June 2018 Paper 1 Higher

(a) Whole numbers: $2 + 1 = 3$. Fractions: $\frac{1}{7} + \frac{1}{4} = \frac{4}{28} + \frac{7}{28} = \frac{11}{28}$

Answer $3\frac{11}{28}$

(b) $1\frac{1}{5} = \frac{6}{5}$, then $\frac{6}{5} \times \frac{4}{3} = \frac{24}{15} = \frac{8}{5} = 1\frac{3}{5}$

Make the flats match: houses : flats $= 7 : 4 = 14 : 8$, so houses : flats : bungalows $= 14 : 8 : 5$.

5 parts $= 50$, so 1 part $= 10$.

Houses $= 14 \times 10 = 140$

$5\text{ kg} = 5000\text{ g}$, so there are $5000 \div 250 = 20$ bags.

Takings: $20 \times £0.65 = £13$

Profit $= £13 - £10 = £3$

Percentage profit $= \frac{3}{10} \times 100 = 30\%$

(a) Round: distance $\approx 3000$ miles, speed $\approx 15$ mph.

Hours $\approx 3000 \div 15 = 200$

Days $\approx 200 \div 8 = 25$

(b) He rides faster, so each day covers more miles and the race takes fewer days.

(a) From the front you see a triangle: base 6 cm, height 4 cm, apex above the middle of the base.

(b) Each sloping face is a triangle with base 6 and slant height 5: $\frac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$

Base $= 6 \times 6 = 36\text{ cm}^2$

Total $= 4 \times 15 + 36 = 96\text{ cm}^2$

Total width $= 38 - 6 = 32$. The four squares sit side by side across, so each side is $32 \div 4 = 8$.

C is two squares to the right of A: $x = 6 + 2 \times 8 = 22$

C is two squares below the top of B: $y = 36 - 2 \times 8 = 20$

So C is $(22, 20)$.

Reflect T in $x = -1$ to get R, then reflect R in $y = -2$ to get S.

Two reflections in perpendicular lines make a half turn about the point where the lines cross.

The lines meet at $(-1, -2)$, so T to S is a rotation of $180^\circ$ about $(-1, -2)$.

$3 + 4 + 5 = 12$ parts, so 1 part $= 72 \div 12 = 6$ cm.

Sides: $18$, $24$ and $30$ cm. The right angle is between 18 and 24.

Area $= \frac{1}{2} \times 18 \times 24 = 216\text{ cm}^2$

(a) $36^{\frac{1}{2}} = \sqrt{36} = 6$

(b) Any non-zero number to the power 0 is 1, so $2^0 = 1$

(c) $27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{27})^2} = \frac{1}{3^2} = \frac{1}{9}$

(a) Draw whiskers from 133 to 145 and from 157 to 170; draw a box from 145 to 157 with a line at 151.

(b) From the minimum to the upper quartile covers three quarters of the data.

$\frac{3}{4} \times 80 = 60$ girls

$OA = OB$ (radii), so triangle $OAB$ is isosceles: angle $BAO = x$ (base angles of an isosceles triangle are equal).

Angle $OBC = 90^\circ$ (a tangent meets a radius at $90^\circ$), so angle $ABC = 90 + x$.

In triangle $ABC$ the angles add to $180^\circ$: angle $ACB = 180 - x - (90 + x) = 90 - 2x$

Let an odd number be $2n + 1$, where $n$ is an integer.

$(2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1$

$4(n^2 + n)$ is a multiple of 4, so the square is 1 more than a multiple of 4.

$\sqrt{8} = 2\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$, so $\sqrt{8} + \sqrt{18} = 5\sqrt{2}$

$\sqrt{5} \times 5\sqrt{2} = 5\sqrt{10}$

So $a = 5$

$y = \frac{K}{d^2}$: $4 = \frac{K}{100}$, so $K = 400$

$d = kx^2$: $24 = k \times 4$, so $k = 6$ and $d = 6x^2$

$y = \frac{400}{(6x^2)^2} = \frac{400}{36x^4} = \frac{100}{9x^4}$

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

(b) Let $a = x^2 + 4$ and $b = x^2 - 2$.

$a + b = 2x^2 + 2$ and $a - b = 6$

$(a + b)(a - b) = 6(2x^2 + 2) = 12x^2 + 12 = 12(x^2 + 1)$

Red or blue: $1 - 0.2 = 0.8$

Red : blue $= 3 : 17$, so 20 parts share 0.8, and 1 part $= 0.8 \div 20 = 0.04$

P(red) $= 3 \times 0.04 = 0.12$

Numerator: $3x^2 - 8x - 3 = (3x + 1)(x - 3)$

Denominator: $2x^2 - 6x = 2x(x - 3)$

Cancel $(x - 3)$: $\frac{3x + 1}{2x}$

Subtracting 2 moves every point of $y = \sin x$ down by 2.

Key points: $(-180, -2)$, $(-90, -3)$, $(0, -2)$, $(90, -1)$, $(180, -2)$

Join them with a smooth wave.

$3x + 2y = 7 \Rightarrow y = -\frac{3}{2}x + \frac{7}{2}$, gradient $-\frac{3}{2}$

PQ is perpendicular, so its gradient is $\frac{2}{3}$

$\frac{b - 4}{a - 3} = \frac{2}{3} \Rightarrow b - 4 = \frac{2}{3}(a - 3) = \frac{2}{3}a - 2$

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

$3n + 2 \le 14 \Rightarrow 3n \le 12 \Rightarrow n \le 4$

$n^2 + 5$ is always positive, so $\frac{6n}{n^2 + 5} > 1 \Rightarrow 6n > n^2 + 5 \Rightarrow n^2 - 6n + 5 < 0$

$(n - 1)(n - 5) < 0 \Rightarrow 1 < n < 5$

Both together: $1 < n \le 4$, so $n = 2, 3, 4$