November 2019 Paper 2 Higher

Check each point against the table at the class midpoints.

The point at 15 is plotted at 8, but the frequency is 6.

The first and last points are joined; a frequency polygon should not be closed.

The vertical axis also has no 'Frequency' label.

Nearest millimetre means half a millimetre either side.

Lower bound: $128 - 0.5 = 127.5$

Upper bound: $128 + 0.5 = 128.5$

$127.5 \le \text{length} < 128.5$

Before: $240 \div 10 = 24$, so Tom has $3 \times 24 = 72$

After: $240 \div 8 = 30$, so Tom has $3 \times 30 = 90$

Tom buys $90 - 72 = 18$ stamps

(i) $\frac{17}{50} \times 700 = 238$ sports bags

(ii) I assumed the 50 people are representative of all 700. If they are not, 238 could be too many or too few.

(a) $y = x^3$ passes through the origin, is negative for negative $x$ and rises steeply: graph F.

(b) $y = \frac{1}{x}$ never touches the axes, with branches in the top-right and bottom-left: graph D.

First sequence $2n^2 - 1$: 1, 7, 17, 31, 49, 71, ... (increasing)

Second sequence $40 - n^2$: 39, 36, 31, 24, 15, 4, −9, ... (decreasing)

31 appears in both. After that the first sequence keeps rising above 31 and the second keeps falling below it, so no other number can be in both.

$3.42 \div 7.5 = 0.456$

$10^{-7} \div 10^{-6} = 10^{-1}$

$0.456 \times 10^{-1} = 0.0456 = 4.56 \times 10^{-2}$

Rearrange: $n = \frac{720}{d}$

Ali: $720 \div 40 = 18$ workers

Hayley: $720 \div 30 = 24$ workers

Difference: $24 - 18 = 6$ more workers.

Cuboid surface area: $2(6\times8 + 6\times18 + 8\times18) = 2(48+108+144) = 600\text{ cm}^2$

Each cube face: $600 \div 6 = 100$, so the side is $\sqrt{100} = 10$ cm

Cube volume $= 10^3 = 1000\text{ cm}^3$; cuboid volume $= 6 \times 8 \times 18 = 864\text{ cm}^3$

$1000 \ne 864$, so Janet is not correct.

$y = \sqrt{2m - k}$

Square both sides: $y^2 = 2m - k$

Add $k$ and subtract $y^2$: $k = 2m - y^2$

(a) The line inside Megan's box is at 57 g. This is the median, so half her potatoes weigh less than 57 g, not 50 g. She is wrong.

(b) Medians: Megan 57 g, Amy 42 g, so Megan's potatoes are generally heavier.

IQRs: Megan $65 - 39 = 26$ g, Amy $51 - 35 = 16$ g, so Megan's weights are more varied.

In triangle BDC: $BD = 8.6 \sin 40° = 5.528$ cm

E is the midpoint, so $ED = 5.528 \div 2 = 2.764$ cm

In triangle EAD: $\tan(EAD) = \frac{2.764}{4.4}$

$EAD = \tan^{-1}(0.6282) = 32.1°$

After 2 years: $3550 \times 1.026^2 = £3736.9998$

Multiplier for year 3: $3819.21 \div 3736.9998 = 1.02199...$

So $R = 2.199... \approx 2.2$

Total choices $= 19 \times r \times 25$

$19 \times 25 = 475$

$r = 3325 \div 475 = 7$ roof colours.

$(3x + 2)(2x + 1) = 6x^2 + 7x + 2$

$(6x^2 + 7x + 2)(x - 5) = 6x^3 - 30x^2 + 7x^2 - 35x + 2x - 10$

$= 6x^3 - 23x^2 - 33x - 10$

The product is odd only if both cards are odd. There are 5 odd cards out of 9.

$P(\text{both odd}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}$

$P(\text{even}) = 1 - \frac{20}{72} = \frac{52}{72} = \frac{13}{18}$

The radius meets the tangent at 90°, so angle $OAD = 90°$.

