AMC 10 · 2010 · #4
Grade 6 algebraFor a real number x, define ♡(x) to be the average of x and x2. What is ♡(1)+♡(2)+♡(3)?
Pick an answer.
AMC 10 2010 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 rule $\heartsuit(x)$ takes a number $x$ and gives the average of $x$ and $x^2$. Compute $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$.
Givens: $\heartsuit(x)$ is defined as the average of the two numbers $x$ and $x^2$; We must evaluate the rule at $x=1$, $x=2$, and $x=3$, then add the three results; Answer choices: (A) $3$, (B) $6$, (C) $10$, (D) $12$, (E) $20$
Unknowns: The single number equal to $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$
Understand
Restated: A rule $\heartsuit(x)$ takes a number $x$ and gives the average of $x$ and $x^2$. Compute $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$.
Givens: $\heartsuit(x)$ is defined as the average of the two numbers $x$ and $x^2$; We must evaluate the rule at $x=1$, $x=2$, and $x=3$, then add the three results; Answer choices: (A) $3$, (B) $6$, (C) $10$, (D) $12$, (E) $20$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The question is really three small identical problems glued together, so Tool #7 (Identify Subproblems) fits: find $\heartsuit(1)$, $\heartsuit(2)$, $\heartsuit(3)$ one at a time, then add. Tool #4 (Introduce a Variable) turns the unfamiliar $\heartsuit$ symbol into a formula $\tfrac{x+x^2}{2}$ so plugging in is mechanical. Tool #3 (Eliminate Possibilities) catches the built-in trap: forgetting to divide by $2$ gives $20$, which is exactly choice (E).
Execute — Answer: C
6.SP.A.3 Step 1 Turn the heart rule into a formula
- The symbol $\heartsuit$ is just a recipe: take the number $x$ and the number $x^2$, then average them.
- The average of two numbers is their sum divided by $2$.
- Writing that with the letter $x$ gives a formula you can plug numbers into.
💡 An unfamiliar symbol is just a rule in disguise; rewrite it as a formula you can compute with.
6.EE.A.2 Step 2 Compute the value at 1
- Substitute $x=1$.
- First $1^2=1$, so we average $1$ and $1$: their sum is $2$, divided by $2$ gives $1$.
- So $\heartsuit(1)=1$.
💡 Plug the number into the formula, square it, then average the two numbers.
6.EE.A.1 Step 3 Compute the value at 2
- Substitute $x=2$.
- Here $2^2=4$, so we average $2$ and $4$: their sum is $6$, divided by $2$ gives $3$.
- So $\heartsuit(2)=3$.
💡 The average of $2$ and $4$ sits exactly halfway between them, at $3$.
6.EE.A.1 Step 4 Compute the value at 3
- Substitute $x=3$.
- Here $3^2=9$, so we average $3$ and $9$: their sum is $12$, divided by $2$ gives $6$.
- So $\heartsuit(3)=6$.
💡 Halfway between $3$ and $9$ is $6$, so the average lands there.
1.OA.C.6 Step 5 Add the three values
- Add the three results: $1+3+6=10$.
- That is choice (C).
- The trap value $20$ in (E) is what you get if you forget to divide by $2$ and just add $x+x^2$ each time ($2+6+12=20$), so a careful average kills it.
💡 Once each piece is found, the answer is just their sum.
6.SP.A.3 The symbol $\heartsuit$ is just a recipe: take the number $x$ and the number $x^ 6.EE.A.2 Substitute $x=1$. First $1^2=1$, so we average $1$ and $1$: their sum is $2$, di 6.EE.A.1 Substitute $x=2$. Here $2^2=4$, so we average $2$ and $4$: their sum is $6$, div 6.EE.A.1 Substitute $x=3$. Here $3^2=9$, so we average $3$ and $9$: their sum is $12$, di 1.OA.C.6 Add the three results: $1+3+6=10$. That is choice (C). The trap value $20$ in (E Review
Reasonableness: A genuine average of two numbers always lands between them, so $\heartsuit(1)=1$, $\heartsuit(2)=3$ (between $2$ and $4$), and $\heartsuit(3)=6$ (between $3$ and $9$) all look right. Their sum $10$ is a modest number in the middle of the choices. The tempting $20$ is exactly the sum of the $x+x^2$ pieces with no halving, which matches choice (E) and confirms it is the forget-to-average trap, not the true answer.
Alternative: Combine before halving. Add all the numerators first: $(1+2+3)+(1^2+2^2+3^2)=6+14=20$, then divide the whole thing by $2$ once to get $\tfrac{20}{2}=10$. This does the averaging a single time at the end and lands on (C), while showing plainly where the trap value $20$ comes from.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Reading "average of $x$ and $x^2$" as the mean $\tfrac{x+x^2}{2}$ of two numbers.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the squares $2^2=4$ and $3^2=9$ inside each $\heartsuit$ value.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting $x=1,2,3$ into the formula $\tfrac{x+x^2}{2}$.)1.OA.C.6Add and subtract within 20 using strategies (Adding the three results $1+3+6=10$.)
⭐ The average of two numbers is their sum divided by $2$ — skip the divide-by-$2$ and you land on double the real answer.
⭐ The average of two numbers is their sum divided by $2$ — skip the divide-by-$2$ and you land on double the real answer.
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