Bikes and Trikes

Generated March 24, 2026

Standard: 4.OA.A.3 — Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted.

Submissions: 15 students

Class average: 47%

Gap summary: 9 of 15 students got Question 2 wrong

Misconception: Students divided or split the total wheel count (30) without accounting for the fact that bikes contribute 2 wheels each and trikes contribute 3 wheels each — most commonly dividing 30 by 2 to get 15 of each vehicle type.

Comparison type: Contrast Type B (correct vs. common error) — the 60% error rate on Q2 and the clear, teachable nature of the equal-split misconception make this ideal for a correct-vs-error comparison.

Teaching point: When two things contribute different amounts to a total, you cannot just split the total evenly — you have to account for how much each one contributes.

Rationale: Cupid's Q2 error is especially instructive because their Q1 work demonstrates strong multiplication (7x2=14, 4x3=12) — they CAN think multiplicatively but lost that structure when working backward from a total.

Student Monitoring Table

	Student	Score	Lookfors
🔴	Bert	50%	Check: uses multiplication (x2 and x3) separately for each vehicle type, not division of total
🟢	Big Bird	100%	Check: can explain WHY 15+15 doesn't work even though 15+15=30
🔴	Blitzen	0%	Check: recognizes that bikes and trikes have DIFFERENT wheel counts before computing
🔴 ⭐	Cupid	50%	Check: applies the same multiplication structure from Q1 (x2, x3) to the new problem
🔴	Dancer	50%	Check: re-reads Q2 as a NEW problem — not reusing Q1 numbers
🔴	Dasher	0%	Check: can set up any organized approach (drawing, table, or multiplication) for the problem
🔴	Donnor	0%	Check: can explain what "same number of bikes as trikes" means and starts from that constraint
🟢	Elmo	100%	Check: can articulate why keeping the two rates separate matters
🟢	Ernie	100%	Check: can explain the pair-based approach (2+3=5 wheels per pair)
🟢 ⭐	Gonzo	100%	Check: can verify a solution by multiplying each vehicle count by its wheel count
🟢	Kermit	100%	Check: can explain their systematic trial approach clearly to a partner
🟢	Miss Piggie	100%	Check: can connect their skip-counting strategy to the multiplication check
🔴	Prancer	50%	Check: recognizes that 30÷2 and 30÷3 don't work because both types SHARE the 30 wheels
🔴	Rudolph	0%	Check: can draw a picture model with pairs (bikes) and triples (trikes) to represent the problem
🔴	Snuffelupagus	0%	Check: attempts any organized strategy (drawing, listing, or multiplication) rather than guessing

Presenter Notes

Slide 1: Shout-Out

Shout-outs: Cupid (multiplication structure: 7x2=14, 4x3=12) and Ernie (circle drawing model)
Both demonstrate the skill that is the teaching point: keeping different group sizes separate
Data: 10/15 correct on Q1, 14/15 showed work

Slide 2: Frame the Task

Cold call (read-aloud): Bert
- ZPD: Mastered Q1 with strong multiplication, no reading difficulty. In the gap group for Q2, so engaging with the question text primes him for the comparison.
Display the full Q2 text verbatim, including the setup about the cycle shop
Performance split: 6 of 15 got it right

Slide 3: Side-by-Side Display + Think Time

Student 1 (left): Gonzo's Q2 work — 6x2=12, 6x3=18, 18+12=30
Student 2 (right): Cupid's Q2 work — 30÷2=15 with multiple crossed-out attempts
Observational prompt: "Pay attention to what each student did with the number of wheels"
35-second silent think time with timer

Slide 4: Finger Check

Evaluative question: "Which student found the right answer?"
Count: "Show me on three... one, two, three"
Scan for students who show 2 — those students may share the equal-split misconception
Expect most to show 1 after think time

Slide 5: Compare — Turn and Talk + Debrief

Phase 1: Pose both questions, set up turn and talk
Phase 2: 90-second partner talk with timer
Phase 3 — Debrief:
- Q1 "What's the same?" — Cold call Elmo
  - ZPD: Mastery group, solved Q2 with systematic trial. Can articulate shared understanding.
  - Exemplar: "Both students knew there were 30 wheels and that the number of bikes had to equal the number of trikes."
  - Scaffolds: "What information from the problem did both students use?" / "Did both students know bikes and trikes were involved?"
- Q2 "What's different?" — Cold call Dancer
  - ZPD: Gap group, mastered Q1 multiplication but copied Q1 values for Q2. Ideal for noticing the structural difference.
  - Exemplar: "Student 1 multiplied 6 times 2 and 6 times 3 separately, then added. Student 2 just divided 30 by 2 — they treated all the cycles the same instead of keeping bikes and trikes separate."
  - Scaffolds: "Why did Student 1 multiply by 2 AND by 3?" / "What does Student 2's division by 2 assume about the wheels?"

