November 2023 Paper 1 Higher

Work out $63 \times 24$: $63 \times 20 = 1260$ and $63 \times 4 = 252$, so $1260 + 252 = 1512$.

There are two decimal places in the question, so $6.3 \times 2.4 = 15.12$

(a)(i) Any non-zero number to the power 0 is 1.

(a)(ii) $5^{-2} = \frac{1}{5^2} = \frac{1}{25}$

(b) Add powers when multiplying, subtract when dividing: $2^{5+4-3} = 2^6$

(a) $156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$, so $156 = 2^2 \times 3 \times 13$

(b) $130 = 2 \times 5 \times 13$

Shared primes: $2 \times 13 = 26$

(a) Total length: $4.2 \times 5 = 21$ cm

Other 4 sticks: $21 - 7 = 14$ cm, so mean $= 14 \div 4 = 3.5$ cm

(b) The other 4 sticks now total $21 - 17 = 4$ cm, so their mean goes down to $4 \div 4 = 1$ cm.

With compasses centred on P, mark two points on AB the same distance either side of P.

Widen the compasses. From each of those two points draw arcs above the line that cross.

Join P to the crossing point with a ruler. This line meets AB at $90^\circ$. Leave all arcs showing.

BA = BD, so the angles at A and D are equal: both are $x$.

Angles in the triangle: $x + x + y = 180$. With $x = 2y$: $5y = 180$, so $y = 36$.

Angles on a straight line: $w = 180 - 36 = 144$

$x + 3x + 1 + 2x - 5 = 44$, so $6x - 4 = 44$, $6x = 48$, $x = 8$

Shelf B: $3 \times 8 + 1 = 25$ books, so one book is $7500 \div 25 = 300$ g

Shelf A: $8 \times 300 = 2400$ g

The sale price is $100\% - 40\% = 60\%$ of the normal price.

$60\%$ is £660, so $1\%$ is £11 and $100\%$ is £1100

(a) 4 cups of rice need 5 cups of water, and 0 need 0. Plot $(0, 0)$, $(4, 5)$ and $(8, 10)$ and join with a ruler.

(b)(i) Gradient $= \frac{\text{rise}}{\text{run}} = \frac{5}{4} = 1.25$

(b)(ii) It is the number of cups of water used for every cup of rice.

$2\pi r = 10$, so $r = \frac{5}{\pi}$

Area $= \pi r^2 = \pi \times \frac{25}{\pi^2} = \frac{25}{\pi}$ m²

(a) Greatest value $= 300 + 1000 = 1300$. Draw whiskers from the box to 300 and 1300, and a line in the box at 900.

Reading the box: lower quartile $= 780$, upper quartile $= 1200$

(b) 1200 is the upper quartile, so about $\frac{3}{4}$ of the days were below it: $\frac{3}{4} \times 80 = 60$

Rearrange: $y = \frac{3}{2}x - \frac{7}{2}$, so the gradient of $L$ is $\frac{3}{2}$

Perpendicular gradient: $-\frac{2}{3}$

$-5 = -\frac{2}{3} \times 6 + c = -4 + c$, so $c = -1$

$y = -\frac{2}{3}x - 1$

Length scale factor $= \frac{5}{2} = 2.5$, so the volume scale factor is $2.5^3 = \frac{125}{8}$

Volume of B $= 12 \times \frac{125}{8} = 1.5 \times 125 = 187.5$ cm³

$27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$

$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$

$9 + 8 = 17$

Draw a tangent touching the curve at $t = 3$ (where $d \approx 42$).

Pick two points on the tangent, for example $(2, 17)$ and $(4, 67)$.

Gradient $\approx \frac{67 - 17}{4 - 2} = \frac{50}{2} = 25$ (metres per second)

(a) $P_2 = kP_1$: $6 = 8k$, so $k = 0.75$

$P_3 = 0.75 \times 6 = 4.5$ million

(b) New $k = 0.75 + 0.3 = 1.05$. Multiplying by a number greater than 1 each year makes the population grow, so Louise is correct.

(a) $6 \times (-4) = -24$; use $-8$ and $3$: $6x^2 - 8x + 3x - 4 = 2x(3x - 4) + (3x - 4) = (2x + 1)(3x - 4)$

(b) Critical values: $x = -\frac{1}{2}$ and $x = \frac{4}{3}$

The graph is a U shape, so it is below zero between the roots: $-\frac{1}{2} < x < \frac{4}{3}$

$\frac{1}{4} \times P(B) = \frac{1}{24}$, so $P(B) = \frac{1}{6}$

A red, B not: $\frac{1}{4} \times \frac{5}{6} = \frac{5}{24}$

A not, B red: $\frac{3}{4} \times \frac{1}{6} = \frac{3}{24}$

Total: $\frac{5}{24} + \frac{3}{24} = \frac{8}{24} = \frac{1}{3}$

(a) Draw $y = 0.3$ across the graph. It meets the curve twice, at about $x = 18$ and $x = 180 - 18 = 162$.

(b) $\sin\theta = 0$ when $\theta = 0$ or $180$. So $x + 20 = 0$ gives $x = -20$ (or $x + 20 = 180$ gives $x = 160$).

$(5\sqrt{7})^2 = 25 \times 7 = 175$

$\cos B = \frac{10^2 + 5^2 - 175}{2 \times 10 \times 5} = \frac{125 - 175}{100} = -\frac{1}{2}$

$\cos 60° = \frac{1}{2}$, so $B = 180° - 60° = 120°$

(a) $x^2 + y^2 = 169$ is a circle centred at the origin with radius $\sqrt{169} = 13$.

(b) Draw $y = 1.5x$ through $(0, 0)$, $(4, 6)$ and $(8, 12)$.

It crosses the circle at about $(7.2, 10.8)$ and $(-7.2, -10.8)$.

Check: $x^2(1 + 2.25) = 169$ gives $x^2 = 52$, $x \approx 7.21$

$r = \frac{13 + 9\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}$

Numerator: $39 - 26\sqrt{2} + 27\sqrt{2} - 36 = 3 + \sqrt{2}$

Denominator: $9 - 8 = 1$

So $r = 3 + \sqrt{2}$