11 6.EE.A.3 Add all three equations. Look at how often each variable appears on the left sid
[3] #11 6.EE.A.3 Combine like terms on the left and add on the right. The left side becomes $2x +
[4] #4 6.EE.B.7 Divide both sides by $2$ to isolate the combined weight.

Review

Reasonableness: Subtract pairs to recover each box and double-check. From $x + z = 125$ and $x + y = 122$: $z - y = 3$. Combined with $y + z = 127$, we get $z = 65$ and $y = 62$, then $x = 60$. Sum: $60 + 62 + 65 = 187$. The pair totals are $60 + 62 = 122$, $60 + 65 = 125$, $62 + 65 = 127$ — all match. Also, every box weighs between $60$ and $65$ pounds, so any single box is under $100$ pounds — that is why Mr. Harman had to weigh them in pairs in the first place.

Alternative: Tool #7 (Break into Subproblems): solve the full system step by step. Subtract $x + y = 122$ from $x + z = 125$ to get $z - y = 3$. Add this to $y + z = 127$ to get $2z = 130$, so $z = 65$, then $y = 62$ and $x = 60$. Sum to $187$. This works, but it does more arithmetic than necessary — Tool #11 skipped straight to the total.

CCSS standards used (min grade 6)

  • 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world problem (Naming the three unknown box weights $x$, $y$, $z$ and writing each pair weighing as an equation.)
  • 6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Combining like terms in $(x + y) + (x + z) + (y + z)$ and factoring out the $2$ to expose $2(x + y + z)$.)
  • 6.EE.B.7 Solve real-world and mathematical problems by writing and solving one-variable equations of the form $x + p = q$ and $px = q$ (Solving $2(x + y + z) = 374$ by dividing both sides by $2$ to get the combined weight.)

⭐ When every pair gets weighed, add all the pair totals — each box was counted twice, so half the sum is the answer.

⭐ When every pair gets weighed, add all the pair totals — each box was counted twice, so half the sum is the answer.

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