Slide 6: Takeaway Stamp

Phase 1 — Teacher call: Mr. Bravo selects 1-2 students to articulate the big idea
- Nudge: "Mr. Bravo, who wants to try to put the big idea into their own words?"
- Target response: something about keeping the different wheel counts separate, not just dividing evenly
Phase 2: Teaching point + bar diagram visual
- Stamp the correct approach (Student 1): "Six bikes means six times two — twelve wheels. Six trikes means six times three — eighteen wheels. Twelve plus eighteen is thirty."
- Do NOT reference Student 2's error again. Stick the landing with the correct approach.

Slide 7: Release to IP

Students receive personalized worksheets (Worksheet A for "Needs Teaching Point" group, Worksheet B for "Mastery" group)
Primary lookfor: Does the student multiply each vehicle/item type by its rate SEPARATELY before combining?
Use the student monitoring table above as a checklist — visit 🔴 students first

Answer Keys

Worksheet A (Needs Teaching Point)

Problem 1 (for all Worksheet A students): Kermit's Pet Shop sells hamsters and parrots. Hamsters have 4 legs. Parrots have 2 legs. Kermit counted 5 hamsters and 3 parrots. How many legs in all? Answer: 5 × 4 = 20 hamster legs, 3 × 2 = 6 parrot legs, 20 + 6 = 26 legs

Problem 2 (for all Worksheet A students): Big Bird counted 24 legs total at the pet shop. There were the same number of hamsters as parrots. How many hamsters? How many parrots? Answer: Same number means try equal counts: if 4 of each, 4 × 4 = 16 hamster legs + 4 × 2 = 8 parrot legs = 24. 4 hamsters and 4 parrots.

Problem 3 (for all Worksheet A students): Ernie says there are 3 hamsters and 3 parrots, and that makes 24 legs because 3 + 3 = 6 and 6 × 4 = 24. Is Ernie right? Explain why or why not. Answer: Ernie is wrong. He multiplied the total number of animals (6) by 4, but only hamsters have 4 legs. Parrots have 2 legs. The correct calculation: 3 × 4 = 12 hamster legs + 3 × 2 = 6 parrot legs = 18 legs, not 24.

Finished Early? Bonus: Gonzo's Go-Kart Track has go-karts (4 wheels) and motorcycles (2 wheels). There are 18 wheels total. Same number of go-karts as motorcycles. How many of each? Can you find a DIFFERENT combination of go-karts and motorcycles that also makes 18 wheels? Answer: Same number: each pair = 4 + 2 = 6 wheels. 18 ÷ 6 = 3. 3 go-karts and 3 motorcycles. Check: 3 × 4 = 12 + 3 × 2 = 6 = 18. Other combinations: 1 go-kart + 7 motorcycles (4 + 14 = 18), or 2 go-karts + 5 motorcycles (8 + 10 = 18).

Worksheet B (Demonstrated Mastery)

Problem 1 (for all Worksheet B students): Miss Piggie's Farm has chickens (2 legs) and cows (4 legs). There are 40 legs total. There are twice as many chickens as cows. How many chickens? How many cows? Answer: Let cows = C, chickens = 2C. Then 4C + 2(2C) = 4C + 4C = 8C = 40, so C = 5. 5 cows and 10 chickens. Check: 5 × 4 = 20 + 10 × 2 = 20 = 40 legs.

Problem 2 (for all Worksheet B students): Dancer's Parking Garage has cars (4 wheels) and motorcycles (2 wheels). There are 30 wheels total. There are 3 more cars than motorcycles. How many cars? How many motorcycles? Answer: Try systematically: if 3 motorcycles and 6 cars, that's 3 × 2 = 6 + 6 × 4 = 24 = 30. 3 motorcycles and 6 cars. Check: 6 + 24 = 30.

Problem 3 (for all Worksheet B students): At Snuffelupagus Stadium, adult tickets cost $5 and kid tickets cost $3. The stadium sold the same number of adult and kid tickets and made $96 total. How many of each ticket? Answer: Same number means each pair = $5 + $3 = $8. Number of pairs: 96 ÷ 8 = 12. 12 adult tickets and 12 kid tickets. Check: 12 × 5 = 60 + 12 × 3 = 36 = 96.

Finished Early? Bonus: Invent your own "different rates" problem. Write a story where two types of items contribute different amounts to a total. Make sure it has a whole-number answer. Trade with a partner and solve each other's problem. Answer: Open-ended — student creates and solves their own problem.