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Primary 6 Method and Model Drawing

Build on P5 methods to solve linked fractions, changing ratios and multi-step word problems. Track what changes, explain the model and check every condition.

Choose a method

Choose what stays unchanged, what is transferred, or which quantities are linked. These P6 examples combine ratio, fractions, percentages and average.

Important phrases in fraction and ratio word problems

Phrases such as “of the whole”, “more than”, “less than” and “as many” tell you which quantities to compare. Open each example to identify the whole or comparison amount, draw its equal parts, and see how the wording affects the ratio.

A quick way to start
  1. Write the ratio labels in the requested order.
  2. Identify the whole or comparison amount that the fraction is taken from.
  3. Give that base the denominator’s number of parts, then find the other amount.
  4. Check whether an extra number, such as “9 more”, is an amount rather than a fraction of a part.
01“74 as many”Mia has 74 as many pencils as Ben.

Start with the base: Ben’s pencils are the comparison amount: 4 parts.

Each box is one equal ratio part.
Work it out
  1. Ben’s pencils are the amount being compared.
  2. Mia has 7 equal parts for every 4 parts Ben has.

Mia : Ben = 7 : 4

Watch for: Keep the names in the asked order: Mia : Ben.

02“Of the whole … the rest”512 of the balls are red. The rest are blue.

Start with the base: All the balls together are 12 parts.

Each box is one equal ratio part.
Work it out
  1. Red balls take 5 of the 12 total parts.
  2. Blue balls take the remaining 12 − 5 = 7 parts.

Red : Blue = 5 : 7

Watch for: The 12 is the total number of balls, so Red : Blue is not 5 : 12.

03“A fraction fewer than”There are 38 fewer blue pencils than red pencils.

Start with the base: Red pencils are the comparison amount: 8 parts.

Each box is one equal ratio part.
Work it out
  1. The difference is 3 of the red pencils’ 8 parts.
  2. Blue pencils have 8 − 3 = 5 parts.

Blue : Red = 5 : 8

Watch for: The fraction is taken from the number of red pencils, so red is the base.

04“A fraction more than”Amy has 23 more stickers than Leo.

Start with the base: Leo’s stickers are the comparison amount: 3 parts.

Each box is one equal ratio part.
Work it out
  1. The extra stickers equal 2 parts of Leo’s 3 parts.
  2. Amy has 3 + 2 = 5 parts.

Amy : Leo = 5 : 3

Watch for: Add the extra fraction to Leo’s whole amount.

05“A fraction of the whole”Emma coloured 710 of the squares.

Start with the base: All the squares together are 10 parts.

Each box is one equal ratio part.
Work it out
  1. Coloured squares take 7 of the 10 parts.
  2. Uncoloured squares take 10 − 7 = 3 parts.

Coloured : Uncoloured = 7 : 3

Watch for: Compare coloured with uncoloured, not with all the squares.

06“More than a fraction of” plus a numberThere are 9 more blue cubes than 25 of the red cubes.

Start with the base: Red cubes are the comparison amount: 5 equal units.

The blue and red unit boxes have equal value. The yellow +9 box is a fixed number of cubes.
Work it out
  1. Two fifths of the red cubes = 2 units.
  2. Blue cubes = 2 units + 9 cubes.
  3. Blue : Red = (2 units + 9) : 5 units.

Blue : Red = (2 units + 9) : 5 units

Special hint: The value of 1 unit is unknown, so there is no fixed numerical ratio yet. Keep the 9 cubes separate from the units.

Watch for: The 9 is a number of cubes, not 9 units. Do not add it to the 2 units.

07The requested labels change the ratioLily ate 49 of a pizza.

Start with the base: The whole pizza is 9 parts.

Each box is one equal ratio part.
Work it out
  1. The amount eaten is 4 parts of the 9-part pizza.
  2. Eaten : Whole pizza = 4 : 9.
  3. The amount left is 9 − 4 = 5 parts, so Eaten : Left = 4 : 5.

Eaten : Whole pizza = 4 : 9

Eaten : Left = 4 : 5

Watch for: Check whether the second label says the whole pizza or the part left.

08A mixed number “times as many”Sara has 3 12 times as many books as Tom.

Start with the base: Tom’s books are the comparison amount: 2 parts.

Each box is one equal ratio part.
Work it out
  1. Convert the mixed number: 3½ = 7/2.
  2. Sara has 7 parts for every 2 parts Tom has.

Sara : Tom = 7 : 2

Watch for: Convert the whole-number part too: 3½ is 7/2, not 1/2.

Ask at home

“Which amount is the fraction taken from or compared with? What are the two labels in our ratio?”

Common mistake

Using the denominator as the second ratio number every time. If the ratio compares a part with the rest, subtract the part from the whole first.

Model drawing · equal concept

What it means

Make the stated equal fractions cover the same number of small units, then compare the complete bars.

When to use itUse when “a fraction of A is the same as a fraction of B” and a difference or total is given.
Keywords & phrases to notice
a fraction of A is the same as a fraction of Bequal tomore than / less than

Work through a new example

of Amira’s savings equals of Ben’s savings. Ben has $12 more than Amira. How much do they have altogether?

Read the storyStep 0 of 4

Read the two fractions carefully. The equal portions belong to different whole amounts.

The fraction bars will appear in the first model step.

