AMC 8 · 2008 · #12
Grade 6 arithmeticProblem
A ball is dropped from a height of 3 meters. On its first bounce it rises to a height of 2 meters. It keeps falling and bouncing to 32 of the height it reached in the previous bounce. On which bounce will it not rise to a height of 0.5 meters?
Pick an answer.
AMC 8 2008 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: A ball is dropped from $3$ m. Its first bounce rises to $2$ m. After that, each bounce rises to $\tfrac{2}{3}$ of the previous bounce's height. Which numbered bounce is the first one that does not reach $0.5$ m?
Givens: Drop height: $3$ m (this is the drop, not a bounce); First bounce height: $2$ m; Each later bounce reaches $\tfrac{2}{3}$ of the bounce before it; Threshold height to compare against: $0.5$ m; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The number $n$ of the first bounce whose height is less than $0.5$ m
Understand
Restated: A ball is dropped from $3$ m. Its first bounce rises to $2$ m. After that, each bounce rises to $\tfrac{2}{3}$ of the previous bounce's height. Which numbered bounce is the first one that does not reach $0.5$ m?
Givens: Drop height: $3$ m (this is the drop, not a bounce); First bounce height: $2$ m; Each later bounce reaches $\tfrac{2}{3}$ of the bounce before it; Threshold height to compare against: $0.5$ m; Answer choices: (A) $3$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #5 Spot a Pattern
Secondary: #2 Make a List
Each bounce is a fixed $\tfrac{2}{3}$ of the one before, so the heights form a geometric pattern. Tool #5 (Spot a Pattern) names that rule and tells us we can keep multiplying by $\tfrac{2}{3}$. Tool #2 (Make a List) is the cleanest way to use that rule: write out the bounce heights as fractions one by one and stop the first time the value drops below $0.5$ m. With only five or six steps to check, a list is faster than setting up an inequality.
Execute — Answer: C
4.OA.C.5 Step 1 - Name the pattern.
- Let $h_n$ be the height of bounce $n$.
- The rule "each bounce is $\tfrac{2}{3}$ of the previous bounce" means $h_{n+1} = \tfrac{2}{3} h_n$, starting from $h_1 = 2$.
💡 Grade 4 "generate a pattern that follows a given rule" — the rule here is multiply by $\tfrac{2}{3}$ each step.
5.NF.B.4 Step 2 - List the heights, multiplying by $\tfrac{2}{3}$ each time.
- Keep the fractions exact so we can compare to $0.5 = \tfrac{1}{2}$ without rounding mistakes.
💡 Grade 5 "multiply a fraction by a fraction": numerator times $2$, denominator times $3$, each step.
4.NF.A.2 Step 3 - Compare each height to $\tfrac{1}{2}$ using a common denominator.
- A fraction $\tfrac{a}{b}$ is less than $\tfrac{1}{2}$ exactly when $2a < b$.
💡 Grade 4 "compare two fractions with different denominators" using cross-multiplication.
6.NS.C.7 Step 4 - Pick the first bounce that fails.
- Bounces $1$ through $4$ all rise to at least $0.5$ m, but bounce $5$ rises only to $\tfrac{32}{81} \approx 0.395$ m, which is below $0.5$ m.
- So bounce $5$ is the first one that does not reach the threshold.
💡 Grade 6 "order rational numbers" — read off the first bounce in the list that lands below $0.5$.
4.OA.C.5 Name the pattern. Let $h_n$ be the height of bounce $n$. The rule "each bounce i 5.NF.B.4 List the heights, multiplying by $\tfrac{2}{3}$ each time. Keep the fractions ex 4.NF.A.2 Compare each height to $\tfrac{1}{2}$ using a common denominator. A fraction $\t 6.NS.C.7 Pick the first bounce that fails. Bounces $1$ through $4$ all rise to at least $ Review
Reasonableness: Decimal check: $h_1 = 2$, $h_2 \approx 1.333$, $h_3 \approx 0.889$, $h_4 \approx 0.593$, $h_5 \approx 0.395$. The heights shrink by about $33\%$ each bounce, so going from $h_4 \approx 0.593$ to $h_5 \approx 0.395$ crossing under $0.5$ matches what we expect. The bounce numbering also matches: the $3$-m drop is not counted, and bounce $1$ is the $2$-m rise, so bounce $5$ is the fifth rebound, which is choice (C).
Alternative: Tool #3 (Set Up an Equation): the closed form is $h_n = 2 \cdot \left(\tfrac{2}{3}\right)^{n-1}$. We want the smallest $n$ with $2 \cdot \left(\tfrac{2}{3}\right)^{n-1} < \tfrac{1}{2}$, i.e. $\left(\tfrac{2}{3}\right)^{n-1} < \tfrac{1}{4}$. Powers of $\tfrac{2}{3}$: $\tfrac{2}{3}, \tfrac{4}{9}, \tfrac{8}{27}, \tfrac{16}{81}$. The fourth power $\tfrac{16}{81}$ is the first one less than $\tfrac{1}{4} = \tfrac{20.25}{81}$, so $n - 1 = 4$ and $n = 5$, again (C).
CCSS standards used (min grade 6)
4.OA.C.5Generate a number pattern that follows a given rule (Reading the problem's "each bounce is $\tfrac{2}{3}$ of the previous bounce" as the rule $h_{n+1} = \tfrac{2}{3} h_n$ for generating successive bounce heights.)5.NF.B.4Multiply a fraction or whole number by a fraction (Computing each bounce height by multiplying the previous height by $\tfrac{2}{3}$, keeping numerator and denominator as exact fractions.)4.NF.A.2Compare two fractions with different numerators and different denominators (Checking whether each $h_n$ is above or below $\tfrac{1}{2}$ by cross-multiplying, which avoids rounding errors.)6.NS.C.7Understand ordering and absolute value of rational numbers (Identifying the first bounce in the list whose height drops below the $0.5$-m threshold.)
⭐ When each step shrinks by the same fraction, listing the values is faster than algebra — multiply by $\tfrac{2}{3}$, compare to $\tfrac{1}{2}$, and stop at the first one that's too small.
⭐ When each step shrinks by the same fraction, listing the values is faster than algebra — multiply by $\tfrac{2}{3}$, compare to $\tfrac{1}{2}$, and stop at the first one that's too small.
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