AMC 8 · 2006 · #3
Grade 6 rate-ratioProblem
Elisa swims laps in the pool. When she first started, she completed 10 laps in 25 minutes. Now, she can finish 12 laps in 24 minutes. By how many minutes has she improved her lap time?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Elisa used to swim $10$ laps in $25$ minutes. Now she swims $12$ laps in $24$ minutes. How many minutes per lap has she shaved off?
Givens: Old pace: $10$ laps in $25$ minutes; New pace: $12$ laps in $24$ minutes; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{3}{4}$, (C) $1$, (D) $2$, (E) $3$
Unknowns: The improvement in minutes per lap (old time per lap $-$ new time per lap)
Understand
Restated: Elisa used to swim $10$ laps in $25$ minutes. Now she swims $12$ laps in $24$ minutes. How many minutes per lap has she shaved off?
Givens: Old pace: $10$ laps in $25$ minutes; New pace: $12$ laps in $24$ minutes; Answer choices: (A) $\tfrac{1}{2}$, (B) $\tfrac{3}{4}$, (C) $1$, (D) $2$, (E) $3$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The question asks for an improvement in "minutes per lap," but the data is given as totals (laps and minutes mixed together). Tool #8 (Analyze the Units) tells us to convert each pace to the requested unit first — minutes $\div$ laps gives minutes per lap. Tool #7 (Identify Subproblems) splits the work into three small steps: find the old per-lap time, find the new per-lap time, and subtract. Each subproblem is a single division or subtraction.
Execute — Answer: A
6.RP.A.2 Step 1 - Find Elisa's old time per lap.
- Divide total minutes by total laps.
💡 Dividing a total by the count gives the per-unit rate — a Grade 6 unit-rate move.
6.RP.A.2 Step 2 Find Elisa's new time per lap the same way.
💡 Same unit-rate formula, new numbers — $24 \div 12 = 2$ is exact.
5.NBT.B.7 Step 3 - Subtract the new per-lap time from the old per-lap time.
- Both are in minutes per lap, so the subtraction is valid and the answer keeps the same unit.
💡 Improvement is old minus new because Elisa got faster, so the old pace is the larger number.
6.RP.A.2 Find Elisa's old time per lap. Divide total minutes by total laps. 6.RP.A.2 Find Elisa's new time per lap the same way. 5.NBT.B.7 Subtract the new per-lap time from the old per-lap time. Both are in minutes per Review
Reasonableness: Sanity check: Elisa went from $10$ laps to $12$ laps in roughly the same time window, so her per-lap time should drop only a little — half a minute fits that picture. The other choices are too big: an improvement of $1$, $2$, or $3$ minutes per lap would mean the old time was at least $3$ minutes per lap, but $25 \div 10 = 2.5$ rules those out. Choice (B) $\tfrac{3}{4}$ would require the new time to be $2.5 - 0.75 = 1.75$ min/lap, but $24 \div 12 = 2$ exactly, so (A) is the only consistent answer.
Alternative: Tool #3 (Eliminate Possibilities): the old per-lap time is $25/10 = 2.5$ min, and the new per-lap time must be a number between $0$ and $2.5$. Test each choice as the improvement: (A) $0.5 \Rightarrow$ new time $= 2$, and $2 \times 12 = 24$ minutes — matches the given $24$ minutes. No other choice produces a whole-minute total when multiplied by $12$ laps, so (A) is forced.
CCSS standards used (min grade 6)
6.RP.A.2Understand the concept of a unit rate $a/b$ associated with a ratio $a:b$ (Converting each total (laps and minutes) into a unit rate of minutes per lap by dividing minutes by laps.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Subtracting $2.5 - 2 = 0.5$ to get the improvement in minutes per lap.)
⭐ When a problem asks "per lap" or "per minute," convert both rates to that unit first — once the units match, subtraction tells the whole story.
⭐ When a problem asks "per lap" or "per minute," convert both rates to that unit first — once the units match, subtraction tells the whole story.