AMC 10 · 2003 · #12
Grade 7 algebraAl, Betty, and Clare split \textdollar1000 among them to be invested in different ways. Each begins with a different amount. At the end of one year, they have a total of \textdollar1500 dollars. Betty and Clare have both doubled their money, whereas Al has managed to lose \textdollar100 dollars. What was Al's original portion?
Pick an answer.
AMC 10 2003 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: Al, Betty, and Clare split $\textdollar 1000$, each getting a different amount. After one year Betty and Clare have each doubled their money, while Al has lost $\textdollar 100$, and together the three now hold $\textdollar 1500$. Find how much Al started with.
Givens: The three original amounts add up to $\textdollar 1000$; Betty's and Clare's amounts each double after one year; Al's amount drops by $\textdollar 100$ after one year; The three year-end amounts add up to $\textdollar 1500$; Answer choices: (A) $\textdollar 250$, (B) $\textdollar 350$, (C) $\textdollar 400$, (D) $\textdollar 450$, (E) $\textdollar 500$
Unknowns: Al's original portion in dollars
Understand
Restated: Al, Betty, and Clare split $\textdollar 1000$, each getting a different amount. After one year Betty and Clare have each doubled their money, while Al has lost $\textdollar 100$, and together the three now hold $\textdollar 1500$. Find how much Al started with.
Givens: The three original amounts add up to $\textdollar 1000$; Betty's and Clare's amounts each double after one year; Al's amount drops by $\textdollar 100$ after one year; The three year-end amounts add up to $\textdollar 1500$; Answer choices: (A) $\textdollar 250$, (B) $\textdollar 350$, (C) $\textdollar 400$, (D) $\textdollar 450$, (E) $\textdollar 500$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #16 Change Focus / Count the Complement
The problem hides three unknown starting amounts, so tool #4 (Introduce a Variable) gives each a letter and turns "they split $\textdollar 1000$" into one equation. Tool #13 (Convert to Algebra) then rewrites the year-end facts — doubling, losing $\textdollar 100$, new total $\textdollar 1500$ — 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 $A$.
Execute — Answer: C
6.EE.B.6 Step 1 Name the three starting shares
- Let $A$, $B$, and $C$ be the number of dollars Al, Betty, and Clare each put in at the start.
- They split $\textdollar 1000$ among them, so the three shares add up to $1000$.
💡 Giving each unknown share a letter turns the plain sentence "they split $\textdollar 1000$" into one exact equation.
7.EE.B.4 Step 2 Write the year-end total
- After a year Betty and Clare have each doubled, so they now hold $2B$ and $2C$.
- Al lost $\textdollar 100$, so he now holds $A - 100$.
- All three together come to $\textdollar 1500$, which gives a second equation.
💡 "Doubled," "lost $\textdollar 100$," and "total $\textdollar 1500$" each translate straight into a piece of one equation.
6.EE.A.3 Step 3 Trade Betty and Clare for Al
- We only want $A$, and $B$ and $C$ always travel together as $B + C$.
- From the first equation, $B + C = 1000 - A$, so $2B + 2C = 2(B + C) = 2(1000 - A)$.
- Substitute that into the year-end equation so only $A$ is left.
💡 Since the two doubled shares only ever appear as a pair, replacing their sum keeps every unknown but $A$ out of the way.
7.EE.B.4 Step 4 Solve for Al's share
- Distribute the $2$ and combine like terms: $A - 100 + 2000 - 2A = 1500$, which simplifies to $1900 - A = 1500$.
- So $A = 1900 - 1500 = 400$.
- Al's original portion was $\textdollar 400$, which is choice (C).
💡 With every unknown but $A$ gone, one line of tidy-up hands you Al's share.
6.EE.B.6 Let $A$, $B$, and $C$ be the number of dollars Al, Betty, and Clare each put in 7.EE.B.4 After a year Betty and Clare have each doubled, so they now hold $2B$ and $2C$. 6.EE.A.3 We only want $A$, and $B$ and $C$ always travel together as $B + C$. From the fi 7.EE.B.4 Distribute the $2$ and combine like terms: $A - 100 + 2000 - 2A = 1500$, which s Review
Reasonableness: Check the answer by rebuilding the year. If Al started with $\textdollar 400$, Betty and Clare together started with $1000 - 400 = \textdollar 600$, which doubles to $\textdollar 1200$. Al ends with $400 - 100 = \textdollar 300$. The year-end total is $1200 + 300 = \textdollar 1500$, exactly as stated, so (C) checks out. It is also sensible that the group grew by $\textdollar 500$: the doubling added $\textdollar 600$ while Al's loss took back $\textdollar 100$.
Alternative: Reason with the total instead of solving an equation. If Al had not lost his $\textdollar 100$, the year-end total would be $1500 + 100 = \textdollar 1600$, made up of Al's unchanged share plus the doubled shares of Betty and Clare. Betty and Clare doubling means their part of the total grew by exactly their starting amount, and the whole picture grew from $\textdollar 1000$ to $\textdollar 1600$, a gain of $\textdollar 600$. That $\textdollar 600$ gain is Betty and Clare's starting sum, leaving $1000 - 600 = \textdollar 400$ for Al — again (C).
CCSS standards used (min grade 7)
6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the three starting shares $A$, $B$, $C$ and writing $A + B + C = 1000$.)7.EE.B.4Use variables to represent quantities and construct simple equations to solve problems (Turning the doubling, the $\textdollar 100$ loss, and the $\textdollar 1500$ total into an equation and solving it for $A$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Replacing $2B + 2C$ with $2(1000 - A)$ and distributing so only $A$ remains.)
⭐ 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.
⭐ 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.
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