Notes

Here you go — all the working in plain, copy‑and‑paste‑ready text.

--------------------------------
YEAR 8 COMMON HOMEWORK 05
--------------------------------

Question 1
Work out 3.67 × 4.2

Step 1: Ignore decimals first.
Compute 367 × 42.

367 × 40 = 14,680
367 × 2 = 734
Total = 14,680 + 734 = 15,414

Step 2: Put decimals back.
3.67 has 2 decimal places, 4.2 has 1 decimal place → total 3 decimal places.

So 3.67 × 4.2 = 15.414

Answer: 15.414


Question 2
The diagram shows five nested squares with side lengths 1, 2, 3, 4, 5 cm.
What percentage of the area of the outer square is shaded?

Outer square side = 5 cm
Area of outer square = 5² = 25 cm²

Areas of the squares:
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25

Method:
1) Use the diagram to see which regions are shaded (they are made from differences between these squares).
2) Add the areas of all shaded regions = A_shaded.
3) Percentage shaded = (A_shaded / 25) × 100.

In this question, A_shaded = 10 cm², so:
Percentage shaded = (10 / 25) × 100 = 40%

Answer: 40%


Question 3
On a farm there are chickens, ducks and pigs.
Chickens : Ducks = 7 : 2
Ducks : Pigs = 5 : 9
There are 36 pigs.
Work out the number of chickens.

From Ducks : Pigs = 5 : 9,
9 corresponds to 36, so scale factor = 36 ÷ 9 = 4.
Ducks = 5 × 4 = 20.

From Chickens : Ducks = 7 : 2,
2 corresponds to 20, so scale factor = 20 ÷ 2 = 10.
Chickens = 7 × 10 = 70.

Answer: 70 chickens


Question 4
Right-angled triangle with perimeter 126 cm.
Work out the area.

Let the two perpendicular sides be a and b, and the hypotenuse be c.

Perimeter: a + b + c = 126
Right angle → Pythagoras: a² + b² = c²

Method:
1) Use the side lengths given in the diagram to form equations.
2) Solve for a and b.
3) Area = 1/2 × a × b = 630 cm².

Answer: 630 cm²


Question 5
Expand and simplify: 4(2d + 3) − 2(3d − 5)

First expand each bracket:

4(2d + 3) = 8d + 12
−2(3d − 5) = −6d + 10

Now combine like terms:

8d + 12 − 6d + 10
= (8d − 6d) + (12 + 10)
= 2d + 22

Answer: 2d + 22


Question 6
On the grid, draw the line with equation x = 4.

The equation x = 4 means:
All points on the line have x-coordinate 4, and y can be any value.

So it is a vertical line crossing the x-axis at x = 4.

Answer: A vertical line through x = 4


Question 7
Sorting triangles by isosceles/not and right/obtuse/acute.

Method:
1) For each triangle, check if it has two equal sides → isosceles or not.
2) Check its largest angle:
- 90° → right angle
- greater than 90° → obtuse angle
- less than 90° → acute angle
3) Place each triangle’s letter in the correct region of the sorting diagram.
4) Any empty region gets an N.

From your completed table:
Triangle A: isosceles and has a right angle.
Triangle C: isosceles and has an obtuse angle.
Triangle D: not isosceles and has a right angle.
All other regions: N.


Question 8
L-shaped garden. Lanza charges £16 per square metre.
Find the total cost.

Method:
1) Split the L-shape into two rectangles.
2) Find each rectangle’s area:
A1 = length1 × width1
A2 = length2 × width2
3) Total area A = A1 + A2.
4) Cost = 16 × A.

In this question, the total area works out to 54.5 m², so:
Cost = 16 × 54.5 = 872

Answer: £872


Question 9
Change 32.4 m³ into cm³.

1 m = 100 cm
So 1 m³ = 100³ cm³ = 1,000,000 cm³

32.4 m³ = 32.4 × 1,000,000 = 32,400,000 cm³

Answer: 32,400,000 cm³


Question 10
Ravina counts matches in 40 boxes.
Table:

Number of matches: 21, 22, 23, 24, 25
Frequency: 13, 8, 8, 6, 5

Work out the mean number of matches.

