AMC 8 · 2004 · #12

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

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

📏 Medium solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Call a full battery 1. Idle empties it in 24 hours, so $\frac{1}{24}$ per hour; constant use empties it in 3 hours, so $\frac{1}{3}$ per hour.

r_idle =

\frac{1}{24}

batt/h, r_use =

\frac{1}{3}

batt/h

2STEP 2

Turn 60 minutes into 1 hour of use, so the 9 hours splits into 1 hour active and 8 hours idle.

t_use = 1 h, t_idle = 9 - 1 = 8 h

3STEP 3

Rate × time gives the battery spent: idle 8 × $\frac{1}{24}$ = $\frac{1}{3}$ , active 1 × $\frac{1}{3}$ = $\frac{1}{3}$ .

idle: 8 ×

\frac{1}{24}

\frac{8}{24}

\frac{1}{3}

batt ; use: 1 ×

\frac{1}{3}

\frac{1}{3}

batt

4STEP 4

Add the pieces: $\frac{1}{3}$ + $\frac{1}{3}$ = $\frac{2}{3}$ spent, so 1 − $\frac{2}{3}$ = $\frac{1}{3}$ of the battery is left.

spent =

\frac{1}{3}

\frac{1}{3}

\frac{2}{3}

→ left = 1 -

\frac{2}{3}

\frac{1}{3}

batt

5STEP 5

The $\frac{1}{3}$ left at idle rate $\frac{1}{24}$ lasts ( $\frac{1}{3}$ ) ÷ ( $\frac{1}{24}$ ) = $\frac{1}{3}$ × 24 = 8 hours → (B).

t = (

\frac{1}{3}

batt)/(

\frac{1}{24}

batt/h) =

\frac{1}{3}

× 24 = 8 h → (B)

Answer

Sanity check:

\frac{1}{3}

of the battery is left, and a full battery in idle mode lasts 24 hours, so

\frac{1}{3}

× 24 = 8 more hours — that matches (B). It also matches a quick scaling argument: spending

\frac{2}{3}

of the battery in 9 hours means

\frac{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

\frac{1}{3}

— so they're too large.

💡Key takeaway

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.

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