AMC 10 Step by Step

2021 AMC 10A Problem 4

P1-10~2 minPrint

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?

Log in to save your progress.
Show solution without answering

Key insight

The distances 5, 12, 19, ... grow by 7 each second. Find the distance in second 30, then add all thirty by pairing the first with the last.

Solution

  1. Write out the first few seconds to see the pattern:
second1234
inches5121926
  1. To reach second 44 we added 77 three times: 5+3×7=265 + 3 \times 7 = 26. So in second nn the cart has had n−1n - 1 increases of 77.
  2. In second 3030 it travels 5+29×7=5+203=2085 + 29 \times 7 = 5 + 203 = 208 inches.
  3. Now add 5+12+19+⋯+2085 + 12 + 19 + \cdots + 208. Pair the first with the last: 5+208=2135 + 208 = 213. Pair the second with the second-to-last: 12+201=21312 + 201 = 213. Every pair adds to 213213, because one number goes up by 77 exactly when the other goes down by 77.
  4. Thirty numbers make 1515 pairs, so the total is 15×213=319515 \times 213 = 3195 inches.

The answer is (D) 3195\boxed{\textbf{(D)}\ 3195}.

Why this works

A list that grows by the same amount each step is called an arithmetic sequence. Its total is the number of terms times the average of the first and last terms, which is exactly what the pairing shows. Remember that the nnth term has only n−1n - 1 jumps; choice (A), 215215, is what you get if you count 3030 jumps.

Alternative approach

Estimate as a check. The cart goes about halfway between 55 and 208208 inches per second on average, roughly 106106, for 3030 seconds: about 32003200 inches. That rules out 215215, 360360 and 29922992 at once, and the exact pairing above decides between 31953195 and 32423242.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The distance covered in each second is an arithmetic sequence with first term 55 and common difference 77. The 3030th term is
5+29⋅7=5+203=208.5 + 29 \cdot 7 = 5 + 203 = 208.
The total distance is the sum of these 3030 terms, which equals the number of terms times the average of the first and last:
30⋅5+2082=15⋅213=3195.30 \cdot \frac{5 + 208}{2} = 15 \cdot 213 = 3195.

The answer is (D) 3195\boxed{\textbf{(D)}\ 3195}.

Review in the AoPS books

Similar problems

Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.