The first model step will show both fractions with equal shaded lengths.
View the complete worked solution
The equal blue lengths each contain 6 identical small units
Amira
6 small units
Ben
6 small units
  1. The blue portions match, so Amira has 9 small units and Ben has 10 small units. Their difference is 1 small unit.
    Amira = 9 small units
    Ben = 10 small units
    1 small unit = $12
    Together = 19 × $12 = $228
  2. Check both equal portions and the stated difference.
    Amira = 9 × $12 = $108
    Ben = 10 × $12 = $120
    2 of Amira’s 3 parts = $72
    3 of Ben’s 5 parts = $72
    $120 − $108 = $12

Answer: Amira and Ben have $228 altogether.

Alternative method · equal numerators

Start with the equal portions in the question:

of Amira’s savings of Ben’s savings
  1. Match the portions. Two of Amira’s larger parts equal three of Ben’s smaller parts. The individual parts have different values.
  2. Make the numerators equal. The lowest common multiple of 2 and 3 is 6. Multiply each fraction’s numerator and denominator by the same number:
    of Amira’s savings of Ben’s savings
  3. Read the new whole bars. Simplify the products without changing either fraction:
    of Amira’s savings of Ben’s savings

    The equal six-part portions can now use the same small unit. Amira’s whole is 9 units; Ben’s whole is 10 units.

    The equal blue lengths each contain 6 identical small units
    Amira
    6 small units
    Ben
    6 small units
    The first 6 small units align. Ben has 1 extra small unit.
  4. Use the $12 difference.
    10 units − 9 units = 1 unit
    1 unit = $12
    Amira = 9 × $12 = $108
    Ben = 10 × $12 = $120
    Together = $108 + $120 = $228
  5. Check. Two-thirds of $108 = $72; three-fifths of $120 = $72. Ben has $120 − $108 = $12 more.

Answer: Amira and Ben have $228 altogether.

Ask at home

“Which portions are said to be equal? Do the whole bars need to be equal?”

Common mistake

Setting the two whole amounts equal instead of only the stated fractions.

Try together · Primary 6

3/4 of Asha’s savings equals 2/3 of Ben’s savings. Ben has $20 more than Asha. How much do they have altogether?

Reveal the worked answer

$340.

Make the shared fraction amount 6 units: Asha has 8 units and Ben has 9 units. The one-unit difference is $20. Their total is 17 × $20 = $340. Check: 3/4 of $160 and 2/3 of $180 both equal $120.

Model drawing · more than / less than

What it means

Draw the known extra part at the stage where “more than” is stated. Show the later transfer on the same bars before applying the new ratio.

When to use itUse when the problem gives one amount as a fixed number more or less at one stage and a ratio at another stage.
Keywords & phrases to notice
more than / less thangave … totimes as many after

Work through a new example

Kai had 18 more stickers than Leo. Leo gave 6 stickers to Kai. Kai then had twice as many as Leo. How many stickers did Leo have at first?

Read the storyStep 0 of 6

Read the whole question. Notice which clue describes the beginning and which describes the end.

Read the clues first. Draw the starting difference before moving the stickers.

Draw the 18-sticker starting gap, then show the transfer and ending ratio.
View the complete worked solution
At first: Kai has 18 more
Leo
36 at first
Kai
same part18 more
Move 6 stickers from Leo to Kai
Leo
30 left−6
Kai
same as Leo first18 more+6
After the transfer: Kai has twice as many
Kai
30 left618 more+62 units
Leo
30 stickers1 unit
Leo gives 6Kai receives 6
  1. Both sides of the transfer enlarge the gap. The 2-unit bar is 1 unit longer than the 1-unit bar.
    Ending gap = 18 + 6 + 6 = 30 stickers
    2 units − 1 unit = 1 unit
    1 unit = 30 stickers
  2. Leo’s ending bar is 1 unit. Put back the 6 stickers he gave Kai.
    Leo after = 1 unit = 30 stickers
    Leo at first = 30 + 6 = 36 stickers
  3. Use the original 18-sticker gap, then check the transfer and ending ratio.
    Kai at first = 36 + 18 = 54 stickers
    Leo after = 36 − 6 = 30 stickers
    Kai after = 54 + 6 = 60 stickers
    60 = 2 × 30 ✓

Answer: Leo had 36 stickers at first.

Ask at home

“Which stage does ‘18 more’ describe? What happens to the gap when 6 moves?”

Common mistake

Keeping the gap at 18 after a transfer between the two people.

Try together · Primary 6

Amy had 20 more stickers than Ben. Ben gave 10 stickers to Amy. Amy then had twice as many as Ben. How many did Ben have at first?

Reveal the worked answer

50 stickers.

The gap grows by 10 + 10 = 20, from 20 to 40. The ending 2 : 1 ratio has a one-part gap, so Ben has 40 at the end. Add back his 10: 50. Check: Amy starts with 70, and ends with 80 while Ben has 40.

Model drawing · constant difference

What it means

Draw the known starting amounts first. Mark the same amount given away at the front of both bars, then compare only the amounts left. Equal subtractions keep the difference unchanged.

When to use itUse when both quantities gain or lose the same amount and their ratio changes.
Keywords & phrases to notice
each gave away the same numberboth gained / both losttimes as many later

Work through a new example

Ravi had 230 stickers and Zoe had 110 stickers. They each gave away the same number of stickers. Ravi then had three times as many stickers as Zoe. How many stickers did each give away?

Read the storyStep 0 of 6

Read the whole question. Draw the known starting amounts first, then show what both children gave away.

Draw the known starting amounts, mark equal amounts given away, then compare the stickers left.

Draw 230 and 110 first, then mark equal giveaways at the front of both bars.
View the complete worked solution
At first: Ravi has 230 stickers; Zoe has 110
Ravi
230 stickers
Zoe
110 stickers
After equal giveaways: compare stickers left
Ravi
50 given1 unit1 unit1 unit3 units left
Zoe
50 given60 left1 unit left

The difference stays 120 stickers: Ravi’s 3 units left − Zoe’s 1 unit left = 2 units.

