AMC 10 · 2011 · #15
Grade 6 arithmeticLet @ denote the "averaged with" operation: a@b=2a+b. Which of the following distributive laws hold for all numbers x,y, and z? I. x @ (y + z) = (x @ y) + (x @ z) II. x + (y @ z) = (x + y) @ (x + z) III. x @ (y @ z) = (x @ y) @ (x @ z)
Pick an answer.
AMC 10 2011 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: A new operation $@$ means "average with": $a @ b = \dfrac{a+b}{2}$. Three candidate distributive laws are listed. Decide which of the three hold for every choice of numbers $x$, $y$, and $z$.
Givens: $a @ b = \dfrac{a+b}{2}$ for all numbers $a$ and $b$; Law I: $x @ (y + z) = (x @ y) + (x @ z)$; Law II: $x + (y @ z) = (x + y) @ (x + z)$; Law III: $x @ (y @ z) = (x @ y) @ (x @ z)$; Answer choices name subsets of $\{\text{I}, \text{II}, \text{III}\}$
Unknowns: Which of the three laws are true for all numbers $x$, $y$, and $z$
Understand
Restated: A new operation $@$ means "average with": $a @ b = \dfrac{a+b}{2}$. Three candidate distributive laws are listed. Decide which of the three hold for every choice of numbers $x$, $y$, and $z$.
Givens: $a @ b = \dfrac{a+b}{2}$ for all numbers $a$ and $b$; Law I: $x @ (y + z) = (x @ y) + (x @ z)$; Law II: $x + (y @ z) = (x + y) @ (x + z)$; Law III: $x @ (y @ z) = (x @ y) @ (x @ z)$; Answer choices name subsets of $\{\text{I}, \text{II}, \text{III}\}$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The symbol $@$ is just shorthand, so Tool #13 (Convert to Algebra) replaces every $@$ with its definition $\dfrac{a+b}{2}$ and turns each law into a plain algebra statement. Tool #7 (Identify Subproblems) splits the job into three independent checks — one per law. For each, simplify both sides and see if they are the same expression; Tool #3 (Eliminate Possibilities) then crosses off any law whose two sides differ, leaving only the true ones.
Execute — Answer: E
6.EE.A.2 Step 1 Rewrite the operation
- Every time you see $@$, swap in the definition: $a @ b$ becomes $\dfrac{a+b}{2}$.
- To test a law, simplify its left side and its right side separately into a single fraction, then compare.
- If the two match no matter what $x$, $y$, $z$ are, the law holds; if they can differ, it fails.
💡 A made-up symbol carries no magic — replace it with its plain meaning and the problem becomes ordinary algebra.
6.EE.A.4 Step 2 Test Law I
- Left side: $x @ (y+z) = \dfrac{x + (y+z)}{2} = \dfrac{x+y+z}{2}$.
- Right side: $(x @ y) + (x @ z) = \dfrac{x+y}{2} + \dfrac{x+z}{2} = \dfrac{2x+y+z}{2}$.
- The two numerators are $x+y+z$ versus $2x+y+z$, which differ whenever $x \neq 0$.
- So Law I fails.
💡 Adding $x$ into the average once is not the same as folding it into two separate averages, which sneaks in an extra $x$.
6.EE.A.3 Step 3 Test Law II
- Left side: $x + (y @ z) = x + \dfrac{y+z}{2} = \dfrac{2x + y + z}{2}$.
- Right side: $(x+y) @ (x+z) = \dfrac{(x+y) + (x+z)}{2} = \dfrac{2x+y+z}{2}$.
- Both sides simplify to the exact same fraction, so Law II holds for all $x$, $y$, $z$.
💡 Adding $x$ to two numbers first and then averaging shifts the average up by exactly $x$ — the same as adding $x$ after averaging.
6.EE.A.3 Step 4 Test Law III
- Left side: $x @ (y @ z) = x @ \dfrac{y+z}{2} = \dfrac{x + \frac{y+z}{2}}{2} = \dfrac{2x + y + z}{4}$.
- Right side: $(x @ y) @ (x @ z) = \dfrac{x+y}{2} @ \dfrac{x+z}{2} = \dfrac{\frac{x+y}{2} + \frac{x+z}{2}}{2} = \dfrac{2x+y+z}{4}$.
- Both sides equal $\dfrac{2x+y+z}{4}$, so Law III holds.
💡 Averaging is symmetric in $x$, $y$, $z$ once you flatten it, so nesting averages one way matches nesting them the other way.
6.EE.A.4 Step 5 Collect the survivors
- Law I failed, while Laws II and III each simplified to matching expressions on both sides.
- Crossing off Law I leaves "II and III only," which is choice (E).
💡 Keep only the laws whose two sides are truly the same expression, and the answer names exactly those.
6.EE.A.2 Every time you see $@$, swap in the definition: $a @ b$ becomes $\dfrac{a+b}{2}$ 6.EE.A.4 Left side: $x @ (y+z) = \dfrac{x + (y+z)}{2} = \dfrac{x+y+z}{2}$. Right side: $( 6.EE.A.3 Left side: $x + (y @ z) = x + \dfrac{y+z}{2} = \dfrac{2x + y + z}{2}$. Right sid 6.EE.A.3 Left side: $x @ (y @ z) = x @ \dfrac{y+z}{2} = \dfrac{x + \frac{y+z}{2}}{2} = \d 6.EE.A.4 Law I failed, while Laws II and III each simplified to matching expressions on b Review
Reasonableness: Plug in easy numbers to confirm. Take $x=2$, $y=4$, $z=6$. Law I: $2 @ 10 = 6$ but $(2@4)+(2@6)=3+4=7$ — unequal, so I really fails. Law II: $2 + (4@6) = 2+5 = 7$ and $(2+4)@(2+6)=6@8=7$ — equal. Law III: $2 @ (4@6) = 2@5 = 3.5$ and $(2@4)@(2@6)=3@4=3.5$ — equal. Numbers agree with the algebra: only II and III hold, so (E) is right.
Alternative: Skip the general algebra and go straight to a single numeric test with distinct values like $x=2,\;y=4,\;z=6$ (or $x=1,\;y=0,\;z=0$). Checking each law once is enough to eliminate Law I and to see II and III survive; because these operations are all linear, one well-chosen test that separates the sides is decisive, again pointing to (E).
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Replacing the symbol $@$ with the literal expression $\dfrac{a+b}{2}$ so each law becomes an algebra statement.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Simplifying both sides of Laws II and III into the same single fraction to confirm they are equivalent.)6.EE.A.4Identify when two expressions are equivalent (Comparing the simplified left and right sides to decide each law — matching means true, differing means false.)
⭐ When a problem uses a strange new symbol, swap in its definition and simplify — two laws survive only if both sides turn into the exact same expression.
⭐ When a problem uses a strange new symbol, swap in its definition and simplify — two laws survive only if both sides turn into the exact same expression.
More like this
Same archetype — closest grade level first.