AMC 8 · 2000 · #10
Grade 6 arithmeticrate-ratioPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The sentence packs three small ideas: a percent, a comparison, and a sum. Tool #7 (Break Into Subproblems) splits the work into three short steps — (a) recover Shea's original height from her current 60 inches, (b) find how many inches Shea grew, (c) take half of that for Ara's growth and add it to Ara's original height. Step (a) is itself a Tool #9 (Work Backwards) move: we know the post-growth value and the growth rate, and we undo the × 1.2 by dividing. No variables or equations are needed — each subproblem is one line of arithmetic.
Undo the growth: 60 is 120% of Shea's old height, so divide by 1.2 to recover her original 50 inches.
If 120% of the old height is 60, then 100% of the old height is = 50 — divide by the growth factor to step back.
6.RP.A.3Solve An Easier Related ProblemSubtract old from new to see how much Shea grew: 60 - 50 = 10 inches.
Growth equals the difference between the new and old heights — a single subtraction once both heights are known.
4.OA.A.3Identify SubproblemsAra grew half as many inches as Shea, so halve the growth: half of 10 is 5 inches.
"Half as many" is a multiplicative comparison — multiply Shea's growth by to get Ara's growth.
4.OA.A.2Identify SubproblemsAra began at the same 50 inches as Shea and grew 5, so Ara is now 55 inches tall.
New height = old height + growth — the three subproblems come together in one addition.
4.OA.A.3Identify SubproblemsA 20% growth means the new value is 1.2 times the old, so dividing by 1.2 undoes it. Once you know Shea grew 10 inches, Ara's "half as many" is just 5 inches on top of the same 50-inch starting height — answer 55, choice (E).