AMC 10 · 2021 · #4

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

Pick an answer.

(A)
~215
(B)
~360
(C)
~2992
(D)
~3195
(E)
~3242

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

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, …, 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?".

1STEP 1

List the first few: 5, then 5 + 7 = 12, then 19, 26 — each second adds the same 7, a steady ladder.

5, 12, 19, 26, … (gap = 7)
2STEP 2

From the 1st second to the 30th the +7 jump happens 29 times, so the last second is 5 + 29 × 7 = 208 inches.

a₃0 = 5 + 29 × 7 = 5 + 203 = 208
3STEP 3

Pair first with last, second with second-to-last: each pair sums to 5 + 208 = 213, and 30 terms make exactly 15 pairs.

15 pairs, each summing to 5 + 208 = 213
4STEP 4

Add all 15 pair sums: 15 × 213 = 3,195 inches — choice (D).

15 × 213 = 3,195 → (D)
Answer
~3195
Average per-second distance is roughly 5+2082\frac{5 + 208}{2} ≈ 106.5 inches, so over 30 seconds the total should be about 106.5 × 30 ≈ 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.
💡Key takeaway

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