  1. Equal giveaways leave the difference unchanged. Find the value of one ending unit.
    Starting difference = 230 − 110 = 120 stickers
    Ending difference = 3 units − 1 unit = 2 units
    1 unit = 120 ÷ 2 = 60 stickers
  2. Zoe has 1 unit left. Subtract it from the number she had at first.
    Zoe gave away = 110 − 60 = 50 stickers
    Ravi has 3 × 60 = 180 stickers left
    Ravi gave away = 230 − 180 = 50 stickers
  3. Check the equal giveaways and the ending comparison.
    Ravi: 230 − 50 = 180 stickers
    Zoe: 110 − 50 = 60 stickers
    180 = 3 × 60 ✓

Answer: Ravi and Zoe each gave away 50 stickers.

Ask at home

“What happens to the 120-sticker gap if both give away the same number? Which parts show the stickers left?”

Common mistake

Including the shaded stickers given away in the 3 : 1 ending ratio.

Try together · Primary 6

Asha had 150 cards and Ben had 90. Each gave away the same number. Asha then had three times as many as Ben. How many did each give away?

Reveal the worked answer

60 cards.

The difference stays 150 − 90 = 60. In the ending 3 : 1 ratio, two parts equal 60, so Ben has 30 left. Each gave away 90 − 30 = 60. Asha has 90 left.

Model drawing · remainder concept

What it means

Draw the $45 toy expense and the money remaining after it as two attached parts of one whole bar. Divide only that remaining part into four equal parts: one for food and three left. The three left parts are one-third of the original whole.

When to use itUse when a number of dollars or objects is removed first, followed by a fraction of the remainder.
Keywords & phrases to notice
spent a fixed amount firstof the remainingof the originalleft

Work through a new example

Priya spent $45 on a toy and of the remaining money on food. The money left was of what she had at first. How much did she have at first?

Read the storyStep 0 of 6

Read the question. The two fractions refer to different amounts: the original money and the money after the toy.

The model will show which whole each fraction refers to.

First draw $45 and the uncut remainder as parts of one whole.
View the complete worked solution
$81
of the original (total)
left = 3 units
Toy
5 units = $45Food
1 unitLeft
3 units
After toy: 4 units
Food = of this
Find the whole from the three units left
of total = 3 units
of total = 3 × 3 units = 9 units
  1. The 3 units left are one-third of the total. Find the whole before subtracting the 4 units after the toy.
    of total = 3 units
    of total = 3 × 3 units = 9 units
    9 units − 4 units = 5 units (toy)
    5 units = $45
    1 unit = $45 ÷ 5 = $9
  2. All 9 small units make up the original money.
    Original money = 9 × $9 = $81
    After toy = 4 × $9 = $36
    Food = 1 × $9 = $9; left = 3 × $9 = $27
  3. Check both fractions against the correct whole.
    $81 − $45 = $36 after the toy
    Food = $36 ÷ 4 = $9
    $36 − $9 = $27 left
    $81 ÷ 3 = $27 ✓

Answer: Priya had $81 at first.

Ask at home

“Which three units show one-third of the original? Which four units were left after the toy?”

Common mistake

Taking one-quarter of the starting amount instead of the amount after buying the toy.

Try together · Primary 6

Jia spent $30, then 1/3 of the remaining money. She had half of her original money left. How much did she have at first?

Reveal the worked answer

$120.

The final 1/2 of the original is 2/3 of the remainder. Therefore the remainder is 3/4 of the original. The $30 spent is 1/4 of the original, so the original is $30 × 4 = $120. Check: $90 remains, then $30 is spent and $60 is left.

Model drawing · pull-down concept

What it means

Draw the first whole bar, pull its remainder into a second bar below, and divide that new bar for the next fraction and extra amount.

When to use itUse when the first step removes a fraction and an extra quantity, and the next step removes a fraction and another quantity from the remainder.
Keywords & phrases to notice
a fraction and another amountof the remainingthenleft

Work through a new example

Rina spent of her money and another $12 on a book. She then spent of the remaining money and another $6 on a game. She had $30 left. How much did she have at first?

Read the storyStep 0 of 7

Read the question. The first fraction is of the original money; the second is of the money remaining after the book.

Read the question first. The lower bar will represent only the money left after the book.

Draw the original amount, then pull only its remainder down.
View the complete worked solution
Original money: 3 equal units = $126
Book spent: 1 unit + $12
1 unit1 unit1 unit
$42 $12 $72
1 part1 part
$36$6$30 left
Game spent: 1 part + $6

After the book: 2 parts = $72

  1. Start with the lower bar. The $30 left and extra $6 together make 1 part of the money after the book.
    1 part = $30 + $6 = $36
    2 parts = $36 × 2 = $72 (after the book)
  2. Return to the upper bar. The other 2 original units contain the extra $12 and the $72 remainder.
    2 units = $72 + $12 = $84
  3. Find 1 original unit, then all 3 units of Rina’s original money.
    1 unit = $84 ÷ 2 = $42
    3 units = $42 × 3 = $126
  4. Check both purchases against the question.
    Book: 1 unit + $12 = $42 + $12 = $54; $126 − $54 = $72
    Game: 1 part + $6 = $36 + $6 = $42; $72 − $42 = $30

Answer: Rina had $126 at first.

Ask at home

“Can you pull the first remainder into a fresh bar? What is the new fraction based on?”

Common mistake

Treating half of the remainder as half of the original money.

