November 2020 Paper 2 Higher

(a) $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$

(b) $60 = 2^2 \times 3 \times 5$ and $84 = 2^2 \times 3 \times 7$

LCM $= 2^2 \times 3 \times 5 \times 7 = 420$

(a) Even: 2, 4, 6, 8, 10. Factors of 10: 1, 2, 5, 10.

Both: 2, 10. Only $A$: 4, 6, 8. Only $B$: 1, 5. Neither: 3, 7, 9.

(b) $A \cap B$ has 2 of the 10 numbers, so the probability is $\frac{2}{10} = \frac{1}{5}$.

One part: $3000 \div 5 = 600$ tins.

Small boxes: $2 \times 600 = 1200$ tins, $1200 \div 6 = 200$ boxes.

Large boxes: $3 \times 600 = 1800$ tins, $1800 \div 20 = 90$ boxes.

$\frac{90}{290} \times 100 = 31.0\%$, which is more than 30%, so Carlo is not correct.

(a) $x = -2$: $5 - (-8) = 13$; $x = 0$: $5$; $x = 1$: $5 - 1 = 4$; $x = 2$: $5 - 8 = -3$

(b) Plot the five points and join them with a smooth curve, not straight lines.

$x$ is opposite $34^\circ$ and 178 is the hypotenuse, so use sine.

$x = 178 \times \sin 34^\circ = 99.536\ldots$

$x = 99.5$ mm (1 d.p.)

$2\mathbf{a} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$, $3\mathbf{b} = \begin{pmatrix} 15 \\ -6 \end{pmatrix}$

$2\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} 6 - 15 \\ 8 - (-6) \end{pmatrix} = \begin{pmatrix} -9 \\ 14 \end{pmatrix}$

$CB^2 = 9^2 - 6^2 = 81 - 36 = 45$, so $CB = \sqrt{45}$

Quarter circle area $= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \pi \times 45 = 35.34\ldots$

$= 35.3\text{ m}^2$ (3 s.f.)

(a) £551 is 95% of the normal price: $551 \div 0.95 = 580$, so £580.

(b) After year 1: $6000 \times 1.024 = 6144$.

Years 2 and 3: $6144 \times 1.017^2 = 6354.67$.

The investment is worth £6354.67.

Check each value against the table.

Least height 142 and lower quartile 154 are plotted correctly. Greatest height $= 142 + 40 = 182$, which is correct.

Median should be 162, but it is drawn at 161.

Upper quartile $= 154 + 17 = 171$, but it is drawn at 172.

So she should move the median line to 162 and the right end of the box to 171.

(a) Anything (non-zero) to the power 0 is 1.

(b) Cancel one factor of $(x - 4)$ from top and bottom: $\frac{8(x-4)}{(x-4)^2} = \frac{8}{x-4}$.

(c) Cube each part: $3^3 = 27$, $(n^4)^3 = n^{12}$, $(w^2)^3 = w^6$, giving $27n^{12}w^6$.

Starter and main combinations: $5 \times 8 = 40$.

$240 \div 40 = 6$, which is a whole number.

So Jack could be correct, if there are 6 desserts.

(a) The line starts at 27 litres when $d = 0$ and reaches 0 litres at $d = 300$.

Gradient $= \frac{0 - 27}{300 - 0} = -0.09$.

(b) The car uses 0.09 litres of petrol for each kilometre it travels (the rate of fuel consumption).

Angle $A = 180° - 34° - 26° = 120°$, which is opposite the 23.8 cm side.

$\frac{AB}{\sin 34°} = \frac{23.8}{\sin 120°}$

$AB = \frac{23.8 \times \sin 34°}{\sin 120°} = 15.367...$

$AB = 15.4$ cm (1 d.p.)

Let square A have side $x$ cm, so square B has side $x + 4$.

$(x + 4)^2 = x^2 + 70$

$x^2 + 8x + 16 = x^2 + 70$, so $8x = 54$ and $x = 6.75$.

Side of B $= 10.75$ cm, so area $= 10.75^2 = 115.5625$.

Area of B $= 116$ cm² (3 s.f.)

B is upside down and on the opposite side, and its sides are longer: A's short sides are 2, B's are 3. So it is an enlargement with scale factor $-\frac{3}{2} = -1.5$.

Join matching corners, e.g. $(5, 5)$ to $(-5, -5)$ and $(3, 3)$ to $(-2, -2)$. The lines cross at $(1, 1)$.

Enlargement, scale factor $-1.5$, centre $(1, 1)$.

First differences: 11, 17, 23, 29. Second differences: 6, so the $n^2$ term is $\frac{6}{2}n^2 = 3n^2$.

Subtract $3n^2$ (3, 12, 27, 48, 75): leaves 7, 9, 11, 13, 15, which is $2n + 5$.

nth term $= 3n^2 + 2n + 5$.

For $y = (x + a)^2 + b$ the turning point is $(-a, b)$.

Here $a = 12$ and $b = -7$, so the turning point is $(-12, -7)$.

Whole cone: $r = 10$, $l = 25$, so curved surface area $= \pi \times 10 \times 25 = 250\pi$.

The grey part is a smaller similar cone with slant height $25 - 10 = 15$.

Its radius is $10 \times \frac{15}{25} = 6$, so its area is $\pi \times 6 \times 15 = 90\pi$.

Unpainted area $= 250\pi - 90\pi = 160\pi$ cm².

$1040 = K \times 1200 + 20$, so $K = \frac{1020}{1200} = 0.85$.

At 09 17: $0.85 \times 1040 + 20 = 904$.

At 09 18: $0.85 \times 904 + 20 = 788.4$ m.

(a) $\frac{n}{n + 8} = \frac{7}{10}$ gives $10n = 7n + 56$, so $3n = 56$ and $n = 18.67$, not a whole number. So the probability cannot be $\frac{7}{10}$.

(b) $\frac{n}{n+8} \times \frac{n-1}{n+7} = \frac{3}{5}$

$5n(n - 1) = 3(n + 8)(n + 7)$

$5n^2 - 5n = 3n^2 + 45n + 168$, so $n^2 - 25n - 84 = 0$.

$(n - 28)(n + 3) = 0$, so $n = 28$ (it cannot be negative).

(a) $f(-x)$ reflects the curve in the $y$-axis: each point $(a, b)$ moves to $(-a, b)$. So $(-4, 2)$, $(-3, 0)$, $(-1, 2)$, $(0, 0)$ become $(4, 2)$, $(3, 0)$, $(1, 2)$, $(0, 0)$.

(b) The curve moves 3 to the right, so replace $x$ with $x - 3$:

$y = 5 + 2(x - 3) - (x - 3)^2$

which simplifies to $y = -x^2 + 8x - 10$.

Tangent gradient $= \frac{10 - 0}{0 - (-20)} = \frac{1}{2}$, so the tangent is $y = \frac{1}{2}x + 10$.

The radius to the point of contact is perpendicular, gradient $-2$, through the origin: $y = -2x$.

$\frac{1}{2}x + 10 = -2x$ gives $x = -4$, $y = 8$.

$r^2 = (-4)^2 + 8^2 = 80$, so the circle is $x^2 + y^2 = 80$.