Step 1: Multiply each value by its frequency.
21 × 13 = 273
22 × 8 = 176
23 × 8 = 184
24 × 6 = 144
25 × 5 = 125

Step 2: Add these totals.
273 + 176 = 449
449 + 184 = 633
633 + 144 = 777
777 + 125 = 902

Step 3: Total frequency = 40.
Mean = 902 ÷ 40 = 22.55

Answer: 22.55


Question 11
Three girls’ marks: 92%, 80%, 71%
Mean of all 12 marks = 84%
Find the mean of the boys’ marks.

Step 1: Total of all 12 marks.
84 × 12 = 1008

Step 2: Total of girls’ marks.
92 + 80 + 71 = 243

Step 3: Total of boys’ marks.
1008 − 243 = 765

Step 4: There are 9 boys.
Mean of boys’ marks = 765 ÷ 9 = 85

Answer: 85%


Question 12
Pie chart: Trekking 90°, City break 72°, Walking 54°, Beach 144°.
48 people chose beach.
How many chose walking?

Step 1: Fraction for beach.
Beach angle = 144°
Total angle = 360°
Fraction = 144 / 360 = 2/5

Step 2: Let total number of people be T.
(2/5) × T = 48
T = 48 × (5/2) = 48 × 2.5 = 120

Step 3: Fraction for walking.
Walking angle = 54°
Fraction = 54 / 360 = 3/20

Step 4: Number who like walking.
(3/20) × 120 = 3 × 6 = 18

Answer: 18 people


--------------------------------
YEAR 8 COMMON HOMEWORK 06
--------------------------------

Question 1
Four different positive integers have a product of 110.
What is the sum of the four integers?

Prime factorise 110:
110 = 2 × 5 × 11

Use 1 as the fourth integer:
1, 2, 5, 11

Check product:
1 × 2 × 5 × 11 = 110

Sum:
1 + 2 + 5 + 11 = 19

Answer: 19


Question 2
Gabi wants to buy a flat costing £175,000.
Mortgage formula: M = 4.625A, where A is annual income.
Gabi’s annual income A = £34,000.
Deposit = cost of flat − mortgage.
Work out the deposit.

Step 1: Calculate mortgage M.
M = 4.625 × 34,000

Break 4.625 into 4 + 0.625:
4 × 34,000 = 136,000
0.625 × 34,000 = 21,250

Total M = 136,000 + 21,250 = 157,250

Step 2: Deposit = 175,000 − 157,250 = 17,750

Answer: £17,750


Question 3
First two terms of a sequence are 2/3 and 4/5.
Each term after the second is the mean of the previous two terms.
Find the fifth term.

Let T1 = 2/3, T2 = 4/5.
Each new term: Tn = (Tn−1 + Tn−2) ÷ 2.

Find T3:
T3 = (2/3 + 4/5) ÷ 2

Convert to a common denominator (15):
2/3 = 10/15
4/5 = 12/15

10/15 + 12/15 = 22/15

T3 = (22/15) ÷ 2 = 22/15 × 1/2 = 22/30 = 11/15

Find T4:
T4 = (T2 + T3) ÷ 2 = (4/5 + 11/15) ÷ 2

4/5 = 12/15
12/15 + 11/15 = 23/15

T4 = (23/15) ÷ 2 = 23/15 × 1/2 = 23/30

Find T5:
T5 = (T3 + T4) ÷ 2 = (11/15 + 23/30) ÷ 2

11/15 = 22/30
22/30 + 23/30 = 45/30 = 3/2

T5 = (3/2) ÷ 2 = 3/2 × 1/2 = 3/4

Answer: 3/4


Question 4
Factorise fully: 16m³g³ + 24m²g⁵

Step 1: Find the common factor.
Numbers: HCF(16, 24) = 8
m: smallest power is m²
g: smallest power is g³

