November 2022 Paper 3 Higher

$p = 3a - 9$

Add 9: $p + 9 = 3a$

Divide by 3: $a = \frac{p + 9}{3}$

The ratio 3 : 5 has $3 + 5 = 8$ parts.

One part $= 120 \div 8 = 15$, giving $45$ and $75$.

Rob divided by 3 and by 5 separately instead of by 8.

German: $200 - 104 - 70 = 26$, so boys choosing German $= 26 - 18 = 8$

Boys: 90, of which 60 chose French and 8 chose German.

Boys choosing Spanish $= 90 - 60 - 8 = 22$

One tank: $\pi \times 40^2 \times 160 \approx 804\,248\text{ cm}^3$

Four tanks: $\approx 3\,216\,991\text{ cm}^3 \approx 3217$ litres

Fertiliser is 1 part in 101: $3217 \div 101 \approx 31.85$ litres

$31.85 < 32$, so yes, she has enough.

AC = 5 matches DF = 20, so the scale factor is $20 \div 5 = 4$.

(a) EF matches BC: $4 \times 4 = 16$ cm

(b) AB matches DE: $22 \div 4 = 5.5$ cm

(a) Not winning the sports quiz: $1 - 0.3 = 0.7$. Not winning the music quiz: $1 - 0.35 = 0.65$ on both pairs of branches.

(b) $P(\text{both}) = 0.3 \times 0.35 = 0.105$

(a) $1\text{ m}^3 = 100^3 = 1\,000\,000\text{ cm}^3$, so $8000 \div 1\,000\,000 = 0.008\text{ m}^3$

(b) $180$ km $= 180\,000$ m, and 1 hour $= 3600$ s

$180\,000 \div 3600 = 50$ m/s

Total for everyone: $50 \times 167.6 = 8380$ cm

Total for men: $20 \times 182 = 3640$ cm

Women: $(8380 - 3640) \div 30 = 4740 \div 30 = 158$ cm

(a) Move the decimal point 4 places left: 0.000675

(b) $2.56 \times 10^6 \times 4.12 \times 10^{-3} = 10\,547.2$

$10\,547.2 \div (1.6 \times 10^{-2}) = 659\,200 = 6.592 \times 10^5$

Subtracting a bracket changes the sign of every term inside it:

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

So the answer should be $x^2 + 3x + 5 - x^2 + 2x + 4 = 5x + 9$. Peter kept $-2x - 4$.

Possible x: 4, 5, 6, 7. Possible y: 5, 6, 7, 8, 9

$y = 14 - x$: x = 4 gives y = 10 (not allowed); x = 5, 6, 7 give y = 9, 8, 7 (all allowed)

So x = 5, 6 or 7

Martin wrote down the first two digits, so he truncated rather than rounded.

Any number from 1.2 up to (but not including) 1.3 would show as 1.2

$1.2 \le P < 1.3$

Check the axes: the time axis has no label (it should say time in seconds).

Check the points: cumulative frequency is plotted at the upper end of each class (10, 20, 30, …), but these are plotted at 5, 15, 25, …

(a) $3^4 = 81$, $(x^5)^4 = x^{20}$, $(y^6)^4 = y^{24}$, so $81x^{20}y^{24}$

(b) $(x+2)(x-3) = x^2 - x - 6$

$(x^2 - x - 6)(x + 4) = x^3 + 4x^2 - x^2 - 4x - 6x - 24 = x^3 + 3x^2 - 10x - 24$

Guppy and angel: $7 \times 5 = 35$

Tetra and angel: $13 \times 5 = 65$

Guppy, tetra and angel: $7 \times 13 \times 5 = 455$

Total: $35 + 65 + 455 = 555$

Let angle $BCD = x$, so angle $EAB = 2x$.

Opposite angles of a cyclic quadrilateral add to $180^\circ$, so angle $BDE = 180 - 2x$, and angle $BDC = 180 - (180 - 2x) = 2x$ (angles on a straight line).

