November 2018 Paper 3 Higher

(a) The 4th figure is 7, so round up: $7357 \to 7360$

(b) Top: $\sqrt{17 + 16} = \sqrt{33} = 5.7445\ldots$

Bottom: $7.3^2 = 53.29$

$5.7445\ldots \div 53.29 = 0.1077981356$

Increase: $883 - 245 = 638$

Percentage increase: $\frac{638}{245} \times 100 = 260.4\%$

(a) $x = -3$: $9 - 3 - 4 = 2$; $x = -1$: $1 - 1 - 4 = -4$; $x = 2$: $4 + 2 - 4 = 2$; $x = 3$: $9 + 3 - 4 = 8$

(b) Plot all seven points and join with a smooth curve, dipping just below $-4$ between $x = -1$ and $x = 0$.

(c) The roots are where the curve crosses the $x$-axis: about $1.6$ or $-2.6$

(a) 2 students bought 20 to 24 books: $\frac{2}{40} \times 100 = 5\%$

(b) Frequencies 11, 8, 13, 6, 2 and midpoints 2, 7, 12, 17, 22

$11 \times 2 + 8 \times 7 + 13 \times 12 + 6 \times 17 + 2 \times 22 = 22 + 56 + 156 + 102 + 44 = 380$

Mean $\approx 380 \div 40 = 9.5$

(a) 1 min 54 s $= 114$ s

Time $= \frac{700}{475} \times 114 = 168$ s

$168$ s $= 2$ minutes $48$ seconds

(b) If she speeds up over a longer race, she will finish more quickly, so the real time is less than 2 min 48 s.

Angles A and B share $180 - 90 = 90°$ in the ratio $3 : 2$ (B : A)

One part $= 90 \div 5 = 18°$, so $A = 36°$ and $B = 54°$

AC is next to angle A and AB is the hypotenuse: $\cos 36° = \frac{14}{AB}$

$AB = \frac{14}{\cos 36°} = 17.3$ cm

Plot each frequency above the middle of its class: 5, 15, 25 and 35.

Points: $(5, 14)$, $(15, 18)$, $(25, 26)$, $(35, 12)$

Join them in order with straight lines. Do not join the last point back to the first.

Face areas 15, 27, 45 need edges with pair products 15, 27, 45: the edges are 3, 5 and 9 cm.

Volume of cuboid $= 3 \times 5 \times 9 = 135$ cm$^3$

Volume of one cube $= 2.5^3 = 15.625$ cm$^3$

$135 \div 15.625 = 8.64$, so at most 8 whole cubes.

(a) $(x - 2)(2x + 3) = 2x^2 - x - 6$

$(2x^2 - x - 6)(x + 1) = 2x^3 + 2x^2 - x^2 - x - 6x - 6 = 2x^3 + x^2 - 7x - 6$

(b) $4 + n - 2 = -3$, so $n = -5$

(c) $x = \frac{4 \pm \sqrt{16 + 60}}{10} = \frac{4 \pm \sqrt{76}}{10}$, so $x = 1.27$ or $x = -0.472$

(a) $f(23) = 4\sin 23° = 1.56$

(b) $g(34) = 2 \times 34 - 3 = 65$, then $f(65) = 4\sin 65° = 3.63$

(c) $25$ has two square roots, $5$ and $-5$. So $x + 4 = -5$ is also possible, giving a second answer $x = -9$.

$\tan x°$ is 0 at 0°, 180° and 360°.

It rises towards $+\infty$ as $x$ approaches 90° and 270°, and comes up from $-\infty$ just after them.

Draw vertical dashed asymptotes at 90 and 270, and three increasing branches through $(0,0)$, $(180,0)$, $(360,0)$.

Diagonal $AC = \sqrt{5^2 + 5^2} = \sqrt{50} = 7.071$ m, so half of it is $3.536$ m.

The triangle from A to the centre to T is right-angled:

$\tan TAC = \frac{12}{3.536}$

$TAC = \tan^{-1}(3.394) = 73.6°$

$P_1 = 400$

$P_2 = 1.01 \times 400 = 404$

$P_3 = 1.01 \times 404 = 408.04$, so about $408$ animals.

$y = \frac{k}{x^3}$

$44 = \frac{k}{a^3}$, so $k = 44a^3$

When $x = 2a$: $y = \frac{44a^3}{(2a)^3} = \frac{44a^3}{8a^3} = 5.5$

Let the odd numbers be $2n - 1$ and $2n + 1$, where $n$ is an integer.

$(2n+1)^2 - (2n-1)^2 = (4n^2 + 4n + 1) - (4n^2 - 4n + 1)$

$= 8n$

$8n$ is 8 times an integer, so it is always a multiple of 8.

The sector's outer arc uses $360 - 140 = 220°$:

Arc $= \frac{220}{360} \times 2\pi \times 6 = 23.04$ cm

Sine rule in triangle AOD: $\frac{OD}{\sin 24°} = \frac{14}{\sin 140°}$, so $OD = 8.859$ cm

$CD = OD - 6 = 2.859$ cm

Perimeter $= 23.04 + 14 + 2.859 = 39.9$ cm

(a) Frequency density = frequency ÷ class width:

$120 \div 20 = 6$, $90 \div 30 = 3$, $120 \div 30 = 4$, $140 \div 70 = 2$, $100 \div 50 = 2$

(b) The median is the $285$th value. $120 + 90 = 210$, so it is $75$ into the 50–80 class.

$50 + \frac{75}{120} \times 30 = 68.75 \approx 69$ miles

Time: 3 hours $= 180$ minutes, bounds $179.5$ and $180.5$.

Distance bounds: $486.5$ and $487.5$ km.

Upper bound of speed $= \frac{487.5}{179.5} = 2.7158\ldots$

Lower bound of speed $= \frac{486.5}{180.5} = 2.6953\ldots$

Both round to $2.7$ (1 d.p.) so the speed is $2.7$ km/minute.

From the linear equation: $x = 1 - 2y$

$2(1 - 2y)^2 - y^2 = 17$

$2 - 8y + 8y^2 - y^2 = 17 \Rightarrow 7y^2 - 8y - 15 = 0$

$(7y - 15)(y + 1) = 0$, so $y = \frac{15}{7}$ or $y = -1$

$y = -1$: $x = 3$. $\quad y = \frac{15}{7}$: $x = 1 - \frac{30}{7} = -\frac{23}{7}$

Triangle A has corners $(-4, 1)$, $(-3, 1)$, $(-3.5, 4)$.

Rotating 180° about $(-2, 0)$ sends $(x, y)$ to $(-4 - x, -y)$. Then translating by $\binom{-3}{2}$ gives $(-7 - x, 2 - y)$.

The corners go to $(-3, 1)$, $(-4, 1)$, $(-3.5, -2)$. The base maps onto itself reversed, so its midpoint stays put.

Check: $-7 - (-3.5) = -3.5$ and $2 - 1 = 1$, so the point is $(-3.5, 1)$.