After giving, Jason = 3 × Don.
Let Don (after) = 1 unit, so Jason (after) = 3 units.

Use the bar model to see the transfer, then find the difference:

So:  2u = − − =

From Step 1: 2u = 72

1u = 72 ÷ 2 =

Jason after = 3u = 3 × = $

Jason gave away $24, so he had $24 more before:

Jason (before) = 3u + 24 = + 24 =

✅ Jason had $ at first.

After giving, Elly = 4 × Jenny.
Let Jenny (after) = 1 unit, so Elly (after) = 4 units.

The total does not change — stickers are only moved, not added or removed.
Before giving: Jenny had 48 more → (1u + 48). Elly had 48 fewer → (4u − 48).

Use the bar model to see the transfer, then find the difference:

So:  3u = + + =

From Step 1: 3u = 84

1u = 84 ÷ 3 =

Jenny after = 1u = stickers

Jenny gave away 48 stickers, so she had 48 more before:

Jenny (before) = 1u + 48 = + 48 =

✅ Jenny had stickers at first.

At end of day: Red = 2 × Blue left.
Let Blue (after) = 1 unit, so Red = 2 units.

Red files are not sold — the count stays the same all day.
So Red = 2u in the morning too.
Morning: Blue = 2 × Red = 2 × 2u = 4u.

Use the bar model to see the files sold, then find how many units were sold:

Blue sold = 4u − 1u = 3u =

3u = 45

1u = 45 ÷ 3 =

Blue files in morning = 4u = 4 × =

✅ There were blue files at first.

After eating: Muffins = 2 × Cookies left.
Let Cookies (after) = 1 unit, so Muffins = 2 units.

Muffins are not eaten — the count stays the same throughout.
So Muffins = 2u always. Before eating: Cookies = 4 × Muffins = 4 × 2u = 8u.

Use the bar model to see the cookies eaten, then solve:

Eaten = Cookies(before) − Cookies(after) = 8u − 1u = 7u
So:  7u =

7u = 84

1u = 84 ÷ 7 =

Cookies made = 8u = 8 × =

Muffins made = 2u = 2 × =

Total = + =

✅ Ivy made cookies and muffins in all.

After shopping: Tina = 3 × Kim left.
Let Kim (after) = 1 unit, so Tina (after) = 3 units.

Since they started with the same amount:
Kim (before) = 1u + 180    Tina (before) = 3u + 70

Use the bar model to see the spending, then find the difference:

Since they started equal: 1u + 180 = 3u + 70
So: 2u = − =

2u = 110

1u = 110 ÷ 2 =

Each girl's amount = 1u + 180 = 3u + 70

= + 180 = $

✅ Each girl had $ at first.

Each shirt costs ½ as much as a blouse.
Let 1 shirt = 1 unit, so 1 blouse = 2 units.

Use the bar model to count all the units:

Total units = 3u + 4 × 2u = 3u + 8u = u

11u = $176

1u = $176 ÷ 11 = $

1 shirt = 1u = $

✅ Each shirt cost $.

Let 1 pair of shorts = 1 unit.
Then 1 pair of shoes = 1 unit + $12.

Use the bar model to count all the units and extras:

Total = 7u + 48 = 181
7u = 181 − 48 =

7u = 133

1u = 133 ÷ 7 = $

1 pair of shoes = 1u + $12 = $ + $12 = $

✅ Each pair of shoes cost $.

Let Day 1 = 1 unit. Each day adds 3 more:
Day 1 = 1u   Day 2 = 1u+3   Day 3 = 1u+6   Day 4 = 1u+9   Day 5 = 1u+12

Use the bar model to see the pattern, then find the totals:

Total extras = 0+3+6+9+12 =
5u = 55 − =

5u = 25

1u = 25 ÷ 5 =

Day 5 = 1u + 12 = + 12 =

✅ Angela made keychains on the last day.

Let sandals = 1u.
Shoes = 3 × sandals = 3u.
Boots = shoes + $43 = 3u + $43.

Use the bar model to see all three items, then find the total units:

7u = $211 − $43 = $

7u = $168

1u = $168 ÷ 7 = $

Boots = 3u + $43
= 3 × $ + $43
= $ + $43
= $

✅ A pair of boots costs $.