Triangle $BCD$: $60 + x + 2x = 180$, so $3x = 120$ and $x = 40^\circ$

$A : B = 2 : 3$ and $C : D = 3 : 4$ (75% = $\frac{3}{4}$)

Let $A = 2k$, $B = 3k$, $C = 3m$, $D = 4m$. Then $5k = 3 \times 7m$, so $5k = 21m$: take $k = 21$, $m = 5$

$A : B : C : D = 42 : 63 : 15 : 20$

Check: $105 = 3 \times 35$

Reflecting in two parallel lines gives a translation.

The distance moved is twice the gap between the lines: $2 \times (6 - 2) = 8$ units to the right.

Translation by $\begin{pmatrix} 8 \\ 0 \end{pmatrix}$

P(red or green) $= 1 - 0.4 = 0.6$

Green is $\frac{8}{15}$ of that: $0.6 \times \frac{8}{15} = 0.32$

Expected greens: $0.32 \times 50 = 16$

$AE = DE$ (given).

Triangle $ADE$ is isosceles, so angle $EAC$ = angle $EDB$ (base angles).

Using the ratio, $AC = 1 + 2 = 3$ parts and $DB = 1 + 2 = 3$ parts, so $AC = DB$.

Two sides and the included angle match, so the triangles are congruent (SAS).

$y = 4(x^2 - 14x)$

$x^2 - 14x = (x - 7)^2 - 49$

$y = 4\left((x - 7)^2 - 49\right) = 4(x - 7)^2 - 196$

The minimum is when $x - 7 = 0$, so the turning point is $(7, -196)$

Common denominator: $(x - 5)(x + 5) = x^2 - 25$

Numerator: $(2x + 3)(x + 5) + (x - 4)(x - 5) - 3(x^2 - 25)$

$= 2x^2 + 13x + 15 + x^2 - 9x + 20 - 3x^2 + 75 = 4x + 110$

So $a = 4$ and $b = 110$

(a) $f(x + 2)$ moves the graph 2 units to the left. The roots $-1, 1, 2$ become $-3, -1, 0$ and the peak at $(0, 2)$ moves to $(-2, 2)$.

(b) Graph B is graph A flipped in the x-axis, so each y-value changes sign: $y = -g(x)$

$\overrightarrow{CE} = \overrightarrow{CD} + \overrightarrow{DE} = 2\mathbf{a} - \mathbf{b}$, so $\overrightarrow{EP} = 2\mathbf{a} - \mathbf{b}$

$\overrightarrow{CF} = \overrightarrow{CE} + \overrightarrow{EF} = 2\mathbf{a} - \mathbf{b} - \mathbf{a} = \mathbf{a} - \mathbf{b}$

$\overrightarrow{DP} = \overrightarrow{DE} + \overrightarrow{EP} = -\mathbf{b} + 2\mathbf{a} - \mathbf{b} = 2\mathbf{a} - 2\mathbf{b}$

$\overrightarrow{DP} = 2(\mathbf{a} - \mathbf{b}) = 2\overrightarrow{CF}$, so CF is parallel to DP

Top volume: $92.8 \div 2.9 = 32\text{ cm}^3$

Total mass: $92.8 + 972.8 = 1065.6$ g

Total volume: $1065.6 \div 4.7 \approx 226.72\text{ cm}^3$

Percentage: $32 \div 226.72 \times 100 \approx 14.1\%$

Interior angle: $180 - 360 \div 7 \approx 128.57^\circ$

Area: $\frac{1}{2}s^2 \sin 128.57^\circ = 30$, so $s = \sqrt{60 \div \sin 128.57^\circ} \approx 8.760$ cm

Cosine rule: $GB^2 = 2s^2 - 2s^2 \cos 128.57^\circ \approx 249.2$

$GB \approx 15.8$ cm