November 2020 Paper 1 Higher

The terms go up by 3, so start with $3n$: 3, 6, 9, ...

Each term is 2 less than that, so the nth term is $3n - 2$.

Change to improper fractions: $2\frac{1}{3} = \frac{7}{3}$ and $3\frac{3}{4} = \frac{15}{4}$.

Multiply: $\frac{7 \times 15}{3 \times 4} = \frac{105}{12} = \frac{35}{4}$.

$35 \div 4 = 8$ remainder 3, so $\frac{35}{4} = 8\frac{3}{4}$.

$y = x^2$ is the U shape: D.

$y = \frac{1}{x}$ has two separate branches and never touches the axes: A.

$y = x^3$ rises from bottom left to top right: C.

$y = -x^3$ is its reflection, falling from top left: B.

Triangle A: the 10 cm side lies between angles of 55° and 45°.

Triangle D: the third angle is $180 - 80 - 45 = 55°$, and its 10 cm side joins the 45° corner to that 55° corner.

So both have a 10 cm side between angles of 55° and 45° (ASA), and A and D are congruent.

Selling price: $24 \times 50$p $= 1200$p $= £12$.

Profit: $12 - 10 = £2$.

Percentage profit: $\frac{2}{10} \times 100 = 20\%$.

Angle $AEB = 63°$ (corresponding angles are equal).

Angle $ABE = 180 - 148 = 32°$ (angles on a straight line add up to $180°$).

Angle $EAB = 180 - 63 - 32 = 85°$ (angles in a triangle add up to $180°$).

Girls: median 165 cm, range $170 - 150 = 20$ cm.

Boys: 15 values, so the median is the 8th: 168 cm. Range $182 - 158 = 24$ cm.

On average the boys are taller (168 > 165), and their heights are more spread out (24 > 20).

Base area $= \frac{18}{3} = 6\text{ m}^2$.

Force $=$ pressure $\times$ area $= 75 \times 6 = 450$ N.

Write each as an ordinary number:

$67.2 \times 10^{-4} = 0.00672$, $6.72 \times 10^5 = 672\,000$, $672 \times 10^4 = 6\,720\,000$.

So: 0.000672, $67.2 \times 10^{-4}$, $6.72 \times 10^5$, $672 \times 10^4$.

$\frac{a}{b} = \frac{2}{5}$ gives $a : b = 2 : 5$, and $\frac{b}{c} = \frac{3}{4}$ gives $b : c = 3 : 4$.

Make b the same in both (15): $a : b = 6 : 15$ and $b : c = 15 : 20$.

So $a : b : c = 6 : 15 : 20$.

(a) $\sqrt[4]{81} = 3$ and $\sqrt[4]{10^8} = 10^2$, so the answer is $3 \times 100 = 300$

(b) $64^{\frac{1}{2}} = 8$, and the negative power means the reciprocal: $\frac{1}{8}$

(c) $9^{n-1} = 3^{2(n-1)} = 3^{2n-2}$, so $\frac{3^n}{3^{2n-2}} = 3^{n-(2n-2)} = 3^{2-n}$

(a) Add the frequencies as you go: 5, 15, 35, 55, 70, 80

(b) Plot each total at the top of its class (250, 300, ..., 500) and join the points.

(c) 60% of 80 $= 0.6 \times 80 = 48$ people.

Reading the graph at £360 gives about 40 people, which is $\frac{40}{80} = 50\%$.

So Juan is wrong: only about 50% earn £360 or less.

Volume of A $= \frac{1400}{70} = 20\text{ m}^3$

Mass of B $= 280 \times 30 = 8400$ kg

Density of C $= \frac{1400 + 8400}{20 + 30} = \frac{9800}{50} = 196\text{ kg/m}^3$

P(not win) $= 1 - 0.3 = 0.7$

Win then not win: $0.3 \times 0.7 = 0.21$

Not win then win: $0.7 \times 0.3 = 0.21$

Total: $0.21 + 0.21 = 0.42$

Gradient of $L_1$ is 3, so the perpendicular gradient is $-\frac{1}{3}$.

$y = -\frac{1}{3}x + c$ through $(9, 5)$: $5 = -3 + c$, so $c = 8$.

$L_2$: $y = -\frac{1}{3}x + 8$

(a) In the second sample $\frac{20}{120} = \frac{1}{6}$ are marked.

So the 90 marked bees are about $\frac{1}{6}$ of the hive: $90 \times 6 = 540$

(b) If marks rubbed off, fewer marked bees would be counted, which makes the estimate bigger than it should be. So 540 would be an overestimate.

Multiply both sides by $(f - 4)$: $d(f - 4) = 3(1 - f)$

Expand: $df - 4d = 3 - 3f$

Collect the $f$ terms: $df + 3f = 3 + 4d$

Factorise: $f(d + 3) = 3 + 4d$

Divide: $f = \frac{3 + 4d}{d + 3}$

$x = k\sqrt{y}$

New $y$ is $1.44y$, so new $x = k\sqrt{1.44y} = 1.2k\sqrt{y} = 1.2x$

So $x$ rises by $20\%$.

(a) $g(5) = 3(2 \times 5 + 1) = 3 \times 11 = 33$

(b) $f(9) = \frac{12}{\sqrt{9}} = 4$, then $g(4) = 3(9) = 27$

(c) Solve $3(2x + 1) = 6$: $2x + 1 = 2$, so $x = \frac{1}{2}$

$\sqrt{180} = 6\sqrt{5}$, so the top is $6\sqrt{5} - 2\sqrt{5} = 4\sqrt{5}$

Multiply top and bottom by $5\sqrt{5} + 5$:

Top: $4\sqrt{5}(5\sqrt{5} + 5) = 100 + 20\sqrt{5}$

Bottom: $(5\sqrt{5})^2 - 5^2 = 125 - 25 = 100$

So the fraction is $\frac{100 + 20\sqrt{5}}{100} = 1 + \frac{\sqrt{5}}{5}$

$\overrightarrow{DE} = -\mathbf{a} + \mathbf{b}$, so $\overrightarrow{DQ} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$

$\overrightarrow{PD} = \frac{1}{2}\mathbf{a}$

$\overrightarrow{PQ} = \frac{1}{2}\mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{b}$

$\overrightarrow{FE} = \mathbf{b} = 2\overrightarrow{PQ}$, so PQ is parallel to FE.

Outer radius $= (3x - 1) + 2x = 5x - 1$

Area A $= \frac{45}{360}\pi[(5x - 1)^2 - (3x - 1)^2] = \frac{\pi}{8}(16x^2 - 4x)$

Area B $= \pi(1 - x)^2 = \pi(x^2 - 2x + 1)$

Equate and cancel $\pi$: $2x^2 - \frac{x}{2} = x^2 - 2x + 1$, so $2x^2 + 3x - 2 = 0$

$(2x - 1)(x + 2) = 0$, so $x = 0.5$ or $x = -2$

$x = -2$ gives negative lengths, so $x = 0.5$

Black cards are $\frac{3}{8}$ of all cards; triangles are $\frac{7}{9}$ of all cards.

Use a total of 72 (LCM of 8 and 9): black $= 27$, triangles $= 56$

So black as a fraction of triangles $= \frac{27}{56}$