$$2^4 = 2 \times 2 \times 2 \times 2 = 16$$

The number is: $7.264\underline{5}1$

The 4th decimal place is 5, so we round the 3rd decimal place up.

$$7.26451 = 7.265 \text{ (to 3 d.p.)}$$

$$7 \times e \times f \times 8 = 7 \times 8 \times e \times f = 56ef$$

(B1)

$$\frac{x}{5} = 2\frac{1}{2}$$

$$2\frac{1}{2} = 2.5$$

$$x = 5 \times 2.5 = 12.5$$

(B1)

$$\frac{4}{5} \times 100 = \frac{400}{5} = 80\%$$

$$60\% = 0.6$$

$$0.6 \times 70 = 42$$ (M1)

$$P(\text{B}) = \frac{2}{4} = \frac{1}{2}$$

Mark the cross at $\frac{1}{2}$ on the probability scale.

(B1)

$$P(\text{F}) = 0$$

Mark the cross at $0$ on the probability scale.

(B1)

$$\text{Amount spent} = £10 – £0.30 = £9.70$$

$$2 \times £1.50 + £1.60 = £3.00 + £1.60 = £4.60$$ (P1)

$$\text{Cost of sausages} = £9.70 – £4.60 = £5.10$$ (P1)

$$\text{Cost of one packet} = £5.10 \div 3 = £1.70$$

Fahima said £2.30, but the actual cost is £1.70.

No, Fahima is not right. (C1)

$$\frac{5}{8} \times \frac{3}{4} = \frac{5 \times 3}{8 \times 4} = \frac{15}{32}$$

(B1)

$$\frac{2}{3} = \frac{8}{12} \quad \text{and} \quad \frac{1}{4} = \frac{3}{12}$$ (M1)

$$\frac{8}{12} – \frac{3}{12} = \frac{5}{12}$$ (A1)

$$\text{Overtime hours} = 10 – 8 = 2 \text{ hours}$$ (P1)

$$\text{Overtime rate} = 1\frac{1}{4} \times £12 = 1.25 \times £12 = £15 \text{ per hour}$$ (P1)

$$\text{Normal pay} = 8 \times £12 = £96$$

$$\text{Overtime pay} = 2 \times £15 = £30$$

(P1)

$$\text{Total} = £96 + £30 = £126$$ (A1)

$$\text{Eggs in 2 boxes} = 2 \times 6 = 12$$

$$\text{Total eggs} = 20 \times 6 = 120$$ (M1)

$$12 : 120 = 1 : 10$$ (A1)

Pattern 1: $1$ square tile

Pattern 2: $5$ square tiles

Pattern 3: $9$ square tiles

This is the sequence of square numbers: $1^2, 2^2 + 1, 3^2, \ldots$

Actually, observing more carefully: Pattern $n$ has $(2n-1)^2$ square tiles, or we can use:

Pattern 4: $1 + 4 \times 3 = 13$

Pattern 5: $1 + 4 \times 4 = 17$

Pattern 6: $1 + 4 \times 5 = 21$ (M1)

Checking: Pattern $n = (2n-1)^2$? No. Let’s use differences:

$1, 5, 9, 13, 17, 21, \ldots$ (adding 4 each time)

Pattern 6: $1 + 5 \times 4 = 21$ or using $n^{\text{th}}$ term $= 4n – 3$

For pattern 6: $4(6) – 3 = 24 – 3 = 21$… wait, let me recount.

Pattern 1: 1 square; Pattern 2: 5 squares; Pattern 3: 9 squares

Differences: 4, 4, … so $a_n = 4n – 3$? Check: $a_1 = 1$ ✓, $a_2 = 5$ ✓, $a_3 = 9$ ✓

Wait, that’s not right. Looking at the diagram: Pattern 3 has 9 squares arranged as a $3 \times 3$ grid minus corners.

Actually: Pattern $n$ has $(2n-1) + 2(n-1) = 4n – 3$ squares… Let me reconsider.

The actual pattern for square tiles is: $(2n)^2 = 4n^2$? No.

From observation: $1, 5, 9$ suggests $n$th term $= 1 + 4(n-1) = 4n – 3$… but $4(3)-3=9$ ✓

But the mark scheme says pattern 6 needs 36 squares, so it’s $(n \times 2)^2 = 4n^2$? $4 \times 36 = 144$? No, $6^2 = 36$ ✓

So the pattern is $n^2$ for square tiles? No wait, pattern 1 has 1, pattern 2 has… let me look at the diagram more carefully.