Common factor = 8m²g³

Step 2: Divide each term by the common factor.
16m³g³ ÷ 8m²g³ = 2m
24m²g⁵ ÷ 8m²g³ = 3g²

So:
16m³g³ + 24m²g⁵ = 8m²g³(2m + 3g²)

Answer: 8m²g³(2m + 3g²)


Question 5
Solve the inequality and list integer values of x:
−4 < 3x + 5 ≤ 11

Step 1: Subtract 5 from all three parts.
−4 − 5 < 3x + 5 − 5 ≤ 11 − 5
−9 < 3x ≤ 6

Step 2: Divide all parts by 3.
−9 ÷ 3 < x ≤ 6 ÷ 3
−3 < x ≤ 2

Step 3: List integer values of x.
Integers greater than −3 and less than or equal to 2:
x = −2, −1, 0, 1, 2

Answer: −2, −1, 0, 1, 2


Question 6
Find the nth term of the sequence: 7, 4, 1, −2, …

Step 1: Find the common difference.
4 − 7 = −3
1 − 4 = −3
−2 − 1 = −3

Common difference d = −3.

Step 2: Use the arithmetic sequence formula:
a_n = a_1 + (n − 1)d

Here a_1 = 7, d = −3:
a_n = 7 + (n − 1)(−3)
= 7 − 3n + 3
= 10 − 3n

Answer: nth term = −3n + 10


Question 7
Volume of a cuboid from its net (face areas given).

Let the three different edge lengths be a, b, c.
Then the three pairs of opposite faces have areas:
ab, bc, ac

Suppose the net shows these three areas as A, B, C.
So:
ab = A
bc = B
ac = C

Multiply all three equations:
(ab)(bc)(ac) = A × B × C
a²b²c² = ABC
(abc)² = ABC

So:
abc = √(ABC)

But abc is the volume of the cuboid.

In your question, ABC = 3600, so:
Volume = √3600 = 60 cm³

Answer: 60 cm³


Question 8
Angle x in quadrilateral ABCD and isosceles triangle ADE, with AE = AD and EDC a straight line.

Method:
1) Use the fact that EDC is a straight line → angles on a straight line sum to 180°.
2) Use that triangle ADE is isosceles (AE = AD) → base angles are equal.
3) Use angles in a triangle sum to 180° to find missing angles in triangle ADE.
4) Use angles in a quadrilateral sum to 360° to solve for x.

Carrying this through with the given angles leads to:
x = 62°

Answer: 62°


Question 9
Sophia’s gym floor is a trapezium with parallel sides 7 m and 4 m, height 6 m.
Each tin of paint covers 20 m².
Prices: 1 tin for $13, 4 tins for $40.
Find the least amount of money she must pay.

Step 1: Area of the trapezium.
Area = 1/2 × (sum of parallel sides) × height
= 1/2 × (7 + 4) × 6
= 1/2 × 11 × 6
= 33 m²

Step 2: Number of tins needed.
Each tin covers 20 m².
Tins needed = 33 ÷ 20 = 1.65 → must buy 2 tins.

Step 3: Compare cost options.
Option 1: 2 single tins → 2 × 13 = 26
Option 2: 4-tin offer → 40

Cheapest is $26.

Answer: $26


Question 10
(Not visible in your screenshots, so no working shown.)


Question 11
Two fair spinners:
Spinner A: 1, 2, 3
Spinner B: 1, 2, 3, 4
Chanthira spins both 84 times.
Estimate how many times they land on the same number.

Step 1: Total number of equally likely outcomes.
3 options on A × 4 options on B = 12 outcomes.

Step 2: Outcomes where numbers are the same:
(1,1), (2,2), (3,3) → 3 outcomes.

Probability(same) = 3/12 = 1/4

Step 3: Expected number in 84 spins.
(1/4) × 84 = 21

Answer: 21


Question 12
Spinner with 6, 7, 8 and a fair dice 1–6.
Score = spinner − dice.
Complete the table of all possible scores.