Let key pouch = 1u.
Wallet = 1u + $12.
Bag = 3 × wallet = 3 × (1u + $12) = 3u + $36.

Use the bar model to see all three items:

5u = $258 − $48 = $

5u = $210

1u = $210 ÷ 5 = $

Key pouch = 1u = $

✅ The key pouch costs $.

Let hair dryer = 1u.
Rice cooker = 3 × hair dryer = 3u.
Oven = rice cooker − $15 = 3u − $15.

Use the bar model to see all three items:

7u = $272 + $15 = $

7u = $287

1u = $287 ÷ 7 = $

Rice cooker = 3u = 3 × $ = $

✅ The rice cooker costs $.

Both situations have the same box. Subtracting removes the box:
(Box + 8 books) − (Box + 5 books) = 3 books

Use the bar model to see the difference:

3 books = 3410 − 2300 = g
1 book = ÷ 3 = g

Using Box + 5 books = 2300 g:

5 books = 5 × = g

Box = 2300 − = g

✅ The mass of the box is g.

Let apples = 1u, so oranges = 3u.

Box + 3u = 25 kg
Box + 1u = 11 kg

Subtracting removes the box — use the bar model to see the difference:

2u = 25 − 11 = kg
1u = ÷ 2 = kg

Using Box + 1u = 11 kg:

Box = 11 − = kg

✅ The mass of the box is kg.

Let apples = 1u, so oranges = 3u.

Box + 1u = 560 g
Box + 3u = 1240 g

Subtracting removes the box — use the bar model to see the difference:

2u = 1240 − 560 = g
1u = ÷ 2 = g

Using Box + 1u = 560 g:

Box = 560 − = g

✅ The mass of the box is g.

Since there are equal numbers, let adults = children = 1u (people).

Total collected from adults = 1u × $2 = $2u
Total collected from children = 1u × $1 = $1u

Use the bar model to see the total collected:

3u = $228
1u = $228 ÷ 3 =

So there were adults and children.

Total people = adults + children

= + =

✅ There were people at the carnival.

Since equal numbers, let egg tarts = cheese tarts = 1u.

Total from egg tarts  = 1u × $2 = $2u
Total from cheese tarts = 1u × $3 = $3u

Use the bar model to see the total spent:

5u = $60
1u = $60 ÷ 5 =

Number of egg tarts = 1u =

✅ Peggy bought egg tarts.

1 file = $6
1 notebook = 2 × $6 = $

Since equal numbers, let notebooks = files = 1u.

Total from notebooks = 1u × $12 = $12u
Total from files     = 1u × $6 = $6u

Use the bar model to see the total spent:

18u = $90
1u = $90 ÷ 18 =

Number of files = 1u =

✅ Helen bought files.

20 ÷ 5 = groups of 5

Total cost = 4 × $100 = $
Total discount = 4 × $6 = $

Amount to pay = $400 − $ = $

✅ They had to pay $.

32 ÷ 5 = groups remainder

Total cost = 32 × $20 = $
Total discount = 6 × $6 = $

Amount to pay = $640 − $ = $

✅ They had to pay $.

Each group of 5 people pays $100 but gets $6 off:
1 group of 5 = $100 − $6 = $

The bar model shows 1 group — how many such groups fit into $812?

Number of full groups = $812 ÷ $94 = remainder $

Leftover amount $60 ÷ $20 = more people

Total people = 8 × 5 + =

✅ Mr Samad paid for people.

62 ÷ 3 = remainder kg

Number of 3 kg packets =

✅ packets of rice can be packed.

Mass of remaining rice = kg

✅ The remaining rice has a mass of kg.

257 ÷ 42 = remainder pupils

6 full buses + 1 bus for the remaining 5 pupils

Total buses needed = 6 + 1 =

✅ buses were needed.

980 ÷ 30 = remainder m

The 20 m remainder cannot make another 30 m piece, so it is not counted.

Greatest number of 30 m pieces =

✅ The greatest number of pieces that can be cut is .

Equation 1: 2f + 3p = $20
Equation 2: 4f + 7p = $42

Multiply Equation 1 by :
4f + p = $  …(3)

Eqn 2 − Eqn 3: (4f + 7p) − (4f + 6p) = $42 − $40

1p = $

✅ 1 pen costs $.

