November 2018 Paper 1 Higher

$3^7 \times 3^{-2} = 3^{7 + (-2)} = 3^5$

$3^5 \div 3^3 = 3^{5 - 3} = 3^2 = 9$

(a) $v^2 = 12^2 + 2 \times (-3) \times 18 = 144 - 108 = 36$

$v = \sqrt{36} = 6$

(b) $v^2 - u^2 = 2as$

$s = \frac{v^2 - u^2}{2a}$

Managers get $0.4 \times 2100 = £840$, so salesmen get $2100 - 840 = £1260$.

Each salesman: $1260 \div 7 = £180$.

Shared equally: $2100 \div 10 = £210$.

25% more than £180 is $180 \times 1.25 = £225$.

$£210 \ne £225$, so the salesman is wrong (it is only about 16.7% more).

(a) One tap alone would take $120 \times 5 = 600$ minutes.

Three taps: $600 \div 3 = 200$ minutes.

(b) Assume each tap gives water at the same rate, and that rate stays constant.

(a) Round 213 to 200 mph.

1 hour $= 60 \times 60 = 3600$ seconds.

Time for 1 mile $= 3600 \div 200 = 18$ seconds.

(b) Overestimate: 200 is slower than the real speed, so the real time is shorter.

Multiply the first equation by 3: $15x + 3y = 63$

Add $x - 3y = 9$: $16x = 72$, so $x = 4.5$

$5(4.5) + y = 21 \Rightarrow y = 21 - 22.5 = -1.5$

Check: $4.5 - 3(-1.5) = 9$ ✓

Quarter circle, radius 20: $\frac{1}{4}\pi \times 20^2 = 100\pi$

Semicircle, radius 10: $\frac{1}{2}\pi \times 10^2 = 50\pi$

Shaded $= 100\pi - 50\pi = 50\pi$

Square $= 20 \times 20 = 400$

$\frac{50\pi}{400} = \frac{\pi}{8}$

(a) $\tan 45° = 1$

(b) The 4 cm side is next to the 60° angle and $x$ is the hypotenuse.

$\cos 60° = \frac{4}{x}$, so $0.5 = \frac{4}{x}$

$x = 4 \div 0.5 = 8$

(a) There are 48 trains. Read across at 12, 24 and 36: LQ = 4, median = 10, UQ = 20. Draw the box from 4 to 20 with a line at 10 and whiskers to 0 and 42.

(b) From Tuesday's box plot: median 9, LQ 6, UQ 17 (IQR 11). Monday: median 10, IQR 16. Monday's delays were longer on average and more spread out.

(c) A box plot only tells us that 12 delays lie between 17 and 33. They could all avoid 25–30, so Mary is not right.

(a) Cancel one factor of $(x-1)$ from top and bottom: $\dfrac{1}{5(x-1)}$

(b) Take out the common factor 2: $2(25 - y^2)$

$25 - y^2$ is a difference of two squares: $2(5+y)(5-y)$

Jack: 125 g for the price of 100 g, so 1 g costs $\frac{100}{125} = 0.8$ of the old price.

Sadia needs the price of the normal box to be 80% of the original.

Reduction $= 100\% - 80\% = 20\%$

Tangent meets radius at 90°, so angle OAB $= 90° - 56° = 34°$.

OA = OB, so angle OBA $= 34°$ and angle ABC $= 34° + 35° = 69°$.

Alternate segment theorem: angle ACB $= 56°$.

Angle CAB $= 180° - 69° - 56° = 55°$

Angle CAO $= 55° - 34° = 21°$

B is smaller and upside down compared with A, so it is an enlargement with a negative scale factor.

A's side of length 6 becomes 2, so the scale factor is $-\frac{1}{3}$.

Join matching corners, e.g. $(-4,-4)$ to $(4,4)$ and $(2,-4)$ to $(2,4)$: the lines cross at $(2, 2)$.

