AMC 10 · 2009 · #3
Grade 6 rate-ratioPick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The whole problem hinges on one hidden rate: how much paint one room takes, measured in cans per room. Tool #8 (Analyze the Units) puts that rate at the center — once we know cans per room, the 25-room total is one multiplication. Tool #7 (Identify Subproblems) sets it up in stages: first turn the lost cans into a number of lost rooms, then find the per-room rate, then scale up to 25 rooms.
Turn the lost cans into lost rooms
The lost paint is the gap between the supplies: 30-25=5, so the 3 cans are worth 5 rooms of paint.
Whatever was lost is just the gap between what she could paint before and after.
4.OA.A.3Identify SubproblemsFind the paint per room
Share those 3 cans evenly over the 5 rooms they cover: one room takes 3/5 of a can.
Splitting the cans evenly among the rooms they cover gives the cost of a single room.
Splitting the cans evenly among the rooms they cover gives the paint one room costs.
▸ Why?
Each room takes one of the equal shares, so the share is the whole amount cut by the room count.
▸ Why?
Every room takes the same fixed amount, so one rate describes them all.
Scale up to 25 rooms
25 rooms need 25 times that rate: 25×3/5=15 cans, which is choice (C).
Once you know the cost of one room, any number of rooms is just repeated copies of that cost.
5.NF.B.4Analyze The UnitsFind how much paint one room needs first; then any number of rooms is just that amount multiplied up.
- Turn the lost cans into lost rooms
- Find the paint per room
- Scale up to 25 rooms