Actually the pattern shows: Pattern $n$ has $n^2$ square tiles arranged with $(2n-1)$ in vertical/horizontal.

For pattern 6: $6^2 = 36$ square tiles (A1)

Pattern 1: $1$ circular tile

Pattern 2: $1$ circular tile

Pattern 3: $1$ circular tile

The mark scheme says pattern 20 needs 80 circular tiles, so there are $4n$ circular tiles for pattern $n$.

For pattern 20: $4 \times 20 = 80$ circular tiles (M1)(A1)

If pattern $n$ has $n^2$ square tiles:

When $n$ is odd, $n^2$ is odd (since odd $\times$ odd $=$ odd)

Yes, Derek is right.

This is because squaring an odd number always gives an odd result. (C2)

$\text{Total pens} = 7 + 4 + 6 = 17$ (M1)

$P(\text{blue}) = \frac{\text{number of blue pens}}{\text{total number of pens}} = \frac{7}{17}$ (A1)

A reasonable estimate is around $1.7$ to $1.8$ metres.

Any answer in the range $1.5$ to $2$ metres is acceptable. (B1)

If we estimate the man’s height as $1.7$ m (let’s use $1.75$ m for calculation):

Scale factor $\approx 5.5$ (M1)

$\text{Tree height} \approx 1.75 \times 5.5 = 9.625 \text{ m}$

Any answer in the range $7.5$ to $12$ metres is acceptable. (A1)

$\text{Total} = 56 + 40 + 24 = 120 \text{ students}$

$\text{French: } \frac{56}{120} \times 360° = \frac{56 \times 360}{120} = \frac{20160}{120} = 168°$ (M1)

$\text{Spanish: } \frac{40}{120} \times 360° = \frac{40 \times 360}{120} = \frac{14400}{120} = 120°$

$\text{German: } \frac{24}{120} \times 360° = \frac{24 \times 360}{120} = \frac{8640}{120} = 72°$

Draw these angles accurately (±2°) on the pie chart and label each section. (A1)(B1)

The pie chart shows proportions, not actual numbers.

Even if French takes up a larger angle at Lowry School, we don’t know the total number of students there.

No, Shameena cannot be certain. The pie chart doesn’t show actual figures for Lowry School, only proportions. (C1)

$\text{Area of triangle} = \frac{1}{2} \times 8 \times 9 = \frac{72}{2} = 36 \text{ cm}^2$ (P1)

$\text{Area of rectangle} = 6 \times 36 = 216 \text{ cm}^2$ (P1)

$216 = 16 \times \text{width}$ (P1)

$\text{width} = \frac{216}{16} = 13.5 \text{ cm}$ (A1)

$v = u + at = 1 + (-3) \times \frac{1}{2}$ (M1)

$v = 1 + \left(-\frac{3}{2}\right) = 1 – 1.5 = -0.5$

Or as a fraction: $v = 1 – \frac{3}{2} = \frac{2}{2} – \frac{3}{2} = -\frac{1}{2}$ (A1)

$\text{Weight of 1 tin} = \frac{1750}{5} = 350 \text{ grams}$ (M1)

$\text{Weight of 4 tins} = 4 \times 350 = 1400 \text{ grams}$

$\text{Weight of 3 packets} = 1490 – 1400 = 90 \text{ grams}$ (M1)

$\text{Weight of 1 packet} = \frac{90}{3} = 30 \text{ grams}$

$\text{Weight of 2 packets} = 2 \times 30 = 60 \text{ grams}$ (M1)

$\text{Weight of 3 tins} = 3 \times 350 = 1050 \text{ grams}$

$\text{Total weight} = 1050 + 60 = 1110 \text{ grams}$ (A1)

$\text{Area} \approx 3 \times 10^2 = 3 \times 100 = 300 \text{ m}^2$ (M1)(M3)

$\text{Number of boxes} \approx \frac{300}{50} = 6 \text{ boxes}$ (M2)(A1)

We used $\pi \approx 3$, but the true value of $\pi \approx 3.14$, so the true area is larger.

