AMC 8 · 2000 · #10
Grade 6 arithmeticrate-ratioProblem
Ara and Shea were once the same height. Since then Shea has grown 20% while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?
Pick an answer.
AMC 8 2000 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: Ara and Shea used to be the same height. Since then Shea has grown $20\%$, and Ara has grown half as many inches as Shea. Shea is now $60$ inches tall. How tall, in inches, is Ara now?
Givens: Ara and Shea once had the same height; Shea has grown $20\%$ since then; Ara has grown $\dfrac{1}{2}$ as many inches as Shea did; Shea is now $60$ inches tall; Answer choices: (A) $48$, (B) $51$, (C) $52$, (D) $54$, (E) $55$
Unknowns: Ara's current height in inches
Understand
Restated: Ara and Shea used to be the same height. Since then Shea has grown $20\%$, and Ara has grown half as many inches as Shea. Shea is now $60$ inches tall. How tall, in inches, is Ara now?
Givens: Ara and Shea once had the same height; Shea has grown $20\%$ since then; Ara has grown $\dfrac{1}{2}$ as many inches as Shea did; Shea is now $60$ inches tall; Answer choices: (A) $48$, (B) $51$, (C) $52$, (D) $54$, (E) $55$
Plan
Primary tool: #7 Break Into Subproblems
Secondary: #9 Work Backwards
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 $\times 1.2$ by dividing. No variables or equations are needed — each subproblem is one line of arithmetic.
Execute — Answer: E
6.RP.A.3 Step 1 - Subproblem 1: recover Shea's original height.
- Growing by $20\%$ means the new height is $100\% + 20\% = 120\%$ of the old height, so $60 = 1.2 \times \text{(old height)}$.
- Divide both sides by $1.2$ to undo the growth.
💡 If $120\%$ of the old height is $60$, then $100\%$ of the old height is $\tfrac{60}{1.2} = 50$ — divide by the growth factor to step back.
4.OA.A.3 Step 2 - Subproblem 2: find Shea's growth in inches.
- Subtract her old height from her new height.
💡 Growth equals the difference between the new and old heights — a single subtraction once both heights are known.
4.OA.A.2 Step 3 - Subproblem 3: find Ara's growth.
- Ara grew $\dfrac{1}{2}$ as many inches as Shea, so take half of Shea's $10$ inches.
💡 "Half as many" is a multiplicative comparison — multiply Shea's growth by $\tfrac{1}{2}$ to get Ara's growth.
4.OA.A.3 Step 4 Combine: Ara started at the same $50$ inches as Shea and grew $5$ inches, so add them.
💡 New height $=$ old height $+$ growth — the three subproblems come together in one addition.
6.RP.A.3 Subproblem 1: recover Shea's original height. Growing by $20\%$ means the new he 4.OA.A.3 Subproblem 2: find Shea's growth in inches. Subtract her old height from her new 4.OA.A.2 Subproblem 3: find Ara's growth. Ara grew $\dfrac{1}{2}$ as many inches as Shea, 4.OA.A.3 Combine: Ara started at the same $50$ inches as Shea and grew $5$ inches, so add Review
Reasonableness: Sanity check the size: Ara grew less than Shea, so Ara should be shorter than $60$ but taller than $50$. Our answer $55$ sits exactly between $50$ and $60$, which fits. Also verify the percent step by going forward: $1.2 \times 50 = 60$, matching Shea's current height. The trap choices line up with common slips. (A) $48$ treats Shea's $20\%$ growth as a $20\%$ drop from $60$. (D) $54$ takes $20\%$ of Shea's $60$ ($= 12$) and then halves it ($= 6$), forgetting that $20\%$ is measured off the old height. Choosing (E) means the percent was undone correctly before halving.
Alternative: Tool #6 (Guess and Check) on the original height. Since Shea grew $20\%$, the original was a "nice" number whose $1.2$ multiple lands on $60$. Try $50$: $50 + 0.2 \times 50 = 50 + 10 = 60$. That matches, so the original is $50$ and the growth is $10$. Half of $10$ is $5$, giving Ara $50 + 5 = 55$ — same answer (E) with no division at all.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world problems, including finding a percent of a quantity (Reading a $20\%$ growth as a $1.2$ multiplier and recovering the original height by dividing $60 \div 1.2 = 50$.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Turning "half as many inches" into the multiplicative step $\tfrac{1}{2} \times 10 = 5$ for Ara's growth.)4.OA.A.3Solve multistep word problems with the four operations (Subtracting to find Shea's growth ($60 - 50 = 10$) and adding to find Ara's new height ($50 + 5 = 55$).)
⭐ A $20\%$ 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).
⭐ A $20\%$ 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).
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