November 2024 Paper 1 Higher

Multiply both numbers by 10: $8184 \div 12$

$81 \div 12 = 6$ r 9, $98 \div 12 = 8$ r 2, $24 \div 12 = 2$

So the answer is $682$.

(a) $0.10 + 0.30 + 0.05 + 0.25 = 0.70$, so $P(1) + P(2) = 0.30$

$P(2) = 0.30 \div 2 = 0.15$

Estimate $= 0.15 \times 500 = 75$

(b) P(2) is really more than 0.15, so the number of 2s would be more than 75.

(a) $3\frac{1}{2} - 1\frac{1}{6} = 3\frac{3}{6} - 1\frac{1}{6} = 2\frac{2}{6} = 2\frac{1}{3}$

(b) $5\frac{1}{4} = \frac{21}{4}$ and $2\frac{1}{3} = \frac{7}{3}$

$\frac{21}{4} \div \frac{7}{3} = \frac{21}{4} \times \frac{3}{7} = \frac{63}{28} = \frac{9}{4} = 2\frac{1}{4}$

AB is parallel to DC, so angle $ACD = e$ (alternate angles are equal).

Angles in triangle $ACD$ add up to $180°$:

angle $CAD = 180 - 3e - e = 180 - 4e$

(a) Round: $4.96 \approx 5$ miles, $30.4 \approx 30$ mph

Time $= 5 \div 30 = \frac{1}{6}$ hour $= 10$ minutes

(b) The distance was rounded up and the speed rounded down, so the estimate is too big: an overestimate.

Angle sum $= (5 - 2) \times 180 = 540°$

$c + 155 + c + 3c + 2c = 540$

$7c = 385$, so $c = 55$

Angle $a = 55°$

The gradient is change in volume ÷ change in time.

Here $-20 \div 50 = -0.4$, so water leaves the tank at 0.4 litres per second: the rate of change of volume.

Area $= \text{force} \div \text{pressure} = 720 \div 80 = 9\text{ m}^2$

Length $= 9 \div 2 = 4.5$ m

(a) Read across at cf 15, 30 and 45:

LQ $\approx 25$, median $\approx 34$, UQ $\approx 42$. Plot these with the lowest mark 8 and highest 74.

(b) At mark 40 the graph reads a cumulative frequency of about 42.

Passed: $60 - 42 = 18$, and $\frac{18}{60} \times 100 = 30\%$. So Sian is correct.

(a) $25^{\frac{1}{2}} = \sqrt{25} = 5$ and $8^{\frac{1}{3}} = \sqrt[3]{8} = 2$, so the product is $10$.

(b) $\left(\frac{1}{32}\right)^{\frac{1}{5}} = \frac{1}{2}$, then $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$

(a) Expanding $(x + a)(x + b)$ gives $x^2 + (a + b)x + ab$, so $a + b = 5$ and $ab = 6$. Kate has them the wrong way round.

(b) $2m^2 - 2 = 2(m^2 - 1) = 2(m - 1)(m + 1)$

(c) $ax + bx - ay - by = x(a + b) - y(a + b) = (a + b)(x - y)$

Spheres are similar. Volume ratio $A : B = 64 : 125$, so radius ratio $= \sqrt[3]{64} : \sqrt[3]{125} = 4 : 5$

Radius of C is half of B's: $4 : 2.5$ for A : C

Area ratio $= 4^2 : 2.5^2 = 16 : 6.25 = 64 : 25$

Volume A $= (x + 2)(2x - 1)(x - 1) = 2x^3 + x^2 - 5x + 2$

Volume B $= 2x(x + 3)(x - 3) = 2x^3 - 18x$

$2x^3 + x^2 - 5x + 2 = 2x^3 - 18x + 142$

$x^2 + 13x - 140 = 0$, so $(x + 20)(x - 7) = 0$

Lengths must be positive, so $x = 7$.

Sine rule: $\frac{AB}{\sin 45°} = \frac{3\sqrt{2}}{\sin 30°}$

$\sin 45° = \frac{\sqrt{2}}{2}$, $\sin 30° = \frac{1}{2}$

$AB = 3\sqrt{2} \times \frac{\sqrt{2}}{2} \div \frac{1}{2} = 3\sqrt{2} \times \sqrt{2} = 6$ cm

$\overrightarrow{OM} = \frac{1}{4}\mathbf{a}$, $\overrightarrow{ON} = \frac{1}{3}\mathbf{b}$, $\overrightarrow{OX} = \overrightarrow{ZY} = \mathbf{a} - \mathbf{b}$

$\overrightarrow{XN} = -(\mathbf{a} - \mathbf{b}) + \frac{1}{3}\mathbf{b} = \frac{4}{3}\mathbf{b} - \mathbf{a} = \frac{1}{3}(4\mathbf{b} - 3\mathbf{a})$

$\overrightarrow{MN} = -\frac{1}{4}\mathbf{a} + \frac{1}{3}\mathbf{b} = \frac{1}{12}(4\mathbf{b} - 3\mathbf{a})$

$XN : MN = \frac{1}{3} : \frac{1}{12} = 4 : 1$

(a) $\frac{15}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}$

(b) $\sqrt{75} = 5\sqrt{3}$. Multiply top and bottom by $1 - 2\sqrt{3}$:

Top: $(5\sqrt{3} - 2)(1 - 2\sqrt{3}) = 5\sqrt{3} - 30 - 2 + 4\sqrt{3} = 9\sqrt{3} - 32$

Bottom: $1 - 12 = -11$

$\frac{9\sqrt{3} - 32}{-11} = \frac{32 - 9\sqrt{3}}{11}$

Alternate segment theorem: angle $BAC =$ angle $BCD = 40°$

Angle $OAB = 3 \times$ angle $OAC$ and they add to $40°$, so angle $OAC = 10°$ and angle $OAB = 30°$

Triangle $OAB$ is isosceles, so angle $AOB = 180 - 30 - 30 = 120°$

Angle at the centre is twice the angle at the circumference: angle $ACB = 60°$

Angle $ACD = 60 + 40 = 100°$

(a) Let $y = \frac{5x - 3}{4}$. Then $4y = 5x - 3$, $5x = 4y + 3$, $x = \frac{4y + 3}{5}$

So $f^{-1}(x) = \frac{4x + 3}{5}$

(b) $h(5) = 1 - 10 = -9$

$g(-9) = (-9 - 1)^2 = 100$

A must beat B: $0.6$

Route 1: C reaches the final ($0.2$) and A beats C ($0.5$): $0.6 \times 0.2 \times 0.5 = 0.06$

Route 2: D reaches the final ($0.8$) and A beats D ($0.3$): $0.6 \times 0.8 \times 0.3 = 0.144$

Total: $0.06 + 0.144 = 0.204$

$p^2 + 1^2 = 4$, so $p = \sqrt{3}$ (as $p > 0$)

Radius gradient $= \frac{1}{\sqrt{3}}$, so tangent gradient $= -\sqrt{3}$

$y - 1 = -\sqrt{3}(x - \sqrt{3})$, so $y = -\sqrt{3}x + 3 + 1$

$y + \sqrt{3}x = 4$