AMC 10 · 2012 · #19

Grade 8 rate-ratio

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

ratesystems-of-equationsunit-conversion convert-to-algebradimensional-analysis ↑ Prerequisites: ratesystems-of-equations

📏 Medium solution 💡 3 insights

Problem

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2012 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: Paula and her two helpers paint at constant but different rates and take the same lunch break each day. Monday all three worked 8:00 AM to 4:00 PM and painted $50\%$ of a house. Tuesday the two helpers alone worked 8:00 AM to 2:12 PM and painted $24\%$. Wednesday Paula alone worked 8:00 AM to 7:12 PM and finished the rest. Find the length, in minutes, of each day's lunch break.

Givens: All three rates are constant; the lunch break has the same length $L$ every day; Monday (all three, 8:00 AM to 4:00 PM): $50\%$ painted; Tuesday (two helpers only, 8:00 AM to 2:12 PM): $24\%$ painted; Wednesday (Paula only, 8:00 AM to 7:12 PM): the house is finished; Answer choices in minutes: (A) $30$, (B) $36$, (C) $42$, (D) $48$, (E) $60$

Unknowns: The lunch break length $L$, in minutes

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #8 Analyze the Units, #16 Change Focus / Count the Complement, #13 Convert to Algebra

Three days, three unknowns — Paula's rate $p$, the helpers' combined rate $h$, and the lunch length $L$ — so tool #4 (Introduce a Variable) is the backbone: name the three rates and the break, then write one equation per day using rate $\times$ working time $=$ fraction painted. Tool #8 (Analyze the Units) keeps the bookkeeping honest: rates are in house/hour, so every clock span must be turned into hours and the break subtracted before multiplying. The clever shortcut is tool #16 (Change Focus): notice Tuesday's $24\%$ plus Wednesday's $26\%$ equals Monday's $50\%$, so adding the Tuesday and Wednesday equations must reproduce the Monday equation — that cancels the products and hands over a clean rate ratio $h = \tfrac{16}{9}p$. Tool #13 (Convert to Algebra) then finishes: substitute to get one equation in $L$ alone.

Execute — Answer: D

#8 Analyze the Units 4.MD.A.1 Step 1

Turn clock spans into working hours

Let $p$ be Paula's rate and $h$ the two helpers' combined rate, both in house per hour, and let $L$ be the lunch break in hours.
Convert each day's clock span to hours (12 minutes $= 0.2$ hour), then the actual working time is the span minus $L$.
Monday 8:00 AM to 4:00 PM is $8$ hours; Tuesday 8:00 AM to 2:12 PM is $6.2$ hours; Wednesday 8:00 AM to 7:12 PM is $11.2$ hours.

$$\text{Mon span}=8,\quad \text{Tue span}=6.2,\quad \text{Wed span}=11.2 \text{ (hours)}$$

💡 Rates are per hour, so every time must be measured in hours before it can be multiplied.

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

Write one equation per day

Each day, (rate) $\times$ (working time) $=$ (fraction of house painted).
Monday all three paint together, so their rate is $p+h$ and they paint $0.50$.
Tuesday only the helpers work, painting $0.24$.
Wednesday only Paula works, and since $50\%+24\%=74\%$ is already done, she paints the remaining $0.26$.

$$(p+h)(8-L)=0.50,\quad h(6.2-L)=0.24,\quad p(11.2-L)=0.26$$

💡 Constant rate means work done is just rate multiplied by time on the clock, minus lunch.

#16 Change Focus / Count the Complement 8.EE.C.8 Step 3

Add Tuesday and Wednesday

Change focus: Tuesday's $0.24$ plus Wednesday's $0.26$ equals $0.50$, exactly Monday's amount.
So the sum of the Tuesday and Wednesday equations equals the Monday equation.
Expanding both sides, the $-pL$ and $-hL$ terms cancel, leaving a plain relation between the two rates.

$$h(6.2-L)+p(11.2-L)=(p+h)(8-L)\;\Rightarrow\; 6.2h+11.2p=8h+8p\;\Rightarrow\; 3.2p=1.8h\;\Rightarrow\; h=\tfrac{16}{9}p$$

💡 Because the percentages add up, combining two days must rebuild the third — that cancels the unknown products.