Compute each entry as (spinner value − dice value):

Row for spinner = 6:
6 − 1 = 5
6 − 2 = 4
6 − 3 = 3
6 − 4 = 2
6 − 5 = 1
6 − 6 = 0

Row for spinner = 7:
7 − 1 = 6
7 − 2 = 5
7 − 3 = 4
7 − 4 = 3
7 − 5 = 2
7 − 6 = 1

Row for spinner = 8:
8 − 1 = 7
8 − 2 = 6
8 − 3 = 5
8 − 4 = 4
8 − 5 = 3
8 − 6 = 2

Table:

SpinnerDice: 1 2 3 4 5 6
6 5 4 3 2 1 0
7 6 5 4 3 2 1
8 7 6 5 4 3 2


--------------------------------
EXTRA: COUNTERS IN A BAG
--------------------------------

There are blue, green, red, yellow counters.
P(blue) = 0.32
P(green) = 0.20
P(red) = 5 × P(yellow)
Let P(yellow) = y.
Total probability = 1.
There are 300 counters.
Find the number of yellow counters.

Step 1: Write the probability equation.
0.32 + 0.20 + y + 5y = 1
0.52 + 6y = 1
6y = 1 − 0.52 = 0.48
y = 0.48 ÷ 6 = 0.08

So P(yellow) = 0.08.

Step 2: Convert probability to number of counters.
Number of yellow counters = 0.08 × 300 = 24

Answer: 24 yellow counters

--------------------------------

If you want, you can now just grab any block of this and drop it straight into your notes.Great, let’s put **all the working** in one clean place.

I’ll go:

- **Homework 5 — Q1 to Q12**
- **Homework 6 — Q1 to Q12**
- **Extra counters question**

For a few geometry ones where the **diagram isn’t fully visible**, I’ll show the **method** rather than pretend I can see lengths/angles I don’t.

---

## YEAR 8 COMMON HOMEWORK 05 — FULL WORKING

### Q1 — Work out (3.67 times 4.2)

1. Ignore decimals first:
(367 times 42)

2. Multiply:
(367 times 40 = 14{,}680)
(367 times 2 = 734)
Total (= 14{,}680 + 734 = 15{,}414)

3. Put decimals back:
3.67 has 2 d.p., 4.2 has 1 d.p. → 3 d.p. total.

[
3.67 times 4.2 = 15.414
]

**Answer: (15.414)**

---

### Q2 — Nested squares, shaded percentage

Outer square side = 5 cm → area:

[
A_{text{outer}} = 5^2 = 25text{ cm}^2
]

Areas of squares with sides 1, 2, 3, 4, 5:

- (1^2 = 1)
- (2^2 = 4)
- (3^2 = 9)
- (4^2 = 16)
- (5^2 = 25)

Method:

1. Use the diagram to see which regions are shaded (they’re made from differences between these squares).
2. Add the areas of all shaded regions → call this (A_{text{shaded}}).
3. Percentage shaded:

[
text{Percentage} = frac{A_{text{shaded}}}{25} times 100
]

In your question, this works out to **40%**, so (A_{text{shaded}} = 10text{ cm}^2).

**Answer: (40%)**

---

### Q3 — Farm ratios (chickens, ducks, pigs)

- Chickens : Ducks = (7:2)
- Ducks : Pigs = (5:9)
- Pigs = 36

1. From Ducks : Pigs = (5:9):
(9 to 36) so scale factor = 4.
Ducks (= 5 times 4 = 20).

2. From Chickens : Ducks = (7:2):
(2 to 20) so scale factor = 10.
Chickens (= 7 times 10 = 70).

**Answer: 70 chickens**

---

### Q4 — Right‑angled triangle, perimeter 126 cm, area

Let the two perpendicular sides be (a) and (b), hypotenuse (c).

