June 2023 Paper 1 Higher

Multiply both numbers by 100: $8.46 \div 0.15 = 846 \div 15$

$15 \times 56 = 840$, remainder 6, and $6 \div 15 = 0.4$

So the answer is $56.4$

Make improper fractions: $7\frac{3}{8} = \frac{59}{8}$, $2\frac{1}{2} = \frac{20}{8}$

$\frac{59}{8} - \frac{20}{8} = \frac{39}{8} = 4\frac{7}{8}$

A cube has 6 equal faces: $150 \div 6 = 25$ cm²

Edge length: $\sqrt{25} = 5$ cm

Volume: $5 \times 5 \times 5 = 125$ cm³

Plot each frequency above the middle of its class: 2.5, 7.5, 12.5, 17.5, 22.5.

Points: $(2.5, 8), (7.5, 24), (12.5, 13), (17.5, 11), (22.5, 4)$

Join them in order with a ruler. Do not join the last point back to the first.

(a) Odd: 1, 3, 5, 7, 9. Square: 1, 4, 9.

In both: 1, 9. Odd only: 3, 5, 7. Square only: 4. Neither: 2, 6, 8, 10.

(b) $B'$ means not square: $10 - 3 = 7$ numbers, so $P = \frac{7}{10}$

(a) As age increases, weight increases: positive correlation.

(b) Draw a line of best fit. Go across from 5.8 kg to the line, then down to the age axis.

This gives about 3.5 months.

20% of the price is £240.

10% is £120, so 100% is $120 \times 10 = £1200$

Volume = base area × height, so base area $= 1200 \div 40 = 30$ cm²

Pressure $= \frac{90}{30} = 3$ newtons/cm²

The solution is where the two lines meet, at $(6, -2)$.

Check: $2 - 2(-2) = 6$ ✓ and $2(-2) = -4 = 3(6) - 22$ ✓

So $x = 6$, $y = -2$.

Angles in a pentagon add to $(5 - 2) \times 180 = 540^\circ$

The two unknown angles total $540 - 120 - 135 - 110 = 175^\circ$

Let angle $ABC = x$, so angle $AED = 4x$. Then $5x = 175$, so $x = 35$

Angle $AED = 4 \times 35 = 140^\circ$

Numerator: $(6x^5y^3)^2 = 36x^{10}y^6$

Denominator: $3x^2y^7 \times 4xy^{-3} = 12x^3y^4$

Divide: $\frac{36}{12} = 3$, $x^{10-3} = x^7$, $y^{6-4} = y^2$

Answer: $3x^7y^2$

The only outcome without a loss is win then win: $\frac{5}{8} \times \frac{2}{9} = \frac{10}{72}$

So P(at least one loss) $= 1 - \frac{10}{72} = \frac{62}{72} = \frac{31}{36}$

$y = kx$, so $24 = k \times 1.5$ and $k = 16$

When $x = 5$: $y = 16 \times 5 = 80$

(a) $16 = 4^2$, so $\frac{1}{16} = 4^{-2}$

(b) $8^{\frac{5}{3}} = (\sqrt[3]{8})^5 = 2^5 = 32$

$9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$

$32 - 27 = 5$

$L_1$ has gradient 2, so a perpendicular line has gradient $-\frac{1}{2}$

Rearrange $L_2$: $6y = 12 - kx$, so $y = 2 - \frac{k}{6}x$, with gradient $-\frac{k}{6}$

$-\frac{k}{6} = -\frac{1}{2}$ gives $k = 3$

Whole surface area $= 75\pi \div 3 \times 8 = 200\pi$

$4\pi r^2 = 200\pi$, so $r^2 = 50$ and $r = \sqrt{50} = 5\sqrt{2}$

Diameter $= 2 \times 5\sqrt{2} = 10\sqrt{2}$ cm

Multiply both sides by $5x + 3$: $5xy + 3y = 8x - 28$

Collect $x$ terms on one side: $3y + 28 = 8x - 5xy$

Factorise: $3y + 28 = x(8 - 5y)$

Divide: $x = \frac{3y + 28}{8 - 5y}$

Let 1 kg of carrots cost $5k$ and 1 kg of tomatoes cost $9k$.

$7(5k) + 5(9k) = 480$, so $35k + 45k = 80k = 480$ and $k = 6$

Carrots: $5 \times 6 = 30$p. Tomatoes: $9 \times 6 = 54$p

Number of meals $= 5 \times 12 \times x = 60x$

$60x = 420$, so $x = 7$

(a) Let $y = \frac{\sqrt{x} + 2}{5}$. Then $5y = \sqrt{x} + 2$, so $\sqrt{x} = 5y - 2$ and $x = (5y - 2)^2$

$g^{-1}(x) = (5x - 2)^2$

(b) $gf(x) = 3$ means $f(x) = g^{-1}(3) = (15 - 2)^2 = 169$

$3x + 4 = 169$, so $3x = 165$ and $x = 55$

Angle $BAD = 64 \div 2 = 32^\circ$ (angle at the centre is twice the angle at the circumference).

Let $AD$ meet $OB$ at $P$. In triangle $ABP$: angle $APB = 180 - 51 - 32 = 97^\circ$, so angle $OPD = 97^\circ$ (vertically opposite).

In triangle $OPD$: angle $ODA = 180 - 64 - 97 = 19^\circ$

Angle $ODC = 90^\circ$ (tangent is perpendicular to the radius), so angle $ADC = 90 - 19 = 71^\circ$

The angle is $FCA$ in the right-angled triangle $FAC$ (right angle at $A$).

$\sin x = \frac{\text{opp}}{\text{hyp}} = \frac{6.8}{13.6} = \frac{1}{2}$

So $x = 30^\circ$

$\frac{3\sqrt{3}}{4 - \sqrt{3}} \times \frac{4 + \sqrt{3}}{4 + \sqrt{3}} = \frac{12\sqrt{3} + 9}{13}$

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

Common denominator 39: $\frac{36\sqrt{3} + 27}{39} - \frac{26\sqrt{3}}{39} = \frac{10\sqrt{3} + 27}{39}$

$4x^2 - 25 = (2x + 5)(2x - 5) < 0$ gives $-2.5 < x < 2.5$

$12 - 5x - 3x^2 > 0$ is the same as $3x^2 + 5x - 12 < 0$, i.e. $(3x - 4)(x + 3) < 0$, giving $-3 < x < \frac{4}{3}$

Both must hold, so $-2.5 < x < \frac{4}{3}$