November 2020 Paper 2 Foundation

The 7 is in the hundredths column, so $0.37$ means 37 hundredths.

$0.37 = \frac{37}{100}$

Look at the hundreds digit: 29 381. It is 3, which is less than 5, so round down.

$29\,381 \approx 29\,000$

Combine the numbers in front of $e$: $3 - 1 + 4 = 6$.

So the expression simplifies to $6e$.

$\frac{1}{4} = 1 \div 4 = 0.25$

$0.25 \times 100 = 25\%$

Cube numbers are $1, 8, 27, 64, \ldots$

$3^3 = 3 \times 3 \times 3 = 27$, which is in the list.

105 minutes = 1 hour 45 minutes.

Film ends: 14 30 + 1 h 45 min = 16 15

At the bus stop: 16 15 + 20 min = 16 35

16 35 is before 16 45, so yes, with 10 minutes to spare.

Check the chart against good practice:

- The vertical axis has no label (e.g. 'Mark').
- The scale jumps from 0 straight to 71, so the bar heights mislead.
- Tom's bar is twice as wide as the other bars.

Any two of these earn the marks.

(a) Angles on a straight line add up to $180^\circ$, so $x = 180 - 150 = 30^\circ$.

(b) Angles around a point make $360^\circ$. Here $90 + 280 = 370^\circ$, which is too big (the reflex angle should be $270^\circ$).

(a) Read across from 40 km on the scale: about 25 miles.

(b) $40 \div 10 = 4$, then $4 \times 6 = 24$ miles.

(c) 24 is close to 25, but the rule gives a slightly smaller answer.

(a) Divide both sides by 3: $m = 36 \div 3 = 12$.

(b) $7 - x = 3$, so $x = 7 - 3 = 4$.

Volume = length × width × height

$= 15 \times 10 \times 4 = 600$

Answer: $600\text{ cm}^3$

Primes between 20 and 30: 23 and 29 (21, 25, 27 are odd but not prime).

Pair each letter with each prime:

L23, L29, U23, U29

$4275 \div 28 = 152.67\ldots$, so 152 full bags.

$152 \times 28 = 4256$ sweets used.

$4275 - 4256 = 19$ sweets left.

Total goals: $50 + 45 + 25 = 120$

Each goal is $360 \div 120 = 3^\circ$.

City $50 \times 3 = 150^\circ$, Rovers $45 \times 3 = 135^\circ$, United $25 \times 3 = 75^\circ$

Draw and label each sector with a protractor.

(a) $T = 3(5) + 4(-7) = 15 - 28 = -13$

(b) $38 = 3(6) + 4y = 18 + 4y$

$4y = 20$, so $y = 5$

Total marks: $60 + 90 = 150$

Pass mark: $\frac{2}{3} \times 150 = 100$

Paper 1: $70\% \times 60 = 42$

Needed on Paper 2: $100 - 42 = 58$

(a) 8 litres is $8 \div 2 = 4$ lots of 2 litres.

Oranges needed: $4 \times 30 = 120$

Boxes: $120 \div 24 = 5$

(b) $1260 : 280$. Divide both by 140: $9 : 2$

Triangle B is triangle A turned upside down, so it is a half-turn.

A vertex at $(-5, 5)$ maps to $(3, -5)$. The centre is the midpoint: $\left(\frac{-5+3}{2}, \frac{5-5}{2}\right) = (-1, 0)$.

Rotation $180^\circ$ about $(-1, 0)$.

Let Rytis get 1 part. Linda gets $3 \times 1 = 3$ parts. Adam gets double Linda: 6 parts.

Total: $1 + 3 + 6 = 10$ parts.

Linda gets $\frac{3}{10}$.

One pencil: $£1.80 \div 12 = £0.15$

One ratio part: $£0.15 \div 3 = £0.05$

One pen: $7 \times £0.05 = £0.35$

Five pens: $5 \times £0.35 = £1.75$

(a) $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$

(b) $60 = 2^2 \times 3 \times 5$ and $84 = 2^2 \times 3 \times 7$

LCM $= 2^2 \times 3 \times 5 \times 7 = 420$

(a) Even: 2, 4, 6, 8, 10. Factors of 10: 1, 2, 5, 10.

Both: 2, 10. Only $A$: 4, 6, 8. Only $B$: 1, 5. Neither: 3, 7, 9.

(b) $A \cap B$ has 2 of the 10 numbers, so the probability is $\frac{2}{10} = \frac{1}{5}$.

One part: $3000 \div 5 = 600$ tins.

Small boxes: $2 \times 600 = 1200$ tins, $1200 \div 6 = 200$ boxes.

Large boxes: $3 \times 600 = 1800$ tins, $1800 \div 20 = 90$ boxes.

$\frac{90}{290} \times 100 = 31.0\%$, which is more than 30%, so Carlo is not correct.

(a) $x = -2$: $5 - (-8) = 13$; $x = 0$: $5$; $x = 1$: $5 - 1 = 4$; $x = 2$: $5 - 8 = -3$

(b) Plot the five points and join them with a smooth curve, not straight lines.

$x$ is opposite $34^\circ$ and 178 is the hypotenuse, so use sine.

$x = 178 \times \sin 34^\circ = 99.536\ldots$

$x = 99.5$ mm (1 d.p.)

$2\mathbf{a} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$, $3\mathbf{b} = \begin{pmatrix} 15 \\ -6 \end{pmatrix}$

$2\mathbf{a} - 3\mathbf{b} = \begin{pmatrix} 6 - 15 \\ 8 - (-6) \end{pmatrix} = \begin{pmatrix} -9 \\ 14 \end{pmatrix}$

$CB^2 = 9^2 - 6^2 = 81 - 36 = 45$, so $CB = \sqrt{45}$

Quarter circle area $= \frac{1}{4} \pi r^2 = \frac{1}{4} \times \pi \times 45 = 35.34\ldots$

$= 35.3\text{ m}^2$ (3 s.f.)

Exterior angles of any polygon add up to $360^\circ$.

$360 \div 15 = 24$ sides.

In $y = mx + c$ the gradient is $m$.

For $y = 2x + 3$, the gradient is 2.