Given:

- Perimeter:
[
a + b + c = 126
]
- Right angle → Pythagoras:
[
a^2 + b^2 = c^2
]

Method:

1. Use the side lengths shown in the diagram to set up these equations.
2. Solve for (a) and (b).
3. Area:

[
text{Area} = frac{1}{2}ab = 630text{ cm}^2
]

**Answer: (630text{ cm}^2)**

---

### Q5 — Expand and simplify
[
4(2d + 3) - 2(3d - 5)
]

1. Expand:
(4(2d + 3) = 8d + 12)
(-2(3d - 5) = -6d + 10)

2. Combine like terms:
(8d + 12 - 6d + 10 = (8d - 6d) + (12 + 10) = 2d + 22)

**Answer: (2d + 22)**

---

### Q6 — Draw the line (x = 4)

- Equation (x = 4) means all points with x‑coordinate 4.
- That’s a **vertical line** through (x = 4) on the x‑axis.

**Answer: vertical line through (x = 4)**

---

### Q7 — Triangle sorting (isosceles / not, right / obtuse / acute)

You have triangles A, C, D, N on a grid.

Method:

1. For each triangle, check if it has **two equal sides** → isosceles or not.
2. Check its largest angle:
- 90° → right
- >90° → obtuse
- <90° → acute
3. Place each letter in the correct box of the sorting diagram.
4. Any empty box gets an **N**.

From your completed table:

- A: isosceles & right angle
- C: isosceles & obtuse angle
- D: not isosceles & right angle
- All other regions: N

---

### Q8 — L‑shaped garden, cost of turf

L‑shaped garden with given side lengths; cost = £16 per m².

Method:

1. Split the L‑shape into **two rectangles**.
2. Find each rectangle’s area:
[
A_1 = text{length}_1 times text{width}_1,quad
A_2 = text{length}_2 times text{width}_2
]
3. Total area:
[
A = A_1 + A_2
]
4. Cost:
[
text{Cost} = 16 times A
]

In your question, this gives **£872**, so:

[
A = frac{872}{16} = 54.5text{ m}^2
]

**Answer: £872**

---

### Q9 — Convert (32.4text{ m}^3) into (text{cm}^3)

[
1text{ m} = 100text{ cm} Rightarrow 1text{ m}^3 = 100^3text{ cm}^3 = 1{,}000{,}000text{ cm}^3
]

[
32.4text{ m}^3 = 32.4 times 1{,}000{,}000 = 32{,}400{,}000text{ cm}^3
]

**Answer: (32{,}400{,}000text{ cm}^3)**

---

### Q10 — Mean from frequency table (matches)

Table:

| Matches | Frequency |
|---------|-----------|
| 21 | 13 |
| 22 | 8 |
| 23 | 8 |
| 24 | 6 |
| 25 | 5 |

1. Multiply each value by its frequency:
(21 times 13 = 273)
(22 times 8 = 176)
(23 times 8 = 184)
(24 times 6 = 144)
(25 times 5 = 125)

2. Sum:
(273 + 176 = 449)
(449 + 184 = 633)
(633 + 144 = 777)
(777 + 125 = 902)

3. Total frequency = 40.

[
text{Mean} = frac{902}{40} = 22.55
]

**Answer: 22.55**

---

### Q11 — Mean of boys’ marks

Girls’ marks: 92%, 80%, 71%
Mean of all 12 = 84%.

1. Total of all 12 marks:
[
84 times 12 = 1008
]

2. Total of girls’ marks:
[
92 + 80 + 71 = 243
]

3. Total of boys’ marks:
[
1008 - 243 = 765
]

4. There are 9 boys:
[
text{Mean} = frac{765}{9} = 85
]

**Answer: 85%**

---

### Q12 — Pie chart (Beach = 48 people, find Walking)

Angles:

- Trekking: (90^circ)
- City break: (72^circ)
- Walking: (54^circ)
- Beach: (144^circ)

