T

Multi-step Word Problems

Maths · Y4 OC Prep
Year 4 OC
Lesson Guide
Explanation
OC problems stack layers: a tray holds 6 rows of 4 (layer 1: 24), Josh bakes 3 trays (layer 2: 72), sells 50 (layer 3: 22 left). Peel one layer at a time and WRITE each result — numbers held in your head evaporate under time pressure. And know this: the wrong options are the middle layers! 24 and 72 will be sitting in the choices, dressed up as answers. Only the final layer answers the question.
Your Turn
Layers: a box holds 5 packs of 8 cards. Emma buys 2 boxes and gives away 25 cards. How many left? Check: 40 → 80 → 55.
Watch Out
Finishing early is the danger — if you've only multiplied, ask 'did I answer the QUESTION or just a layer?'
Explanation
OC style: Mia and Tom ate 21 grapes; Tom ate twice as many as Mia. How many did Tom eat? DRAW it as bars: Mia gets 1 part, Tom gets 2 parts — together 3 equal parts make 21, so one part is 7. Tom: 14. The bar model turns 'twice as many' from a word-fog into a picture you can divide. It scales to 'three times as many' (4 parts) and beyond.
Your Turn
Tom and Ali share $20; Tom gets three times as much as Ali. How much does Tom get? Check: 4 parts of $5 → $15.
Watch Out
'Twice as many' makes THREE parts in total (1 + 2), not two. Draw the bars and the parts count themselves.
Explanation
The famous 'together and apart' shape: Mia and Ben have 30 marbles together; Mia has 6 more than Ben. Bars again — Ben's bar plus Mia's longer bar. The move: chop off Mia's extra 6, leaving 24 split between two EQUAL bars: 12 each. So Ben has 12, Mia has 12 + 6 = 18. Verify: 12 + 18 = 30 ✓. Remove the difference, share equally, add it back.
Your Turn
Two shelves hold 40 books; the top shelf has 8 more. How many on each? Check: (40−8)÷2 = 16, and 24.
Watch Out
Halving 30 first (15 and 15, then shifting) invites errors. Remove the extra FIRST, then halve — cleaner every time.
Explanation
The hardest OC money questions stack constraints: change from $10 for a $3.20 purchase ($6.80), containing EXACTLY eight 50c coins ($4.00) — what's the greatest number of 20c coins in the rest? Handle fixed parts first: $6.80 − $4.00 = $2.80 to fill. Maximise the 20c coins: fourteen make exactly $2.80, with nothing left over ✓. Answer: 14. (Fifteen would make $3.00 — too much.) Constraints are just layers: lock in what's forced, maximise what's free, check the leftover works.
Your Turn
Change of $2.60 must include exactly two 50c coins. Max 20c coins in the rest? Check: $1.60 → 8 × 20c exactly.
Watch Out
After maximising, the leftover must be payable with REAL coins. If 5c or 10c can't cover it, drop one coin and re-check.
Explanation
Here's the engine for every OC word problem: READ twice (circle numbers, underline the question), DRAW when there are parts or comparisons, PEEL the layers writing each result, ANSWER the actual question (never a middle layer), and CHECK by plugging your answer back into the story. Five steps, about 70 seconds. It feels slow the first week and becomes automatic by the third — that's the training. Quiz and exercises now!
Your Turn
Run the engine: "3 friends each buy 2 packs of 6 stickers. They combine and share equally among 4 kids. How many each?" Check: 36 ÷ 4 = 9.
Watch Out
Skipping the CHECK step saves 5 seconds and costs whole questions. An answer that breaks the story is always wrong.
Step 1: Introduction
Step 1 of 5
Peel the onion 🧅 "A tray holds 6 rows of 4 muffins. Josh bakes 3 full trays and sells 50 muffins. How many are left?" layer 1 6 × 4 = 24 layer 2 3 × 24 = 72 layer 3 72 − 50 = 22 One layer at a time, each result feeding the next. Write each layer's answer down — held-in-head numbers evaporate. The options will include 24, 72 AND 22 — only the LAST layer wins.
OC style: twice as many "Mia and Tom ate 21 grapes. Tom ate twice as many as Mia." Mia 1 part Tom 2 parts 3 parts = 21 1 part = 7 Bars: Mia = 1 part, Tom = 2 parts → 3 parts share 21. One part = 7, so Tom ate 2 × 7 = 14 ✓ (Mia ate 7.)
Together AND apart "Mia and Ben have 30 marbles together. Mia has 6 MORE than Ben." Ben Mia +6 Remove the extra: 30 − 6 = 24 → two EQUAL bars of 12. Ben = 12, Mia = 12 + 6 = 18. Check: 12 + 18 = 30 ✓
OC style: coins in the change "Stickers $2.30 + card 90c, paid with $10. No $1/$2 coins, no notes in the till. The change contains exactly eight 50c coins. Greatest number of 20c coins?" Layer 1 — the change: 10 − 3.20 = $6.80 Layer 2 — remove the fixed part: eight 50c = $4.00 → $2.80 remains Layer 3 — maximise 20c coins in $2.80: fourteen 20c = $2.80 exactly, leaving nothing over. Answer: 14 ✓ Constraints are LAYERS: handle the fixed parts first, then push the flexible part to its limit — and check leftovers are legal coins!
Your problem-solving engine 🚂 1. READ twice circle numbers, underline the actual question 2. DRAW if there are parts/comparisons bar models turn words into maths 3. PEEL layer by layer write every intermediate result down 4. ANSWER the question asked not a middle layer! reread before choosing 5. CHECK against the story plug your answer back in — does it fit? Same engine, every problem. Quiz and exercises — go!
Was this lesson helpful? Rate this Course
Discussion
Loading comments...