Try together · Primary 6

Rina spent 1/4 of her money and another $10. She then spent half of the remainder and another $5. She had $35 left. How much did she have at first?

Reveal the worked answer

$120.

Reverse the second stage: ($35 + $5) × 2 = $80. Reverse the first fixed amount: $80 + $10 = $90, which is 3/4 of the original. The original is $90 ÷ 3 × 4 = $120.

Unitary method

What it means

Match the units that represent an unchanged quantity across two ratios. Once those units are on the same scale, the change tells you the value of one unit.

When to use itUse it for before-and-after ratios after checking whether one quantity, the total, or the difference stayed constant. It also joins two ratios that share a common person or item.
Keywords & phrases to notice
ratio … became …unchangedjoined / lefttransferred
Useful forms of this method

How to make an equivalent ratio

Multiply every part by the same number. The ratio keeps the same relationship.

Two-part ratio
3:4
↓× 2↓× 2
6:8
Three-part ratio
2:3:5
↓× 2↓× 2↓× 2
4:6:10

For each question, first identify what can be matched between the two ratios or cases.

  1. Repeated identity

    Two ratios share one person or item.

    Ali : Bao = 2 : 3 and Bao : Cara = 4 : 5. Ali and Cara have 46 cards altogether. How many cards does each child have?

    Read the storyStep 0 of 8

    Find the two ratios and the quantity that stays the same.

    The first ratio or case appears when you begin.

    View the complete worked solution

    What stays the same: Bao is the same child in both ratios. The common multiple of 3 and 4 is 12.

    First ratio
    AliBao
    2:3
    ↓× 4↓× 4
    8:12
    Second ratio
    BaoCara
    4:5
    ↓× 3↓× 3
    12:15
    1. Bao is 12 units in both equivalent ratios, so the two ratios can be joined.
      Ali : Bao : Cara = 8 : 12 : 15
    2. The question says Ali and Cara have 46 cards altogether. Add only their units: 8 + 15 = 23.
      23 units = 46 cards
    3. Divide the 46 cards equally among the 23 units.
      1 unit = 46 ÷ 23 = 2 cards
    4. Multiply each child’s units by 2 cards.
      Ali = 8 × 2 = 16; Bao = 12 × 2 = 24; Cara = 15 × 2 = 30
    5. Check both original ratios and the given total.
      16 : 24 = 2 : 3; 24 : 30 = 4 : 5; 16 + 30 = 46

    Answer: Ali has 16 cards, Bao has 24 cards, and Cara has 30 cards.

    Ask at home

    “Who appears in both ratios? How can we make Bao’s parts equal before joining them?”

    Common mistake

    Joining 2 : 3 and 4 : 5 directly. The 3 Bao parts and 4 Bao parts must be scaled to the same amount first.

  2. Constant variable

    One amount does not change.

    The ratio of pens to pencils was 3 : 4. After 5 pens were added, the ratio became 7 : 8. How many pens and pencils were there at first?

    Read the storyStep 0 of 7

    Find the two ratios and the quantity that stays the same.

    The first ratio or case appears when you begin.

    View the complete worked solution

    What stays the same: The number of pencils did not change. Make their 4 parts in the first ratio match the 8 parts in the second.

    First ratio
    PensPencils
    3:4
    ↓× 2↓× 2
    6:8
    Pencils: 4 → 8 units
    Second ratio
    PensPencils
    7:8
    Pencils: 8 units
    1. Pencils are unchanged. Double both parts of the first ratio so pencils are 8 units in both ratios.
      3 : 4 = 6 : 8
    2. Now compare the pen parts. The extra 1 unit is the 5 pens added.
      7 units − 6 units = 1 unit = 5 pens
    3. Use the equivalent first ratio, 6 : 8, to find the starting amounts.
      Pens = 6 × 5 = 30; pencils = 8 × 5 = 40
    4. Check the new ratio after adding the 5 pens.
      30 + 5 = 35; 35 : 40 = 7 : 8

    Answer: There were 30 pens and 40 pencils at first.

    Ask at home

    “Which item count stayed the same? Can we make the pencil parts match before comparing pens?”

    Common mistake

    Subtracting 3 pen parts from 7 before matching the unchanged pencils. The first 4 pencil parts must become 8.

  3. Constant total

    A transfer stays within the group.

    Maya and Noah had tokens in the ratio 4 : 1. Maya gave 10 tokens to Noah. Their new ratio was 2 : 1. How many tokens did each have at first?

    Read the storyStep 0 of 9

    Find the two ratios and the quantity that stays the same.

    The first ratio or case appears when you begin.

    View the complete worked solution

    What stays the same: No tokens enter or leave, so Maya and Noah have the same total before and after. The first ratio has 5 total parts; the second has 3.

    First ratio
    MayaNoahTotal
    4:1:5
    ↓× 3↓× 3↓× 3
    12:3:15
    Total: 5 → 15 units
    Second ratio
    MayaNoahTotal
    2:1:3
    ↓× 5↓× 5↓× 5
    10:5:15
    Total: 3 → 15 units
    1. Add Maya’s and Noah’s parts to write each ratio with its total.
      First: 4 : 1 : 5; second: 2 : 1 : 3
    2. The same total must have the same number of units. The common multiple of 5 and 3 is 15. Multiply every part of the first ratio by 3.
      4 : 1 : 5 = 12 : 3 : 15
    3. Multiply every part of the second ratio by 5. Both totals are now 15 units.
      2 : 1 : 3 = 10 : 5 : 15
    4. Maya’s share falls from 12 to 10 units when she gives Noah 10 tokens.
      12 units − 10 units = 2 units = 10 tokens
    5. Find one unit, then use the equivalent first ratio for the starting amounts.
      1 unit = 10 ÷ 2 = 5 tokens; Maya = 12 × 5 = 60; Noah = 3 × 5 = 15
    6. Check the transfer, the ending ratio, and the unchanged total.
      60 − 10 = 50; 15 + 10 = 25; 50 : 25 = 2 : 1; 60 + 15 = 50 + 25 = 75

    Answer: Maya had 60 tokens and Noah had 15 tokens at first.

