November 2024 Paper 3 Higher

Midpoints: 130, 150, 170, 190, 210

Plot (130, 12), (150, 28), (170, 30), (190, 22), (210, 8).

Join neighbouring points with straight lines. Do not join the last point back to the first.

(a) Move the decimal point 5 places right: $3.402 \times 10^5 = 340\,200$

(b) Move the point 1 place right to get 8.026, so the power is $-1$: $8.026 \times 10^{-1}$

Put the compass point on B and draw an arc crossing BA and BC.

From each crossing point, draw equal arcs that meet inside the angle.

Rule a line from B through where the arcs meet. Leave all arcs visible.

(a) Not winning: $1 - 0.3 = 0.7$. Each pair of branches is 0.3 (wins) and 0.7 (does not win).

(b) $0.7 \times 0.7 = 0.49$

One part: $240 \div 8 = 30$, so 90 pens and 150 pencils.

Cost: $90 \times 9 + 150 \times 6 = 1710$p

Income: $90 \times 11 + 150 \times 10 = 2490$p

Profit: $2490 - 1710 = 780$p

$\frac{780}{1710} \times 100 = 45.6\%$

School A has 23 scores, so the median is the 12th: 57.

Range: $79 - 30 = 49$

Median: A (57) is higher than B (56), so A did slightly better on average.

Range: A (49) is higher than B (47), so A's scores were more spread out.

The calculator display began 10.2, so the number was cut off (truncated), not rounded.

It could be anything from 10.2 up to but not including 10.3:

$10.2 \le t < 10.3$

AB is opposite the $62°$ angle and BC (7.6 cm) is adjacent.

$\tan 62° = \frac{AB}{7.6}$

$AB = 7.6 \times \tan 62° = 14.29... \approx 14.3$ cm

(a) Multiply the numbers: $2 \times 7 = 14$. Add the powers: $x^{3+2} = x^5$, $y^{5+1} = y^6$.

So $14x^5y^6$

(b) Multiply the powers: $2 \times -3 = -6$, so $m^{-6}$

After a 15% reduction the sale price is 85% of the normal price.

Normal price $= 46.75 \div 0.85 = £55$

Arc length $= 34.3 - 2 \times 4.7 = 24.9$ m

Arc $= \frac{\theta}{360} \times 2\pi r$, so

$\theta = \frac{24.9}{2\pi \times 4.7} \times 360 = 303.5... \approx 304^\circ$

(a) $4500 \times 1.024 = 4608$

$4608 \times 1.018^2 = 4775.38...$, so £4775.38

(b) Multiplier over 2 years $= 4107 \div 7500 = 0.5476$

Yearly multiplier $= \sqrt{0.5476} = 0.74$

$1 - 0.74 = 0.26$, so $x = 26$

Triangles ABE and ACD are similar with scale factor $\frac{4.2}{6.3} = \frac{2}{3}$.

$AB = \frac{2}{3} \times 10.8 = 7.2$, so $BC = 10.8 - 7.2 = 3.6$ cm

Area $= \frac{4.2 + 6.3}{2} \times 3.6 = 18.9$ cm²

Let $x = 0.4626262...$

$1000x = 462.626262...$

$10x = 4.626262...$

Subtract: $990x = 458$

$x = \frac{458}{990} = \frac{229}{495}$

$t(5 - 2p) = 2(2p - 3)$

$5t - 2pt = 4p - 6$

$5t + 6 = 4p + 2pt$

$5t + 6 = p(4 + 2t)$

$p = \frac{5t + 6}{4 + 2t}$

Gradient of OB $= \frac{-4}{2} = -2$

DE is parallel to OB through $(12, -6.5)$: $y = -2x + c$, $-6.5 = -24 + c$, so $y = -2x + 17.5$

OD is perpendicular to OB through O: gradient $\frac{1}{2}$, so $y = \frac{1}{2}x$

$\frac{1}{2}x = -2x + 17.5 \Rightarrow 2.5x = 17.5 \Rightarrow x = 7$, $y = 3.5$

$D = (7, 3.5)$

Angle A is between 14.6 and 7.9 cm:

$\cos A = \frac{14.6^2 + 7.9^2 - 18.2^2}{2 \times 14.6 \times 7.9} = -0.2413...$, so $A = 103.97^\circ$

Area $= \frac{1}{2} \times 14.6 \times 7.9 \times \sin 103.97^\circ = 55.97 \approx 56.0$ cm²

(a) In the second sample $\frac{11}{55} = \frac{1}{5}$ are tagged.

So the 40 tagged frogs are about $\frac{1}{5}$ of all frogs:

$N = \frac{40 \times 55}{11} = 200$

(b) For example: the number of frogs stayed the same (none born, died, arrived or left).

Let height $= h$, so $r = \frac{3}{4}h$.

$l = \sqrt{h^2 + \frac{9}{16}h^2} = \frac{5}{4}h$

Total area $= \pi r l + \pi r^2 = \pi\left(\frac{15}{16}h^2 + \frac{9}{16}h^2\right) = \frac{24}{16}\pi h^2$

$\frac{3}{2}h^2 = 54 \Rightarrow h^2 = 36 \Rightarrow h = 6$ cm

$y = 7 - 2x$, so $(7 - 2x)^2 = 3x^2 + 4$

$49 - 28x + 4x^2 = 3x^2 + 4$

$x^2 - 28x + 45 = 0$

$x = 14 \pm \sqrt{151}$, so $x = 26.3$ or $x = 1.71$

Then $y = 7 - 2x$: $y = -45.6$ or $y = 3.58$

(a) Each point $(x, y)$ moves to $(x, -y)$: reflect the graph in the x-axis.

(b) P is the minimum of $\sin x$ at $(270, -1)$.

$\sin(x + 180) + 4$ moves the graph 180 left and 4 up.

$R = (270 - 180,\ -1 + 4) = (90, 3)$

Split 0–30 s into strips of width 10. Read velocities: 0, 54, 76, 92 m/s.

Triangle: $\frac{1}{2} \times 10 \times 54 = 270$

Trapezium: $\frac{1}{2} \times 10 \times (54 + 76) = 650$

Trapezium: $\frac{1}{2} \times 10 \times (76 + 92) = 840$

Distance $\approx 270 + 650 + 840 = 1760$ m

$y = x^2 - 6px - 7 = (x - 3p)^2 - 9p^2 - 7$

Turning point (minimum): $(3p,\ -9p^2 - 7)$

When $x = 0$, $y = -7$, so the y-intercept is $(0, -7)$.

Since $p > 0$, $3p > 0$ and $-9p^2 - 7 < 0$: the minimum is in the fourth quadrant. Sketch a U-shape through $(0, -7)$ with its lowest point there.