1 pair of pants = 3 ties, so 4 pairs of pants = ties

Rewrite Equation 1: 2s + t = $144  …(3)

Equation 2: 3s + 6t = $126
Multiply Equation 2 by :
s + 12t = $  …(4)

Eqn 4 − Eqn 3: (6s + 12t) − (2s + 12t) = $252 − $144

4s = $
1s = $

✅ 1 shirt costs $.

1 bowl = $6.40 − $1.70 = $

Let bowls = 1u. Then plates = 1u + 12.

Cost of 12 extra plates = 12 × $6.40 = $

Remaining cost (for u bowls + u plates) = $143.40 − $ = $

Cost per unit pair = $4.70 + $6.40 = $
Number of units = $66.60 ÷ $ =

Bowls = 1u =
Plates = 1u + 12 = + 12 =

✅ Mrs Tan bought plates.

Age gap = 47 − 8 = years

This gap never changes — both age by the same amount each year.

In x years: Mr Ng = 4 × Son → Son = 1u, Mr Ng = 4u.

Age gap = 4u − 1u = 3u =

1u = 39 ÷ 3 =

Son's future age = 1u = years old

Years from now = − 8 =

✅ In years' time, Mr Ng will be 4 times as old as his son.

Son's age now = 40 − 24 = years old

In 3 years, both age by 3:

Mr Lim in 3 years = 40 + 3 =
Son in 3 years = 16 + 3 =

Combined age = + =

✅ Their combined age three years later will be .

Let Alice = 1u, so Simon = 3u.

In 7 years: Alice = 1u + 7, Simon = 3u + 7.
Total = (1u + 7) + (3u + 7) = 4u + 14 = 78

Use the bar model to see the future ages, then solve:

Remove the two +7 strips: 4u = 78 − =
1u = ÷ 4 =

Alice's age now = 1u = years old

✅ Alice is years old now.

Denominators are 5 and 4. LCM =

Let total = 20 units.
Guppies  = 15 × 20 = units
Goldfish = 14 × 20 = units

Angelfish = 20 − − = units

Fraction of angelfish =

✅ of the fish are angelfish.

Denominators are 10 and 4. LCM =

Let total = 20 units.   1u = 400 ÷ 20 =

Red  = 310 × 20 = units
Blue = 14 × 20 = units

Green = 20 − − = units

Green buttons = 9u = 9 × = buttons

✅ The seamstress has green buttons.

Denominators are 8 and 4. LCM =

Let total = 8 units.   1u = 208 ÷ 8 =

Friends   = 18 × 8 = unit(s)
Neighbour = 14 × 8 = unit(s)

Left = 8 − − = units

Marbles left = 5u = 5 × =

✅ Harry had marbles left.

After giving to neighbours, the remainder is split:
14 of remainder → friends  = 1u
34 of remainder → kept     = 3u = 24

Use the bar model to see the remainder split:

3u = 24 → 1u = 24 ÷ 3 =

Remainder = 4u = 4 × = cookies

Total cookies baked = 35 + =

✅ Grace baked cookies.

After giving to neighbours, the remainder is split:
14 of remainder → friends = 1u
34 of remainder → kept    = 3u

Kept = 13 of total baked → total = 3 × kept = 3 × 3u = 9u

Neighbours = total − remainder = 9u − 4u = u = 35
1u = 35 ÷ =

Total = 9u = 9 × =

✅ Gracia baked cookies.

Boys = 45 × 400 =
Girls = 15 × 400 =

Boys who left = 34 × 320 =
Boys remained = 320 − =

Girls who left = 15 × 80 =
Girls remained = 80 − =

More boys than girls = − =

✅ There were more boys than girls remaining.

Cloth for dress = 12 × 5 m = m

Total used = dress + bag = + = m

Cloth left = 5 − = m

✅ Fiona had m of cloth left.

Remaining = 34 − 18 = kg

Muffins = 310 × remaining = 310 × = kg

✅ Mrs Lim used kg of sugar for muffins.

Total = 112 ℓ = ℓ

Remaining = 32 − 110 = ℓ

Drank = 14 of remaining = 14 × = ℓ

✅ Siti drank ℓ of juice.

