A cart rolls down a hill, travelling 5 inches the first second and accelerating so that during each successive 1-second time interval, it travels 7 inches more than during the previous 1-second interval. The cart takes 30 seconds to reach the bottom of the hill. How far, in inches, does it travel?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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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.3Step 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.2Step 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.
💡 Multiplying a two-digit and a three-digit number — Grade 5 "fluently multiply multi-digit whole numbers".
[1]
#5 5.OA.B.3Write the first few distances to see the pattern. Second $1: 5$. Second $2: 5 +
[2]
#7 5.OA.A.2Find the last second's distance ($30$th term). From the $1$st second to the $30$
[3]
#9 5.OA.B.3Pair first with last, second with second-to-last, and so on. Each pair sums to t
[4]
#7 5.NBT.B.5Total 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.