AMC 8 · 2002 · #18

Grade 6 rate-ratioalgebra

Archetype Arithmetic Expression Decomposition

mean-median-mode-rangeunit-conversionlinear-equations-one-var convert-to-algebraidentify-subproblems ↑ Prerequisites: multi-digit-arithmeticunit-conversion

📏 Short solution 💡 2 insights

Problem

Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?

Pick an answer.

(A)

1 hr

(B)

1 hr 10 min

(C)

1 hr 20 min

(D)

1 hr 40 min

(E)

2 hr

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Toolkit + CCSS Solution

Understand

Restated: Gage skated $1$ hr $15$ min on each of $5$ days and $1$ hr $30$ min on each of $3$ days. How long must he skate on day $9$ so that his average over all $9$ days is exactly $85$ minutes per day?

Givens: $5$ days at $1$ hr $15$ min $= 75$ minutes per day; $3$ days at $1$ hr $30$ min $= 90$ minutes per day; Target average over $9$ days $= 85$ minutes per day; Answer choices: (A) $1$ hr, (B) $1$ hr $10$ min, (C) $1$ hr $20$ min, (D) $1$ hr $40$ min, (E) $2$ hr

Unknowns: The number of minutes Gage must skate on day $9$, expressed in hours and minutes

Understand

Plan

Primary tool: #4 Introduce a Variable

Secondary: #7 Identify Subproblems

The unknown day-$9$ time is the only mystery quantity, so Tool #4 (Introduce a Variable) names it $x$ and turns the average into a single equation $\dfrac{75 \cdot 5 + 90 \cdot 3 + x}{9} = 85$. Tool #7 (Identify Subproblems) splits the bookkeeping into clean pieces — required total, known total, missing piece — so the arithmetic stays simple. Together they reduce a word problem about averages to one subtraction.

Execute — Answer: E

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

Convert every time to minutes.
$1$ hr $15$ min $= 75$ min and $1$ hr $30$ min $= 90$ min.
Working in one unit removes a common stumble.

$1$ hr $15$ min $= 75$ min, $\;1$ hr $30$ min $= 90$ min

💡 Hours-to-minutes is the Grade 4 unit-conversion move: $1$ hr $= 60$ min, then add the extra minutes.

#7 Identify Subproblems 6.SP.B.5 Step 2

Subproblem 1: required $9$-day total.
By the mean formula, a $9$-day average of $85$ minutes needs a total of $9 \times 85$ minutes.

$$\text{Required total} = 9 \times 85 = 765 \text{ minutes}$$

💡 Average $\times$ count $=$ total — the Grade 6 mean formula run in reverse to find the target sum.

#7 Identify Subproblems 5.NBT.B.5 Step 3

Subproblem 2: minutes already skated on days $1$–$8$.
Add the five $75$s and the three $90$s.

$$5 \times 75 + 3 \times 90 = 375 + 270 = 645 \text{ minutes}$$

💡 Two multiplications plus an addition give the running total after $8$ days.

#4 Introduce a Variable 6.EE.B.7 Step 4

Introduce the variable and solve.
Let $x$ be the minutes Gage skates on day $9$.
The required total equals the known total plus $x$.

$$645 + x = 765 \;\Rightarrow\; x = 765 - 645 = 120 \text{ minutes}$$

💡 Naming the unknown and balancing the equation is the Grade 6 "solve $p + x = q$" move.

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

Convert $120$ minutes back to hours and minutes and match a choice.

$$120 \text{ min} = 2 \text{ hr} \;\Rightarrow\; \textbf{(E)}$$

💡 Dividing $120$ by $60$ gives $2$ hours exactly — the reverse of the conversion in step $1$.

[1] #7 4.MD.A.1 Convert every time to minutes. $1$ hr $15$ min $= 75$ min and $1$ hr $30$ min $=

[2] #7 6.SP.B.5 Subproblem 1: required $9$-day total. By the mean formula, a $9$-day average of

[3] #7 5.NBT.B.5 Subproblem 2: minutes already skated on days $1$–$8$. Add the five $75$s and the

[4] #4 6.EE.B.7 Introduce the variable and solve. Let $x$ be the minutes Gage skates on day $9$.

[5] #7 4.MD.A.1 Convert $120$ minutes back to hours and minutes and match a choice.

Review

Reasonableness: Plug $x = 120$ back into the average. Total minutes $= 375 + 270 + 120 = 765$, and $765 \div 9 = 85$ — exactly the target. The answer also passes a quick sanity check: on the first $5$ days Gage was $85 - 75 = 10$ minutes short each day ($50$ minutes short total), and on the next $3$ days he was $90 - 85 = 5$ minutes over each day ($15$ minutes over total). Net deficit after $8$ days is $50 - 15 = 35$ minutes, so day $9$ must run $85 + 35 = 120$ minutes — the same $2$ hours.

Alternative: Tool #11 (Find an Invariant): the sum of daily deviations from the target average is always $0$. Days $1$–$5$ contribute $5 \times (75 - 85) = -50$ and days $6$–$8$ contribute $3 \times (90 - 85) = +15$, so day $9$ must contribute $+35$ to zero out the running total. That makes day $9$'s time $85 + 35 = 120$ minutes $= 2$ hours, confirming (E).

CCSS standards used (min grade 6)

4.MD.A.1 Know relative sizes of measurement units; convert from a larger unit to a smaller unit (Converting $1$ hr $15$ min to $75$ min, $1$ hr $30$ min to $90$ min, and $120$ min back to $2$ hr.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $5 \times 75 = 375$, $3 \times 90 = 270$, and $9 \times 85 = 765$.)
6.SP.B.5 Summarize numerical data sets, including reporting the number of observations and measures of center (Using mean $\times$ count $=$ total to turn the $85$-minute target average into a required $9$-day total of $765$ minutes.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Letting $x$ be the day-$9$ minutes and solving $645 + x = 765$ to get $x = 120$.)

⭐ A target average sets the required total: $85 \times 9 = 765$ minutes. Subtract the $645$ minutes already skated and only $120$ minutes — exactly $2$ hours — are left for day $9$, answer (E).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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