Enlargement, scale factor $-\frac{1}{3}$, centre $(2, 2)$.

(a) $\left(\frac{16}{81}\right)^{\frac{3}{4}}$: fourth root first gives $\frac{2}{3}$, then cube: $\frac{8}{27}$

(b) $\frac{1}{9} = 3^{-2}$, so $a = -2$

$9\sqrt{3} = 3^2 \times 3^{\frac{1}{2}} = 3^{2.5}$, so $b = 2.5$

$\frac{1}{\sqrt{3}} = 3^{-0.5}$, so $c = -0.5$

$a + b + c = -2 + 2.5 - 0.5 = 0$

Area ratio A:B = 4:25, so length ratio A:B $= \sqrt{4}:\sqrt{25} = 2:5$

Volume ratio B:C = 27:64, so length ratio B:C $= \sqrt[3]{27}:\sqrt[3]{64} = 3:4$

Make B match: A:B = 6:15 and B:C = 15:20, so A:B:C = 6:15:20

A:C = 6:20 = 3:10

Let $x = 0.2565656\ldots$

$10x = 2.565656\ldots$ and $1000x = 256.565656\ldots$

Subtract: $990x = 254$

$x = \frac{254}{990} = \frac{127}{495}$

At $(0, -5)$: $b = -5$

At $(5, 0)$: $0 = 25 + 5a - 5$, so $a = -4$

$y = x^2 - 4x - 5 = (x - 2)^2 - 9$

Turning point: $(2, -9)$

(a) $f(x - 2)$ moves the graph 2 units to the right. The roots 0 and 2 become 2 and 4, and the minimum $(1, -1)$ becomes $(3, -1)$.

(b) Graph B is graph A reflected in the y-axis, so its equation is $y = g(-x)$.

(a) $gf(x) = g\big((x+1)^2\big) = 2\big((x+1)^2 - 1\big)$

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

(b) Solve $g(x) = 7$: $2(x - 1) = 7$, so $x - 1 = 3.5$ and $x = 4.5$

$\sqrt{18} + \sqrt{2} = 3\sqrt{2} + \sqrt{2} = 4\sqrt{2}$, so the numerator is $(4\sqrt{2})^2 = 32$

$\sqrt{8} - 2 = 2\sqrt{2} - 2$

$\dfrac{32}{2\sqrt{2} - 2} \times \dfrac{2\sqrt{2} + 2}{2\sqrt{2} + 2} = \dfrac{32(2\sqrt{2} + 2)}{8 - 4} = 8(2\sqrt{2} + 2)$

$= 16\sqrt{2} + 16 = 16(1 + \sqrt{2})$

$\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$, so $\overrightarrow{OP} = \frac{3}{5} \times \frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{3}{10}(\mathbf{a} + \mathbf{b})$

$\overrightarrow{AP} = -\mathbf{a} + \frac{3}{10}(\mathbf{a} + \mathbf{b}) = \frac{1}{10}(-7\mathbf{a} + 3\mathbf{b})$

N is on OB, so $\overrightarrow{AN} = -\mathbf{a} + k\mathbf{b}$ must be a multiple of $-7\mathbf{a} + 3\mathbf{b}$: $k = \frac{3}{7}$

$ON = \frac{3}{7}OB$, so $ON : NB = 3 : 4$

Let $x$ = green pens, so there are $x + 3$ blue and $2x + 3$ in total.

$\dfrac{x(x-1) + (x+3)(x+2)}{(2x+3)(2x+2)} = \dfrac{27}{55}$

$55(2x^2 + 4x + 6) = 27(4x^2 + 10x + 6)$

$2x^2 - 50x + 168 = 0$, i.e. $x^2 - 25x + 84 = 0$

$(x - 21)(x - 4) = 0$, so $x = 21$ or $x = 4$

There are more than 12 pens, so $x = 4$ (11 pens) is rejected: 21 green pens.