AMC 8 · 2003 · #4
Grade 4 arithmeticProblem
A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A group of $7$ children rode past on bicycles and tricycles, and Billy Bob counted $19$ wheels total. How many of the children were riding tricycles?
Givens: There are $7$ children in all; Each child rides exactly one bike or one trike; A bicycle has $2$ wheels; a tricycle has $3$ wheels; The total wheel count is $19$; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Unknowns: The number of tricycles among the $7$ vehicles
Understand
Restated: A group of $7$ children rode past on bicycles and tricycles, and Billy Bob counted $19$ wheels total. How many of the children were riding tricycles?
Givens: There are $7$ children in all; Each child rides exactly one bike or one trike; A bicycle has $2$ wheels; a tricycle has $3$ wheels; The total wheel count is $19$; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $6$, (E) $7$
Plan
Primary tool: #6 Guess and Check
Secondary: #5 Look for a Pattern
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.
Execute — Answer: C
3.OA.A.3 Step 1 - Start from the all-bicycles case as a baseline.
- If every one of the $7$ children rode a bicycle, the wheel count would be $7 \times 2$.
💡 Picking the simplest guess first gives a number to compare $19$ against.
4.OA.A.3 Step 2 - See how much more wheel count we need.
- The target is $19$ wheels, and the all-bikes baseline gives $14$, so we are short by $19 - 14$.
💡 The gap between the guess and the goal is what each "swap" has to close.
4.OA.A.3 Step 3 - Notice the per-swap pattern.
- Replace one bicycle ($2$ wheels) with one tricycle ($3$ wheels), and the total wheel count rises by $3 - 2 = 1$.
- So each swap closes the gap by exactly $1$ wheel.
💡 One swap, one extra wheel — a clean rate that turns the rest of the problem into one division.
3.OA.A.3 Step 4 - Make $5$ swaps to add the missing $5$ wheels.
- That turns $5$ of the $7$ bicycles into tricycles, leaving $7 - 5 = 2$ bicycles.
💡 Closing a gap of $5$ at $1$ wheel per swap takes exactly $5$ swaps.
3.OA.A.3 Start from the all-bicycles case as a baseline. If every one of the $7$ children 4.OA.A.3 See how much more wheel count we need. The target is $19$ wheels, and the all-bi 4.OA.A.3 Notice the per-swap pattern. Replace one bicycle ($2$ wheels) with one tricycle 3.OA.A.3 Make $5$ swaps to add the missing $5$ wheels. That turns $5$ of the $7$ bicycles Review
Reasonableness: Check the totals directly. With $5$ tricycles and $2$ bicycles: children $= 5 + 2 = 7$ (matches), and wheels $= 5 \times 3 + 2 \times 2 = 15 + 4 = 19$ (matches). The answer also has to be between the all-bike total of $14$ and the all-trike total of $21$, and $19$ falls inside that range — so a valid mix must exist. The trap choices $4$, $6$, and $7$ each fail the wheel count: $4$ trikes give $4 \times 3 + 3 \times 2 = 18$, $6$ trikes give $6 \times 3 + 1 \times 2 = 20$, and $7$ trikes give $21$.
Alternative: Tool #2 (Make a Systematic List): try every value of $t$ from $0$ to $7$, computing $3t + 2(7-t) = t + 14$ each time. The list is $14, 15, 16, 17, 18, 19, 20, 21$ — the first to hit $19$ is $t = 5$. Same answer (C), with the bonus that the formula $t + 14$ is exactly the "$+1$ per swap" pattern in compact form.
CCSS standards used (min grade 4)
3.OA.A.3Use multiplication and division within 100 to solve word problems involving equal groups, arrays, and measurement quantities (Computing the baseline wheel count $7 \times 2 = 14$ and the final $5 \times 3 + 2 \times 2 = 19$ check as equal-group multiplications.)4.OA.A.3Solve multistep word problems with whole numbers using the four operations, including problems with remainders (Finding the wheel gap $19 - 14 = 5$ and dividing by the per-swap rate of $1$ wheel to count the swaps needed.)
⭐ Start 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.
⭐ Start 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.
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