November 2023 Paper 3 Higher

(a) Put the decimal point after the first digit: 4.68. The point moves 5 places, so $4.68 \times 10^5$.

(b) A power of $-4$ means move the point 4 places left: $0.0005037$

Expected number $=$ probability × number of spins

$0.4 \times 200 = 80$

(a) Midpoints 12.5, 17.5, 22.5, 27.5, 32.5.

$\sum fx = 25 + 140 + 292.5 + 577.5 + 195 = 1230$

Mean $\approx 1230 \div 50 = 24.6$ °C

(b) No. The median is about the 25.5th value. $2 + 8 + 13 = 23$, so it is in the $25 < T \le 30$ group, not the middle group.

(a) $-3 < x$ means $x$ is bigger than $-3$ but not equal to it, so the line should start at $-3$ with an open circle. Jenna started her line at $-2$.

$x \le 4$ includes 4, so the circle at 4 should be filled in. Jenna drew an open circle there.

(b) $5y - 7 < 16$

$5y < 23$, so $y < 4.6$

The greatest integer less than 4.6 is 4.

Multiples of 30: 30, 60, 90, 120

Multiples of 24: 24, 48, 72, 96, 120

LCM $= 120$, so $120 \div 30 = 4$ packs and $120 \div 24 = 5$ boxes.

Total work $= 30 \times 4 = 120$ machine-hours.

In 6 hours: $120 \div 6 = 20$ machines.

Car time $= 143 \div 55 = 2.6$ hours $= 2$ hours 36 minutes ($0.6 \times 60 = 36$).

Total $= 2$ h 36 min $+ 5$ h 20 min $= 7$ hours 56 minutes

Area of face on the table $= \text{force} \div \text{pressure} = 504 \div 3.5 = 144$ cm²

A cube has 6 equal faces: $6 \times 144 = 864$ cm²

$864 < 900$, so the surface area is less than 900 cm².

The line crosses the $y$-axis at 3, so $c = 3$.

From (0, 3) to (1, 1): down 2 for every 1 across, so gradient $= -2$.

Equation: $y = -2x + 3$

$k = p + \dfrac{2m}{5}$

Subtract $p$: $k - p = \dfrac{2m}{5}$

Multiply by 5: $5(k - p) = 2m$

Divide by 2: $m = \dfrac{5(k - p)}{2}$

Length scale factor 50, so area scale factor $50^2 = 2500$.

Real area $= 48 \times 2500 = 120\,000\text{ cm}^2$

$1\text{ m}^2 = 10\,000\text{ cm}^2$, so $120\,000 \div 10\,000 = 12\text{ m}^2$

$\dfrac{x}{360} \times \pi \times 6.2^2 = 82.6$

$x = \dfrac{82.6 \times 360}{\pi \times 6.2^2} = \dfrac{29\,736}{120.76...} = 246.2...$

$x = 246°$ (3 s.f.)

(a) At 90 cm the cumulative frequency is about 74, so $80 - 74 = 6$ plants are taller. Probability $\approx \frac{6}{80}$.

(b) Median at $80 \div 2 = 40$th plant: about 60 cm.

(c) LQ (20th plant) $\approx 40$ cm, UQ (60th plant) $\approx 75$ cm. IQR $= 75 - 40 = 35$ cm.

(d) $35 > 30$, so the heights of the outside plants are more varied than those grown inside.

First differences: 6, 10, 14, 18, 22. Second difference: 4, so the sequence starts $2n^2$.

$2n^2$: 2, 8, 18, 32, ...

Sequence minus $2n^2$: 3, 3, 3, 3, ... so the nth term is $2n^2 + 3$.

$AC = 21.6 + 58.4 = 80$ cm and $AB = 32 + 22 = 54$ cm.

In triangle $AED$, $AD = 32$ matches $AC = 80$, and $AE = 21.6$ matches $AB = 54$.

$80 \div 32 = 2.5$ and $54 \div 21.6 = 2.5$

The ratios are equal and angle $A$ is common, so the triangles are similar.

