June 2024 Paper 1 Higher

The terms go up by 4, so start with $4n$.

$4n$ gives 4, 8, 12, ... Each term is 3 less, so the nth term is $4n - 3$.

(a) $3\frac{4}{5} - 1\frac{2}{3} = \frac{57}{15} - \frac{25}{15} = \frac{32}{15} = 2\frac{2}{15}$

(b) $35 - 24 = 11$, so $\frac{35}{24} = 1\frac{11}{24}$. Kevin wrote 9 instead of 11 in the numerator.

(a) Area $= 10 \times 8 - 4 \times 3 = 80 - 12 = 68\text{ m}^2$

Paint: $3 \times 2.5 = 7.5$ litres, covering $7.5 \times 10 = 75\text{ m}^2$

$75 > 68$, so yes, she has enough.

(b) With 11 m² per litre it covers $82.5\text{ m}^2$, even more, so the answer does not change.

(a) $P'$ is everything not in $P$: $10, 11, 13, 14, 16, 17$

(b) $P \cup Q = \{10, 12, 14, 15, 18\}$, which is 5 of the 9 numbers.

Probability $= \frac{5}{9}$

(a) Round: $513 \approx 500$ km and $0.81 \approx 0.8$ euros.

$500 \div 10 = 50$ lots of 10 km, $50 \times 0.8 = 40$ euros.

(b) Underestimate: both values were rounded down.

(a) The line crosses the $y$-axis at 3.

From $(0, 3)$ to $(2, 6)$: gradient $= \frac{6-3}{2-0} = \frac{3}{2}$

$y = \frac{3}{2}x + 3$

(b) Parallel lines have the same gradient, e.g. $y = 5x + 1$.

Empty jars: $\frac{3}{8} \times 400 = 150$

S : M $= 3 : 4$ and M : L $= 1 : 2 = 4 : 8$, so S : M : L $= 3 : 4 : 8$

Small empty: $\frac{3}{15} \times 150 = 30$

Percentage: $\frac{30}{400} \times 100 = 7.5\%$

Total weight of 8 parcels: $8 \times 2.5 = 20$ kg

Total of the 3 parcels: $3 \times 2 = 6$ kg

The other 5 weigh $20 - 6 = 14$ kg, so their mean is $14 \div 5 = 2.8$ kg

A multiplier of 0.7 means the sale price is 70% of the normal price.

The reduction is $100\% - 70\% = 30\%$, so $R = 30$

Multiply the first equation by 3 and the second by 2:

$15x - 6y = 69$ and $4x - 6y = 36$

Subtract: $11x = 33$, so $x = 3$

Substitute: $5(3) - 2y = 23 \Rightarrow -2y = 8 \Rightarrow y = -4$

A has vertices $(-3, 2), (-2, 2), (-2, 4)$.

Translate by $\binom{6}{-4}$: B is $(3, -2), (4, -2), (4, 0)$.

Rotate B $90^\circ$ clockwise about $(1, 2)$: C is $(-3, 0), (-3, -1), (-1, -1)$.

Comparing A and C, C is A turned $90^\circ$ clockwise; tracing paper or perpendicular bisectors give the centre $(-4, 1)$.

Single transformation: **rotation $90^\circ$ clockwise about $(-4, 1)$**

(i) $y = x^2 - 4$ is a U-shaped parabola crossing the $y$-axis at $-4$: H

(ii) $y = -x^3$ is a cubic that falls from top left to bottom right through the origin: F

(iii) $y = -\frac{5}{x}$ is a reciprocal graph in the second and fourth quadrants: J

(a) Frequency density = frequency ÷ class width:

$20 \div 10 = 2$, $70 \div 20 = 3.5$, $22 \div 5 = 4.4$, $30 \div 15 = 2$, $8 \div 10 = 0.8$

Draw bars of these heights with no gaps; label the vertical axis 'frequency density'.