From the grid (each square = 1 cm × 1 cm):

Width UW = cm
Height UY = cm

Area of rectangle UWXY = × = cm²

Subtract 3 triangles from the rectangle:

△UZV: base = UV = cm, height = UZ = cm
Area = ½ × × = cm²

△VWX: base = VW = cm, height = WX = cm
Area = ½ × × = cm²

△ZYX: base = YX = cm, height = YZ = cm
Area = ½ × × = cm²

Area of △VXZ = Rectangle − △UZV − △VWX − △ZYX

= − − − = cm²

✅ The area of triangle VXZ is cm².

The base of the shaded triangle is AB = side of rectangle = cm

Height of triangle = EF = side of square = cm

Area = ½ × base × height

= ½ × × = cm²

✅ The area of the shaded part is cm².

GF = CH = HD = cm (equal, shown by tick marks)

Left side AF = AG + GF = 12 + = cm
Right side CE = DE + BC = 18 + = cm

Parallel sides: AF = cm, CE = cm, base FE = cm

Area = ½ × ( + ) × = cm²

△ABG: base AB = cm, height AG = cm
Area = ½ × × = cm²

△CHD: base CH = cm, height HD = cm
Area = ½ × × = cm²

Total = trapezoid + △ABG + △CHD

= + + = cm²

✅ The area of the shaded figure is cm².

BC = BF − CF = 17 − = cm

Big rectangle (30 × 17):
Area = (GF + FE) × GF = ( + ) × 17 = × 17 = cm²

△ABC: ½ × AB × BC = ½ × × = cm²

Rect BD: BC × CD = × = cm²

△CDE: ½ × CD × DE = ½ × × = cm²

△AGE: ½ × GE × AG = ½ × × = cm²

Area △ACE = Big rect − △ABC − Rect BD − △CDE − △AGE

= − − − − = cm²

✅ The area of triangle ACE is cm².

Initial water height = 56 × 18 cm = cm

Volume = length × width × height

= × × = cm³

Volume = × × = cm³

Volume poured out = − = cm³

Since 1 cm³ = 1 ml:
✅ Mr Lee poured out ml of water.

By symmetry, △FGB ≅ △CEG (congruent triangles).

Area of △FGB = cm²

The base and height of △CEF are the same as the length and breadth of the rectangle.
Area of △CEF = ½ × base × height = ½ × length × breadth = ½ × area of rectangle
= ½ × = cm²

G lies on FC, so △CEF = △CGE + △GFE:
Area of △GFE = 80 − = cm²

By symmetry △FBG ≅ △CEG:
Area of △FBG = cm²

Shaded area = Area of rectangle − △CGE − △GFE − △FBG

= − − − = cm²

✅ The area of the shaded parts is cm².

Since the right angle is at X, QX and XR are the base and height.

Area of △QXR = 12 × QX × XR
= 12 × × = cm²

QR, which is the length of the rectangle, can also be the base of △QXR and the corresponding height will be equal to SR which is the breadth of the rectangle.
Therefore, Area of △QXR = 12 × base × height = 12 × length × breadth = 12 × area of rectangle.
Therefore, Area of rectangle = 2 × area of △QXR:

Area of rectangle PQRS = 2 × area of △QXR
= 2 × = cm²

Area of rectangle = QR × SR
= × SR

SR = ÷ = cm

✅ SR = cm.

A cuboid has 3 pairs of faces:

Top & bottom (10 × 5): 2 × = cm²
Front & back (10 × 4): 2 × = cm²
Left & right (5 × 4): 2 × = cm²

Total surface area = + + = cm²

Unpainted cubes are fully inside the cuboid — not touching any face.
Remove 1 cube from each side along every dimension:

Inner length = 10 − 2 =
Inner width  = 5 − 2 =
Inner height = 4 − 2 =

Number of unpainted cubes = × × =

Cubes with 2 painted faces lie along the edges (not at corners).
A cuboid has 12 edges — 4 of each length.

4 edges of length 10: 4 × (10 − 2) = 4 × =
4 edges of length 5:  4 × (5 − 2) = 4 × =
4 edges of length 4:  4 × (4 − 2) = 4 × =

Total = + + =