November 2020 Paper 3 Higher

(a) Add the powers: $n^{3+5} = n^8$

(b) Subtract the powers: $c^{3-2}d^{4-1} = cd^3$

(c) $5x > 14$, so $x > \frac{14}{5} = 2.8$

Cycling: $30 \div 24 = 1.25$ h

Running: $12 \div 8 = 1.5$ h

Total: $2.75$ h $= 2$ hours $45$ minutes

Half of 0.1 is 0.05.

Lower bound $9.4 - 0.05 = 9.35$, upper bound $9.4 + 0.05 = 9.45$

$9.35 \le m < 9.45$

(a) 3 kg covers $5 \times 9 = 45$ m², so 1 kg covers $15$ m².

New lawn: $10 \times 14 = 140$ m² needs $140 \div 15 = 9.33$ kg.

She has $5 \times 2 = 10$ kg, so yes.

(b) Now she has $4 \times 2 + 1 = 9$ kg, less than 9.33 kg, so she does not have enough.

(a) Each spinner has 3 equal sections, one of which is 2: $P(2) = \frac{1}{3}$, $P(\text{not } 2) = \frac{2}{3}$

(b) $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$

(a) The lines cross at $(-2, 4)$, so $x = -2$, $y = 4$.

Check: $5(-2) - 9(4) = -46$ ✓

(b) Read where the curve meets the $x$-axis: $x \approx 0.6$ and $x \approx 3.4$

Boys' total $= 18 \times 16.2 = 291.6$

Girls' total $= 27 \times 16.7 = 450.9$

Mean $= \frac{291.6 + 450.9}{45} = \frac{742.5}{45} = 16.5$ years

Red and yellow together: $1 - 0.32 - 0.20 = 0.48$

Red $= 5 \times$ yellow, so $6 \times$ P(yellow) $= 0.48$, giving P(yellow) $= 0.08$

Yellow counters $= 0.08 \times 300 = 24$

The triangle is isosceles, so split it into two right-angled triangles with base $5$ cm.

Height $h = 5\tan 40^\circ \approx 4.195$ cm

Cross-section $= 10 \times 12 + \frac{1}{2} \times 10 \times 4.195 \approx 140.98$ cm²

Volume $= 140.98 \times 20 \approx 2819.5 \approx 2820$ cm³

(a) $10^5 \times 365 \times 81 = 2\,956\,500\,000 \approx 3.0 \times 10^9$

(b) $90 \div (2 \times 10^{12}) = 45 \times 10^{-12} = 4.5 \times 10^{-11}$ grams

(a) P has vertices $(1,2), (1,5), (2,5)$.

Rotating $180^\circ$ about O gives Q: $(-1,-2), (-1,-5), (-2,-5)$.

Translating by $\binom{5}{-2}$ gives R: $(4,-4), (4,-7), (3,-7)$.

R is upside down relative to P, so it is a rotation of $180^\circ$. The centre is the midpoint of $(1,2)$ and its image $(4,-4)$: $(2.5, -1)$.

(b) The centre of rotation does not move, so $A = (2.5, -1)$.

(a) $\frac{x(x-4) + 2x(x+2)}{(x+2)(x-4)} = \frac{x^2 - 4x + 2x^2 + 4x}{(x+2)(x-4)} = \frac{3x^2}{(x+2)(x-4)}$

(b) $(x-3)(2x+3) = 2x^2 - 3x - 9$

$(2x^2 - 3x - 9)(4x + 5) = 8x^3 + 10x^2 - 12x^2 - 15x - 36x - 45$

$= 8x^3 - 2x^2 - 51x - 45$

(a) Draw $x = 2$ (vertical), $y = x + 3$ (through $(0,3)$ and $(2,5)$) and $2x + 3y = 6$ (through $(0,2)$ and $(3,0)$).

The region is between $x = 0$ and $x = 2$, below $y = x + 3$ and above $2x + 3y = 6$. Label it R.

(b) At $(2, 4)$: $y = 4 \le 4x = 8$, $y = 4 \ge \frac{1}{2}x = 1$, and $x + y = 6 \le 6$. All are satisfied because the point is on the boundary, so Geoffrey is wrong.

$BDEF$ is a cyclic quadrilateral, so angle $DBF = 180^\circ - 100^\circ = 80^\circ$ (opposite angles of a cyclic quadrilateral add to $180^\circ$).

In triangle $BDF$: angle $BFD = 180^\circ - 80^\circ - 40^\circ = 60^\circ$ (angles in a triangle add to $180^\circ$).

Angle $ABD$ = angle $BFD = 60^\circ$ (alternate segment theorem).

Let $x = 0.7333\ldots$

$10x = 7.333\ldots$ and $100x = 73.333\ldots$

$100x - 10x = 90x = 66$

$x = \frac{66}{90} = \frac{11}{15}$

(a) At 30 s the speed is about 9 m/s. Treat the area as a triangle: $\frac{1}{2} \times 30 \times 9 = 135$ m

(b) The curve bulges above the straight line from $(0,0)$ to $(30,9)$, so the triangle misses some area: an underestimate.

(c) Speed ÷ time gives the average acceleration from 0 to 60 s. The acceleration at 60 s is the gradient of the tangent to the curve at that point.

Frequencies = width × density:

$0.8 \times 10 = 8$, $0.4 \times 50 = 20$, $0.4 \times 100 = 40$, $0.2 \times 60 = 12$ (total 80)

Mean $= \frac{6.8 \times 8 + 7.4 \times 20 + 7.8 \times 40 + 8.1 \times 12}{80} = \frac{611.6}{80} = 7.645$ m

Lower bound of $AH = 11.25$ cm

For a cube of edge $a$, the space diagonal satisfies $a^2 + a^2 + a^2 = AH^2$

$3a^2 = 11.25^2$, so $a = \frac{11.25}{\sqrt{3}} \approx 6.495$ cm

A regular hexagon of side $x$ is six equilateral triangles.

One triangle $= \frac{1}{2}x^2\sin 60^\circ = \frac{\sqrt{3}}{4}x^2$, so hexagon $= \frac{3\sqrt{3}}{2}x^2$

The enlarged hexagon has side $px$, area $\frac{3\sqrt{3}}{2}p^2x^2$

Shaded $= \frac{3\sqrt{3}}{2}p^2x^2 - \frac{3\sqrt{3}}{2}x^2 = \frac{3\sqrt{3}}{2}(p^2 - 1)x^2$

Every number in the list is a power of 98.

Each one divides the largest, $98^{91}$, so the lowest common multiple is $98^{91}$.

(a) $5c - c = 4d - d$, so $4c = 3d$, giving $c : d = 3 : 4$

(b) $6x^2 - 7xy - 20y^2 = 0$

$(2x - 5y)(3x + 4y) = 0$

Since $x, y > 0$, $3x + 4y \ne 0$, so $2x = 5y$ and $x : y = 5 : 2$