AMC 10 · 2010 · #3
Grade 6 arithmeticPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Instead of tracking the messy transfer, focus on the ending totals. The marbles are just shared between two boys, so the pile stays the same size the whole time. Naming Eric's final count as a variable turns 'twice as many' into one short equation, and once the ending amounts are known, a single subtraction recovers how many marbles actually moved.
Find the unchanging total
Marbles only move between the boys, so the two piles always add to the same total: 97+11=108.
Moving marbles around never changes how many there are in total.
Moving marbles from one boy to the other never changes how many there are in total.
▸ Why?
What one loses the other gains, so the two changes cancel exactly.
▸ Why?
The two piles together are the whole collection, so their sum is fixed from the start.
Name the ending amounts
Let E be Eric's ending count; Tyrone ends with twice that, 2E, and together they still hold 108: 2E+E=108.
'Twice as many' just means one share for Eric and two equal shares for Tyrone.
6.EE.B.6Introduce A VariableSolve for the ending amounts
Combine like terms into 3E=108, divide by 3: Eric ends with E=36 and Tyrone with 2E=72.
Three equal shares make 108, so each share is a third of 108.
6.EE.B.7Introduce A VariableAnswer the real question
The ask is the transfer, not the totals: Eric rose 36-11=25, and Tyrone fell 97-72=25 too, so choice (D).
To find what was handed over, compare each boy's before and after.
4.NBT.B.4Work BackwardsWhen things are only shared back and forth, the total stays fixed, so split that total into equal shares and read off the answer.
- Find the unchanging total
- Name the ending amounts
- Solve for the ending amounts
- Answer the real question