AMC 8 · 2018 · #17

Easy mode Grade 5

Problem

Bella and Ella are friends who live $2$ miles apart. That is the same as $10{,}560$ feet.

At the exact same moment, they leave their houses and head toward each other. Bella walks, and Ella rides her bicycle.

Picture them moving on the path between their houses. Each one moves at a steady speed the whole way. In any chunk of time, Ella covers $5$ times as much ground as Bella does.

Bella's step is $2 \tfrac{1}{2}$ feet long. So each time Bella's foot hits the ground, she has moved forward $2 \tfrac{1}{2}$ feet.

How many steps will Bella take before she meets Ella on the path?

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

Pick an answer.

(A)

704

(B)

845

(C)

1056

(D)

1760

(E)

3520

View mode:

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

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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 5)

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