November 2024 Paper 2 Foundation

Per cent means out of 100, so $17\% = \frac{17}{100}$.

There are 60 minutes in an hour.

$140 = 60 + 60 + 20$, so it is 2 hours 20 minutes.

Give each number three decimal places: 0.570, 0.507, 0.500, 0.050.

Smallest first: 0.05, 0.5, 0.507, 0.57

Count the sides: there are 6.

A polygon with 6 sides is a hexagon.

Add the two numbers and halve: $\frac{-2 + 8}{2} = \frac{6}{2} = 3$

Red: $52 \div 5 = 10.4$, so enough for 10 bracelets.

White: $80 \div 7 = 11.4...$, so enough for 11.

The red beads run out first, so the greatest number is 10.

(a) Total $= 14+10+10+13+15+9+15+10 = 96$; mean $= 96 \div 8 = 12$ years.

(b) Range $= 15 - 9 = 6$ years.

(c) Two of the 8 children are 15, so $P = \frac{2}{8} = \frac{1}{4}$: mark halfway between 0 and $\frac{1}{2}$.

(a) No. $10^2 = 10 \times 10 = 100$, not 20. He has worked out $10 \times 2$. The squares near 20 are 16 and 25.

(b) $12 \div 4 = 3$, and 3 is odd, so the claim is false.

Red : yellow $= 90 : 150$.

Divide both by 30 (their HCF): $3 : 5$.

One side of a square is $40 \div 4 = 10$ cm.

The cross has 12 outside edges.

Perimeter $= 12 \times 10 = 120$ cm.

(a) $2 \times 3 = 6$ and $x \times y = xy$, giving $6xy$.

(b) $3d + 2d = 5d$ and $-4e + e = -3e$, so $5d - 3e$.

Child tickets: $60\%$ of $180 = 108$.

Child income: $108 \times 8 = £864$.

Adult tickets: $180 - 108 = 72$; adult income: $1944 - 864 = £1080$.

Each adult ticket: $1080 \div 72 = £15$

Angle $BAC = 360 - 310 = 50^\circ$ (angles around a point add to $360^\circ$).

Angle $ACB = 180 - 115 = 65^\circ$ (angles on a straight line add to $180^\circ$).

Angle $ABC = 180 - 50 - 65 = 65^\circ$ (angles in a triangle add to $180^\circ$).

Angles $ACB$ and $ABC$ are both $65^\circ$, so the triangle is isosceles.

(a) The friend's house is the first flat part at 3 km; the skate park is at 5 km. $5 - 3 = 2$ km.

(b) She arrives at 3pm, stays 2 hours: flat line at 5 km until 5pm.

She then walks home in $1\frac{1}{2}$ hours: straight line from 5 km at 5pm to 0 km at 6:30pm.

Real length $= 14 \times 25\,000 = 350\,000$ cm.

There are $100\,000$ cm in 1 km, so $350\,000 \div 100\,000 = 3.5$ km.

Box A: $P(\text{red}) = \frac{10}{40} = 0.25$

Box B: $P(\text{red}) = \frac{7}{25} = 0.28$

$0.28 > 0.25$, so box B gives the greater chance.

(a) $285 \times 0.93 = £265.05$

(b) The bonus is $2100 - 1700 = £400$.

$5\% = 400$, so $1\% = 80$ and $100\% = 80 \times 100 = £8000$.

The triangle is bigger, so it is an enlargement.

Sides of A double in B (e.g. height 2 becomes 4), so the scale factor is 2.

Lines through matching corners, e.g. $(3, 1)$ to $(6, 2)$, meet at the origin, so the centre is $(0, 0)$.

(a) The HCF of $15w^2$ and $5w$ is $5w$.

$15w^2 - 5w = 5w(3w - 1)$

(b) $-2$ is not included: open circle. $4$ is included: filled circle. Join them with a line.

Top: $208.3 - 15.7 = 192.6$

Bottom: $5.694 + 1.8^2 = 5.694 + 3.24 = 8.934$

$\sqrt{192.6 \div 8.934} = \sqrt{21.558...} = 4.643069317...$

(a) Longer ships tend to be wider, so the correlation is positive.

(b) Draw one straight line through the middle of the crosses, with roughly as many points above as below.

(c) Go up from 194 m on the length axis to the line, then across: about 28 m.

Change the London price to francs: $3.50 \times 1.25 = 4.375$ francs for 200 g.

Zurich price for 200 g: $7.20 \div 360 \times 200 = 4.00$ francs.

$4.00 < 4.375$, so Zurich is better value.

The universal set is 1, 3, 5, ..., 19.

17 is in both A and B, so it belongs only in the overlap; Tom also wrote it in A only.

1 is odd but in neither A nor B, so it must go inside the rectangle, outside both circles.

(a) $x = -2$: $4 + 4 - 3 = 5$; $x = 0$: $-3$; $x = 1$: $1 - 2 - 3 = -4$; $x = 3$: $9 - 6 - 3 = 0$; $x = 4$: $16 - 8 - 3 = 5$.

(b) Plot $(-2, 5), (-1, 0), (0, -3), (1, -4), (2, -3), (3, 0), (4, 5)$ and join them with a smooth curve.

First class: $\frac{85 - 76}{76} \times 100 = 11.84\%$

Second class: $\frac{66 - 65}{65} \times 100 = 1.54\%$

$7 \times 1.54 = 10.77\%$, and $11.84 > 10.77$, so Filip is correct.

$AC = BC$: $2x + 11 = 4x - 4$, so $2x = 15$ and $x = 7.5$.

Sides: $AC = BC = 26$, $AB = 20$ (perimeter $72$ ✓).

Height from $C$ to the midpoint of $AB$: $\sqrt{26^2 - 10^2} = \sqrt{576} = 24$.

Area $= \frac{1}{2} \times 20 \times 24 = 240\text{ cm}^2$

$k = \frac{1.25 \times 10^{-12}}{4 \times 10^{-20}}$

$1.25 \div 4 = 0.3125$ and $10^{-12} \div 10^{-20} = 10^{8}$

$k = 0.3125 \times 10^{8} = 3.125 \times 10^{7}$

Volume $= \pi \times 3^2 \times 10 = 90\pi = 282.74...\text{ cm}^3$

Density $= 2250 \div 282.74... = 7.957...$

$= 7.96\text{ g/cm}^3$ (3 s.f.)