June 2018 Paper 2 Higher

(a) Add the powers: $m^{3+4} = m^7$

(b) Cube each part: $5^3 \times n^3 \times (p^3)^3 = 125n^3p^9$

(c) Divide the numbers and subtract the powers: $\frac{32}{4} = 8$, $q^{9-3} = q^6$, $r^{4-1} = r^3$, giving $8q^6r^3$

(a) $40 = 2^3 \times 5$ and $56 = 2^3 \times 7$

LCM $= 2^3 \times 5 \times 7 = 280$

(b) Take the lowest power of each shared prime: $2^2 \times 3 \times 5 = 60$

The line crosses the $y$-axis at $-6$, so $c = -6$.

From $(0, -6)$ to $(2, 0)$: up 6, across 2, so gradient $= 6 \div 2 = 3$.

$y = 3x - 6$

Price with VAT: $8500 \times 1.2 = £10\,200$

Monthly payments: $12 \times 531.25 = £6375$

Deposit: $10\,200 - 6375 = £3825$

Ratio: $3825 : 6375$. Divide both by 1275: $3 : 5$

(a) e.g. $x = -2$: $4 + 2 - 6 = 0$; $x = 1$: $1 - 1 - 6 = -6$; $x = 2$: $4 - 2 - 6 = -4$

(b) Plot $(-3, 6), (-2, 0), (-1, -4), (0, -6), (1, -6), (2, -4), (3, 0)$ and join with a smooth curve.

(c) Draw the line $y = -2$ and read off where it crosses the curve: $x \approx 2.6$ and $x \approx -1.6$

Original pressure: $70 \div 20 = 3.5$ N/cm²

New pressure: $80 \div 30 = 2.67$ N/cm²

Decrease: $3.5 - 2.67 = 0.83$

Percentage decrease: $\frac{0.83}{3.5} \times 100 \approx 24\%$

This is more than 20%, so Helen is wrong.

Measure each vertex from the centre $(0, 1)$ and multiply by $\frac{1}{3}$.

$(3, 4)$ is $(3, 3)$ from the centre → $(1, 1)$ → image $(1, 2)$

$(3, 10)$ → $(1, 3)$ → $(1, 4)$; $(6, 10)$ → $(2, 3)$ → $(2, 4)$; $(6, 7)$ → $(2, 2)$ → $(2, 3)$

Females: $60 - 38 = 22$

Males choosing Britain: $32 - 11 = 21$

Males choosing Spain: $38 - 21 - 8 = 9$, so females choosing Spain: $12 - 9 = 3$

Probability $= \frac{3}{22}$

Multiplier for year 1: $12\,336 \div 12\,000 = 1.028$, so $x = 2.8$

Year 2 rate: $2.8 \div 2 = 1.4\%$

$12\,336 \times 1.014 = £12\,508.70$

From the grid, $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$

(a) $-2\mathbf{a} = \begin{pmatrix} -2 \\ -4 \end{pmatrix}$: draw 2 left, 4 down.

(b) $\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 1 + 2 \\ 2 - 6 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$

$f(x) = \frac{2}{x^2}$ and $g(x) = 4x^3$

(a) $f(-5) = \frac{2}{(-5)^2} = \frac{2}{25}$

(b) $g(1) = 4$, then $f(4) = \frac{2}{16} = \frac{1}{8}$

$y \propto x$ is a straight line through the origin: B

$y \propto x^2$ curves upwards more and more steeply: D

$y \propto \sqrt{x}$ rises quickly then levels off: A

$y \propto \frac{1}{x}$ falls towards the axes: C

(a) Angle $ODE = 90°$ (tangent meets radius at right angles), so angle $BDE = 90 + x$.

By the alternate segment theorem, angle $BAD =$ angle $BDE$, so $y = 90 + x$.

Therefore $y - x = 90$.

(b) No. $y$ is an angle of triangle $ABD$, so it is less than 180°; it cannot be 200.

Draw a tangent to the curve at $t = 5$.

Pick two points far apart on the tangent and find rise ÷ run, e.g. a rise of about 60 m over 4 s.

Speed $\approx 60 \div 4 = 15$ m/s

(a) $1 - 0.45 = 0.55$, $1 - 0.67 = 0.33$, $1 - 0.35 = 0.65$

(b) Win then lose: $0.45 \times 0.33 = 0.1485$

Lose then win: $0.55 \times 0.35 = 0.1925$

Total: $0.1485 + 0.1925 = 0.341$

(a) $x^2 + y^2 = 12.25$ is a circle centre $O$ with radius $\sqrt{12.25} = 3.5$

(b) Draw $y = 1 - 2x$ (through $(0, 1)$ and $(2, -3)$).

Read off the crossing points: about $(2.0, -2.9)$ and $(-1.2, 3.3)$

(Check: $5x^2 - 4x - 11.25 = 0$ gives $x = 1.95$ or $-1.15$.)

(a) Frequency = class width × frequency density:

$10 \times 0.4 = 4$, $10 \times 0.6 = 6$, $5 \times 1 = 5$, $20 \times 0.2 = 4$

(b) Total $= 4 + 4 + 6 + 5 + 4 = 23$, so the lower quartile is the $\frac{23+1}{4} = 6$th value.

The first class holds 4 values, so we need 2 more from the $5 < t \le 15$ class, which holds 4 values over 10 minutes.

Lower quartile $= 5 + \frac{2}{4} \times 10 = 10$ minutes

In right-angled triangle $ABC$: $AC = \frac{7.3}{\sin 48°} = 9.823$ cm

In right-angled triangle $ACH$: $\tan(CAH) = \frac{8.1}{9.823}$

Angle $= \tan^{-1}(0.8246) = 39.5°$

$\frac{1}{4} \times \frac{4}{3}\pi r^3 = 576\pi$, so $r^3 = 1728$ and $r = 12$

Curved part: $\frac{1}{4} \times 4\pi \times 12^2 = 144\pi$

Two flat semicircles: $2 \times \frac{1}{2}\pi \times 12^2 = 144\pi$

Total $= 288\pi = 904.8 \approx 905\text{ cm}^2$ (3 s.f.)

(a) $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$. Martin wrote $+3$ for $\sqrt{3} \times -\sqrt{3}$. The answer should be $28 - 14\sqrt{3}$.

(b) $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$. Sian wrote $3\sqrt{2}$. The answer should be $\frac{10\sqrt{3}}{12} = \frac{5\sqrt{3}}{6}$.

Lengths are to the nearest mm, mass to the nearest 5 g.

Lower bound: $\frac{1967.5}{13.25 \times 16.05 \times 21.75} = 0.4254$

Upper bound: $\frac{1972.5}{13.15 \times 15.95 \times 21.65} = 0.4344$

Both round to $0.43$ (2 d.p.), so the density is $0.43\text{ g/cm}^3$.