Options for each digit (1 to 9):

- cube: 1, 8 → 2
- prime: 2, 3, 5, 7 → 4
- greater than 6: 7, 8, 9 → 3
- odd: 1, 3, 5, 7, 9 → 5

Number of passcodes $= 2 \times 4 \times 3 \times 5 = 120$

Probability $= \frac{1}{120}$

(a)(i) $x^2 - 8x + 3 = (x - 4)^2 - 16 + 3 = (x - 4)^2 - 13$

(ii) Turning point $(4, -13)$

(b) $x = \dfrac{-8 \pm \sqrt{8^2 - 4(7)(-5)}}{2(7)} = \dfrac{-8 \pm \sqrt{204}}{14}$

$x = 0.449$ or $x = -1.59$ (3 s.f.)

(c) $3k - 2 = 0$ gives $k = \frac{2}{3}$. Alex forgot to divide by 3.

(a) $P$ has vertices $(-8, 1)$, $(-8, 5)$, $(-6, 1)$. From the centre $(-2, -1)$ these are $(-6, 2)$, $(-6, 6)$, $(-4, 2)$ away.

Multiply by $-1.5$: $(9, -3)$, $(9, -9)$, $(6, -3)$. Add back to $(-2, -1)$: $Q$ is $(7, -4)$, $(7, -10)$, $(4, -4)$.

(b) Rotating $90°$ anticlockwise about $O$ sends $(x, y)$ to $(-y, x)$, so $(-8, 1) \to (-1, -8)$.

To send it back to $(-8, 1)$, translate by $\begin{pmatrix} -7 \\ 9 \end{pmatrix}$.

Small cone: height 10 is $\frac{2}{3}$ of 15, so its radius is $\frac{2}{3} \times 12 = 8$ cm.

Large cone $= \frac{1}{3}\pi(12^2)(15) = 720\pi$; small cone $= \frac{1}{3}\pi(8^2)(10) = \frac{640}{3}\pi$

Frustum $= 720\pi - \frac{640}{3}\pi = \frac{1520}{3}\pi$

Hemisphere $= \frac{2}{3}\pi(12^3) = 1152\pi$

$S = 1152\pi - \frac{1520}{3}\pi = 2027.37... \approx 2030\text{ cm}^3$

(a) $g(x - 7)$ moves the graph 7 to the right: $(-3 + 7, 6) = (4, 6)$.

(b) $f(-x)$ reflects the curve in the $y$-axis: maximum $(-2, 3) \to (2, 3)$, ends $(-6, -7) \to (6, -7)$ and $(2, -7) \to (-2, -7)$.

$+3$ moves it up 3: maximum $(2, 6)$, ends $(6, -4)$ and $(-2, -4)$.

Angle $CBD =$ angle $CAD = 35°$ (angles in the same segment are equal).

In triangle $BDE$: angle $BDE = 180 - 35 - 28 = 117°$.

Angle $ADC = 90°$ (angle in a semicircle), so angle $CDE = 180 - 90 = 90°$.

Angle $BDC = 117 - 90 = 27°$

To find the least possible rate, use the smallest takings, the largest costs and the longest time.

Takings LB $= 872.5$, costs UB $= 192.5$, time UB $= 72.5$

LB of rate $= \dfrac{872.5 - 192.5}{72.5} = \dfrac{680}{72.5} = 9.379...$

Even the lowest possible rate, £9.379..., is more than £8.20, so her hourly rate was definitely more than £8.20.

$11(2x^2 + y^2) = 43(4x^2 - y^2)$

$22x^2 + 11y^2 = 172x^2 - 43y^2$

$54y^2 = 150x^2$, so $\dfrac{x^2}{y^2} = \dfrac{54}{150} = \dfrac{9}{25}$

$x$ and $y$ are positive, so $\dfrac{x}{y} = \dfrac{3}{5}$ and $x : y = 3 : 5$

$AP = \frac{5}{7} \times 14.7 = 10.5$ cm and $AM = 3.8 \div 2 = 1.9$ cm.

Angle $MAP = 90°$, so $MP = \sqrt{10.5^2 + 1.9^2} = 10.67...$ cm

T is above M, so the angle is $TPM$:

$\tan TPM = \dfrac{2.3}{10.67...}$, so angle $= 12.16...° \approx 12.2°$