November 2021 Paper 1 Foundation

Percent means out of 100.

$\frac{3}{10} = \frac{30}{100} = 30\%$

The most negative number is the smallest.

Negatives: $-10 < -7 < -2$, then $0$, then $1 < 8$.

Order: $-10, -7, -2, 0, 1, 8$

Hundredths are the second decimal place.

$\frac{9}{100} = 0.09$

327 lies between 320 and 330.

The units digit is 7, which is 5 or more, so round up: $330$.

$7^2 = 7 \times 7 = 49$

(a) The arrows show exactly one pair of parallel sides, so it is a trapezium.

(b) A solid with two equal circular faces joined by a curved surface is a cylinder.

Share equally means divide.

$£42 \div 3 = £14$

Check each bar against the table.

Blue: 10 and 10. Green: 4 and 4. Brown: the table says 15 but the bar reaches 16.

So the brown bar is one too tall.

Cheese: $2 \times £2.30 = £4.60$

Total: $£1.20 + £0.70 + £4.60 = £6.50$

Change: $£10 - £6.50 = £3.50$

Danny should get £3.50, so he is not correct.

Monday: $5$ °C

Tuesday: $5 - 10 = -5$ °C

Wednesday: $-5 + 3 = -2$ °C

Difference: $5 - (-2) = 7$ °C

Each symbol is 8 games, so a quarter symbol is 2.

(a) Monday has 2 symbols: $2 \times 8 = 16$

(b) Tuesday: $8 + 8 + 6 = 22$. Wednesday: $8 + 2 = 10$. Difference: $22 - 10 = 12$

(c) Thursday: $\frac{1}{4} \times 32 = 8$, so draw 1 symbol. Friday: $32 - 8 = 24$, so draw 3 symbols.

Group 1: $36 : 48$. Divide both by 12 to get $3 : 4$.

Group 2: $\frac{3}{7}$ are boys, so $\frac{4}{7}$ are girls. Boys : girls $= 3 : 4$.

Both are $3 : 4$, so Ann is correct.

(a) $1 \to 2$ is $+1$, $2 \to 4$ is $+2$, so the next step could be $+3$: $4 + 3 = 7$.

(b) The gaps grow by 1: $10 + 5 = 15$, $15 + 6 = 21$, $21 + 7 = 28$, $28 + 8 = 36$.

The 8th term is $36$.

Carrots: $£1.80 \div 3 = £0.60$ per kg

2 kg carrots cost $£1.20$, so 5 kg potatoes cost $£3.45 - £1.20 = £2.25$

Potatoes: $£2.25 \div 5 = £0.45$ per kg

Total: $4 \times £0.60 + 2 \times £0.45 = £2.40 + £0.90 = £3.30$

(a) Multiply each term by 2: $2(a + d) = 2a + 2d$

(b) Both terms contain $y$: $6y^2 - 5y = y(6y - 5)$

(c) $4x - 7 = 37$

$4x = 44$

$x = 11$

(a) $4 \times 0.5 - 2 = 0$, so AB would be 0 cm, which is impossible.

(b) A kite has two pairs of equal sides, so the perimeter is $2(AB + AD) = 2(AB + 3AB) = 8AB$.

$8AB = 64$, so $AB = 8$.

$4x - 2 = 8$, so $4x = 10$ and $x = 2.5$

(a) Batches each ingredient allows (12 biscuits per batch):

Butter: $500 \div 125 = 4$. Flour: $700 \div 200 = 3.5$. Sugar: $250 \div 50 = 5$

Flour runs out first: $3.5 \times 12 = 42$ biscuits.

(b) No. Flour is the limit, so extra sugar makes no difference.

Make a table using $y = 2x - 2$:

| $x$ | −2 | −1 | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| $y$ | −6 | −4 | −2 | 0 | 2 | 4 |

Plot the points and join them with a ruler from $x = -2$ to $x = 3$.

Loss: $£80 - £56 = £24$

Percentage loss: $\frac{24}{80} \times 100 = 30\%$

(a) $367 \times 42 = 14680 + 734 = 15414$. There are 3 decimal places in total, so $3.67 \times 4.2 = 15.414$

(b) Multiply both by 10: $598.4 \div 16$

$598.4 \div 16 = 37.4$ (since $16 \times 37 = 592$ and $6.4 \div 16 = 0.4$)

$\mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}$

In both A and B: 6, 18

A only: 12. B only: 2, 14

In neither: 4, 8, 10, 16

$4\frac{1}{5} = \frac{21}{5} = \frac{63}{15}$ and $2\frac{2}{3} = \frac{8}{3} = \frac{40}{15}$

$\frac{63}{15} - \frac{40}{15} = \frac{23}{15} = 1\frac{8}{15}$

Tamara: $220\,000 \times 0.8 = £176\,000$

Rahim: $160\,000 \times 1.3 = £208\,000$

$208\,000 > 176\,000$, so Rahim's house has the greater value.

Ibrahim minus Matilda: $15 - 7 = 8$ parts $= 24$, so 1 part $= 3$.

Ibrahim minus Rosie: $15 - 4 = 11$ parts.

$11 \times 3 = 33$

Cross-section area: $750 \div 25 = 30$ cm²

Triangle area: $\frac{1}{2} \times 5 \times h = 30$

$2.5h = 30$, so $h = 12$ cm

Cube surface area: $6x^2$

Sphere surface area: $4\pi \times 3^2 = 36\pi$

$6x^2 = 36\pi$

$x^2 = 6\pi$

$x = \sqrt{6\pi}$, so $k = 6$, an integer.

To 1 decimal place, the length could be up to half of 0.1 either side.

$7.2 - 0.05 = 7.15$ and $7.2 + 0.05 = 7.25$

$7.15 \le p < 7.25$

Rewrite as $y = -4x + 3$, which is $y = mx + c$.

(i) Gradient $m = -4$

(ii) The intercept $c = 3$, so it crosses at $(0, 3)$