November 2021 Paper 3 Higher

(a) The point far from the others is at $(100, 18)$.

(b) Draw a line of best fit through the main trend and read up from 370: about 13.8 mg/m³.

(c) No. One outlier does not remove the pattern; the other points still show positive correlation.

Total parts: $9 + 2 + 1 = 12$

Cheese: $6000 \div 12 \times 2 = 1000$ g

$1000 \div 175 = 5.714...$ lots of 175 g

$5.714... \times 2.25 = £12.857...$, so about £12.86

(a) Move the digits 5 places: $4.5 \times 10^5 = 450\,000$

(b) $0.007 = 7 \times 10^{-3}$

(c) $4200 + 530 = 4730 = 4.73 \times 10^3$

Company A: 1 min 40 s $= \frac{5}{3}$ min, so the rate is $8 \div \frac{5}{3} = 4.8$ litres per minute.

Time A $= 2400 \div 4.8 = 500$ min.

Company B: $2.2 \times 4.54 = 9.988$ litres per minute.

Time B $= 2400 \div 9.988 = 240.29$ min.

Difference $= 500 - 240.29 = 259.7$, about 260 minutes.

Each term is the sum of the two before, so the fifth term is $3a + 5a = 8a$.

$a + 2a + 3a + 5a + 8a = 19a$

$19a = 228$, so $a = 12$.

(a) The probabilities add to 1: $1 - 0.05 - 0.15 = 0.8$ for green and pink together.

Green is 0.2 more than pink: $2p + 0.2 = 0.8$, so pink $= 0.3$ and green $= 0.5$.

(b) $0.15$ of the counters are blue: $18 \div 0.15 = 120$.

Quarter circle: $\frac{1}{4} \times \pi \times 8^2 = 50.265...$ cm²

Triangle $OPR$: $\frac{1}{2} \times 8 \times 8 = 32$ cm²

Segment $= 50.265... - 32 = 18.265...$

So the area is $18.3$ cm² (3 s.f.).

(a) $y = kx$: a straight line starting at the origin and going up to the right.

(b) $y = \frac{k}{x}$: a curve that is high near the $y$-axis, falls as $x$ grows and gets closer to the $x$-axis, never touching either axis.

(a) Total work $= 12 \times 5 = 60$ person-hours. $60 \div 15 = 4$ hours.

(b) If the people work at different rates, it could take longer (if some are slower) or less time (if some are faster).

$EB = \sqrt{6^2 - 2^2} = \sqrt{32} = 5.656...$

Triangles $DEB$ and $ACB$ are similar with scale factor $\frac{AB}{DB} = \frac{9 + 6}{6} = 2.5$

$CB = 2.5 \times \sqrt{32} = 14.142...$, so $14.14$ cm

Rounding to 1 decimal place means $x$ is within $0.05$ of $6.4$.

Lower bound $= 6.4 - 0.05 = 6.35$, upper bound $= 6.4 + 0.05 = 6.45$

$6.35 \le x < 6.45$

Powers of $x$: $x^{6 \times \frac{1}{n}} = x^3$, so $\frac{6}{n} = 3$ and $n = 2$.

Numbers: $a^{\frac{1}{2}} = 7$, so $a = 7^2 = 49$.

Let each rectangle be $l$ long and $w$ wide.

Across from A to B: $11 - 3 = 8$, so $2l - w = 8$

Up from A to B: $20 - 4 = 16$, so $2l + w = 16$

Adding: $4l = 24$, $l = 6$, then $w = 4$.

C is one length right and one width up from A: $(3 + 6, 4 + 4) = (9, 8)$

Fran and Gary share 5 parts; Olivia and Jessica share 10 parts.

Olivia and Jessica have half as many, so Fran and Gary's total must be 20 parts: multiply $2 : 3$ by 4 to get $8 : 12$.

Olivia and Jessica: $9 : 1$ (total 10, which is half of 20).

$w : x : y : z = 8 : 12 : 9 : 1$

Draw a tangent to the curve at $t = 12$ (where the volume is about 17 litres).

Pick two points far apart on the tangent, e.g. the tangent drops about 28 litres over 30 seconds.

Gradient $\approx 28 \div 30 \approx 0.93$, so water flows out at about $0.9$ litres per second.

From the line: $y = 5x - 4$

Set equal: $x^2 + 3x - 3 = 5x - 4$

$x^2 - 2x + 1 = 0$

$(x - 1)^2 = 0$, so $x = 1$ is the only solution.

One value of $x$ means only one point in common, $(1, 1)$.

$x = ky^2$ and $y = cz^3$, so $x = k(cz^3)^2 = mz^6$

Substitute $z = 2$, $x = 32$: $32 = m \times 64$, so $m = \frac{1}{2}$

$x = \frac{1}{2}z^6$

$\vec{OB} = \mathbf{a} + \mathbf{b}$, so $\vec{DB} = \frac{3}{5}(\mathbf{a} + \mathbf{b})$

$\vec{BC} = -\mathbf{b} - \mathbf{a} + 3\mathbf{b} = 2\mathbf{b} - \mathbf{a}$, so $\vec{BE} = \frac{1}{5}(2\mathbf{b} - \mathbf{a})$

$\vec{DE} = \frac{3}{5}\mathbf{a} + \frac{3}{5}\mathbf{b} + \frac{2}{5}\mathbf{b} - \frac{1}{5}\mathbf{a} = \frac{2}{5}\mathbf{a} + \mathbf{b}$

2017 to 2019 is two years, so $P_{2019} = k^2 P_{2017}$.

$3610 = 4000k^2$

$k^2 = 0.9025$, so $k = \sqrt{0.9025} = 0.95$

The only outcomes that fail are all heads or all tails.

$P(\text{all heads}) = \left(\frac{1}{2}\right)^n$ and $P(\text{all tails}) = \left(\frac{1}{2}\right)^n$

So the probability $= 1 - 2\left(\frac{1}{2}\right)^n = 1 - \left(\frac{1}{2}\right)^{n-1}$

(a) Read the speeds: at $t = 1, 2, 3, 4$ they are $4, 6, 7.2, 7.8$.

$\frac{1}{2}(4 + 6) + \frac{1}{2}(6 + 7.2) + \frac{1}{2}(7.2 + 7.8) = 5 + 6.6 + 7.5 = 19.1$

(b) Area under a speed–time graph is the distance travelled, so this is the distance in metres between 1 and 4 seconds.

$\frac{6x^3}{9x^2 - 144} \times \frac{3(x - 4)}{2x^4}$

$9x^2 - 144 = 9(x - 4)(x + 4)$

$= \frac{18x^3(x - 4)}{18x^4(x - 4)(x + 4)} = \frac{1}{x(x + 4)}$, so $r = 4$

Let angle $ADB = y$, so angle $ADC = 180^\circ - y$.

Sine rule in $ABD$: $\frac{AB}{\sin y} = \frac{BD}{\sin x}$, so $\frac{AB}{BD} = \frac{\sin y}{\sin x}$

Sine rule in $ADC$: $\frac{AC}{DC} = \frac{\sin(180^\circ - y)}{\sin x}$

Since $\sin(180^\circ - y) = \sin y$, both ratios equal $\frac{\sin y}{\sin x}$, so $\frac{AB}{BD} = \frac{AC}{DC}$.