November 2023 Paper 1 Foundation

Range = largest − smallest.

$8 - 2 = 6$

$120 - 89$: take away 90 to get 30, then add back 1.

$120 - 89 = 31$

Multiply the numbers: $3 \times 4 = 12$.

Keep the letter: $12a$.

Place the centre of the protractor on the vertex and line zero up with the vertical arm.

Read from the scale that starts at 0 on that arm: about $40^\circ$.

To find $\frac{1}{5}$ of 300, divide by 5.

$300 \div 5 = 60$

1 litre = 1000 ml, so 3 litres = 3000 ml.

$3000 - 700 = 2300$ ml

(a) Each cm is 5 m, so $3 \times 5 = 15$ m.

(b) $20 \div 5 = 4$ cm.

(a) Count each place: castle 11, farm 7, museum 6. Check: $11 + 7 + 6 = 24$.

(b) The mode is the most common choice: castle.

(c) Label the horizontal axis with the three places and the vertical axis 0 to 12 in equal steps. Draw bars of height 11, 7 and 6 with equal widths.

(i) 9 red out of 22: $\frac{9}{22}$

(ii) Not pink: $22 - 8 = 14$, so $\frac{14}{22} = \frac{7}{11}$

(iii) There are no white counters, so the probability is 0.

60 cookies is $60 \div 20 = 3$ times the recipe.

Peanut butter needed: $3 \times 250 = 750$ g

$750 < 900$, so yes, he has enough (150 g spare).

The triangle has base 2 squares and height 3 squares, with its top above the middle of the base.

Multiply every length by 3: base $2 \times 3 = 6$, height $3 \times 3 = 9$.

Draw a base of 6 squares and put the top 9 squares above its midpoint.

(a)(i) $P = 2 \times 3 + 4 \times 5 = 6 + 20 = 26$

(ii) $38 = 2g + 12$, so $2g = 26$ and $g = 13$

(b) $V = 3 \times (-3) - 2 = -9 - 2 = -11$

Number of scrunchies: $100 \div 5 = 20$

Wool per scrunchy: $300 \div 20 = 15$p

Total cost each: $15 + 8 = 23$p

Make a table of values using $y = 4x - 1$:

| $x$ | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $y$ | −9 | −5 | −1 | 3 | 7 |

Plot the points and join them with one straight ruled line from $x = -2$ to $x = 2$.

Deposit: $25\%$ of £12 000 $= 12000 \div 4 = £3000$

Left to pay: $12000 - 3000 = £9000$

Each payment: $9000 \div 20 = £450$

Pass mark: $10\%$ of 60 is 6, so $70\%$ is $7 \times 6 = 42$ marks.

Shah scored 45, and $45 \ge 42$, so he passes.

Or: $\frac{45}{60} = \frac{3}{4} = 75\%$, which is more than $70\%$.

Keep, change, flip: $\frac{3}{5} \div \frac{1}{6} = \frac{3}{5} \times \frac{6}{1} = \frac{18}{5}$

$18 \div 5 = 3$ remainder 3, so $\frac{18}{5} = 3\frac{3}{5}$

Work out $63 \times 24$: $63 \times 20 = 1260$ and $63 \times 4 = 252$, so $1260 + 252 = 1512$.

There are two decimal places in the question, so $6.3 \times 2.4 = 15.12$

(a)(i) Any non-zero number to the power 0 is 1.

(ii) $5^{-2} = \frac{1}{5^2} = \frac{1}{25}$

(b) Add the powers when multiplying: $2^5 \times 2^4 = 2^9$. Subtract when dividing: $2^9 \div 2^3 = 2^6$.

(a) $156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$, so $156 = 2^2 \times 3 \times 13$

(b) $130 = 2 \times 5 \times 13$

Shared primes: $2 \times 13 = 26$

(a) Total length: $4.2 \times 5 = 21$ cm

Other 4 sticks: $21 - 7 = 14$ cm, so mean $= 14 \div 4 = 3.5$ cm

(b) The other 4 sticks now total $21 - 17 = 4$ cm, so their mean goes down to $4 \div 4 = 1$ cm.

With compasses centred on P, mark two points on AB the same distance either side of P.

Widen the compasses. From each of those two points draw arcs above the line that cross.

Join P to the crossing point with a ruler. This line meets AB at $90^\circ$. Leave all arcs showing.

BA = BD, so the angles at A and D are equal: both are $x$.

Angles in the triangle: $x + x + y = 180$. With $x = 2y$: $5y = 180$, so $y = 36$.

Angles on a straight line: $w = 180 - 36 = 144$

$x + 3x + 1 + 2x - 5 = 44$, so $6x - 4 = 44$, $6x = 48$, $x = 8$

Shelf B: $3 \times 8 + 1 = 25$ books, so one book is $7500 \div 25 = 300$ g

Shelf A: $8 \times 300 = 2400$ g

Density = mass ÷ volume

$27 \div 10 = 2.7$ g/cm³

Round: $5.7 \approx 6$, $8.2 \approx 8$, $0.26 \approx 0.3$

$6 \times 8 = 48$

$48 \div 0.3 = 480 \div 3 = 160$

(a) $(3x + 2)(2x - 5) = 6x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10$

(b) $x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4)$