November 2019 Paper 3 Higher

(a) $x \times x = x^2$, $x \times (-9) = -9x$, $5 \times x = 5x$, $5 \times (-9) = -45$

$x^2 - 9x + 5x - 45 = x^2 - 4x - 45$

(b) The highest common factor of $9x^2$ and $6x$ is $3x$: $3x(3x + 2)$

(a) Top: $29^2 - 4.6 = 836.4$. Bottom: $\sqrt{35 - 6.859} = \sqrt{28.141} = 5.30481...$

$836.4 \div 5.30481... = 157.668255...$

(b) The 4th significant figure is 6 and the next digit is 6, so round up: $157.7$

Draw a line of best fit through the middle of the crosses.

Go up from 34 on the Science axis to the line, then across to the Maths axis. This reads about 40.

Midpoints: 7.5, 12.5, 17.5, 22.5

$\sum fx = 7.5 + 25 + 122.5 + 180 = 335$

Mean $\approx 335 \div 18 = 18.61... = 18.6$ seconds

1 cm $= 10$ mm, so $1\text{ cm}^3 = 10 \times 10 \times 10 = 1000\text{ mm}^3$.

$37 \times 1000 = 37\,000\text{ mm}^3$

Time: 13 30 to 14 48 is 1 hour 18 minutes $= 78$ minutes.

In hours: $78 \div 60 = 1.3$

Speed $= 65 \div 1.3 = 50$ mph

(a) Put the decimal point after the first digit: $3.246$. It moves 7 places, so $3.246 \times 10^7$.

(b) Move the point 3 places left: $0.00496$

(c) No. $A = 621\,200\,000$ and $B = 4\,730\,000\,000$, so $B$ is bigger. Compare the powers first.

Co-interior angles in the parallelogram add to 180°: $180 - 117 = 63°$

Interior angle of a regular pentagon: $(5-2) \times 180 \div 5 = 108°$

$x = 108 - 63 = 45°$

Centre $(0, 1)$. Triangle A has vertices $(1, 1)$, $(1, 3)$ and $(2, 3)$.

From the centre, these are $(1, 0)$, $(1, 2)$ and $(2, 2)$ away. Multiply by 2.5: $(2.5, 0)$, $(2.5, 5)$, $(5, 5)$.

Add back to the centre: $(2.5, 1)$, $(2.5, 6)$, $(5, 6)$.

(a) $9 + x = 7(11 - x) = 77 - 7x$

$8x = 68$, so $x = 8.5$

(b) Cancel $(y + 3)^2$ from top and bottom: $\dfrac{4(y+3)^3}{(y+3)^2} = 4(y + 3)$

Late Monday, not Tuesday: $0.07 \times 0.98 = 0.0686$

Not late Monday, late Tuesday: $0.93 \times 0.11 = 0.1023$

$0.0686 + 0.1023 = 0.1709$

There are 23 values, so the median is the 12th: 197 cm.

LQ is the 6th value: 188 cm. UQ is the 18th value: 209 cm.

Smallest 173 cm, largest 219 cm.

Draw the box from 188 to 209 with a line at 197, and whiskers to 173 and 219.

Volume $= \pi \times 3^2 \times 25 = 225\pi = 706.86$ cm³

A: $\frac{2}{15} \times 225\pi = 30\pi$ cm³, mass $= 30\pi \times 1.21 = 114.04$ g

B: $\frac{13}{15} \times 225\pi = 195\pi$ cm³, mass $= 195\pi \times 1.02 = 624.86$ g

Total $= 738.90... \approx 739$ g

Soup $= \frac{2}{5}$, prawns $= \frac{3}{5}$

Soup then curry: $\frac{2}{5} \times \frac{3}{8} = \frac{3}{20}$

Prawns then curry: $\frac{3}{5} \times \frac{5}{6} = \frac{1}{2}$

Curry total $= \frac{3}{20} + \frac{10}{20} = \frac{13}{20}$

Let the numbers be $2n$ and $2n + 2$.

$(2n)^2 + (2n + 2)^2 = 4n^2 + 4n^2 + 8n + 4 = 8n^2 + 8n + 4$

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

$2n^2 + 2n + 1$ is an integer, so the sum is a multiple of 4.

$y = \dfrac{k}{x^2}$, so $8 = \dfrac{k}{2.5^2}$ and $k = 8 \times 6.25 = 50$

When $y = \frac{8}{9}$: $x^2 = 50 \div \frac{8}{9} = 56.25$

$x = \pm 7.5$, so the negative value is $-7.5$

$x^2 - 2x - 2 = 0$ is the same as $x^2 - 3 = 2x - 1$.

Draw $y = 2x - 1$ on the graph of $y = x^2 - 3$.

The line and curve cross at about $x = 2.7$ and $x = -0.7$.

Cosine rule: $\cos C = \dfrac{6.1^2 + 6.2^2 - 3.4^2}{2 \times 6.1 \times 6.2}$, so $C = 32.08°$

Angle DAC $= \frac{2}{5} \times 32.08 = 12.83°$

Angle ADC $= 180 - 32.08 - 12.83 = 135.09°$

Sine rule: $DC = \dfrac{6.2 \sin 12.83°}{\sin 135.09°} = 1.95$ cm

Draw a tangent touching the curve at $t = 6$ (where the distance is about 10 m).

Pick two points on the tangent, e.g. it rises about 20 m over about 5.8 s.

Speed $= 20 \div 5.8 \approx 3.4$ m/s

First differences: 1, 3, 5, 7. Second difference 2, so start with $n^2$: 1, 4, 9, 16, 25.

Sequence minus $n^2$: −2, −4, −6, −8, −10, which is $-2n$.

nth term $= n^2 - 2n$

P(head)$^4 = \frac{16}{81}$, so P(head) $= \sqrt[4]{\frac{16}{81}} = \frac{2}{3}$

P(tail) $= 1 - \frac{2}{3} = \frac{1}{3}$

P(4 tails) $= \left(\frac{1}{3}\right)^4 = \frac{1}{81}$

$7x - 14 = 7(x - 2)$ and $x^2 + 4x - 12 = (x - 2)(x + 6)$

$x^3 - 36x = x(x - 6)(x + 6)$

$\dfrac{7(x-2)}{(x-2)(x+6)} \times \dfrac{x(x-6)(x+6)}{x-6}$

Cancel $(x - 2)$, $(x + 6)$ and $(x - 6)$ to leave $7x$, so $a = 7$.

$56.8 = \frac{1}{3}\pi r^2 \times 3.6$, so $r^2 = \dfrac{56.8}{1.2\pi} = 15.07$ and $r = 3.882$ cm

Slant height $l = \sqrt{3.882^2 + 3.6^2} = 5.294$ cm (the sector's radius)

Arc length = base circumference: $\frac{\theta}{360} \times 2\pi l = 2\pi r$

$\theta = 360 \times \dfrac{r}{l} = 360 \times \dfrac{3.882}{5.294} = 264°$

$\overrightarrow{OZ} = \overrightarrow{XY} = \mathbf{b} - \mathbf{a}$

$\overrightarrow{ZP} = \frac{1}{3}\mathbf{a} - (\mathbf{b} - \mathbf{a}) = \frac{4}{3}\mathbf{a} - \mathbf{b} = \frac{4}{3}(\mathbf{a} - \frac{3}{4}\mathbf{b})$

$\overrightarrow{ZR} = \frac{1}{4}\mathbf{b} - (\mathbf{b} - \mathbf{a}) = \mathbf{a} - \frac{3}{4}\mathbf{b}$

So $ZP = \frac{4}{3}ZR$ and $ZP : ZR = 4 : 3$