    Ask at home

    “Where did Maya’s 10 tokens go? What total do the two ratios share?”

    Common mistake

    Treating 5 total parts and 3 total parts as equal-sized units. Scale both ratios until their total columns match.

  4. Constant difference

    Both amounts rise or fall by the same amount.

    Lena and Sam had stickers in the ratio 5 : 3 at first. They each received the same number of stickers. After that, the ratio became 3 : 2. Lena had 30 more stickers than Sam throughout. How many stickers did each receive?

    Read the storyStep 0 of 9

    Find the two ratios and the quantity that stays the same.

    The first ratio or case appears when you begin.

    View the complete worked solution

    What stays the same: Their difference stays 30 stickers because both receive the same number. Compare Lena : Sam : Difference in each ratio.

    First ratio
    LenaSamDifference
    5:3:2
    Difference: 2 units
    Second ratio
    LenaSamDifference
    3:2:1
    ↓× 2↓× 2↓× 2
    6:4:2
    Difference: 1 → 2 units
    1. Subtract Sam’s parts from Lena’s parts to add Difference as the third column.
      First: 5 : 3 : 2; second: 3 : 2 : 1
    2. The difference is always 30 stickers. Double every part of the second ratio so its difference is 2 units too.
      3 : 2 : 1 = 6 : 4 : 2
    3. In the first ratio, the 2-unit difference is 30 stickers. Find one unit.
      2 units = 30 stickers; 1 unit = 30 ÷ 2 = 15 stickers
    4. Compare each child’s before and after units. Both gained 1 unit.
      Lena: 6 − 5 = 1 unit; Sam: 4 − 3 = 1 unit
    5. Each child received the value of 1 unit.
      Each received 1 × 15 = 15 stickers
    6. Check both ratios and the unchanged 30-sticker difference.
      At first 75 : 45 = 5 : 3; after 90 : 60 = 3 : 2; 75 − 45 = 90 − 60 = 30

    Answer: Lena and Sam each received 15 stickers.

    Ask at home

    “If both children receive the same number, what happens to the gap between them?”

    Common mistake

    Matching the growing total instead of the unchanged difference, or comparing the two ratios before their difference columns match.

  5. Double-if cases

    Compare two possible scenarios.

    If each child gets 3 stickers, 8 stickers are left. If each child gets 5 stickers, there are 6 stickers too few. How many children and stickers are there?

    Read the storyStep 0 of 8

    Find the two cases and the quantity that stays the same.

    The first ratio or case appears when you begin.

    View the complete worked solution

    What stays the same: Both “if” cases use the same children and the same stock of stickers.

    First if case3 stickers per child8 left
    Second if case5 stickers per child6 short
    1. Giving 5 instead of 3 stickers needs 2 extra stickers for each child.
      5 − 3 = 2 extra stickers per child
    2. The extra needed uses the 8 stickers left in the first case and fills the shortage of 6 in the second.
      8 + 6 = 14 extra stickers
    3. Divide the 14 extra stickers by 2 per child.
      Number of children = 14 ÷ 2 = 7
    4. Use the first case to find the stock of stickers.
      Stickers = 7 × 3 + 8 = 29
    5. Check both possible distributions.
      29 − 7 × 3 = 8 left; 7 × 5 − 29 = 6 short

    Answer: There are 7 children and 29 stickers.

    Ask at home

    “How many more stickers does each child get in the second case? How do the 8 left and 6 short fit together?”

    Common mistake

    Subtracting 6 from 8. Moving from 8 left to 6 short requires 8 + 6 = 14 extra stickers.

Try together · Primary 6

Asha and Ben had tokens in the ratio 4 : 1. Asha gave 12 tokens to Ben and the ratio became 3 : 2. How many tokens did Asha have at first?

Reveal the worked answer

48 tokens.

The total stays at five parts in both ratios. Asha drops from four parts to three, so one part equals 12 tokens. Her original four parts equal 48. Check: 48 : 12 becomes 36 : 24 = 3 : 2.

Working backwards

What it means

Write the events in their original order, tracking both people after each transfer. Use the total and ending comparison to find the final amounts, then reverse the transfers, starting with the last one. A fraction belongs to the amount at the moment it was transferred.

When to use itUse it when a total and ending relationship are given, several changes happened in sequence, and the question asks for the amounts at first. For an internal transfer, one person loses exactly what the other gains.
Keywords & phrases to notice
at firstthenin the endhow many to begin with

Work through a new example

Mei and Ravi had a total of 144 marbles at first. Mei gave 15 marbles to Ravi. Ravi then gave of the marbles he had at that moment to Mei. In the end, Mei had twice as many marbles as Ravi. How many marbles did each have at first?

Read the storyStep 0 of 7

Read the whole story. The ending is described by a comparison, not by two given amounts.

The two-person table will appear as you follow the question.

Record each change for both people, then work upward.
View the complete worked solution
Together throughout: 144 marbles
EventMeiRavi
At first8757
Mei gave 15 to Ravi↓ −15↑ reverse: +15↓ +15↑ reverse: −15
After Mei gave 157272
Ravi gave to Mei↓ receives of Ravi’s amount then↑ reverse: −24↓ gives of his amount then↑ reverse: +24
At the end9648

Read down ↓ for the events. Solve upward ↑ from the ending.

