AMC 10 · 2003 · #12

Grade 7 algebra

Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known

systems-of-equationslinear-equations-one-var convert-to-algebra ↑ Prerequisites: systems-of-equations

📏 Medium solution 💡 2 insights

Problem

Al, Betty, and Clare split 1000 dollars among them, each getting a different amount. After one year Betty and Clare have each doubled their money, while Al has lost 100 dollars, and the three together now hold 1500 dollars. Find how much Al started with.

Pick an answer.

(A)

$\textdollar 250$

(B)

$\textdollar 350$

(C)

$\textdollar 400$

(D)

$\textdollar 450$

(E)

$\textdollar 500$

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The problem hides three unknown starting amounts, so tool #4 (Introduce a Variable) gives each a letter and turns "they split $1000" into one equation. Tool #13 (Convert to Algebra) then rewrites the year-end facts — doubling, losing$ 100, new total $ 1500 — as a second equation. The key move is tool #16 (Change Focus): only Al's share is asked for, and Betty and Clare always appear together as B+C, so we never solve for them individually. Replacing B+C with 1000-A collapses everything to a single equation in A.

1STEP 1

Name the three starting shares

Let A, B, C be Al's, Betty's, and Clare's starting dollars; the three shares split the whole pot, so A + B + C = 1000.

A + B + C = 1000

2STEP 2

Write the year-end total

A year later Betty and Clare hold 2B and 2C, Al holds A - 100, and the three amounts add to 1500.

(A - 100) + 2B + 2C = 1500

3STEP 3

Trade Betty and Clare for Al

Only A is wanted, and B and C always travel together, so B + C = 1000 - A turns 2B + 2C into 2(1000 - A).

(A - 100) + 2(1000 - A) = 1500

4STEP 4

Solve for Al's share

Expanding leaves 1900 - A = 1500, so Al's original portion was 400 dollars — choice (C).

A - 100 + 2000 - 2A = 1500 → 1900 - A = 1500 → A = 400 → (C)

Answer

$ 400

Check the answer by rebuilding the year. If Al started with 400 dollars, Betty and Clare together started with 1000 - 400 = 600 dollars, which doubles to 1200 dollars. Al ends with 400 - 100 = 300 dollars. The year-end total is 1200 + 300 = 1500 dollars, exactly as stated, so (C) checks out. It is also sensible that the group grew by 500 dollars: the doubling added 600 dollars while Al's loss took back 100 dollars.

💡Key takeaway

When a problem only asks about one unknown, group the others together and swap them out, so you solve just one equation instead of chasing every letter.

Name the three starting shares
Write the year-end total
Trade Betty and Clare for Al
Solve for Al's share

Name the three starting shares

Write the year-end total

Trade Betty and Clare for Al

Solve for Al's share

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