June 2022 Paper 1 Higher

$7x - 27 < 8$

Add 27: $7x < 35$

Divide by 7: $x < 5$

$124 = 2 \times 62$

$62 = 2 \times 31$, and 31 is prime.

So $124 = 2 \times 2 \times 31 = 2^2 \times 31$

Cars: $160 \div 10 = 16$, $16 \times 3 = 48$

Electric: $48 \div 8 = 6$

Diesel: $25\%$ of 48 $= 12$

Petrol: $48 - 6 - 12 = 30$

(a) Move the decimal point 3 places left: $0.00163$

(b) $438\,000 = 4.38 \times 100\,000 = 4.38 \times 10^5$

(c) $4 \times 6 = 24$ and $10^3 \times 10^{-5} = 10^{-2}$

$24 \times 10^{-2} = 2.4 \times 10^{-1}$

Hexagon interior angle: $180 - 360 \div 6 = 120^\circ$

Pentagon interior angle: $180 - 360 \div 5 = 108^\circ$

Angles round a point add to $360^\circ$: $x = 360 - 120 - 108 = 132^\circ$

(a) $x = -1$: $1 + 3 + 1 = 5$; $x = 2$: $4 - 6 + 1 = -1$; $x = 3$: $9 - 9 + 1 = 1$; $x = 4$: $16 - 12 + 1 = 5$

(b) Plot $(-1,5), (0,1), (1,-1), (2,-1), (3,1), (4,5)$ and join with a smooth curve.

(c) The curve crosses $y = 0$ at about $x = 0.4$ and $x = 2.6$

Volumes: $3^3 = 27$ cm³ and $4^3 = 64$ cm³

Densities: $81 \div 27 = 3$ g/cm³ and $128 \div 64 = 2$ g/cm³

Ratio: $3 : 2$

Midpoints: 5, 15, 25, 35, 45

$\sum fx = 40 + 150 + 175 + 70 + 135 = 570$

Mean $\approx 570 \div 30 = 19$ cm

Cuboid faces: $2(7\times5) + 2(6\times5) + 2(7\times6) = 70 + 60 + 84 = 214$

Cube faces: $6 \times 16 = 96$

The join hides one $4 \times 4$ face on each solid: $214 + 96 - 2 \times 16 = 278\text{ cm}^2$

(a) Running totals: 10, 25, 50, 80, 85, 100

(b) Plot each total at the top of its class (50, 100, …, 300) and join.

(c) Read up from £125: about 37 days.

(d) Lower quartile (25th value) $= £100$; upper quartile (75th value) $\approx £190$

IQR $\approx 190 - 100 = £90$

$\frac{3}{7} = \frac{9}{21}$, so the ratio has 21 parts in total.

$9 + 4 + x = 21$

$x = 8$

Let $x = 0.1171717\ldots$

$1000x = 117.171717\ldots$ and $10x = 1.171717\ldots$

$990x = 116$

$x = \frac{116}{990} = \frac{58}{495}$

Let the diameters be $a$ (hypotenuse), $b$ and $c$. Pythagoras: $a^2 = b^2 + c^2$

A semicircle with diameter $d$ has area $\frac{1}{2}\pi\left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$

Multiply Pythagoras by $\frac{\pi}{8}$: $\frac{\pi a^2}{8} = \frac{\pi b^2}{8} + \frac{\pi c^2}{8}$

So area A = area B + area C.

(a) Draw a tangent touching the curve at $t = 2$. Pick two points on it, e.g. $(0, 1)$ and $(3, 3.7)$.

Gradient $\approx \frac{3.7 - 1}{3 - 0} = \frac{2.7}{3} = 0.9$

(b) Area under a speed-time graph gives the distance travelled.

(a) $\overrightarrow{AC} = 15\mathbf{a} + 20\mathbf{b} = 5(3\mathbf{a} + 4\mathbf{b}) = 5\overrightarrow{AB}$

So AC is parallel to AB, and both pass through A, so A, B, C are on one line.

(b) $\overrightarrow{DF} = (3\mathbf{e} + 6\mathbf{f}) + (-10.5\mathbf{e} - 21\mathbf{f}) = -7.5\mathbf{e} - 15\mathbf{f} = -2.5\overrightarrow{DE}$

Length DF : length DE $= 2.5 : 1 = 5 : 2$

Let $p$ = P(pass practical).

Only one passed: $0.75(1 - p) + 0.25p = 0.36$

$0.75 - 0.5p = 0.36$

$0.5p = 0.39$, so $p = 0.78$

$y = k\sqrt{t}$: $15 = k \times 3$, so $k = 5$

$t = \frac{K}{x^3}$: $8 = \frac{K}{8}$, so $K = 64$

$y = 5\sqrt{\frac{64}{x^3}} = \frac{5 \times 8}{x^{3/2}} = \frac{40}{x^{3/2}}$

$\left(5\frac{4}{9}\right)^{-\frac{1}{2}} = \left(\frac{49}{9}\right)^{-\frac{1}{2}} = \frac{3}{7}$

Numerator: $\frac{3}{7} \times 4\frac{2}{3} = \frac{3}{7} \times \frac{14}{3} = 2$

$2^{-3} = \frac{1}{8}$, so the value is $2 \div \frac{1}{8} = 16$

Multiply by $(2x-1)(x-1)$: $(x - 1) + 3(2x - 1) = (2x - 1)(x - 1)$

$7x - 4 = 2x^2 - 3x + 1$, so $2x^2 - 10x + 5 = 0$

$x = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2}$

Gradient of radius from $(-1, 3)$ to $(6, 8)$: $\frac{8-3}{6+1} = \frac{5}{7}$

Tangent is perpendicular: gradient $-\frac{7}{5}$

$y - 8 = -\frac{7}{5}(x - 6)$

$5y - 40 = -7x + 42$, so $7x + 5y - 82 = 0$

Join A and B to a crossing point of their circles: all three sides are 4 cm, so the triangle is equilateral and the overlap of circles A and B uses a $120^\circ$ angle at A.

One segment $= \frac{120}{360}\pi(4^2) - \frac{1}{2}(4)(4)\sin 120^\circ = \frac{16\pi}{3} - 4\sqrt{3}$

Each overlap (lens) is 2 segments, and circle B has 2 lenses, so the lenses total $4\left(\frac{16\pi}{3} - 4\sqrt{3}\right)$

Shaded = circle B minus the two lenses $= 16\pi - \frac{64\pi}{3} + 16\sqrt{3} = 16\sqrt{3} - \frac{16\pi}{3}\text{ cm}^2$