1. Beach fraction:
[
frac{144}{360} = frac{2}{5}
]

2. Let total people = (T):
[
frac{2}{5}T = 48 Rightarrow T = 48 times frac{5}{2} = 120
]

3. Walking fraction:
[
frac{54}{360} = frac{3}{20}
]

4. Number who like walking:
[
frac{3}{20} times 120 = 18
]

**Answer: 18 people**

---

## YEAR 8 COMMON HOMEWORK 06 — FULL WORKING

### Q1 — Four integers with product 110, find sum

Prime factorise 110:

[
110 = 2 times 5 times 11
]

Use 1 as the fourth integer:

[
1, 2, 5, 11
]

Check product:

[
1 times 2 times 5 times 11 = 110
]

Sum:

[
1 + 2 + 5 + 11 = 19
]

**Answer: 19**

---

### Q2 — Gabi’s mortgage and deposit

Flat price = £175,000
Formula: (M = 4.625A), (A = 34{,}000)

1. Mortgage:
[
M = 4.625 times 34{,}000
]

Break 4.625:

- (4 times 34{,}000 = 136{,}000)
- (0.625 times 34{,}000 = 21{,}250)

Total:
[
M = 136{,}000 + 21{,}250 = 157{,}250
]

2. Deposit = cost − mortgage:
[
175{,}000 - 157{,}250 = 17{,}750
]

**Answer: £17,750**

---

### Q3 — Sequence where each term is mean of previous two

Given:

[
T_1 = frac{2}{3},quad T_2 = frac{4}{5}
]

Each new term:
[
T_n = frac{T_{n-1} + T_{n-2}}{2}
]

**Find (T_3):**

[
T_3 = frac{frac{2}{3} + frac{4}{5}}{2}
]

Common denominator 15:

[
frac{2}{3} = frac{10}{15},quad frac{4}{5} = frac{12}{15}
]

[
frac{10}{15} + frac{12}{15} = frac{22}{15}
]

[
T_3 = frac{22}{15} div 2 = frac{22}{30} = frac{11}{15}
]

**Find (T_4):**

[
T_4 = frac{T_2 + T_3}{2} = frac{frac{4}{5} + frac{11}{15}}{2}
]

(frac{4}{5} = frac{12}{15})

[
frac{12}{15} + frac{11}{15} = frac{23}{15}
]

[
T_4 = frac{23}{15} div 2 = frac{23}{30}
]

**Find (T_5):**

[
T_5 = frac{T_3 + T_4}{2} = frac{frac{11}{15} + frac{23}{30}}{2}
]

(frac{11}{15} = frac{22}{30})

[
frac{22}{30} + frac{23}{30} = frac{45}{30} = frac{3}{2}
]

[
T_5 = frac{3}{2} div 2 = frac{3}{4}
]

**Answer: (frac{3}{4})**

---

### Q4 — Factorise
[
16m^3g^3 + 24m^2g^5
]

1. Common factor:
- Numbers: HCF(16, 24) = 8
- (m): smallest power = (m^2)
- (g): smallest power = (g^3)

Factor out (8m^2g^3).

2. Divide each term:

[
16m^3g^3 div 8m^2g^3 = 2m
]
[
24m^2g^5 div 8m^2g^3 = 3g^2
]

So:

[
16m^3g^3 + 24m^2g^5 = 8m^2g^3(2m + 3g^2)
]

**Answer: (8m^2g^3(2m + 3g^2))**

---

### Q5 — Solve inequality (-4 < 3x + 5 le 11)

1. Subtract 5 from all parts:

[
-4 - 5 < 3x + 5 - 5 le 11 - 5
]

[
-9 < 3x le 6
]

2. Divide all parts by 3:

[
frac{-9}{3} < x le frac{6}{3}
]

[
-3 < x le 2
]

3. Integer values of (x):

[
x = -2, -1, 0, 1, 2
]

**Answer: (-2, -1, 0, 1, 2)**

---

### Q6 — nth term of (7, 4, 1, -2, dots)

Sequence: (7, 4, 1, -2,dots)

1. Common difference:
(4 - 7 = -3)
(1 - 4 = -3)
(-2 - 1 = -3)

So (d = -3).

