AMC 8 · 2011 · #10

Easy mode Grade 6

Problem

In Gotham City, a taxi works like this. The first half mile costs $\$2.40$ no matter what. After that, every extra $0.1$ mile adds another $\$0.20$ .

You have $\$10$ to spend in total. You plan to give the driver a $\$2$ tip out of that.

How many miles can you ride with the money you have left for the fare?

$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

Pick an answer.

(A)

3.0

(B)

3.25

(C)

3.3

(D)

3.5

(E)

3.75

View mode:

Toolkit + CCSS Solution

Understand

Restated: A Gotham City taxi charges $\$2.40$ for the first $\tfrac{1}{2}$ mile, then $\$0.20$ for each additional $0.1$ mile. You have $\$10$ total and want to leave a $\$2$ tip. How many miles can you ride?

Givens: Flat fare = $\$2.40$ covers the first $0.5$ mile; After $0.5$ mile, the rate is $\$0.20$ per $0.1$ mile; Total money available = $\$10$; Planned tip = $\$2$; Answer choices: (A) $3.0$, (B) $3.25$, (C) $3.3$, (D) $3.5$, (E) $3.75$ (miles)

Unknowns: The total number of miles the ride can cover under the $\$10$ budget

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #8 Analyze the Units

The $\$10$ has to do three different jobs — tip, flat fare for the first $0.5$ mile, and the per-$0.1$-mile charge after that — so Tool #7 (Identify Subproblems) cleanly peels them off one at a time: take out the tip, take out the flat fare, then ask "how far does the leftover money buy?" Tool #8 (Analyze the Units) turns the awkward $\$0.20$ per $0.1$ mile rate into the friendlier $\$2$ per mile, which makes the final division a one-line decimal computation rather than an algebra problem.

Execute — Answer: C

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

Subproblem 1 — set aside the tip.
You will hand the driver $\$2$ at the end, so the meter can only use what is left.

$\$10.00 - \$2.00 = \$8.00 \text{ for the meter}$

💡 Splitting the $\$10$ into "tip" and "fare" is the Tool #7 move — handle one job at a time.

#7 Identify Subproblems 5.NBT.B.7 Step 2

Subproblem 2 — pay the flat fare for the first $\tfrac{1}{2}$ mile.
Subtract it from the meter budget; whatever remains has to cover the rest of the trip.

$\$8.00 - \$2.40 = \$5.60 \text{ left for extra distance}$

💡 Subtracting decimals to the hundredths place is a Grade 5 arithmetic move.

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

Convert the meter's rate into a friendlier unit.
$\$0.20$ per $0.1$ mile means every full mile costs ten times $\$0.20$, i.e.
$\$2.00$ per mile.

$\dfrac{\$0.20}{0.1 \text{ mile}} = \dfrac{\$0.20 \times 10}{0.1 \times 10 \text{ mile}} = \dfrac{\$2.00}{1 \text{ mile}}$

💡 Scaling the top and bottom of the rate by $10$ turns "per $0.1$ mile" into "per mile" — a Grade 6 unit-rate move.

#7 Identify Subproblems 5.NBT.B.7 Step 4

Subproblem 3 — figure out how many extra miles the $\$5.60$ buys at $\$2.00$ per mile.

$\dfrac{\$5.60}{\$2.00 / \text{mile}} = 2.8 \text{ miles of extra distance}$

💡 Dividing $5.60$ by $2$ is a direct Grade 5 decimal calculation.

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

Add the flat-fare distance back to the extra distance to get the full ride length.

$$0.5 + 2.8 = 3.3 \text{ miles} \;\Rightarrow\; \textbf{(C)}$$

💡 Re-combining the two subproblem pieces (first $0.5$ mile $+$ extra $2.8$ mile) finishes the Tool #7 split.

[1] #7 4.MD.A.2 Subproblem 1 — set aside the tip. You will hand the driver $\$2$ at the end, so

[2] #7 5.NBT.B.7 Subproblem 2 — pay the flat fare for the first $\tfrac{1}{2}$ mile. Subtract it

[3] #8 6.RP.A.3 Convert the meter's rate into a friendlier unit. $\$0.20$ per $0.1$ mile means e

[4] #7 5.NBT.B.7 Subproblem 3 — figure out how many extra miles the $\$5.60$ buys at $\$2.00$ per

[5] #7 5.NBT.B.7 Add the flat-fare distance back to the extra distance to get the full ride lengt

Review

Reasonableness: Check the dollars: first $0.5$ mile costs $\$2.40$, the extra $2.8$ miles cost $2.8 \times \$2.00 = \$5.60$, and the tip is $\$2.00$. Total $= 2.40 + 5.60 + 2.00 = \$10.00$ exactly — every dollar is accounted for, so $3.3$ miles is right. A quick sanity check on size: at $\$2$ per mile after the first half, an extra $\$5.60$ should buy under $3$ extra miles, so the total ride should be a bit under $3.5$ miles. $3.3$ fits perfectly.

Alternative: Tool #6 (Guess and Check) on the choices. For each candidate distance $x$, the total cost is $\$2.40 + (x - 0.5) \times \$2.00 + \$2.00 \text{ tip}$, and we need this to equal $\$10.00$. (A) $3.0$: $2.40 + 2.5 \times 2 + 2 = \$9.40$ (too cheap). (B) $3.25$: $2.40 + 2.75 \times 2 + 2 = \$9.90$ (still cheap). (C) $3.3$: $2.40 + 2.8 \times 2 + 2 = \$10.00$ — exact match. (D) $3.5$ and (E) $3.75$ overshoot $\$10$, so (C) is the answer.

CCSS standards used (min grade 6)

4.MD.A.2 Solve word problems involving distances, time, liquid volumes, masses, and money (Setting aside the $\$2$ tip from the $\$10$ total ($\$10 - \$2 = \$8$) as a money word-problem subtraction.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (The decimal arithmetic in the body of the solution: $\$8.00 - \$2.40 = \$5.60$, $\$5.60 \div \$2.00 = 2.8$, and $0.5 + 2.8 = 3.3$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Rewriting the meter rate $\$0.20$ per $0.1$ mile as the equivalent unit rate $\$2.00$ per mile, then using that rate to convert the remaining $\$5.60$ into a distance.)

⭐ Big AMC 8 word problems often shrink down to Grade 6 rate reasoning — once you split off the tip and the flat fare, only a tidy decimal division is left.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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