AMC 10 · 2002 · #9

Grade 7 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
systems-of-equationsmean-median-mode-range convert-to-algebra ↑ Prerequisites: systems-of-equations
📏 Medium solution 💡 2 insights
Problem
Three numbers A, B, and C satisfy two equations: 1001C-2002A=4004 and 1001B+3003A=5005. Find the average of A, B, and C.

Pick an answer.

(A)
1
(B)
3
(C)
6
(D)
9
(E)
Not uniquely determined

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

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.

1STEP 1

Divide each equation by 1001

Every number here is a multiple of 1001, so dividing both equations by it leaves C-2A=4 and B+3A=5.

1001C-2002A=4004 → C-2A=4 1001B+3003A=5005 → B+3A=5
2STEP 2

Add the two equations

Adding them merges the A terms, since -2A+3A=A, so the left side becomes A+B+C and the right 4+5: A+B+C=9.

(C-2A)+(B+3A)=4+5 → A+B+C=9
3STEP 3

Divide the sum by 3

The average is the sum divided by 3, and 9 divided by 3 is 3 — (E) fails because that sum is fixed.

(A+B+C)/3=9/3=3 → (B)
Answer
3
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.
💡Key takeaway

When you only need a total, add the equations so the extra letters cancel — you can find the average without ever finding each number.

  • Divide each equation by 1001
  • Add the two equations
  • Divide the sum by 3

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