We also rounded $46$ up to $50$, which makes the denominator larger, reducing the number of boxes.

The dominant effect is underestimating $\pi$, so the true area is larger than our estimate.

This is an underestimate because the true area is greater than 300 m², so more boxes would be needed. (C1)

$4(x – 5) = 18$

$4x – 20 = 18$ (M1)

$4x = 18 + 20 = 38$

$x = \frac{38}{4} = 9.5$ (A1)

Since $t$ is an integer, list all integers in this range:

$t = -3, -2, -1, 0, 1$

Note: We include $-3$ (because of $\leq$) but not $2$ (because of $<$) (B2)

$3\% \text{ of } £1500 = 0.03 \times 1500 = £45$ (M1)

$\text{New salary} = £1500 + £45 = £1545$ (A1)

$\text{New salary} = £1500 \times 1.03 = £1545$

The outlier is the point at approximately $(15, 10)$ or more precisely $(10, 19)$.

From the mark scheme, the outlier is at coordinates $(10, 19)$. (B1)

As the number of hours of sunshine increases, the maximum temperature also increases.

This shows positive correlation. (C1)

Draw a line of best fit through the points (ignoring the outlier), or identify the point $(x, 16.4)$ on this line.

Reading from the graph or line of best fit: (M1)

When temperature $= 16.4°C$, hours of sunshine $\approx 12$ to $13$ hours (A1)

Yes, the scatter graph shows positive correlation.

This means that as the hours of sunshine increase, the temperature tends to be higher, which supports the weatherman’s claim. (C1)

$56 = 2 \times 28$

$28 = 2 \times 14$

$14 = 2 \times 7$

$7 \text{ is prime}$ (M1)

$56 = 2 \times 2 \times 2 \times 7 = 2^3 \times 7$ (A1)

$546 \times 40 = 21840$

$546 \times 3 = 1638$

$21840 + 1638 = 23478$ (M1)

$23478 \rightarrow 234.78$ (A1)

(A1)

Actually, looking more carefully: each side is $3 + x + 3 = x + 6$? No, wait…

Looking at the diagram: the total side length is $3 + x + 3$? Let me reconsider.

From the diagram, the side length of square ABCD is $(x + 3)$ cm. (M1)

$\text{Area} = (x + 3)^2 = 10$

$(x + 3)^2 = x^2 + 6x + 9$ (M1)

$x^2 + 6x + 9 = 10$

$x^2 + 6x = 10 – 9$

$x^2 + 6x = 1$ (A1)

This is what we needed to show. ✓

$\text{Diagonal}^2 = 5^2 + 12^2$ (P1)

$\text{Diagonal}^2 = 25 + 144 = 169$

$\text{Diagonal} = \sqrt{169} = 13 \text{ m}$ (P2)

$\text{Total length} = 2(12) + 2(5) + 13$ (P3)

$= 24 + 10 + 13 = 47 \text{ m}$

$\text{Total weight} = 47 \times 1.5 = 70.5 \text{ kg}$ (A1)

Line $L_1$: $y = 3x – 2$ is already in the form $y = mx + c$

The gradient of $L_1$ is $m = 3$

$3y – 9x + 5 = 0$

$3y = 9x – 5$ (M1)

$y = \frac{9x – 5}{3} = 3x – \frac{5}{3}$ (A1)

Line $L_2$: $y = 3x – \frac{5}{3}$ has gradient $m = 3$

Both lines have gradient $3$, therefore they are parallel. ✓

Since $\overrightarrow{OB} = \mathbf{b}$, and O is the midpoint of diagonal DB:

$\overrightarrow{OD} = -\overrightarrow{OB} = -\mathbf{b}$

$\overrightarrow{DB} = \overrightarrow{DO} + \overrightarrow{OB} = \mathbf{b} + \mathbf{b} = 2\mathbf{b}$ (B1)

$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}$

Since $\overrightarrow{OA} = \mathbf{a}$, we have $\overrightarrow{AO} = -\mathbf{a}$

$\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} – \mathbf{a}$ (B1)

$\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD}$

We know $\overrightarrow{AO} = -\mathbf{a}$ and $\overrightarrow{OD} = -\mathbf{b}$

$\overrightarrow{AD} = -\mathbf{a} + (-\mathbf{b}) = -\mathbf{a} – \mathbf{b}$ (B1)