AMC 8 · 2006 · #16

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

ratelinear-equations-one-varlcmfraction-arithmetic convert-to-algebraidentify-subproblems ↑ Prerequisites: ratelinear-equations-one-varlcm

📏 Medium solution 💡 3 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.

Before Chandra and Bob start reading, Alice says she would like to team read with them. If they divide the book into three sections so that each reads for the same length of time, how many seconds will each have to read?

Pick an answer.

(A)

6400

(B)

6600

(C)

6800

(D)

7000

(E)

7200

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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A $760$-page novel is split among Alice, Bob, and Chandra so that all three read for the same total time. Alice needs $20$ seconds per page, Bob needs $45$ seconds per page, and Chandra needs $30$ seconds per page. Find that shared reading time, in seconds.

Givens: Total pages: $760$; Alice's rate: $20$ seconds per page; Bob's rate: $45$ seconds per page; Chandra's rate: $30$ seconds per page; Answer choices: (A) $6400$, (B) $6600$, (C) $6800$, (D) $7000$, (E) $7200$

Unknowns: The common time $T$ (in seconds) that each person reads

Understand

Plan

Primary tool: #3 Set Up an Equation

Secondary: #4 Introduce a Variable

The three readers all share one unknown — the time $T$ each spends reading. Tool #4 (Introduce a Variable) names that unknown so we can describe each person's contribution in pages. Tool #3 (Set Up an Equation) ties everything together: the three page counts must add to $760$. Once $T$ appears in a single linear equation, solving for it is straightforward.

Execute — Answer: E

#4 Introduce a Variable 6.EE.B.6 Step 1

Name the shared time.
Let $T$ be the number of seconds each person reads.
This is what we want.

$$T = \text{seconds each person reads}$$

💡 When a problem says "the same amount of time" without giving the number, that quantity gets a letter.

#4 Introduce a Variable 6.RP.A.3 Step 2

Write each person's pages as a rate $\times$ time expression.
"Seconds per page" inverts to "pages per second," so pages read $=$ $T \div$ (seconds per page).

Alice: $\dfrac{T}{20}$ pages,\quad Bob: $\dfrac{T}{45}$ pages,\quad Chandra: $\dfrac{T}{30}$ pages

💡 Unit rates work both ways. If one page takes $20$ seconds, then $T$ seconds covers $T/20$ pages.

#3 Set Up an Equation 6.EE.B.7 Step 3

Set up the equation.
The three page counts must cover the whole book, which is $760$ pages.

$$\dfrac{T}{20} + \dfrac{T}{45} + \dfrac{T}{30} = 760$$

💡 "Parts add to the whole" is the standard work-problem equation.

#3 Set Up an Equation 6.NS.B.4 Step 4

Clear denominators.
The least common multiple of $20$, $45$, $30$ is $180$, so multiply both sides by $180$.

$180 \cdot \dfrac{T}{20} + 180 \cdot \dfrac{T}{45} + 180 \cdot \dfrac{T}{30} = 180 \cdot 760$\\ $9T + 4T + 6T = 180 \cdot 760$\\ $19T = 180 \cdot 760$

💡 Multiplying by the LCM turns the fraction equation into a tidy whole-number one.

#3 Set Up an Equation 6.EE.B.7 Step 5

Solve for $T$.
Notice $760 = 19 \cdot 40$, so the $19$ cancels cleanly.

$$T = \dfrac{180 \cdot 760}{19} = 180 \cdot \dfrac{760}{19} = 180 \cdot 40 = 7200 \;\Rightarrow\; \textbf{(E)}$$

💡 Spotting that $760$ is a multiple of $19$ avoids the big multiplication.

[1] #4 6.EE.B.6 Name the shared time. Let $T$ be the number of seconds each person reads. This i

[2] #4 6.RP.A.3 Write each person's pages as a rate $\times$ time expression. "Seconds per page"

[3] #3 6.EE.B.7 Set up the equation. The three page counts must cover the whole book, which is $

[4] #3 6.NS.B.4 Clear denominators. The least common multiple of $20$, $45$, $30$ is $180$, so m

[5] #3 6.EE.B.7 Solve for $T$. Notice $760 = 19 \cdot 40$, so the $19$ cancels cleanly.

Review

Reasonableness: Plug $T = 7200$ back in. Alice reads $7200/20 = 360$ pages, Bob reads $7200/45 = 160$ pages, Chandra reads $7200/30 = 240$ pages. Total: $360 + 160 + 240 = 760$ pages — exactly the book length. The fastest reader (Alice) handles the most pages and the slowest (Bob) the fewest, which fits the equal-time setup.

Alternative: Tool #5 (Look for a Pattern via ratios): in equal time, pages read are inversely proportional to seconds per page. The reciprocals $\tfrac{1}{20} : \tfrac{1}{45} : \tfrac{1}{30}$ scale by $180$ to $9 : 4 : 6$, a total of $19$ parts. So one part equals $760/19 = 40$ pages, and Alice's $9$ parts mean $9 \cdot 40 = 360$ pages, which at $20$ seconds per page is $360 \cdot 20 = 7200$ seconds. Same answer (E).

CCSS standards used (min grade 6)

6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem (Naming the shared reading time $T$ so each reader's page count can be written as an expression in $T$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Converting "seconds per page" into pages read in time $T$, namely $T/20$, $T/45$, and $T/30$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Building the equation $T/20 + T/45 + T/30 = 760$ and solving it for $T$.)
6.NS.B.4 Find the greatest common factor and least common multiple (Using $\mathrm{lcm}(20, 45, 30) = 180$ to clear denominators in the equation.)

⭐ When the same time hides behind different speeds, name it once and let one equation do the work.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

More like this