November 2021 Paper 1 Higher

(a) $367 \times 42 = 14680 + 734 = 15414$. There are 3 decimal places in total, so $3.67 \times 4.2 = 15.414$

(b) Multiply both by 10: $598.4 \div 16$

$598.4 \div 16 = 37.4$ (since $16 \times 37 = 592$ and $6.4 \div 16 = 0.4$)

$\mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}$

In both A and B: 6, 18

A only: 12. B only: 2, 14

In neither: 4, 8, 10, 16

$4\frac{1}{5} = \frac{21}{5} = \frac{63}{15}$ and $2\frac{2}{3} = \frac{8}{3} = \frac{40}{15}$

$\frac{63}{15} - \frac{40}{15} = \frac{23}{15} = 1\frac{8}{15}$

Tamara: $220\,000 \times 0.8 = £176\,000$

Rahim: $160\,000 \times 1.3 = £208\,000$

$208\,000 > 176\,000$, so Rahim's house has the greater value.

Ibrahim minus Matilda: $15 - 7 = 8$ parts $= 24$, so 1 part $= 3$.

Ibrahim minus Rosie: $15 - 4 = 11$ parts.

$11 \times 3 = 33$

Cross-section area: $750 \div 25 = 30$ cm²

Triangle area: $\frac{1}{2} \times 5 \times h = 30$

$2.5h = 30$, so $h = 12$ cm

Cube surface area: $6x^2$

Sphere surface area: $4\pi \times 3^2 = 36\pi$

$6x^2 = 36\pi$

$x^2 = 6\pi$

$x = \sqrt{6\pi}$, so $k = 6$, an integer.

$x^2 - 5x - 24 = 0$

Find two numbers that multiply to $-24$ and add to $-5$: $-8$ and $3$

$(x - 8)(x + 3) = 0$

$x = 8$ or $x = -3$

(a) Anything (except 0) to the power 0 is 1, so $7^0 = 1$

(b) $3^6 \times 3^{-6} = 3^0 = 1$, so the value is $3 \times 1 = 3$

(c) $2^{-4} = \frac{1}{2^4} = \frac{1}{16}$

(d) $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$

(a) Round 5406 to 5400. One square: $5400 \div 6 = 900\text{ cm}^2$

Side $= \sqrt{900} = 30$ cm

(b) Underestimate: 5406 was rounded down to 5400, so each square's area and side were made slightly smaller than they really are.

$6(2w + y) = 7w(3y + 6)$

$12w + 6y = 21wy + 42w$

$6y - 21wy = 30w$

$y(6 - 21w) = 30w$

$y = \frac{30w}{6 - 21w}$ (or $\frac{10w}{2 - 7w}$)

(a) Plot each cumulative frequency at the upper end of its interval: (5, 4), (10, 11), (15, 24), (20, 34), (25, 38), (30, 40). Start at (0, 0) and join with a smooth curve.

(b) The median is at half of 40, which is 20. Read across from 20 to the curve and down: about $13.5$ cm.

In $0.\dot{4}\dot{3}$ the two digits 4 and 3 both repeat.

$10x = 4.3434...$ and $x = 0.4343...$ have different digits after the point, so subtracting them does not get rid of the recurring part.

He should use $100x = 43.\dot{4}\dot{3}$, then $99x = 43$, so $x = \frac{43}{99}$.

Split into a top rectangle $(x + 1)$ by $4$ and a bottom rectangle $(2x + 6)$ by $(x + 11) - 4 = x + 7$.

Top: $4(x + 1) = 4x + 4$

Bottom: $(2x + 6)(x + 7) = 2x^2 + 14x + 6x + 42 = 2x^2 + 20x + 42$

$A = 2x^2 + 24x + 46$

Common denominator $10x$:

$\frac{4x + 3}{2x} = \frac{5(4x + 3)}{10x} = \frac{20x + 15}{10x}$ and $\frac{3}{5} = \frac{6x}{10x}$

Add: $\frac{20x + 15 + 6x}{10x} = \frac{26x + 15}{10x}$

So $a = 26$, $b = 15$, $c = 10$.

One order (red, yellow, yellow): $\frac{3}{8} \times \frac{5}{7} \times \frac{4}{6} = \frac{60}{336}$

The red can come first, second or third, and each order gives the same product.

$3 \times \frac{60}{336} = \frac{180}{336} = \frac{15}{28}$

Draw $x = 2y + 4$, i.e. $y = \frac{1}{2}x - 2$ (through (0, −2) and (4, 0)).

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

$2y + 4 < x$ means below the first line; $x < 3$ means left of $x = 3$; $y < 6 - 3x$ means below the third line.

The two sloping lines cross at about $(2.3, -0.9)$ and $y = 6 - 3x$ meets $x = 3$ at $(3, -3)$. Shade the part of the grid below both sloping lines and left of $x = 3$, and label it R.

Height $= 6 \sin 30^\circ = 6 \times \frac{1}{2} = 3$ cm

Area: $\frac{1}{2}(AB + CD) \times 3 = 66$, so $AB + CD = 44$

$AB : CD = 2 : 3$, so $AB = \frac{2}{5} \times 44 = 17.6$ cm

$\sqrt{12} = 2\sqrt{3}$

Multiply top and bottom by $5 - \sqrt{3}$:

Top: $(8 + 2\sqrt{3})(5 - \sqrt{3}) = 40 - 8\sqrt{3} + 10\sqrt{3} - 6 = 34 + 2\sqrt{3}$

Bottom: $25 - 3 = 22$

$\frac{34 + 2\sqrt{3}}{22} = \frac{17 + \sqrt{3}}{11}$

Rearrange: $y = 2x + 1$. Draw it through (0, 1), (2, 5) and (−3, −5).

Read off where it crosses the circle:

$x \approx 2.1, y \approx 5.1$

$x \approx -2.9, y \approx -4.7$

(a) $f(1) = 3 + 1 = 4$, then $g(4) = \frac{4}{16} = \frac{1}{4}$

(b) $fg(x) = 3\left(\frac{4}{x^2}\right)^2 + 1 = \frac{48}{x^4} + 1$

Let $y = \frac{48}{x^4} + 1$, so $x^4 = \frac{48}{y - 1}$

$h(x) = \sqrt[4]{\frac{48}{x - 1}}$

$y = -3x^2 + 18x + 9 = -3(x^2 - 6x) + 9$

$= -3[(x - 3)^2 - 9] + 9$

$= -3(x - 3)^2 + 27 + 9 = -3(x - 3)^2 + 36$

Turning point: $(3, 36)$