Angle $AOD = 180 - 90 - 32 = 58°$, so angle $AOB = 180 - 58 = 122°$.

$OA = OB$ (radii), so angle $OAB = (180 - 122) \div 2 = 29°$.

Angle $CAB = 90 - 29 = 61°$

Frequencies from the bars (width × height): $2.5 \times 1.2 = 3$, $2.5 \times 2 = 5$, $5 \times 2.8 = 14$, $12.5 \times 0.8 = 10$

Known total $= 32$, which is 80% of the trees, so there are $32 \div 0.8 = 40$ trees.

Missing class: 20% of 40 $= 8$ trees.

Frequency density $= 8 \div 2.5 = 3.2$, so draw a bar of height 3.2 from 10 to 12.5.

Radius $= 8.4 \div 2 = 4.2$ cm

Volume $= \frac{1}{2} \times \frac{4}{3}\pi \times 4.2^3 = 155.17...$

$= 155\text{ cm}^3$ (3 s.f.)

Bounds of $c$: $10.85 \le c < 10.95$

Lower bound of $d = \frac{1}{8} \times 10.85^3 = 159.66...$

Upper bound of $d = \frac{1}{8} \times 10.95^3 = 164.11...$

Both round to 160 to 2 significant figures, so $d = 160$.

(a) Distance = area: $\frac{1}{2}(30)(12) + 50(12) + \frac{1}{2}(20)(12) = 180 + 600 + 120 = 900$ m

Half distance $= 450$ m. After 30 s the train has gone 180 m, so it needs $270$ m more at 12 m/s: $270 \div 12 = 22.5$ s.

Time $= 30 + 22.5 = 52.5$ s

(b) First part: $12 \div 30 = 0.4\text{ m/s}^2$. Last part: $-12 \div 20 = -0.6\text{ m/s}^2$. The train slows down more sharply than it speeds up.

(a) April: $1.2 \times 200 - 50 = 190$

May: $1.2 \times 190 - 50 = 178$

June: $1.2 \times 178 - 50 = 163.6$, so about 163 or 164 rabbits.

(b) The population drops each month, so over time it will keep decreasing until there are no rabbits left.

(a) Area $= ab\sin C = (2x - 1)(10 - x) \times \sin 150° = \frac{1}{2}(2x - 1)(10 - x)$

$= \frac{1}{2}(-2x^2 + 21x - 10)$

So $\frac{1}{2}(-2x^2 + 21x - 10) > 15 \Rightarrow -2x^2 + 21x - 10 > 30 \Rightarrow 2x^2 - 21x + 40 < 0$

(b) $(2x - 5)(x - 8) < 0$, critical values $x = 2.5$ and $x = 8$.

The U-shaped graph is below zero between the roots, so $2.5 < x < 8$.

Reflecting in $x = -1$ sends $A(1, 4) \to (-3, 4)$, $B(3, 4) \to (-5, 4)$, $C(3, 2) \to (-5, 2)$, $D(1, 2) \to (-3, 2)$.

To bring $B$ and $D$ back to where they started, rotate 90° clockwise about $(-1, 0)$: $(-5, 4) \to (3, 4)$ and $(-3, 2) \to (1, 2)$.

So B and D are invariant.

L: $2y = 17 - 3x$, so $y = 8.5 - 1.5x$. Gradient $-\frac{3}{2}$ and $B = (0, 8.5)$.

M has gradient $\frac{2}{3}$ through $(0, 2)$: $y = \frac{2}{3}x + 2$.

Solve: $\frac{2}{3}x + 2 = 8.5 - 1.5x \Rightarrow \frac{13}{6}x = 6.5 \Rightarrow x = 3$, so $C = (3, 4)$.

Base $AB = 8.5 - 2 = 6.5$ on the y-axis, height = 3.

Area $= \frac{1}{2} \times 6.5 \times 3 = 9.75$