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AMC 10 · 2021 · #4

Grade 6 rate-ratio
Archetype Arithmetic Expression Decomposition
sequences-arithmeticpattern-recognitionequal-spacingmean-median-mode-range pattern-recognitionidentify-subproblems ↑ Prerequisites: sequences-arithmetic
📏 Medium solution 💡 2 insights

Problem

A cart rolls down a hill, travelling 55 inches the first second and accelerating so that during each successive 11-second time interval, it travels 77 inches more than during the previous 11-second interval. The cart takes 3030 seconds to reach the bottom of the hill. How far, in inches, does it travel?

Pick an answer.

(A)
~215
(B)
~360
(C)
~2992
(D)
~3195
(E)
~3242
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Toolkit + CCSS Solution

Understand

Restated: A cart rolls $5$ inches in the first second, and each later second it goes $7$ more inches than the second before. After $30$ seconds it hits the bottom. Find the total inches traveled.

Givens: Distance in the $1$st second $= 5$ inches; Each next second adds $7$ inches to the previous second's distance; Total time $= 30$ seconds; Answer choices: (A) $215$, (B) $360$, (C) $2992$, (D) $3195$, (E) $3242$

Unknowns: Total inches traveled in $30$ seconds

Understand

Restated: A cart rolls $5$ inches in the first second, and each later second it goes $7$ more inches than the second before. After $30$ seconds it hits the bottom. Find the total inches traveled.

Givens: Distance in the $1$st second $= 5$ inches; Each next second adds $7$ inches to the previous second's distance; Total time $= 30$ seconds; Answer choices: (A) $215$, (B) $360$, (C) $2992$, (D) $3195$, (E) $3242$

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems

Tool #5 (Pattern) spots the steady $+7$ jumps and the equal-spaced list. Tool #9 (Easier Problem) checks the idea on a tiny version — pair $5$ and $208$, $12$ and $201$, $\dots$, each pair sums to $213$. Tool #7 (Subproblems) then splits the work into two pieces: "what is the last-second distance?" and "how many pairs, each $213$?".

Execute — Answer: D

#5 Look for a Pattern 5.OA.B.3 Step 1
  • Write the first few distances to see the pattern.
  • Second $1: 5$.
  • Second $2: 5 + 7 = 12$.
  • Second $3: 12 + 7 = 19$.
  • Second $4: 26$.
  • Each step adds the same $7$ — a regular ladder.
$5, \;12, \;19, \;26, \;\dots$ \quad (gap $= 7$)

💡 Listing the first few terms makes the pattern visible — Grade 5 "generate a numerical pattern from a rule".

#7 Identify Subproblems 5.OA.A.2 Step 2
  • Find the last second's distance ($30$th term).
  • From the $1$st second to the $30$th, we add $7$ a total of $29$ times.
  • So $a_{30} = 5 + 29 \times 7 = 5 + 203 = 208$ inches.
$$a_{30} = 5 + 29 \times 7 = 5 + 203 = 208$$

💡 Counting how many $7$-jumps separate the first and last term — Grade 5 "write expressions that record calculations".

#9 Solve an Easier Related Problem 5.OA.B.3 Step 3
  • Pair first with last, second with second-to-last, and so on.
  • Each pair sums to the same value: $5 + 208 = 213$, $12 + 201 = 213$, $19 + 194 = 213$, $\dots$.
  • With $30$ terms there are exactly $15$ pairs.
$$15 \text{ pairs}, \; \text{each summing to } 5 + 208 = 213$$

💡 When two numbers walk toward each other at the same pace, their sum stays constant — Grade 5 "identify relationships between two patterns".

#7 Identify Subproblems 5.NBT.B.5 Step 4
  • Total distance is the sum of all $15$ pair sums: $15 \times 213 = 3{,}195$ inches.
  • Matches choice (D).
$$15 \times 213 = 3{,}195 \;\Rightarrow\; \textbf{(D)}$$

💡 Multiplying a two-digit and a three-digit number — Grade 5 "fluently multiply multi-digit whole numbers".

[1] #5 5.OA.B.3 Write the first few distances to see the pattern. Second $1: 5$. Second $2: 5 +
[2] #7 5.OA.A.2 Find the last second's distance ($30$th term). From the $1$st second to the $30$
[3] #9 5.OA.B.3 Pair first with last, second with second-to-last, and so on. Each pair sums to t
[4] #7 5.NBT.B.5 Total distance is the sum of all $15$ pair sums: $15 \times 213 = 3{,}195$ inche

Review

Reasonableness: Average per-second distance is roughly $\tfrac{5 + 208}{2} \approx 106.5$ inches, so over $30$ seconds the total should be about $106.5 \times 30 \approx 3{,}195$. That lands exactly on choice (D). Tiny choices (A) and (B) are obviously wrong — even the last second alone is $208$ inches, far more than $215$.

Alternative: Tool #6 (Guess and Check) plus Tool #3 (Eliminate): the average pace argument already eliminates (A), (B). Choice (C) $2{,}992$ is below the average estimate of $3{,}195$, and (E) $3{,}242$ is above it. Choice (D) sits right at the predicted center, so a single multiplication confirms it.

CCSS standards used (min grade 6)

  • 5.OA.A.2 Write simple expressions that record calculations with numbers (Writing $a_{30} = 5 + 29 \times 7$ to compute the $30$th term.)
  • 5.OA.B.3 Generate two numerical patterns using two given rules and identify relationships (Generating the list $5, 12, 19, 26, \dots$ and noticing the pairing property that opposite terms sum to a constant.)
  • 5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing $15 \times 213 = 3{,}195$ for the total distance.)
  • 6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Using the average per-second distance ($\tfrac{5 + 208}{2}$) times the number of seconds as the total — the "mean $\times$ count" identity.)

⭐ This AMC 10 problem only needs Grade 6 "average $\times$ count = total" — pair the first second ($5$) with the last ($208$), every pair sums to $213$, and $15$ pairs give $3{,}195$ inches.

⭐ This AMC 10 problem only needs Grade 6 "average $\times$ count = total" — pair the first second ($5$) with the last ($208$), every pair sums to $213$, and $15$ pairs give $3{,}195$ inches.

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