June 2023 Paper 2 Higher

(a) Top: $25 - \sqrt{43.87} = 18.3765...$

Bottom: $6 + 2.1^2 = 10.41$

$18.3765... \div 10.41 = 1.765279230...$

(b) Reciprocal $= 1 \div 0.625 = 1.6$

$60 = 2 \times 30 = 2 \times 2 \times 15 = 2 \times 2 \times 3 \times 5$

So $60 = 2^2 \times 3 \times 5$

The ratio $1 : 2$ has $1 + 2 = 3$ parts.

$48 \div 3 = 16$, so there are 16 red and 32 blue.

Helen is wrong: 24 red would be half, which is a $1 : 1$ ratio.

(a) $n$ is less than 5, so the greatest integer is $4$.

(b) Filled circle at $-4$ (it is included), empty circle at $1$ (not included), and a line joining them.

(c) $\frac{2}{5}g - 4 < 6$

$\frac{2}{5}g < 10$

$2g < 50$

$g < 25$

Triangle: $\frac{1}{2} \times 8 \times 6x = 24x$

Rectangle: $5(4x - 1) = 20x - 5$

$24x = 20x - 5 + 10$

$4x = 5$, so $x = 1.25$

(Check: triangle 30, rectangle 20.)

Paper and glass: $0.57 \times 800 = 456$ kg

Ratio $12 : 7$ has 19 parts: $456 \div 19 = 24$

Glass: $7 \times 24 = 168$ kg

Rounding to 1 d.p. means the value is within $0.05$ of $12.7$.

Lower bound: $12.7 - 0.05 = 12.65$

Upper bound: $12.7 + 0.05 = 12.75$

$12.65 \le d < 12.75$

Tamsin: $150\,000 \times 1.04^2 = £162\,240$

Rachel: $160\,000 \times 1.015^2 = £164\,836$

$164\,836 > 162\,240$, so Rachel's house is worth more.

(a) Plot each cumulative frequency at the upper end of its class: $(30, 20)$, $(40, 48)$, $(50, 64)$, $(60, 75)$, $(70, 80)$, starting from $(20, 0)$, and join with a smooth curve.

(b) Median is at $80 \div 2 = 40$ people. Read across from 40 to the curve and down: about $37$ years.

(a) P(6) $\approx \frac{15}{60} = \frac{1}{4}$

P(6 both times) $= \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 0.0625$

(b) More throws give a more reliable estimate, so $n$ must be greater than 60.

Multiply the first by 2: $4x + 12y = 10$

Multiply the second by 3: $9x - 12y = -36$

Add: $13x = -26$, so $x = -2$

$2(-2) + 6y = 5 \Rightarrow 6y = 9 \Rightarrow y = 1.5$

The diagonal of the rectangle is a diameter.

$d = \sqrt{8^2 + 10^2} = \sqrt{164} = 12.806...$ cm

Circumference $= \pi d = \pi \times 12.806... = 40.23...$

$= 40.2$ cm (3 s.f.)

Angle $C$ is opposite $AB = 15$; angle $B = 70°$ is opposite $AC = 18$.

$\frac{\sin C}{15} = \frac{\sin 70°}{18}$, so $\sin C = \frac{15 \sin 70°}{18} = 0.7830...$

$C = 51.54...°$

$BAC = 180 - 70 - 51.54... = 58.5°$

Top: $x^2 - x - 6 = (x - 3)(x + 2)$

Bottom: $2x^2 - 5x - 3 = (x - 3)(2x + 1)$

Cancel $(x - 3)$: $\frac{x + 2}{2x + 1}$, so $a = 1, b = 2, c = 2, d = 1$.

Differences: 6, 8, 10; second difference 2, so start with $n^2$.

Subtract $n^2$ (1, 4, 9, 16): leaves 2, 5, 8, 11, which is $3n - 1$.

nth term $= n^2 + 3n - 1$

(a) The 30–40 bar holds 28 students, so its fd is $28 \div 10 = 2.8$: this sets the vertical scale.

For 40–60: fd $= 24 \div 20 = 1.2$, so draw a bar of height 1.2 from 40 to 60.

(b) 0–20: fd 1.6, so $1.6 \times 20 = 32$. 20–30: fd 3.6, so $3.6 \times 10 = 36$.

Total $= 32 + 36 + 28 + 24 = 120$

(a) $x^4 - x^2 - 5 = 0 \Rightarrow x^4 = x^2 + 5 \Rightarrow x = \sqrt[4]{x^2 + 5}$

(b) $x_1 = \sqrt[4]{1.5^2 + 5} = 1.6409...$

$x_2 = \sqrt[4]{1.6409...^2 + 5} = 1.6653...$

$x_3 = \sqrt[4]{1.6653...^2 + 5} = 1.6697...$

$2a : 5c = 6 : 25 \Rightarrow 50a = 30c \Rightarrow a = \frac{3c}{5}$

$4b : 7c = 20 : 21 \Rightarrow 84b = 140c \Rightarrow b = \frac{5c}{3}$

Take $c = 15$: $a = 9$, $b = 25$.

$a + b : b + c = 34 : 40 = 17 : 20$

Bounds: $AB$ from 9.25 to 9.35; $AC$ from 12.55 to 12.65.

$\sin x = \frac{AB}{AC}$ is smallest with the smallest $AB$ and largest $AC$.

$\sin x = \frac{9.25}{12.65} = 0.7312...$

$x = 46.989...°$

$\vec{OR} = \vec{OT} + \vec{TR} = \mathbf{a} - \mathbf{b}$

$\vec{OM} = \frac{2}{5}(\mathbf{a} - \mathbf{b})$

$\vec{MT} = -\vec{OM} + \vec{OT} = -\frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b} + \mathbf{a} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$

(a) $f(x) - 4$ moves every point down 4: $(-3,0) \to (-3,-4)$, $(-2,1) \to (-2,-3)$, $(-1,1) \to (-1,-3)$, $(0,0) \to (0,-4)$.

(b) $f(-x)$ reflects in the $y$-axis: change the sign of each $x$: $(3,0)$, $(2,1)$, $(1,1)$, $(0,0)$.

Blue : red = 4 : 1, so P(blue) $= \frac{4}{5}$, P(red) $= \frac{1}{5}$.

The only ways to fail are all blue or all red.

P(both colours) $= 1 - \left(\frac{4}{5}\right)^n - \left(\frac{1}{5}\right)^n$

$AB : AC : AD = 1 : (1+2) : (1+2+3) = 1 : 3 : 6$

Similar triangles, so areas are in the ratio of the squares: $1 : 9 : 36$.

BCFG $= 9 - 1 = 8$; CDEF $= 36 - 9 = 27$.

So the ratio is $1 : 8 : 27$.

Each octagon angle is 135°, so where two octagons meet at a point the gap is $360 - 2 \times 135 = 90°$.

Each point of the star is a right-angled triangle with legs $a$: area $\frac{1}{2}a^2$, hypotenuse $a\sqrt{2}$.

Each side of the central square is an octagon side, a triangle base and another octagon side: $a + a\sqrt{2} + a = (2 + \sqrt{2})a$.

Area $= (2+\sqrt{2})^2a^2 + 4 \times \frac{1}{2}a^2 = (6 + 4\sqrt{2})a^2 + 2a^2 = (8 + 4\sqrt{2})a^2 = 4(2+\sqrt{2})a^2$