  1. The transfer stays within the pair, so their final total is still 144. Mei has 2 parts and Ravi has 1 part.
    2 units + 1 unit = 3 units = 144 marbles
    1 unit = 144 ÷ 3 = 48 marbles
    At the end: Mei = 2 × 48 = 96; Ravi = 48
  2. Ravi gave of his marbles at that stage to Mei. His 48 marbles after the transfer are of what he had before it. Find , then reverse the transfer for both children.
    Ravi after giving = 48 marbles = of his amount before giving
    of his amount before giving = 48 ÷ 2 = 24 marbles
    Ravi before giving = 48 + 24 = 72 marbles
    Mei before receiving = 96 − 24 = 72 marbles
  3. Reverse Mei’s earlier transfer of 15: add 15 back to Mei and subtract 15 from Ravi.
    Mei at first = 72 + 15 = 87 marbles
    Ravi at first = 72 − 15 = 57 marbles
  4. Check the answer by following both transfers forwards.
    87 + 57 = 144 marbles
    After Mei gives 15: Mei = 72, Ravi = 72
    Ravi gives of 72 = 24; at the end Mei = 96, Ravi = 48
    96 = 2 × 48 ✓

Answer: Mei had 87 marbles and Ravi had 57 marbles at first.

Ask at home

“What is the total at the end? If Mei has twice Ravi’s amount, how many equal parts is that? Which transfer should we reverse first?”

Common mistake

Taking of Ravi’s final amount instead of his amount before that transfer, or forgetting that each internal transfer changes both people’s amounts.

Try together · Primary 6

Mei and Ravi had 120 marbles altogether. Mei gave 12 to Ravi. Ravi then gave 1/4 of his marbles to Mei. They ended with equal numbers. How many did each have at first?

Reveal the worked answer

Mei: 52 marbles; Ravi: 68 marbles.

At the end, each has 60. Ravi’s 60 is 3/4 of what he had before the last transfer, so he had 80 and gave 20 to Mei. Mei then had 40 before receiving 20. Reverse the first transfer: Mei had 40 + 12 = 52 and Ravi had 80 − 12 = 68.

Branching method

What it means

Split the original whole once, then split only the remainder. The second fraction has a new base; it is not automatically a fraction of the starting amount.

When to use itUse it when the words say “of the remainder”, “of what was left”, or “of the rest”, especially after an earlier fraction or percentage has been removed.
Keywords & phrases to notice
of what remainsof the restthenleft

Work through a new example

Lina saved 20% of her allowance, then spent of the remaining money. She had $48 left. What was her allowance?

Read the storyStep 0 of 6

Read the story. The second fraction is of the money remaining after Lina saves.

The branches will appear as each clue is read.

The second split uses the 80% remainder as its whole; the branches also show their share of the original allowance.
View the complete worked solution
Total allowance100% = $80
Saved20%of original
Remaining80%of original
Spent of 80%= 20% of original
Left60% of original= $48
$16 saved + $16 spent + $48 left = $80
  1. Find the percentage remaining after Lina saves 20% of the original allowance.
    100% − 20% = 80% remaining
  2. Take one-quarter of only that 80% remainder to find what she spends.
    × 80% = 20% of the original allowance spent
  3. Find the percentage of the original allowance represented by the $48 left.
    80% − 20% = 60% of the original allowance
    $48 = 60% of the original allowance
  4. Use 60% = $48 to find the whole allowance.
    1% = $48 ÷ 60 = $0.80
    100% = $0.80 × 100 = $80
  5. Check the saving, spending and final amount in the original story.
    20% of $80 = $16 saved
    $80 − $16 = $64 remaining
    of $64 = $16 spent
    $64 − $16 = $48 left ✓

Answer: Lina’s allowance was $80.

Ask at home

“One-quarter of which amount: the first allowance or the money after saving?”

Common mistake

Taking one-quarter of the original allowance without first changing the base to the remainder.

Try together · Primary 6

Lina saved 10% of her allowance, then spent 1/3 of the remainder. She had $54 left. What was her allowance?

Reveal the worked answer

$90.

She kept 2/3 of the remaining 90%: 2/3 × 90% = 60% of the original. If 60% is $54, 10% is $9 and 100% is $90. Check: save $9, then spend $27; $54 remains.

Systematic listing

What it means

Start with the smallest possible number of groups. Use the first condition to generate each possible total, then test that total against the second condition. The first matching row gives the smallest total because no earlier possibility was skipped.

When to use itUse it when a question asks for the smallest possible total and two conditions can be checked in a short, ordered table. It also helps with “how many ways” and “all possible” questions.
Keywords & phrases to notice
smallest numberleftneeds moretwo conditionsall possible

Work through a new example

At a craft fair, Maya packs beads into bags. If she puts 6 beads in each bag, 4 beads are left. If she puts 9 beads in each bag, she needs 8 more beads. (a) What is the smallest number of beads Maya has? (b) How many bags does she use?

Read the storyStep 0 of 9

Read both packing plans. We will list possible bead totals from the smallest number of bags.

The conditions and ordered checking table will appear here.

Compare the two total rows in each bag-count column. Equal totals satisfy both conditions.
View the complete worked solution
Condition 1: 6 in each bag; 4 left

Beads = 6 × number of bags + 4

Condition 2: 9 in each bag; 8 more needed

Beads = 9 × bags − 8

List from the smallest: 1 bag → 2 bags → 3 bags → 4 bags → …

Swipe the table sideways to compare the totals →

Number of bags1234
6 beads in each bag6121824
4 beads leftadd 410162228
9 beads in each bag9182736
8 beads more neededsubtract 81101928

The first match is 4 bags. All smaller bag counts were checked, so 28 is the smallest bead total.

