AMC 10 · 2014 · #1
Grade 3 algebraPick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The clean fact is about a made-up future: after adding one nickel the counts are equal. So Tool #11 (Work Backwards) is natural — first describe that easy equal-split state, then undo the added nickel to recover how many pennies and nickels Leah really has. Tool #8 (Analyze the Units) keeps the money straight at the end: a nickel counts as 5 cents and a penny as 1 cent, so the total must be measured in cents, not in coins.
Split the pretend total in half
Pretend the extra nickel is already there: 13+1 = 14 coins, split evenly into 7 nickels and 7 pennies.
"Same number of each" out of a known total just means split the total into two equal groups.
3.OA.A.3Work BackwardsUndo the extra nickel
Remove the pretend nickel: the 7 pennies stay, so Leah really has 6 nickels and 7 pennies, and 6+7 = 13.
To reverse a step, remove exactly what was added and leave everything else alone.
To reverse a step, remove exactly what was added and leave everything else alone.
▸ Why?
Taking away undoes adding, so the original state comes straight back.
▸ Why?
The pile is exactly the original coins plus the extra one, so removing that one leaves the original.
Add up the money in cents
Count value, not coins: 6 × 5 = 30 cents from nickels plus 7 × 1 = 7 cents from pennies gives 30+7 = 37 cents, choice (C).
Coins of different kinds only add up correctly once you turn each into its cent value first.
2.MD.C.8Analyze The UnitsWhen a condition describes an imagined future, build that easy picture first, then undo the change to see what is really true now.
- Split the pretend total in half
- Undo the extra nickel
- Add up the money in cents