AMC 8 · 2004 · #12

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition

ratefraction-arithmeticlinear-equations-one-var identify-subproblemsconvert-to-algebra ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 2 insights

Problem

Niki usually leaves her cell phone on. If her cell phone is on but
she is not actually using it, the battery will last for $24$ hours. If
she is using it constantly, the battery will last for only $3$ hours.
Since the last recharge, her phone has been on $9$ hours, and during
that time she has used it for $60$ minutes. If she doesn’t use it any
more but leaves the phone on, how many more hours will the battery last?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Niki's phone drains a full battery in $24$ hours if it's on but unused, and in $3$ hours if it's used constantly. Since the last charge, the phone has been on for $9$ hours, with $60$ minutes of actual use. If she stops using it but leaves it on, how many more hours will the battery last?

Givens: Idle (on, not used): full battery lasts $24$ hours; In use: full battery lasts $3$ hours; So far, phone has been on for $9$ hours total; Of that, $60$ minutes $= 1$ hour was active use; After now: idle only (no more use); Answer choices: (A) $7$, (B) $8$, (C) $11$, (D) $14$, (E) $15$

Unknowns: Number of additional hours the battery will last from now

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units

The $9$ hours already past actually contains two different things — $1$ hour of active use and $8$ hours of idle. Tool #7 (Identify Subproblems) splits that mixed interval cleanly, so each piece is handled at its own rate. Tool #8 (Analyze the Units) makes the rates safe to use: "battery per hour" times "hours" gives "battery," which is the unit we want for what's already been spent and what's still left. Once the leftover battery is known, dividing by the idle rate (units again) gives the answer in hours.

Execute — Answer: B

#8 Analyze the Units 6.RP.A.2 Step 1

Set the unit and write each rate.
Call a full battery $1$.
Idle drains it in $24$ hours, so the idle rate is $\tfrac{1}{24}$ battery per hour.
Active use drains it in $3$ hours, so the active rate is $\tfrac{1}{3}$ battery per hour.

$$r_{\text{idle}} = \tfrac{1}{24} \text{ batt/h}, \quad r_{\text{use}} = \tfrac{1}{3} \text{ batt/h}$$

💡 Setting the whole battery to $1$ turns each "lasts $N$ hours" into a Grade 6 unit rate $\tfrac{1}{N}$ per hour. Now the rates can be multiplied or added like ordinary fractions.

#7 Identify Subproblems 4.MD.A.1 Step 2

Split the past $9$ hours into the two subproblems.
Convert $60$ minutes to $1$ hour of active use; the remaining $9 - 1 = 8$ hours were idle.

$$t_{\text{use}} = 1 \text{ h}, \quad t_{\text{idle}} = 9 - 1 = 8 \text{ h}$$

💡 The phone can't be in two states at once, so the $9$ hours has to break into idle hours $+$ use hours. The Grade 4 unit-conversion habit handles the $60$ min $=$ $1$ h step first.

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

Compute the battery spent in each piece.
Idle: $8$ h at $\tfrac{1}{24}$ per hour.
Active: $1$ h at $\tfrac{1}{3}$ per hour.
Units check: $\text{h} \times \tfrac{\text{batt}}{\text{h}} = \text{batt}$.

$$\text{idle: } 8 \times \tfrac{1}{24} = \tfrac{8}{24} = \tfrac{1}{3} \text{ batt} \\ \text{use: } 1 \times \tfrac{1}{3} = \tfrac{1}{3} \text{ batt}$$

💡 Rate $\times$ time $=$ amount used is exactly the Grade 6 rate move. The units cancel cleanly, leaving the answer in "battery," which is what we need.

#7 Identify Subproblems 5.NF.A.1 Step 4

Add the two pieces to get total battery spent, then subtract from $1$ to find what's left.

$$\text{spent} = \tfrac{1}{3} + \tfrac{1}{3} = \tfrac{2}{3} \;\Rightarrow\; \text{left} = 1 - \tfrac{2}{3} = \tfrac{1}{3} \text{ batt}$$

💡 Adding the two same-denominator fractions and then taking the complement from $1$ is straight Grade 5 fraction arithmetic.