4 × 6 + 4 = 28 beads · 4 × 9 − 28 = 8 more needed ✓
  1. Use the first condition to generate bead totals from the number of bags.
    Beads = 6 × bags + 4
  2. The second condition says 9 beads per bag needs 8 more beads.
    Beads = 9 × bags − 8
  3. Start at 1 bag and increase by 1 each row to find the smallest matching total.
    Possible bag counts: 1, 2, 3, 4, …
  4. Check 1 bag against both conditions.
    1 × 6 + 4 = 10; 10 − 1 × 9 = 1 left
  5. Check 2 bags against both conditions.
    2 × 6 + 4 = 16; 2 × 9 − 16 = 2 more needed
  6. Check 3 bags against both conditions.
    3 × 6 + 4 = 22; 3 × 9 − 22 = 5 more needed
  7. Check 4 bags against both conditions.
    4 × 6 + 4 = 28; 4 × 9 − 28 = 8 more needed ✓
  8. This is the first matching row, so its bead total is the smallest.
    Earlier rows did not meet the second condition
    Smallest total = 28 beads
  9. Check the number of beads and bags in both packing plans.
    4 × 6 + 4 = 28 beads
    4 × 9 − 28 = 8 more needed ✓

Answer: (a) 28 beads, the smallest possible total. (b) 4 bags.

Ask at home

“Why start at one bag and move up by one? How does the table prove 28 is the smallest total?”

Common mistake

Trying bag counts out of order or stopping at a row that fits only the first packing plan.

Try together · Primary 6

If beads are packed 5 per bag, 3 beads are left. If they are packed 7 per bag, 9 more beads are needed. How many bags and beads are there?

Reveal the worked answer

6 bags and 33 beads.

Each bag needs 2 extra beads in the second case. The move from 3 left to 9 short needs 3 + 9 = 12 beads. There are 12 ÷ 2 = 6 bags and 6 × 5 + 3 = 33 beads. Check: 6 × 7 − 9 = 33.

Excess and shortage (alternative to systematic listing for two “if” conditions)

What it means

Compare how much more each person would get in the second condition with the change from leftovers to a shortage.

When to use itUse it when a question gives two “if each gets…” conditions for the same items and people. One leaves some items over; the other is short. You can solve it directly instead of checking possibilities in a table.
Keywords & phrases to notice
left overshorteach getssame stock

Work through a new example

If a baker gives 4 cookies to each child, 9 cookies remain. If he gives 6 to each child, he is 3 cookies short. How many children are there?

Read the storyStep 0 of 6

Read the two “if” conditions. Follow each clue to build the comparison.

The two conditions and their change will appear here.

The 9 leftover cookies are used up before 3 more are needed.
View the complete worked solution
First condition4 each + 9 left
Second condition6 each + 3 short
More for each child6 − 4 = 2 cookies
More cookies needed9 left + 3 short = 12 cookies
Number of children12 ÷ 2 = 6 children
Check both conditions4 × 6 + 9 = 33; 6 × 6 − 3 = 33 ✓
  1. Read the first condition: 4 cookies for each child and 9 left.
    Cookies = 4 × children + 9
  2. Read the second condition: 6 cookies for each child and 3 short.
    Cookies = 6 × children − 3
  3. Compare the number of cookies for each child.
    6 − 4 = 2 more cookies per child
  4. The extra cookies needed use up the 9 left, then require 3 more.
    9 + 3 = 12 extra cookies needed
  5. Divide the total extra cookies by the extra cookies per child.
    Children = 12 ÷ 2 = 6
  6. Check the same stock of cookies against both conditions.
    4 × 6 + 9 = 33 cookies
    6 × 6 − 3 = 33 cookies ✓

Answer: There are 6 children.

Ask at home

“Where did the 9 leftover cookies go before the baker became 3 short?”

Common mistake

Subtracting 3 from 9. Moving from 9 left over to 3 short is a change of 12.

Try together · Primary 6

If 3 cookies are given to each child, 8 remain. If 5 are given to each child, there are 6 cookies too few. How many children and cookies are there?

Reveal the worked answer

7 children and 29 cookies.

Two more cookies per child require 8 + 6 = 14 extra cookies. There are 14 ÷ 2 = 7 children. The stock is 7 × 3 + 8 = 29 cookies. Check: 7 × 5 − 6 = 29.

Quantity × value table

What it means

Turn the quantity relationship into comparable parts. It may be given as a ratio, a fraction, or “times as many”. Multiply each item’s parts by its value, add the row values, then scale to the known total.

When to use itUse it when the quantities of different items are linked by a ratio, fraction, or “times as many”, each item has a different value or score, and a combined total is given.
Keywords & phrases to notice
number of eachfraction of the wholetimes as manyvalue of eachtotal value

Work through a new example

The number of 20-cent coins and 50-cent coins is in the ratio 2 : 3. Their total value is $9.50. How many coins are there altogether?

Read the storyStep 0 of 8

Read the ratio and total value. Build a table for one ratio group first.

The coin-count and value table will appear here.

The ratio compares the numbers of coins. Multiply each count by its coin value before adding money.
View the complete worked solution

Swipe the table sideways to see every column →

CoinCount in 1 groupValue eachValue in 1 group
20¢220¢2 × 20¢ = 40¢
50¢350¢3 × 50¢ = 150¢
One ratio group2 + 3 = 5 coins—40¢ + 150¢ = 190¢

How many groups? $9.50 = 950¢; 950¢ ÷ 190¢ = 5 groups.

