AMC 8 · 2006 · #14

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

ratemulti-digit-arithmeticformula-substitution identify-subproblems ↑ Prerequisites: ratemulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Problems 14, 15 and 16 involve Mrs. Reed's English assignment.

A Novel Assignment

The students in Mrs. Reed's English class are reading the same $760$ -page novel. Three friends, Alice, Bob and Chandra, are in the class. Alice reads a page in 20 seconds, Bob reads a page in $45$ seconds and Chandra reads a page in $30$ seconds.

If Bob and Chandra both read the whole book, Bob will spend how many more seconds reading than Chandra?

Pick an answer.

(A)

7,600

(B)

11,400

(C)

12,500

(D)

15,200

(E)

22,800

AMC 8 2006 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.

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Toolkit + CCSS Solution

Understand

Restated: A $760$-page novel is read by both Bob and Chandra. Bob takes $45$ seconds per page; Chandra takes $30$ seconds per page. Find how many more total seconds Bob spends reading the whole book than Chandra does.

Givens: The book has $760$ pages; Bob reads a page in $45$ seconds; Chandra reads a page in $30$ seconds; Answer choices: (A) $7{,}600$, (B) $11{,}400$, (C) $12{,}500$, (D) $15{,}200$, (E) $22{,}800$

Unknowns: Bob's total reading time minus Chandra's total reading time, in seconds

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #7 Identify Subproblems

This is a rate problem, so Tool #8 (Analyze the Units) is the natural anchor: pages $\times$ (seconds / page) cancels to seconds, which tells us how to combine $760$ pages with each reader's per-page rate. Tool #7 (Identify Subproblems) splits the main question into two clean pieces — Bob's total time and Chandra's total time — that we then subtract. The factoring shortcut $760 \cdot 45 - 760 \cdot 30 = 760(45 - 30)$ is just the distributive property applied after the subproblems are written down, so no algebra tool is needed.

Execute — Answer: B

#8 Analyze the Units 6.RP.A.3 Step 1

Track units to set up each reader's total time.
"Pages" times "seconds per page" gives seconds, which is exactly what the question asks for.

$$\text{total seconds} \;=\; \text{pages} \times \dfrac{\text{seconds}}{\text{page}}$$

💡 Grade 6 unit-rate reasoning: the "page" units cancel and only seconds remain, so the setup matches the question.

#7 Identify Subproblems 5.NBT.B.5 Step 2

Split the main question into two subproblems.
Subproblem 1: how long does Bob take?
Subproblem 2: how long does Chandra take?
The final answer is the first minus the second.

$$\text{Bob's time} = 760 \times 45,\quad \text{Chandra's time} = 760 \times 30$$

💡 Naming each subproblem keeps the work organized and exposes the shared factor $760$, which the next step will exploit.

#7 Identify Subproblems 5.OA.A.1 Step 3

Subtract the two times.
Both products share the factor $760$, so factor it out using the distributive property — this avoids computing two large products separately.

$$760 \times 45 - 760 \times 30 \;=\; 760 \times (45 - 30) \;=\; 760 \times 15$$

💡 The distributive property reads $a \cdot c - b \cdot c = (a - b) \cdot c$. Recognizing the common factor turns two multiplications into one.

#8 Analyze the Units 5.NBT.B.5 Step 4

Finish the arithmetic.
$760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 3800 = 11{,}400$.
Matching to the answer choices gives (B).

$$760 \times 15 \;=\; 7600 + 3800 \;=\; 11{,}400 \;\Rightarrow\; \textbf{(B)}$$

💡 Breaking $15$ into $10 + 5$ is the Grade 5 "multiply by parts" strategy, which keeps the arithmetic mental-math friendly.

[1] #8 6.RP.A.3 Track units to set up each reader's total time. "Pages" times "seconds per page"

[2] #7 5.NBT.B.5 Split the main question into two subproblems. Subproblem 1: how long does Bob ta

[3] #7 5.OA.A.1 Subtract the two times. Both products share the factor $760$, so factor it out u

[4] #8 5.NBT.B.5 Finish the arithmetic. $760 \times 15 = 760 \times 10 + 760 \times 5 = 7600 + 38

Review

Reasonableness: Sanity check the magnitude. Bob is slower by $15$ seconds on every page, and there are $760$ pages, so the gap should be roughly $760 \times 15 \approx 11{,}000$ seconds. The exact value $11{,}400$ lines up. Bob's full time is $760 \times 45 = 34{,}200$ seconds and Chandra's is $760 \times 30 = 22{,}800$ seconds; their difference is $34{,}200 - 22{,}800 = 11{,}400$, which agrees with the factored computation. The units are seconds, matching what the problem asks for.

Alternative: Tool #13 (Convert to Algebra): let $p = 760$ be the page count and let $b = 45$, $c = 30$ be the per-page rates. The required difference is $pb - pc = p(b - c) = 760(45 - 30) = 760 \cdot 15 = 11{,}400$. The algebra makes the common-factor structure explicit, but the arithmetic path above arrives at the same place without introducing letters.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Reading $45$ seconds per page and $30$ seconds per page as unit rates and multiplying by $760$ pages to get each reader's total time in seconds.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $760 \times 15 = 11{,}400$ by splitting $15$ into $10 + 5$ and adding $7600 + 3800$.)
5.OA.A.1 Use parentheses, brackets, or braces in numerical expressions and evaluate them (Rewriting $760 \times 45 - 760 \times 30$ as $760 \times (45 - 30)$ via the distributive property so the subtraction happens inside the parentheses first.)

⭐ When two products share a factor like $760$, pull it out first: $760 \times 45 - 760 \times 30 = 760 \times 15$. The hard-looking subtraction becomes one easy multiplication.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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