(b) 20 to 30: half of 70 = 35; 30 to 35: 22; 35 to 40: $\frac{5}{15}$ of 30 = 10

Total $= 35 + 22 + 10 = 67$, so the fraction is $\frac{67}{150}$

$(3x - 1)(2x + 3) = 6x^2 + 9x - 2x - 3 = 6x^2 + 7x - 3$

$(6x^2 + 7x - 3)(x - 5) = 6x^3 - 30x^2 + 7x^2 - 35x - 3x + 15$

$= 6x^3 - 23x^2 - 38x + 15$

Full circumference $= 2\pi \times 6 = 12\pi$

Fraction of the circle: $\frac{5\pi}{12\pi} = \frac{5}{12}$ (angle $= 150^\circ$)

Full area $= \pi \times 6^2 = 36\pi$

Sector area $= \frac{5}{12} \times 36\pi = 15\pi\text{ cm}^2$

There are $n + 1$ sweets, $n$ of them orange.

P(first orange) $= \frac{n}{n+1}$

After eating one, there are $n$ sweets left and $n - 1$ are orange: P(second orange) $= \frac{n-1}{n}$

P(both orange) $= \frac{n}{n+1} \times \frac{n-1}{n} = \frac{n-1}{n+1}$

(a) Multiply top and bottom by $\sqrt{7}$: $\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

(b) $\sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$

$4\sqrt{5} - \sqrt{5} = 3\sqrt{5}$

Let $a = 0.151515\ldots$: $100a - a = 15$, so $a = \frac{15}{99} = \frac{5}{33}$

Let $b = 0.2272727\ldots$: $1000b - 10b = 227.27\ldots - 2.27\ldots = 225$, so $b = \frac{225}{990} = \frac{5}{22}$

$\frac{5}{33} + \frac{5}{22} = \frac{10}{66} + \frac{15}{66} = \frac{25}{66}$, so $m = 25$

Let $BC = 4$ and $CD = 21$, so $BD = 25$ and $AD = BD = 25$.

Triangles ABC and DAB are similar, so $\frac{AB}{BC} = \frac{DB}{AB}$

$AB^2 = 4 \times 25 = 100$, so $AB = 10$

$AB : AD = 10 : 25 = 2 : 5$

$\sqrt[3]{2} = 2^{\frac{1}{3}}$, so $2^x = 2^{n - \frac{1}{3}}$ and $x = n - \frac{1}{3}$

$(\sqrt{2})^5 = (2^{\frac{1}{2}})^5 = 2^{\frac{5}{2}}$, so $y = \frac{5}{2}$

$n - \frac{1}{3} + \frac{5}{2} = 8$

$n = 8 + \frac{1}{3} - \frac{5}{2} = \frac{35}{6}$

Volume: $20wh = 300$, so $wh = 15$

Surface area: $2(20w + 20h + wh) = 370 \Rightarrow 40w + 40h + 30 = 370 \Rightarrow w + h = 8.5$

Substitute $w = \frac{15}{h}$: $\frac{15}{h} + h = 8.5 \Rightarrow 2h^2 - 17h + 30 = 0$

$(2h - 5)(h - 6) = 0$, so $h = 2.5$ or $6$

The width is greater than the height, so $h = 2.5$ cm (and $w = 6$ cm)

(a) Start at 0, rise to a maximum of 1 at $90^\circ$, back to 0 at $180^\circ$, down to $-1$ at $270^\circ$ and back to 0 at $360^\circ$.

(b) $\sin x = -\frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$, the negative values are below the axis:

$x = 180 + 30 = 210$ and $x = 360 - 30 = 330$

Radius$^2 = 5^2 + 10^2 = 125$, so C is $x^2 + y^2 = 125$

At Q, $(-2)^2 + y^2 = 125 \Rightarrow y^2 = 121 \Rightarrow y = \pm 11$

The gradient of L is positive, so Q is $(-2, -11)$ (using $(-2, 11)$ gives a negative gradient).

Gradient $= \frac{10 - (-11)}{5 - (-2)} = \frac{21}{7} = 3$

$10 = 3(5) + k \Rightarrow k = -5$