AMC 8 · 2010 · #11
Grade 6 rate-ratioProblem
The top of one tree is feet higher than the top of another tree. The heights of the two trees are in the ratio . In feet, how tall is the taller tree?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Two trees stand side by side. The taller tree's top is $16$ feet above the shorter tree's top, and the two heights are in the ratio $3:4$ (short to tall). In feet, how tall is the taller tree?
Givens: Height of the shorter tree : height of the taller tree $= 3 : 4$; The taller tree's top is $16$ feet higher than the shorter tree's top; Answer choices: (A) $48$, (B) $64$, (C) $80$, (D) $96$, (E) $112$ (feet)
Unknowns: The height of the taller tree in feet
Understand
Restated: Two trees stand side by side. The taller tree's top is $16$ feet above the shorter tree's top, and the two heights are in the ratio $3:4$ (short to tall). In feet, how tall is the taller tree?
Givens: Height of the shorter tree : height of the taller tree $= 3 : 4$; The taller tree's top is $16$ feet higher than the shorter tree's top; Answer choices: (A) $48$, (B) $64$, (C) $80$, (D) $96$, (E) $112$ (feet)
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #3 Eliminate Possibilities
The ratio $3:4$ already gives us a tiny version of the problem — two "trees" of heights $3$ and $4$, differing by $1$. Tool #9 (Easier Related Problem) says: solve the small case first, then scale it up until the difference matches the real $16$ feet. That avoids algebra entirely. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net: the taller tree must be divisible by $4$ (so the shorter side $\tfrac{3}{4}$ of it is a whole number), which already cuts the choices down.
Execute — Answer: B
6.RP.A.1 Step 1 - Start with the easier version.
- Pretend the trees are just $3$ feet and $4$ feet tall — they satisfy the $3:4$ ratio.
- In this baby problem the taller tree is $4 - 3 = 1$ foot higher.
💡 Replacing the unknown heights with the ratio numbers themselves is the cleanest "easier problem" — a Grade 6 ratio idea.
4.OA.A.2 Step 2 - Scale the small case up to the real problem.
- The real gap is $16$ feet but the small-case gap is only $1$ foot, so every length must be multiplied by $16$ to keep the same $3:4$ ratio.
💡 "Multiply both numbers by the same factor" is exactly Grade 4 multiplicative comparison.
6.RP.A.3 Step 3 - Apply the scale factor to both trees.
- The shorter tree becomes $3 \times 16 = 48$ feet and the taller tree becomes $4 \times 16 = 64$ feet.
- Check: the ratio is still $48:64 = 3:4$ and the gap is $64 - 48 = 16$ feet — both conditions match.
💡 Scaling a ratio by a common factor preserves it — the heart of Grade 6 ratio reasoning.
6.RP.A.3 Step 4 - Confirm against the answer choices.
- The taller tree is $64$ feet, which is choice (B).
- As a sanity sweep, the taller tree must be a multiple of $4$ for the shorter tree $\tfrac{3}{4} \cdot h$ to be a whole number — that eliminates (C) $80$ as the only nondivisible-by-$4$...
- actually (C) $80$ IS divisible by $4$, so we instead check the gap $\tfrac{1}{4}h = 16$ directly: only $h = 64$ works.
💡 Testing each choice against "$\tfrac{1}{4}$ of the taller tree equals the gap" is the Tool #3 elimination move.
6.RP.A.1 Start with the easier version. Pretend the trees are just $3$ feet and $4$ feet 4.OA.A.2 Scale the small case up to the real problem. The real gap is $16$ feet but the s 6.RP.A.3 Apply the scale factor to both trees. The shorter tree becomes $3 \times 16 = 48 6.RP.A.3 Confirm against the answer choices. The taller tree is $64$ feet, which is choic Review
Reasonableness: Heights of $48$ ft and $64$ ft are realistic for medium-sized trees, and $64 - 48 = 16$ ft matches the given gap exactly. The ratio $48 : 64$ simplifies to $3 : 4$ by dividing both by $16$, so both problem conditions are satisfied. The answer (B) is the only choice that pairs with a whole-number partner $\tfrac{3}{4}\cdot 64 = 48$ differing by exactly $16$.
Alternative: Tool #13 (Convert to Algebra) also works: let the shorter tree be $3k$ and the taller be $4k$ for some scale $k$. Then $4k - 3k = k = 16$, so the taller tree is $4k = 64$ ft. This is the same idea as the easier-problem scaling but written with a variable; for a Grade 4-6 learner the scaling story is faster and friendlier.
CCSS standards used (min grade 6)
4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Recognizing that scaling both tree heights by the same factor ($\times 16$) is a multiplicative-comparison move that preserves the ratio.)6.RP.A.1Understand the concept of a ratio and use ratio language (Reading the $3:4$ ratio as a small-case pair of heights ($3$ and $4$) before scaling up to the real trees.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Finding the scale factor $\tfrac{16}{1} = 16$ that turns the ratio gap into the real $16$-foot gap, then computing both heights as $48$ and $64$ feet.)
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — scale a small $3{:}4$ pair up until the gap matches!
⭐ This AMC 8 problem only needs Grade 6 ratio reasoning — scale a small $3{:}4$ pair up until the gap matches!