June 2024 Paper 3 Higher

$63 = 3 \times 3 \times 7$

$105 = 3 \times 5 \times 7$

Common prime factors: $3 \times 7 = 21$

(a)(i) Move the decimal point 4 places right: $53\,000$

(a)(ii) Move the decimal point 5 places left: $0.000\,074$

(b) $9\,700\,000 + 24\,500\,000 = 34\,200\,000 = 3.42 \times 10^7$

(a) Rana drew a 7 cm by 4 cm rectangle. The 4 cm sides of the triangle slope, so the prism is not 4 cm tall. Its height is $\sqrt{4^2 - 3^2} = \sqrt7 \approx 2.6$ cm.

(b) From above you see the 6 cm by 7 cm base. The top edge (ridge) is above the middle of the 6 cm side, so draw a 7 cm by 6 cm rectangle with a line along its length 3 cm from each long side.

Multiplier for a 6% increase: $1.06$

$25\,000 \times 1.06^3 = 29\,775.4$

So about $29\,775$ workers.

Volume of tin $= \frac{\text{mass}}{\text{density}} = 600 \div 0.6 = 1000$ cm³

Volume of salt $= 1000 - 700 = 300$ cm³

Density of salt $= 600 \div 300 = 2$ g/cm³

(a) $1 - 0.6 = 0.4$ and $1 - 0.55 = 0.45$. Each coin B pair of branches is $0.55$ (heads) and $0.45$ (not heads).

(b) $P(\text{both heads}) = 0.6 \times 0.55 = 0.33$

Volume $= \pi r^2 h = \pi \times 100^2 \times 30 = 942\,478$ cm³

Time $= 942\,478 \div 250 = 3769.9$ seconds

$3769.9 \div 60 = 62.8$, so about $63$ minutes.

$2\mathbf{a} = \begin{pmatrix}6\\4\end{pmatrix}$

$2\mathbf{a} + \mathbf{b} = \begin{pmatrix}6 + (-1)\\4 + 4\end{pmatrix} = \begin{pmatrix}5\\8\end{pmatrix}$

Draw a line 5 squares right and 8 squares up, with an arrow at the end, and label it $2\mathbf{a} + \mathbf{b}$.

Cube: $6^3 = 216$ cm³

Pyramid: $\frac13 \times 8 \times 8 \times h = \frac{64}{3}h$

$\frac{64}{3}h = 216 \Rightarrow h = \dfrac{216 \times 3}{64} = 10.125$ cm

Red : yellow $= 3 : 5$, so yellow $= \frac{5}{3}x$.

Bag A $= x + \frac{5}{3}x = \frac{8}{3}x$

Bag B $= \frac12 \times \frac{8}{3}x = \frac{4}{3}x$

Total $= \frac{8}{3}x + \frac{4}{3}x = \frac{12}{3}x = 4x$

(a) UQ $= 35 + 12 = 47$; highest $= 25 + 37 = 62$. Draw the box from 35 to 47 with a line at 40, and whiskers to 25 and 62.

(b) Sunday: median 42, IQR $48 - 33 = 15$, range $66 - 22 = 44$.

The median speed was higher on Sunday, so cars were generally faster. The IQR was larger on Sunday, so speeds were more varied.

T has vertices $(1, 2)$, $(1, 3)$ and $(3, 2)$.

Multiply each coordinate by $-2$: $(-2, -4)$, $(-2, -6)$, $(-6, -4)$.

Plot and join these points.

The first student can be chosen in 30 ways and the second in 29 ways: $30 \times 29 = 870$.

Each pair has been counted twice (AB and BA), so $870 \div 2 = 435$.

Before the 2023 withdrawal: $2937.14 + 1000 = 3937.14$

Start of 2023: $3937.14 \div 1.035 = 3804$

Before the 2022 withdrawal: $3804 + 750 = 4554$

Start of 2022: $4554 \div 1.035 = £4400$

(a) Cancel one factor of $(a-3)$: $\dfrac{a-3}{5}$

(b) $3k^2 + 11k - 4 = (3k - 1)(k + 4)$ (check: $3k^2 + 12k - k - 4$)

(c) $\dfrac{(2-x)(2+x)}{x(x+3)} \times \dfrac{x+3}{x+2} = \dfrac{2-x}{x}$

$f(x) = \dfrac{12}{x+1}$, $g(x) = 5 - 3x$

(a) $f(-3) = \dfrac{12}{-2} = -6$

(b) $g(1) = 2$, so $fg(1) = f(2) = \dfrac{12}{3} = 4$

(c) Solve $5 - 3x = 4$: $x = \frac13$

$h_1 = 8$

$h_2 = 0.55 \times 8 = 4.4$

$h_3 = 0.55 \times 4.4 = 2.42$

$h_4 = 0.55 \times 2.42 = 1.331$ m

$\frac12 \times 17 \times AC \times \sin 35° = 54 \Rightarrow AC = 11.076$ cm

Angle $CAD = 180° - 48° - 57° = 75°$

$CD = \dfrac{11.076 \sin 75°}{\sin 57°} = 12.757$ cm

Area $= \frac12 \times 11.076 \times 12.757 \times \sin 48° = 52.5$ cm²

$P$ lower bound $= 5.875 \times 10^8$

$Q$ upper bound $= 3.65 \times 10^5$

Lower bound of $R = \dfrac{5.875 \times 10^8}{3.65 \times 10^5} = 1609.6 \approx 1610$

(a) $(x+2)-(x-4) = 6$ and $(3x+1)-(x+2) = 2x-1$, so $2x - 1 = 6$, $x = 3.5$

(b) $\dfrac{y+2}{y-4} = \dfrac{3y+1}{y+2}$

$(y+2)^2 = (3y+1)(y-4) \Rightarrow y^2 + 4y + 4 = 3y^2 - 11y - 4$

$2y^2 - 15y - 8 = 0 \Rightarrow (2y+1)(y-8) = 0$, so $y = -0.5$ or $y = 8$

Smaller hexagon: made of 6 equilateral triangles of side $r$, so perimeter $= 6r$.

Larger hexagon: the circle's radius is the height of each equilateral triangle, so half a side $= r\tan 30° = \frac{r}{\sqrt3}$, side $= \frac{2r}{\sqrt3}$, perimeter $= \frac{12r}{\sqrt3} = 4\sqrt3\,r$.

The circle lies between them: $6r < 2\pi r < 4\sqrt3\,r$

Divide by $2r$: $3 < \pi < 2\sqrt3$