#8 Analyze the Units 6.NS.A.1 Step 5

Last subproblem: how long does $\tfrac{1}{3}$ battery last at the idle rate $\tfrac{1}{24}$ batt/h?
Divide remaining battery by the rate; units give hours.

$$t = \dfrac{1/3 \text{ batt}}{1/24 \text{ batt/h}} = \tfrac{1}{3} \times 24 = 8 \text{ h} \;\Rightarrow\; \textbf{(B)}$$

💡 $\tfrac{\text{batt}}{\text{batt/h}}$ leaves hours — exactly what the question asks for. Dividing by $\tfrac{1}{24}$ is the Grade 6 "multiply by the reciprocal" move, giving $\tfrac{1}{3} \times 24 = 8$.

[1] #8 6.RP.A.2 Set the unit and write each rate. Call a full battery $1$. Idle drains it in $24

[2] #7 4.MD.A.1 Split the past $9$ hours into the two subproblems. Convert $60$ minutes to $1$ h

[3] #8 6.RP.A.3 Compute the battery spent in each piece. Idle: $8$ h at $\tfrac{1}{24}$ per hour

[4] #7 5.NF.A.1 Add the two pieces to get total battery spent, then subtract from $1$ to find wh

[5] #8 6.NS.A.1 Last subproblem: how long does $\tfrac{1}{3}$ battery last at the idle rate $\tf

Review

Reasonableness: Sanity check: $\tfrac{1}{3}$ of the battery is left, and a full battery in idle mode lasts $24$ hours, so $\tfrac{1}{3} \times 24 = 8$ more hours — that matches (B). It also matches a quick scaling argument: spending $\tfrac{2}{3}$ of the battery in $9$ hours means $\tfrac{1}{3}$ would last about half of $9$ — but $8$ is slightly less because the remaining time is pure idle (slower drain), so $8$ is in the right ballpark. The largest choices (D) $14$ and (E) $15$ would require less than half the battery used, but $1$ hour of active use alone already drains $\tfrac{1}{3}$ — so they're too large.

Alternative: Tool #13 (Convert to Algebra) gives the same path in one equation. Let $t$ be the extra idle hours. Total battery spent $=$ idle drain $+$ active drain $+$ future idle drain $= 1$: $\tfrac{8}{24} + \tfrac{1}{3} + \tfrac{t}{24} = 1$. Multiply by $24$: $8 + 8 + t = 24$, so $t = 8$. Same answer (B), but with one more line of algebra than the subproblem version.

CCSS standards used (min grade 6)

4.MD.A.1 Know relative sizes of measurement units and convert from a larger unit to a smaller unit (Converting $60$ minutes to $1$ hour so use-time and idle-time share the same unit before splitting the $9$ hours.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Adding $\tfrac{1}{3} + \tfrac{1}{3} = \tfrac{2}{3}$ for total drain, then computing $1 - \tfrac{2}{3} = \tfrac{1}{3}$ for the leftover battery.)
6.RP.A.2 Understand the concept of a unit rate associated with a ratio (Reading "lasts $24$ hours" and "lasts $3$ hours" as unit rates $\tfrac{1}{24}$ and $\tfrac{1}{3}$ battery per hour.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world problems (Multiplying rate $\times$ time to find battery used in each phase ($8 \times \tfrac{1}{24}$ and $1 \times \tfrac{1}{3}$).)
6.NS.A.1 Divide fractions by fractions, including in word problems (Dividing the leftover $\tfrac{1}{3}$ battery by the idle rate $\tfrac{1}{24}$ to get the remaining hours: $\tfrac{1}{3} \div \tfrac{1}{24} = 8$.)

⭐ When a problem mixes two drain rates, split the timeline by state, run each piece at its own rate, then divide what's left by the rate that's still running. Units carry you through every step.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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