November 2019 Paper 2 Foundation

The most negative number is the smallest.

Order: $-7, -4, -2, 1, 8$

8375 lies between 8000 and 9000.

The hundreds digit is 3, which is less than 5, so round down to 8000.

Multiply by 100 to turn a decimal into a percentage.

$0.23 \times 100 = 23\%$

Use the square root key: $\sqrt{17.64} = 4.2$

Check: $4.2 \times 4.2 = 17.64$

$6^5 = 6 \times 6 \times 6 \times 6 \times 6$

$= 7776$

Seats: $14 \times 15 = 210$

Tickets sold: $1274 \div 6.50 = 196$

Not sold: $210 - 196 = 14$

Harry has $20 - 7 = 13$ sweets left.

As a fraction of 20: $\frac{13}{20}$

(a) $6 \times 8 = 48$, then $48 - 5 = 43$

(b) After the first box: $17 + 13 = 30$

The second box must turn 30 into 10: $30 \div 3 = 10$ (or $30 - 20 = 10$)

So the missing box is $\div 3$ (or $-20$).

(a) In order: 3, 4, 4, 6, 8, 9. The middle two are 4 and 6, so the median is $\frac{4+6}{2} = 5$

(b) The odd numbers are 9 and 3, so $P(\text{odd}) = \frac{2}{6} = \frac{1}{3}$

(c) Mode 3 means one hidden card is 3. Total $= 5 \times 5 = 25$, so the other card is $25 - 3 - 3 - 8 - 5 = 6$

Minutes parked: $8.40 \div 0.024 = 350$ minutes

$350$ minutes $= 5$ hours $50$ minutes

$10{:}45 + 5\text{ h }50\text{ min} = 16{:}35$

(a) Go up from 3 stones to the line and across: about 19 kg.

(b) 80 kg is off the graph, so use 40 kg: 40 kg is about 6.3 stones.

$80\text{ kg} = 2 \times 6.3 = 12.6$ stones

$\frac{1}{10} = 0.1$ and $\frac{3}{5} = 0.6$

Halfway: $\frac{0.1 + 0.6}{2} = 0.35$ (or $\frac{7}{20}$)

Count the length of each side of the shape in squares, including how far across and up each sloping side goes.

Multiply every horizontal and vertical step by 3.

For example, a side 2 squares long becomes 6 squares long, and a slope of 3 across, 1 up becomes 9 across, 3 up.

20 litre offer: 2 bags = 40 litres for £3.50, so $3.50 \div 40 = £0.0875$ per litre.

40 litre offer: 3 bags = 120 litres for £9, so $9 \div 120 = £0.075$ per litre.

The 40 litre bags are cheaper per litre, so they are better value.

$19.2\text{ m} = 1920\text{ cm}$

Model length: $1920 \div 24 = 80\text{ cm}$

One year: $4500 \times 0.018 = £81$

Three years: $81 \times 3 = £243$

Triangle ABD is isosceles ($AB = BD$), so $\angle ADB = 64°$ (base angles of an isosceles triangle are equal).

$\angle ABD = 180 - 64 - 64 = 52°$ (angles in a triangle add up to $180°$).

$\angle DBC = 180 - 52 = 128°$ (angles on a straight line add up to $180°$).

Triangle DBC is isosceles ($BD = BC$), so $x = \frac{180 - 128}{2} = 26°$

(a) Ben $= n$, Chloe $= 2n$, Dan $= n - 5$

$T = n + 2n + n - 5 = 4n - 5$

(b) An identity is true for every value of the letter. $5m - 3m$ always equals $2m$, so tick $5m - 3m = 2m$.

Batches of 16 each ingredient allows:

Butter $500 \div 175 = 2.86$, sugar $300 \div 75 = 4$, flour $625 \div 250 = 2.5$

Flour runs out first: $2.5 \times 16 = 40$ biscuits

Estimate: $H = 4 \times 110 + 12 = 452$ m

Difference: $452 - 442 = 10$ m

$5\%$ of 442 $= 22.1$ m

$10 < 22.1$, so the difference is less than 5% of the real height ($\frac{10}{442} \times 100 = 2.26\%$).

Check each point against the table at the class midpoints.

The point at 15 is plotted at 8, but the frequency is 6.

The first and last points are joined; a frequency polygon should not be closed.

The vertical axis also has no 'Frequency' label.

Nearest millimetre means half a millimetre either side.

Lower bound: $128 - 0.5 = 127.5$

Upper bound: $128 + 0.5 = 128.5$

$127.5 \le \text{length} < 128.5$

Before: $240 \div 10 = 24$, so Tom has $3 \times 24 = 72$

After: $240 \div 8 = 30$, so Tom has $3 \times 30 = 90$

Tom buys $90 - 72 = 18$ stamps

(i) $\frac{17}{50} \times 700 = 238$ sports bags

(ii) I assumed the 50 people are representative of all 700. If they are not, 238 could be too many or too few.

(a) $y = x^3$ passes through the origin, is negative for negative $x$ and rises steeply: graph F.

(b) $y = \frac{1}{x}$ never touches the axes, with branches in the top-right and bottom-left: graph D.

First sequence $2n^2 - 1$: 1, 7, 17, 31, 49, 71, ... (increasing)

Second sequence $40 - n^2$: 39, 36, 31, 24, 15, 4, -9, ... (decreasing)

The only number in both is 31. After this the first keeps rising above 31 and the second keeps falling below it, so there can be no other.

$\frac{3.42}{7.5} = 0.456$ and $10^{-7} \div 10^{-6} = 10^{-1}$

$0.456 \times 10^{-1} = 0.0456 = 4.56 \times 10^{-2}$

Rearrange: $n = \frac{720}{d}$

Ali: $720 \div 40 = 18$ workers

Hayley: $720 \div 30 = 24$ workers

$24 - 18 = 6$ more

Cuboid surface area: $2(6\times8 + 6\times18 + 8\times18) = 2(48+108+144) = 600\text{ cm}^2$

Each cube face: $600 \div 6 = 100$, so the side is $\sqrt{100} = 10$ cm

Cube volume $= 10^3 = 1000\text{ cm}^3$; cuboid volume $= 6 \times 8 \times 18 = 864\text{ cm}^3$

$1000 \ne 864$, so Janet is not correct.

$2\mathbf{b} = \begin{pmatrix}6\\-2\end{pmatrix}$

$\mathbf{a} - 2\mathbf{b} = \begin{pmatrix}5-6\\2-(-2)\end{pmatrix} = \begin{pmatrix}-1\\4\end{pmatrix}$

Draw an arrow going 1 square left and 4 squares up, and label it.