June 2023 Paper 3 Higher

(a) Multiply the powers: $(m^2)^3 = m^{2 \times 3} = m^6$

(b) Add the powers: $x^{5+8} = x^{13}$

(c) $4p \times p^2 = 4p^3$ and $4p \times 3p = 12p^2$, so $4p^3 + 12p^2$

(a) $0.68 \times 800 = 544$ people

$544 \times 2 = 1088$ cups

$1088 \times 10.6 = 11532.8$, so $11533$ g

(b) More people drink coffee, so he needs more than in (a).

Angle $CFG = 180 - 125 = 55°$ (angles on a straight line add to 180°).

Angle $ACD = 55°$ (corresponding angles are equal).

Angle $ADC = 180 - 110 = 70°$ (angles on a straight line).

Angle $CAD = 180 - 55 - 70 = 55°$ (angles in a triangle add to 180°).

Two angles are equal, so triangle $ACD$ is isosceles.

Total work: $5 \times 14 = 70$ pump-hours.

With 4 pumps: $70 \div 4 = 17.5$ hours

(a) Take the lowest power of each shared prime: $3^2 \times 7 = 63$

(b) Take the highest power of every prime: $2^2 \times 3^4 \times 7^2 = 4 \times 81 \times 49 = 15876$

Seconds: $67\,205\,600 \div 11.9 = 5\,647\,529.4...$

Seconds in a day: $24 \times 60 \times 60 = 86\,400$

$5\,647\,529.4 \div 86\,400 = 65.36...$, so $65$ days

(a) The lowest point of the curve is at $(1, -3)$.

(b) The roots are where the curve crosses the $x$-axis: about $x = -0.7$ or $x = 2.7$.

Mass = density × volume

$9 \times 72 = 648$ g

$80\%$ of $70\% = 0.8 \times 0.7 = 0.56$

So $56\%$ of the people asked.

In triangle $ABC$: $AC = 6.8 \sin 41° = 4.461...$ cm

In triangle $ACD$: $\tan 55° = \dfrac{AC}{DC}$

$DC = \dfrac{4.461...}{\tan 55°} = 3.12...$, so $DC = 3.1$ cm

(a) Plot the five values on the scale: whisker from 169 to 174, box from 174 to 180 with a line at 177, whisker from 180 to 186.

(b) Teenagers: median 169, IQR $171 - 165 = 6$, range $178 - 157.5 = 20.5$

Adults: median 177, IQR $180 - 174 = 6$, range $186 - 169 = 17$

The adults' median height is greater, so adults are generally taller. Both groups have the same IQR, so the middle half of heights is equally spread.

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

$(x^2 + 2x - 3)(x - 5) = x^3 - 5x^2 + 2x^2 - 10x - 3x + 15$

$= x^3 - 3x^2 - 13x + 15$, so $a = 1$, $b = -3$, $c = -13$, $d = 15$, all integers.

The next triangular number is $\dfrac{(n+1)(n+2)}{2}$.

Sum: $\dfrac{n(n+1) + (n+1)(n+2)}{2} = \dfrac{(n+1)(n + n + 2)}{2}$

$= \dfrac{(n+1)(2n+2)}{2} = (n+1)(n+1) = (n+1)^2$

This is a square number for every $n$.

Cosine rule: $OB^2 = 9^2 + 6^2 - 2 \times 9 \times 6 \cos 35° = 28.53...$

$OB = 5.341...$ cm

Sector area $= \dfrac{80}{360} \times \pi \times 28.53... = 19.918...$

$= 19.9$ cm² (3 s.f.)

(a) $a^2 - b^2 = (a - b)(a + b)$

(b) $2^{40} = (2^{20})^2$, so $2^{40} - 1 = (2^{20} - 1)(2^{20} + 1)$

$2^{20}$ is even, so $2^{20} - 1$ and $2^{20} + 1$ are both odd, and they differ by 2, so they are consecutive odd numbers ($1\,048\,575$ and $1\,048\,577$).

Triangle T has vertices $(-6, -2)$, $(-7, -5)$, $(-5, -5)$.

From the centre $(-2, -2)$ each vertex is $(-4, 0)$, $(-5, -3)$, $(-3, -3)$ away.

Multiply by $-2$: $(8, 0)$, $(10, 6)$, $(6, 6)$.

Add back to the centre: $(6, -2)$, $(8, 4)$, $(4, 4)$. The image is upside down and twice the size.

(a) Draw a straight line touching the curve at $t = 2.5$ (where the distance is about 21 m).

Pick two points on the tangent, e.g. $(1.4, 0)$ and $(4, 48)$.

Gradient $= \dfrac{48 - 0}{4 - 1.4} \approx 18$

(b) Distance ÷ time is speed, so the gradient is the speed at 2.5 seconds, about 18 m/s.

Small cone radius: $\dfrac{6}{10} \times 3 = 1.8$ cm

Curved area of frustum $= \pi \times 3 \times 10 - \pi \times 1.8 \times 6 = 19.2\pi = 60.3...$

Bottom circle: $\pi \times 3^2 = 9\pi$; top circle: $\pi \times 1.8^2 = 3.24\pi$

Total $= (19.2 + 9 + 3.24)\pi = 31.44\pi = 98.77...$, so $98.8$ cm²

When $x = 0$, $y = 3^0 = 1$.

Sana's curve starts at $(0, 0)$, so it is wrong there: it should start at $(0, 1)$.

Let $x = 0.1232323...$

$10x = 1.232323...$ and $1000x = 123.232323...$

$1000x - 10x = 122$, so $990x = 122$

$x = \dfrac{122}{990} = \dfrac{61}{495}$

Multiply every term by $(x + 4)(2 - 2x)$:

$(2 - 2x) + 3(x + 4) = (x + 4)(2 - 2x)$

$x + 14 = -2x^2 - 6x + 8$

$2x^2 + 7x + 6 = 0$

$(2x + 3)(x + 2) = 0$, so $x = -\frac{3}{2}$ or $x = -2$

Parallel means $a\begin{pmatrix}2\\6\end{pmatrix} + b\begin{pmatrix}8\\2\end{pmatrix} = k\begin{pmatrix}13\\6\end{pmatrix}$

$2a + 8b = 13k$ and $6a + 2b = 6k$, so $k = a + \frac{b}{3}$

$2a + 8b = 13a + \frac{13b}{3}$

$\frac{11b}{3} = 11a$, so $b = 3a$

Gradient of radius $OP = \dfrac{4}{-3} = -\dfrac{4}{3}$

Tangent is perpendicular, so its gradient is $\dfrac{3}{4}$.

Tangent: $y - 4 = \frac{3}{4}(x + 3)$, so $y = \frac{3}{4}x + \frac{25}{4}$

At $x = -8$: $y = -6 + \frac{25}{4} = \frac{1}{4} \neq 0$

So the tangent does not pass through $(-8, 0)$: Alex is wrong.

$y = 3x$, so green $= 3x - x - 5 = 2x - 5$

P(both pink) $= \dfrac{x}{3x} \times \dfrac{x-1}{3x-1}$

P(both blue) $= \dfrac{5}{3x} \times \dfrac{4}{3x-1}$

P(both green) $= \dfrac{2x-5}{3x} \times \dfrac{2x-6}{3x-1}$

Sum $= \dfrac{x^2 - x + 20 + 4x^2 - 22x + 30}{3x(3x-1)} = \dfrac{5x^2 - 23x + 50}{9x^2 - 3x}$