November 2022 Paper 2 Higher

(a) The point at $(2, 1)$ is far away from the trend of the others.

(b) As rainfall goes up, sunshine goes down: negative correlation.

(c) Draw a line of best fit, go across from 7 hours to the line and down: about 3.5 mm.

Viewed from the right, the width comes from the depth of the plan (3 squares) and the height from the tallest point of the front elevation (5 squares).

Draw a rectangle 3 wide and 5 high. Visible edges where the sloping faces meet appear as horizontal lines across it.

(a) The terms go up by 6, so start with $6n$: 6, 12, 18, ... Each term is 1 more, so $6n + 1$.

(b) $8 - 6n = -58$

$6n = 66$, so $n = 11$.

11 is a whole number, so yes, $-58$ is the 11th term.

Rectangle ABCO: $11 \times 7 = 77$

Rectangle DEFO: $7 \times 9 = 63$

Triangle CDO: $\frac12 \times 11 \times 9 = 49.5$

Quarter circle: $\frac14 \times \pi \times 7^2 = 38.48...$

Total $= 227.98...\text{ m}^2$, and $227.98 \div 14 = 16.28...$ so 17 bags.

$17 \times 10.95 = £186.15$

$x$ is adjacent to $53°$ and 14.5 cm is the hypotenuse, so use cosine.

$x = 14.5 \cos 53° = 8.726...$

$x = 8.73$ (3 s.f.)

After year 1: $7000 \times 1.03 = 7210$

After year 2: $7210 \times 1.015 = £7318.15$

(a) The curve crosses the y-axis where $x = 0$: $y = 4$.

(b) The lowest point is at $(3, -5)$.

(c) Read where the curve crosses the x-axis: about $x = 0.8$ and $x = 5.2$.

Profit $= 135 - 120 = £15$

Percentage profit $= \frac{15}{120} \times 100 = 12.5\%$

Divide $6y = 3x + 7$ by 6: $y = \frac{1}{2}x + \frac{7}{6}$

The first line is $y = \frac{1}{2}x - 6$.

Both gradients are $\frac{1}{2}$, so the lines are parallel. Oscar is correct.

Multipliers: $1 - 0.23 = 0.77$ and $1 - 0.19 = 0.81$

Original $\times 0.77 \times 0.81 = 10\,914.75$

Original $= 10\,914.75 \div 0.6237 = £17\,500$

There are 60 students.

Lower quartile: read across from 15 → 42 s

Median: read across from 30 → 54 s

Upper quartile: read across from 45 → 64 s

Draw the box from 42 to 64 with a line at 54, and whiskers to 24 and 96.

(a) Try powers of 1.13:

$1.13^5 = 1.84...$ (not yet doubled)

$1.13^6 = 2.08...$ (more than 2)

So it takes 6 years.

(b) If $k$ keeps growing, the population grows faster, so it would double in fewer years.

Horizontal difference: $8 - (-7) = 15$

Vertical difference: $6 - (-5) = 11$

$AB = \sqrt{15^2 + 11^2} = \sqrt{346} = 18.60...$

$AB = 18.6$ cm

Let $x = 1.06222...$

$1000x = 1062.222...$

$100x = 106.222...$

Subtract: $900x = 956$

$x = \frac{956}{900} = \frac{239}{225} = 1\frac{14}{225}$

(a) $\frac{12}{42}$ of the second sample were tagged, so $\frac{20}{N} = \frac{12}{42}$

$N = \frac{20 \times 42}{12} = 70$

(b) He caught 55 different rabbits in one day, so the wood must hold at least 55. An estimate of 50 is too small.

Horizontal line: $y = 6$, region below, so $y \le 6$.

Vertical line: $x = -3$, region to the right, so $x \ge -3$.

Steep line through $(0, 6)$ and $(-2, 0)$: gradient 3, so $y = 3x + 6$; region above, $y \ge 3x + 6$.

Shallow line through $(0, 1)$ and $(2, 0)$: gradient $-\frac{1}{2}$; region above, $y \ge -\frac{1}{2}x + 1$.

Volume scale factor $= 1587.762 \div 58.806 = 27$

Length scale factor $= \sqrt[3]{27} = 3$

Height of B's triangle: $\frac{2 \times 43.74}{8.1} = 10.8$

$h = 10.8 \div 3 = 3.6$

Area $= \frac{1}{2}ab\sin C$

$= \frac{1}{2} \times 11.2 \times 4.3 \times \sin 118° = 21.26...$

Area $= 21.3\text{ cm}^2$ (3 s.f.)

Factorise: $6x^2 + 5x - 6 = (2x + 3)(3x - 2)$

$2x + 3 = 0 \Rightarrow x = -\frac{3}{2}$

$3x - 2 = 0 \Rightarrow x = \frac{2}{3}$

Front face: $AF = \sqrt{13^2 - 9^2} = \sqrt{88} = 9.380...$

Base: $GH = 9$, so $FH = \frac{9}{\cos 49°} = 13.71...$

Triangle $FAH$ is right-angled at $F$: $\tan FAH = \frac{13.71...}{9.380...}$

Angle $FAH = 55.6... \approx 56°$

(a) Draw a tangent touching the curve at $t = 17.5$.

Pick two points on it far apart and work out rise ÷ run, e.g. $\frac{18.5}{17.5} \approx 1.06$

(b) It is the rate the container fills: about 1.06 litres per second at $t = 17.5$.

$h(x) = f(g(x)) = \sqrt[3]{2x + 3}$

Let $y = \sqrt[3]{2x + 3}$, so $y^3 = 2x + 3$ and $x = \frac{y^3 - 3}{2}$

$h^{-1}(x) = \frac{x^3 - 3}{2}$

Distance: $10.55 \le d < 10.65$ km

Time: $1907.5 \le t < 1908.5$ s

Upper bound $= 10.65 \div \frac{1907.5}{3600} = 20.099...$ km/h

Lower bound $= 10.55 \div \frac{1908.5}{3600} = 19.900...$ km/h

Both round to 20 (2 s.f.), so $V = 20$ km/h.

Gradient of radius $OP = \frac{2.8}{2.1} = \frac{4}{3}$

Tangent gradient $= -\frac{3}{4}$

$y - 2.8 = -\frac{3}{4}(x - 2.1)$

Multiply by 4: $4y - 11.2 = -3x + 6.3$, so $3x + 4y = 17.5$

Double for integers: $6x + 8y = 35$