Actual counts: 2 × 5 = 10 twenty-cent coins; 3 × 5 = 15 fifty-cent coins. Total: 10 + 15 = 25 coins.

Check: 10 × 20¢ + 15 × 50¢ = 200¢ + 750¢ = 950¢ = $9.50 ✓

  1. Record the 2 : 3 ratio as one group of 2 twenty-cent coins and 3 fifty-cent coins.
    20¢ coins : 50¢ coins = 2 : 3
  2. Keep each coin’s value separate from its count.
    20¢ each; 50¢ each
  3. Find the value of the twenty-cent coins in one group.
    2 × 20¢ = 40¢
  4. Find the value of the fifty-cent coins in one group.
    3 × 50¢ = 150¢
  5. Add both rows to find the value and coin count of one ratio group.
    40¢ + 150¢ = 190¢ per group
    2 + 3 = 5 coins per group
  6. Convert $9.50 to cents, then find how many ratio groups make the total value.
    $9.50 = 950¢
    950¢ ÷ 190¢ = 5 groups
  7. Scale both coin counts by the number of groups and add them.
    20¢ coins = 2 × 5 = 10
    50¢ coins = 3 × 5 = 15
    Total = 10 + 15 = 25 coins
  8. Check the two actual coin counts against the $9.50 total.
    10 × 20¢ = 200¢
    15 × 50¢ = 750¢
    200¢ + 750¢ = 950¢ = $9.50 ✓

Answer: There are 25 coins altogether.

Ask at home

“What do the numbers count, and what is one item of each type worth?”

Common mistake

Treating the quantities as money or score. Multiply each type’s quantity by its own item value before adding.

Try together · Primary 6

The numbers of 10-cent and 50-cent coins are in the ratio 3 : 2. Their total value is $6.50. How many coins are there altogether?

Reveal the worked answer

25 coins.

One ratio group contains 3 ten-cent coins and 2 fifty-cent coins, worth 30 + 100 = 130 cents. There are 650 ÷ 130 = 5 groups. Each group has 5 coins, giving 25 coins altogether.

Average shift

What it means

Compare the new item with the new average. Its excess supplies the increase needed to lift every original item from the old average to the new one.

When to use itUse it when an old average, a new average and the value of an added or removed item are given, but the original number of items is unknown.
Keywords & phrases to notice
average becomesjoins / leavesnew averagehow many at first

Work through a new example

The average mass of a group of boys was 35 kg. Kenny, who weighed 47 kg, joined them. Their average mass became 37 kg. How many boys were there at first?

Read the storyStep 0 of 5

Read the starting and ending averages. We will draw the boys who were there at first before adding Kenny.

The before-and-after mass model will appear here.

The right-hand bars rebalance Kenny’s extra mass to show the new average; they do not show each boy’s actual mass. Bar lengths are schematic.
View the complete worked solution
37 − 35 = 2 kg more per original boy47 − 37 = 10 kg Kenny has above the new average
Before Kenny joinedEach original boy: 35 kg · number unknown
Boy 1
35 kg
Boy 2
35 kg
Boy 3
35 kg
Boy 4
35 kg
Boy 5
35 kg
Balanced at the new averageEach boy’s model bar: 37 kg
Boy 1
35 kg+2 kg
Boy 2
35 kg+2 kg
Boy 3
35 kg+2 kg
Boy 4
35 kg+2 kg
Boy 5
35 kg+2 kg
Kenny
37 kg

Kenny’s 10 kg above 37 kg is shared into the +2 kg blocks.

Kenny’s extra 10 kg → five 2 kg pieces2 + 2 + 2 + 2 + 2 = 10 kg, so there were 5 boys at first.

Before: 5 × 35 = 175 kg. After: 175 + 47 = 222 kg for 6 boys; 222 ÷ 6 = 37 kg. ✓

  1. Draw one 35 kg bar for each boy at first. The number of boys is still unknown.
    Average at first = 35 kg
  2. Compare the averages and Kenny’s mass with the new average.
    37 − 35 = 2 kg
    47 − 37 = 10 kg
  3. Add Kenny to the model. Extend each original 35 kg bar by 2 kg to show the new average of 37 kg.
    35 + 2 = 37 kg
    37 + 10 = 47 kg
  4. Share Kenny’s 10 kg above 37 kg as 2 kg for each original boy.
    10 kg = 2 kg × number of original boys
  5. Find the number of boys at first.
    10 ÷ 2 = 5 boys
  6. Check the original and new averages.
    5 × 35 = 175 kg
    (175 + 47) ÷ 6 = 37 kg

Answer: There were 5 boys at first.

Ask at home

“How far above the new average is the new pupil? Who receives that extra amount?”

Common mistake

Dividing 47 by the change in average. The useful excess is 47 − 37, not the pupil’s full mass.

Try together · Primary 6

The average mass of some children was 32 kg. A child weighing 44 kg joined them and the average became 34 kg. How many children were there at first?

Reveal the worked answer

5 children.

The newcomer has 44 − 34 = 10 kg above the new average. Each original child needs 34 − 32 = 2 kg to reach it. There were 10 ÷ 2 = 5 original children. Check: (5 × 32 + 44) ÷ 6 = 34.

Put the method into practice

  1. Identify the invariant. State which amount, total or difference stays the same.
  2. Connect the stages. Label each ratio or fraction with its own base before calculating.
  3. Explain and verify. Show a complete chain of working and check the answer against every condition.

For topic practice, revisit P6 ratio, P6 division of fractions or P6 average.

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