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AMC 8 · 2018 · #17

Grade 5 rate-ratio
Archetype Meeting Time with Piecewise Rates
rateratio-proportionunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: ratio-proportionunit-conversion
📏 Medium solution 💡 3 insights
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Problem

Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 55 times as fast as Bella walks. The distance between their houses is 22 miles, which is 10,56010,560 feet, and Bella covers 2122 \tfrac{1}{2} feet with each step. How many steps will Bella take by the time she meets Ella?

(A) 704(B) 845(C) 1056(D) 1760(E) 3520\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520

Pick an answer.

(A)
704
(B)
845
(C)
1056
(D)
1760
(E)
3520
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Toolkit + CCSS Solution

Understand

Restated: Bella walks from her house toward Ella's house while Ella bikes toward Bella's house at the same moment. Ella's biking speed is $5$ times Bella's walking speed, and Bella covers $2 \tfrac{1}{2}$ feet per step. Their houses are $2$ miles ($10{,}560$ feet) apart. How many steps has Bella taken by the time they meet?

Givens: Distance between houses $= 2$ miles $= 10{,}560$ feet; Bella's step length $= 2 \tfrac{1}{2} = 2.5$ feet; Ella's speed $= 5 \times$ Bella's speed; Both move at constant speeds, starting at the same time, toward each other; Answer choices: (A) $704$, (B) $845$, (C) $1056$, (D) $1760$, (E) $3520$

Unknowns: The number of steps Bella takes from her start until she meets Ella

Understand

Restated: Bella walks from her house toward Ella's house while Ella bikes toward Bella's house at the same moment. Ella's biking speed is $5$ times Bella's walking speed, and Bella covers $2 \tfrac{1}{2}$ feet per step. Their houses are $2$ miles ($10{,}560$ feet) apart. How many steps has Bella taken by the time they meet?

Givens: Distance between houses $= 2$ miles $= 10{,}560$ feet; Bella's step length $= 2 \tfrac{1}{2} = 2.5$ feet; Ella's speed $= 5 \times$ Bella's speed; Both move at constant speeds, starting at the same time, toward each other; Answer choices: (A) $704$, (B) $845$, (C) $1056$, (D) $1760$, (E) $3520$

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #8 Analyze the Units

We are never told the actual speed in mph or how long the trip takes, so chasing time as a variable is harder than it needs to be. Tool #9 (Easier Related Problem) lets us replace the awkward unit of "time" with a much friendlier unit: "one Bella-step." In the duration of one Bella-step, Bella moves $2.5$ ft and Ella — being $5$ times faster — moves $5 \times 2.5 = 12.5$ ft, so the gap between them shrinks by a fixed amount per Bella-step. Tool #8 (Analyze the Units) then keeps the bookkeeping clean: dividing total feet by feet-per-step cancels to leave plain "steps," exactly what the question asks for.

Execute — Answer: A

#9 Solve an Easier Related Problem 4.NF.C.6 Step 1

Replace the abstract clock with a simpler unit of time: one "Bella-step." During one Bella-step she moves her step length, $2.5$ feet.

$$\text{Bella per step} = 2.5 \text{ ft}$$

💡 Recognizing $2 \tfrac{1}{2}$ as the decimal $2.5$ is a Grade 4 mixed-number-to-decimal skill.

#9 Solve an Easier Related Problem 4.OA.A.2 Step 2
  • In that same "one Bella-step" duration, Ella — going $5$ times as fast — covers $5$ times as much distance as Bella.
  • "$5$ times as fast" is a multiplicative comparison.
$$\text{Ella per Bella-step} = 5 \times 2.5 = 12.5 \text{ ft}$$

💡 "$N$ times as fast" is the classic Grade 4 multiplicative-comparison setup: same time, $N$ times the distance.

#8 Analyze the Units 5.NBT.B.7 Step 3
  • Because they walk and ride toward each other, the gap between them shrinks by the sum of their per-step distances.
  • Add the two decimals to get the gap-closure per Bella-step.
$$\text{Gap closed per Bella-step} = 2.5 + 12.5 = 15 \text{ ft}$$

💡 Adding two decimals to hundredths to get $15.0$ ft is the Grade 5 decimal-arithmetic standard.

#8 Analyze the Units 5.NBT.B.6 Step 4
  • Bella keeps stepping until the original $10{,}560$-ft gap is fully closed.
  • Divide total feet by feet-per-Bella-step; the "ft" units cancel and "Bella-steps" remains, which is exactly the number the problem asks for.
$$\dfrac{10{,}560 \text{ ft}}{15 \text{ ft/step}} = 704 \text{ steps} \;\Rightarrow\; \textbf{(A)}$$

💡 Dividing a $4$-digit number by a $2$-digit divisor is the Grade 5 long-division standard; tracking units confirms the answer is in steps.

[1] #9 4.NF.C.6 Replace the abstract clock with a simpler unit of time: one "Bella-step." During
[2] #9 4.OA.A.2 In that same "one Bella-step" duration, Ella — going $5$ times as fast — covers
[3] #8 5.NBT.B.7 Because they walk and ride toward each other, the gap between them shrinks by th
[4] #8 5.NBT.B.6 Bella keeps stepping until the original $10{,}560$-ft gap is fully closed. Divid

Review

Reasonableness: Sanity check the magnitude. Between them they close $15$ ft for every one of Bella's $2.5$-ft steps, so Bella personally covers only $\tfrac{2.5}{15} = \tfrac{1}{6}$ of the trip — about $\tfrac{10{,}560}{6} = 1{,}760$ ft, or one-third of a mile. At $2.5$ ft per step that is $1760 / 2.5 = 704$ steps, matching (A). It also matches the ratio intuition: if Ella is $5\times$ faster, Bella covers $\tfrac{1}{1+5} = \tfrac{1}{6}$ of the gap. Choice (D) $1760$ is the trap that confuses "feet Bella walks" with "steps Bella takes," and choice (E) $3520$ is the trap of using half the distance ($1$ mile) without the speed ratio.

Alternative: Tool #13 (Convert to Algebra): let $t$ be the meeting time and $v$ be Bella's speed in ft/sec. Then Bella walks $vt$ ft and Ella rides $5vt$ ft, with $vt + 5vt = 10{,}560$, so $vt = 1{,}760$ ft. Bella's steps $= 1{,}760 / 2.5 = 704$. The algebra works but is overkill — the "Bella-step as time unit" reframing reaches the answer with no variable at all.

CCSS standards used (min grade 5)

  • 4.NF.C.6 Use decimal notation for fractions with denominators 10 or 100 (Converting Bella's step length from the mixed number $2 \tfrac{1}{2}$ ft into the decimal $2.5$ ft so the later arithmetic stays clean.)
  • 4.OA.A.2 Multiply or divide to solve word problems involving multiplicative comparison (Translating "Ella is $5$ times as fast as Bella" into "Ella covers $5 \times 2.5 = 12.5$ ft in the time Bella covers $2.5$ ft" — a same-time, $5$-times-the-distance multiplicative comparison.)
  • 5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Adding the two per-step distances $2.5 + 12.5 = 15$ ft to find how fast the gap closes per Bella-step.)
  • 5.NBT.B.6 Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing the total $10{,}560$ ft of separation by the $15$ ft closed per Bella-step to obtain $704$ steps.)

⭐ This AMC 8 problem only needs Grade 5 decimal arithmetic and "times as fast" comparison you already know!

⭐ This AMC 8 problem only needs Grade 5 decimal arithmetic and "times as fast" comparison you already know!

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