June 2019 Paper 1 Higher

(a) Probabilities add to 1: $1 - 0.2 = 0.8$

Red and yellow are equal: $0.8 \div 2 = 0.4$ each.

(b) 0.2 of the cubes are blue, so $12 \div 0.2 = 60$ cubes.

(a) $60 \div 15 = 4$, so she needs $4 \times 50 = 200$ g of sugar.

Flour: $3 \times 200 = 600$ g

(b) Butter: $2 \times 200 = 400$ g

$400 \div 250 = 1.6$, so she must buy 2 packs.

$72 = 2^3 \times 3^2$ and $90 = 2 \times 3^2 \times 5$

Shared primes: $2 \times 3^2 = 18$

The plan is a circle, so the solid is round when seen from above.

Both elevations are rectangles, so the solid is a cylinder.

From the grid: diameter 4 cm (radius 2 cm) and height 5 cm.

Reflect A in the $x$-axis: $(3, 2) \to (3, -2)$, $(5, 2) \to (5, -2)$, $(3, 5) \to (3, -5)$

B has matching corners at $(-3, -3)$, $(-1, -3)$, $(-3, -6)$.

Move from $(3, -2)$ to $(-3, -3)$: 6 left and 1 down.

So $c = -6$ and $d = -1$

Pens in the ratio $7 \times 2 : 3 \times 5 : 4 \times 6 = 14 : 15 : 24$

Total parts $= 53$, and $212 \div 53 = 4$

Green pens $= 24 \times 4 = 96$

Area of PQRS: $PQ \times 10 = 45$, so $PQ = 4.5$ cm, so $BC = 4.5$ cm.

Perimeter of ABCD: $2 \times AB + 2 \times 4.5 = 26$

$2 \times AB = 17$, so $AB = 8.5$ cm

(a) $\sqrt{63.5 \times 101.7} \approx \sqrt{64 \times 100} = \sqrt{6400} = 80$

(b) $0.23 = 2.3 \div 10$, so $(0.23)^6 = 148 \div 10^6 = 0.000148$

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

$3\frac{1}{2} = \frac{7}{2}$ and $1\frac{3}{5} = \frac{8}{5}$

$\frac{7}{2} \times \frac{8}{5} = \frac{56}{10} = \frac{28}{5} = 5\frac{3}{5}$

The solution is where the two lines cross.

Reading the grid, they cross at about $x = 2.25$ and $y = -1.35$.

(Solving exactly gives $x = \frac{9}{4}$ and $y = -\frac{4}{3}$.)

(a) There are 23 ages, so the median is the 12th value: 59.

The lower quartile is the 6th value: 53. The upper quartile is the 18th value: 66.

(b) Coach A's median, 59, is below coach B's, 70, so yes.

(c) Range: A is $79 - 41 = 38$, B is $85 - 42 = 43$. IQR: A is $66 - 53 = 13$, B is $73 - 54 = 19$. Coach A varies less, so no.

$Q = 1.5P$ and $R = 1.5Q = 1.5^2 P = 2.25P$

$\frac{P}{R} = \frac{1}{2.25} = \frac{4}{9}$

$n^2 - n = n(n - 1)$

$n$ and $n - 1$ are consecutive integers, so one of them is even.

An even number times any integer is even, so $n^2 - n$ is never odd.

$\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2}$

Radius $= 6 \div 2 = 3$ cm.

Cone: $\frac{1}{3} \pi \times 3^2 \times 10 = 30\pi$

Hemisphere: $\frac{1}{2} \times \frac{4}{3} \pi \times 3^3 = 18\pi$

Total $= 48\pi$, so $k = 48$.

(a) Each dial has 5 choices: $5 \times 5 \times 5 = 125$

(b) The first dial has 5 choices, the second 4 left, the third 3 left: $5 \times 4 \times 3 = 60$

$\frac{x^2}{3x + 5} = \frac{1}{2}$, so $2x^2 = 3x + 5$

$2x^2 - 3x - 5 = 0$

$(2x - 5)(x + 1) = 0$

$x = 2.5$ or $x = -1$

(a) $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$, so $\sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$

(b) $(\sqrt{3})^7 = (\sqrt{3})^6 \times \sqrt{3} = 27\sqrt{3}$

$\frac{1}{27\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{81}$

(i) Halve $-6$ to get $(x - 3)^2 = x^2 - 6x + 9$.

$x^2 - 6x + 1 = (x - 3)^2 - 9 + 1 = (x - 3)^2 - 8$, so $a = 3$ and $b = 8$.

(ii) The turning point is $(a, -b) = (3, -8)$.

$h = \frac{k}{p}$: $10 = \frac{k}{6}$, so $k = 60$

$p = K\sqrt{t}$: $6 = K \times 12$, so $K = \frac{1}{2}$

$h = \frac{60}{\frac{1}{2}\sqrt{t}} = \frac{120}{\sqrt{t}}$

(a) $y = 3x - 1 \Rightarrow x = \frac{y + 1}{3}$, so $f^{-1}(x) = \frac{x + 1}{3}$

(b) $fg(x) = 3(x^2 + 4) - 1 = 3x^2 + 11$

$gf(x) = (3x - 1)^2 + 4 = 9x^2 - 6x + 5$

$3x^2 + 11 = 2(9x^2 - 6x + 5) = 18x^2 - 12x + 10$

$0 = 15x^2 - 12x - 1$

Green is $\frac{3}{7}$, so $g = 3x$ and $r = 4x$ for some $x$.

After adding: $\frac{3x + 3}{7x + 5} = \frac{6}{13}$

$39x + 39 = 42x + 30$, so $3x = 9$ and $x = 3$.

Red $= 4 \times 3 = 12$, green $= 3 \times 3 = 9$.