AMC 8 · 2003 · #4
Grade 4 arithmeticPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only 8 possible tricycle counts (0 through 7), so testing values is faster than algebra. Tool #6 (Guess and Check) makes that direct: pick a tricycle count, pair it with the matching bicycle count, and add the wheels. Tool #5 (Look for a Pattern) makes the search even shorter — every time we swap one bicycle for one tricycle, the total wheel count goes up by exactly 1. That "+1 wheel per swap" rule pins down the answer in a single jump instead of trying every option.
Take all bicycles as the baseline: if all 7 children rode bicycles, the wheels would total 7 × 2 = 14.
Picking the simplest guess first gives a number to compare 19 against.
3.OA.A.3Guess And CheckThe target is 19 wheels and the baseline gives 14, so we are short by 19 - 14 = 5 wheels.
The gap between the guess and the goal is what each "swap" has to close.
4.OA.A.3Look For A PatternSwapping one bicycle (2 wheels) for one tricycle (3 wheels) raises the total by 3 - 2 = 1 wheel per swap.
One swap, one extra wheel — a clean rate that turns the rest of the problem into one division.
4.OA.A.3Look For A PatternClosing the gap of 5 wheels takes 5 swaps, so there are 5 tricycles and 7 - 5 = 2 bicycles.
Closing a gap of 5 at 1 wheel per swap takes exactly 5 swaps.
3.OA.A.3Guess And CheckStart with the simplest guess (all bicycles), then notice that each bike-to-trike swap adds exactly one wheel — the missing wheel count tells you the number of tricycles directly. This AMC 8 problem becomes a Grade 4 multistep word problem, no algebra required.