AMC 10 · 2002 · #9
Grade 7 arithmeticThere are 3 numbers A, B, and C, such that 1001C−2002A=4004, and 1001B+3003A=5005. What is the average of A, B, and C?
Pick an answer.
AMC 10 2002 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 $A$, $B$, and $C$ satisfy two equations: $1001C-2002A=4004$ and $1001B+3003A=5005$. Find the average of $A$, $B$, and $C$.
Givens: $1001C-2002A=4004$; $1001B+3003A=5005$; Answer choices: (A) $1$, (B) $3$, (C) $6$, (D) $9$, (E) not uniquely determined
Unknowns: The average $\dfrac{A+B+C}{3}$
Understand
Restated: Three numbers $A$, $B$, and $C$ satisfy two equations: $1001C-2002A=4004$ and $1001B+3003A=5005$. Find the average of $A$, $B$, and $C$.
Givens: $1001C-2002A=4004$; $1001B+3003A=5005$; Answer choices: (A) $1$, (B) $3$, (C) $6$, (D) $9$, (E) not uniquely determined
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities
The average only depends on the sum $A+B+C$, so instead of chasing each of the three numbers, Tool #16 (Change Focus) aims straight at that sum — with two equations and three unknowns, hunting for $A$, $B$, and $C$ one at a time is both harder and unnecessary. Tool #15 (Organize Information in More Ways) makes the numbers manageable first: every coefficient and constant is a multiple of $1001$, so dividing each equation by $1001$ turns a wall of big numbers into two tiny equations. Tool #3 (Eliminate Possibilities) guards against the trap answer (E): the fact that the numbers are not individually fixed does not mean the average is undetermined, and this is exactly the reasoning that rules (E) out.
Execute — Answer: B
6.EE.A.3 Step 1 Divide each equation by 1001
- Every number in both equations is a multiple of $1001$: $2002=2\cdot1001$, $4004=4\cdot1001$, $3003=3\cdot1001$, and $5005=5\cdot1001$.
- Dividing the first equation through by $1001$ gives $C-2A=4$, and dividing the second by $1001$ gives $B+3A=5$.
- The two equations now hold exactly the same information but with single-digit numbers.
💡 Dividing every term of an equation by the same number keeps it true, so pulling out the shared factor $1001$ costs nothing and clears the clutter.
7.EE.A.1 Step 2 Add the two equations
- Add the left sides and the right sides of the two simplified equations.
- On the left, $(C-2A)+(B+3A)$ collects the $A$ terms as $-2A+3A=A$, leaving $A+B+C$.
- On the right, $4+5=9$.
- So $A+B+C=9$.
- Notice this pins down the sum even though $A$, $B$, and $C$ are still free individually — the $A$ terms were chosen to add up to exactly one $A$, which is why the total lands on the plain sum $A+B+C$.
💡 You do not need each number to know their total — adding the equations makes the leftover $A$ terms merge into the sum you actually want.
6.SP.A.3 Step 3 Divide the sum by 3
- The average of three numbers is their sum divided by $3$.
- Since $A+B+C=9$, the average is $\dfrac{9}{3}=3$.
- Choice (E) is a trap: it is true that $A$, $B$, and $C$ are not each determined, but the combination $A+B+C$ — and therefore the average — is fixed at a single value, so the average is uniquely determined.
- The answer is $3$, which is choice (B).
💡 A single, fixed sum forces a single, fixed average, so 'not uniquely determined' is wrong even though the three numbers themselves can wobble.
6.EE.A.3 Every number in both equations is a multiple of $1001$: $2002=2\cdot1001$, $4004 7.EE.A.1 Add the left sides and the right sides of the two simplified equations. On the l 6.SP.A.3 The average of three numbers is their sum divided by $3$. Since $A+B+C=9$, the a Review
Reasonableness: The average came out to $3$, a clean whole number that matches choice (B) and sits sensibly among the choices $1,3,6,9$. A quick independent check: pick any convenient value for one letter. Let $C=0$; then $C-2A=4$ gives $A=-2$, and $B+3A=5$ gives $B=5-3(-2)=11$. Their sum is $0+(-2)+11=9$ and the average is $3$ — the same answer, even from a different starting choice, which is exactly what 'uniquely determined' should look like.
Alternative: Instead of adding, solve the simplified equations for two letters in terms of the third: $C=2A+4$ and $B=5-3A$. Then $A+B+C=A+(5-3A)+(2A+4)=9$, and the $A$ terms cancel for every value of $A$. This shows directly that the sum is always $9$ no matter what $A$ is, so the average is always $3$ and answer (E) cannot be right.
CCSS standards used (min grade 7)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Recognizing that $2002,3003,4004,5005$ are all multiples of $1001$ and dividing each equation by $1001$ to get the equivalent equations $C-2A=4$ and $B+3A=5$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Adding the two linear equations and combining like terms, where $-2A+3A=A$, to obtain $A+B+C=9$ without solving for the individual variables.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Computing the average as the sum divided by the count, $\tfrac{9}{3}=3$, and reasoning that a fixed sum forces a single, uniquely determined average.)
⭐ When you only need a total, add the equations so the extra letters cancel — you can find the average without ever finding each number.
⭐ When you only need a total, add the equations so the extra letters cancel — you can find the average without ever finding each number.
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