2. Arithmetic sequence formula:

[
a_n = a_1 + (n-1)d
]

[
a_n = 7 + (n-1)(-3) = 7 - 3n + 3 = 10 - 3n
]

**Answer: ( text{nth term} = -3n + 10 )**

---

### Q7 — Volume of cuboid from net

Net shows three different face areas (A, B, C).
Let side lengths be (a, b, c):

[
ab = A,quad bc = B,quad ac = C
]

Multiply:

[
(ab)(bc)(ac) = A B C
]

[
a^2 b^2 c^2 = (abc)^2 = A B C
]

So:

[
abc = sqrt{A B C}
]

But (abc) is the **volume**.

In your question, (A B C = 3600), so:

[
text{Volume} = sqrt{3600} = 60text{ cm}^3
]

**Answer: (60text{ cm}^3)**

---

### Q8 — Angle (x) in quadrilateral + isosceles triangle

Given:

- Quadrilateral (ABCD)
- Triangle (ADE) is isosceles with (AE = AD)
- (EDC) is a straight line

Method:

1. Use **angles on a straight line** at D or C → sum to 180°.
2. Use **isosceles triangle** (ADE): base angles equal.
3. Use **angles in a triangle = 180°** to find missing angles in triangle ADE.
4. Use **angles in a quadrilateral = 360°** to solve for (x).

Carrying this through with the given angles (39°, 100°, 41° etc.) gives:

**Answer: (x = 62^circ)**

---

### Q9 — Trapezium gym floor + paint cost

Trapezium with:

- Parallel sides: 7 m and 4 m
- Height: 6 m
- Each tin covers (20text{ m}^2)
- Prices: 1 tin for $13, 4 tins for $40

1. Area of trapezium:

[
text{Area} = frac{1}{2}(a + b)h = frac{1}{2}(7 + 4)times 6 = frac{1}{2} times 11 times 6 = 33text{ m}^2
]

2. Tins needed:

[
frac{33}{20} = 1.65 Rightarrow text{need 2 tins}
]

3. Compare costs:

- 2 single tins: (2 times 13 = 26)
- 4‑tin offer: 40

Cheapest: **2 single tins**.

**Answer: $26**

---

### Q10 — (Not visible in your screenshots)

No question text given, so I can’t show working for this one.

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### Q11 — Two spinners, expected same number (84 spins)

Spinner A: 1, 2, 3
Spinner B: 1, 2, 3, 4

1. Total outcomes:
[
3 times 4 = 12
]

2. Same number outcomes: (1,1), (2,2), (3,3) → 3 outcomes.

[
P(text{same}) = frac{3}{12} = frac{1}{4}
]

3. Expected number in 84 spins:

[
frac{1}{4} times 84 = 21
]

**Answer: 21**

---

### Q12 — Spinner 6/7/8 and dice 1–6 (subtraction table)

Score = spinner − dice.

| Spinner Dice | 1 | 2 | 3 | 4 | 5 | 6 |
|----------------|---|---|---|---|---|---|
| 6 | 5 | 4 | 3 | 2 | 1 | 0 |
| 7 | 6 | 5 | 4 | 3 | 2 | 1 |
| 8 | 7 | 6 | 5 | 4 | 3 | 2 |

Each entry is:

[
text{score} = text{spinner value} - text{dice value}
]

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## EXTRA — Counters in a bag (blue/green/red/yellow)

Given:

- (P(text{blue}) = 0.32)
- (P(text{green}) = 0.20)
- Let (P(text{yellow}) = y)
- (P(text{red}) = 5y)
- Total probability = 1
- Total counters = 300

1. Equation:

[
0.32 + 0.20 + y + 5y = 1
]

[
0.52 + 6y = 1
]

[
6y = 0.48 Rightarrow y = 0.08
]

So (P(text{yellow}) = 0.08).

2. Number of yellow counters:

[
0.08 times 300 = 24
]

**Answer: 24 yellow counters**

---

If any one of these still feels shaky, tell me **which homework and question number** (like “06 Q5”) and I’ll slow that one down even more.