#13 Convert to Algebra 8.EE.C.7 Step 4

Solve for the lunch break

Solve the Wednesday equation for $p$ and, using $h=\tfrac{16}{9}p$, solve the Tuesday equation for $p$ as well.
Setting the two expressions equal eliminates $p$ and leaves one equation in $L$.
Cross-multiplying gives a linear equation.

$$\dfrac{0.26}{11.2-L}=\dfrac{0.135}{6.2-L}\;\Rightarrow\; 0.26(6.2-L)=0.135(11.2-L)\;\Rightarrow\; 0.1=0.125\,L\;\Rightarrow\; L=0.8$$

💡 Two formulas for the same rate must match, and that match pins down the only unknown left.

#8 Analyze the Units 4.MD.A.1 Step 5

Convert back to minutes

The lunch break is $L=0.8$ hour.
Convert to minutes by multiplying by $60$.

$$0.8 \times 60 = 48 \text{ minutes} \;\Rightarrow\; \textbf{(D)}$$

💡 Answer choices are in minutes, so the hour value must be converted back before reading off the letter.

[1] #8 4.MD.A.1 Let $p$ be Paula's rate and $h$ the two helpers' combined rate, both in house pe

[2] #4 6.RP.A.3 Each day, (rate) $\times$ (working time) $=$ (fraction of house painted). Monday

[3] #16 8.EE.C.8 Change focus: Tuesday's $0.24$ plus Wednesday's $0.26$ equals $0.50$, exactly Mo

[4] #13 8.EE.C.7 Solve the Wednesday equation for $p$ and, using $h=\tfrac{16}{9}p$, solve the Tu

[5] #8 4.MD.A.1 The lunch break is $L=0.8$ hour. Convert to minutes by multiplying by $60$.

Review

Reasonableness: Back-substitute $L=0.8$: Paula's rate is $p=0.26/(11.2-0.8)=0.26/10.4=0.025$ and the helpers' rate is $h=0.24/(6.2-0.8)=0.24/5.4=0.0444$, which matches $\tfrac{16}{9}(0.025)=0.0444$. Check Monday: $(0.025+0.0444)(8-0.8)=0.0694\times7.2=0.50$, exactly the $50\%$ given. All three equations hold, so $L=0.8$ hour $=48$ minutes is consistent. Eliminate distractors: $L=48$ falls between the choices $30$ and $60$ as expected for a full lunch break, and choices like (A) $30$ or (E) $60$ would break the Monday check.

Alternative: Tool #6 (Guess and Check): since the answer is one of five given minute values, test each. For a candidate $L$, compute $p$ from Wednesday and $h$ from Tuesday, then check whether Monday's $(p+h)(8-L)=0.50$ holds. Only $L=48$ minutes $=0.8$ hour satisfies it, giving the same answer without the algebraic elimination.

CCSS standards used (min grade 8)

4.MD.A.1 Convert measurement units within a given system (Turning each day's clock span (and the final answer) between hours and minutes, e.g. 12 minutes = 0.2 hour and 0.8 hour = 48 minutes.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world problems (Writing each day's painted fraction as rate times working time, giving the three equations.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Adding the Tuesday and Wednesday equations to match Monday and reduce the system to the rate relation $h=\tfrac{16}{9}p$.)
8.EE.C.7 Solve linear equations in one variable (Cross-multiplying the two rate expressions to get $0.1=0.125L$ and solving $L=0.8$ hour.)

⭐ Write rate times working time for each day, notice Tuesday's $24\%$ plus Wednesday's $26\%$ rebuilds Monday's $50\%$, and the equations collapse to one line giving a $48$-minute lunch.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Turn clock spans into working hours

Write one equation per day

Add Tuesday and Wednesday

Solve for the lunch break

Convert back to minutes

Review

CCSS standards used (min grade 8)

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