June 2022 Paper 2 Higher

(a) Multiply the powers: $(x^3)^5 = x^{15}$

(b) $4x + 12 + 28 - 14x = 40 - 10x$

(c) The HCF of $15x^3$ and $3x^2y$ is $3x^2$.

$15x^3 + 3x^2y = 3x^2(5x + y)$

The shape is the same size and way round, so it is a translation.

Pick a corner: $(1, -5)$ on S moves to $(-4, 1)$ on T.

That is 5 left and 6 up: translation by $\begin{pmatrix}-5\\6\end{pmatrix}$.

Rounded to the nearest metre, so go half a metre either side.

Lower bound: $90 - 0.5 = 89.5$

Upper bound: $90 + 0.5 = 90.5$

$89.5 \le \text{length} < 90.5$

(a) Field B: $700 \times 2000 = 1\,400\,000\text{ m}^2$

B per person: $1\,400\,000 \div 6750 = 207.4...$

A per person: $80\,000 \div 425 = 188.2...$

Difference: $207.4 - 188.2 = 19.17... \approx 19\text{ m}^2$

(b) A square metre is $100\text{ cm} \times 100\text{ cm} = 10\,000\text{ cm}^2$, so $300\text{ cm}^2 = 0.03\text{ m}^2$.

From L to M: $x$ goes up $4 - (-3) = 7$, $y$ goes up $9 - 1 = 8$. This is 2 parts.

One part: $x$ up 3.5, $y$ up 4.

MN is 3 parts: $x$ up 10.5, $y$ up 12.

$N = (4 + 10.5,\ 9 + 12) = (14.5, 21)$

The multiplier for a 4% decrease is $0.96$.

$679 \times 0.96^3 = 600.7357...$

Value after 3 years: £600.74

Spain in £: $27 \times 0.85 = 22.95$, so per litre $22.95 \div 18 = 1.275$.

Wales: $8 \times 4.5 = 36$ litres, so per litre $40.80 \div 36 = 1.133...$

£1.13 < £1.275, so petrol is cheaper in Wales. Sam is not correct.

Top: $1.57^4 + \tan 60^\circ = 6.0757... + 1.7320... = 7.8077...$, then $\sqrt[3]{7.8077...} = 1.9838...$

Bottom: $7.2^{\frac{1}{2}} = \sqrt{7.2} = 2.6832...$

$1.9838... \div 2.6832... = 0.73933...$, so $0.739$ to 3 s.f.

Area $= \text{force} \div \text{pressure} = 180 \div 187.5 = 0.96\text{ m}^2$

Area $= 1.2 \times x$, so $x = 0.96 \div 1.2 = 0.8$

(a) The lower quartile is at 25% and the upper quartile at 75%, so only $75\% - 25\% = 50\%$ of the data lies between them.

(b) Draw whiskers from 30 to 80 and from 260 to 350, a box from 80 to 260 and a line at 170.

(c) Online: median 200, IQR $350 - 160 = 190$. Shop: median 170, IQR $260 - 80 = 180$.

The online median is higher, so it typically sells more each month. The online IQR is bigger, so its monthly sales are less consistent.

Total work $= 6 \times 14.5 = 87$ worker-days.

With 13 workers: $87 \div 13 = 6.69...$ days.

$6.69 < 7$, so 13 workers are enough.

Rearrange $L_2$: $5y = 10x - 4$, so $y = 2x - \frac{4}{5}$

$L_1$: $y = 2x + 3$ has gradient 2. $L_2$ has gradient 2.

Equal gradients, so the lines are parallel.

The shape's vertices are $(-6, -4)$, $(-5, -2)$, $(-3, -2)$, $(-2, -4)$.

Multiply each coordinate by $-2$: $(12, 8)$, $(10, 4)$, $(6, 4)$, $(4, 8)$.

The image is twice as big, upside down, on the opposite side of the origin.

The fraction tagged in the second catch should match the fraction tagged in the whole lake:

$\frac{45}{305} = \frac{180}{N}$

$N = \frac{180 \times 305}{45} = 1220$

Let the pay be $6x$ and $5x$. After the rise: $\frac{6x + 1.5}{5x + 1.5} = \frac{13}{11}$

$11(6x + 1.5) = 13(5x + 1.5)$

$66x + 16.5 = 65x + 19.5$, so $x = 3$

Marta: $6 \times 3 = £18$, Khalid: $5 \times 3 = £15$

(a) All three: 4. A and B only: 8. B and C total 14, so B and C only $= 14 - 4 = 10$.

Nobody uses C alone, so A and C only $= 20 - 4 - 10 - 0 = 6$.

A only $= 62 - 8 - 4 - 6 = 44$. Outside: 16.

B only $= 100 - 16 - 44 - 8 - 4 - 6 - 10 = 12$

(b) Among the 62 A users, $8 + 4 = 12$ also use B: $P = \frac{12}{62} = \frac{6}{31}$

(a) Radius $= 3.5$ cm. Hemisphere: $\frac{2}{3}\pi(3.5)^3 = 28.583...\pi$

Cone volume $= 120\pi - 28.583...\pi = 91.416...\pi$

$\frac{1}{3}\pi(3.5)^2 h = 91.416...\pi$ gives $h = 91.416... \div 4.0833... = 22.387...$

$y = h + 3.5 = 25.887...$, so $y = 25.9$

(b) With a wider base the same volume needs less height, so $y$ would be smaller.

Triangle PQR: $QR^2 = 11^2 + 9.4^2 - 2 \times 11 \times 9.4 \cos 27^\circ = 25.09...$, so $QR = 5.0099...$

Triangle QRS: $\frac{QS}{\sin 88^\circ} = \frac{QR}{\sin 41^\circ}$

$QS = \frac{5.0099... \times \sin 88^\circ}{\sin 41^\circ} = 7.6318...$, so $QS = 7.63$ cm

(a) $g(16) = \sqrt[3]{32 - 5} = \sqrt[3]{27} = 3$

(b) Let $y = \sqrt[3]{2x - 5}$. Cube: $y^3 = 2x - 5$, so $x = \frac{y^3 + 5}{2}$

$g^{-1}(x) = \frac{x^3 + 5}{2}$

$hg^{-1}(x) = \frac{1}{\frac{x^3 + 5}{2}} = \frac{2}{x^3 + 5}$

Angle $BAD = 132 \div 2 = 66^\circ$ (angle at the centre is twice the angle at the circumference).

Angle $BCD = 180 - 66 = 114^\circ$ (opposite angles of a cyclic quadrilateral add to 180°).

Angle $DCE = 180 - 114 = 66^\circ$ (angles on a straight line).

Angle $CDE = 180 - 66 - 16 = 98^\circ$ (angles in a triangle).

(a) $f(-x)$ swaps left and right: reflect every point of the curve in the $y$-axis.

(b) Q is at $(360, 0)$ and R is at $(90, -5)$.

The graph moves 270 left and 5 down, i.e. by $\begin{pmatrix}-270\\-5\end{pmatrix}$

Moving left by 270 means replacing $x$ with $x + 270$; moving down 5 means subtracting 5:

$g(x) = \tan(x + 270)^\circ - 5$

$x^2 - 49 > 0$: $(x + 7)(x - 7) > 0$, so $x < -7$ or $x > 7$

$5x^2 - 31x - 72 > 0$: $(5x + 9)(x - 8) > 0$, so $x < -1.8$ or $x > 8$

Both true: $x < -7$ or $x > 8$