November 2024 Paper 1 Foundation

Per cent means out of 100, so divide by 100.

$25 \div 100 = 0.25$

The even numbers are the 2 times table.

The 10th one is $2 \times 10 = 20$.

There are 10 mm in 1 cm.

$15 \times 10 = 150$ mm

Multiples of 9: 9, 18, 27, 36, …

The only one between 20 and 30 is $27$.

$500 + 145 = 645$

Hire: $10 \times £45 = £450$

Total cost: $£450 + £30 = £480$

Paid: $10 \times £50 = £500$

Change: $£500 - £480 = £20$

(a) Place the ruler's zero on P and read off at Q: about $9.5$ cm.

(b) The angle is smaller than $90°$, so it is acute.

(c) Put the protractor's centre on the vertex and read from the zero line: about $32°$.

(a) A is 3 right and 1 up: $(3, 1)$.

(b) Go 3 right and 4 down and mark B there.

(c) C is $(-3, 3)$. Midpoint $= \left(\frac{3 + (-3)}{2}, \frac{1 + 3}{2}\right) = (0, 2)$

(a) Tennis is 12 and football is 17. Cricket $= 40 - 12 - 17 = 11$, so draw the bar up to 11.

(b) $\frac{12}{40} = \frac{3}{10}$

(c) There are $5 + 4 = 9$ parts and 5 are large: $\frac{5}{9}$

(d) Divide both sides by 4: $5 \div 4 = 1.25$, so $1.25 : 1$

(a) Each vertex keeps its $x$ value and moves to the same distance on the other side of $y = 1$:

$(1,2)\to(1,0)$, $(1,3)\to(1,-1)$, $(3,4)\to(3,-2)$, $(4,2)\to(4,0)$

(b) Every point on the line has $y$-coordinate 1, so it is $y = 1$.

(a) Children $= 500 - 280 = 220$. Children action $= 150 - 100 = 50$.

Thriller total $= 500 - 200 - 150 = 150$, so adult thriller $= 150 - 80 = 70$.

Adult comedy $= 280 - 70 - 100 = 110$, child comedy $= 200 - 110 = 90$.

(b) 100 of the 500 people are adults who like action: $\frac{100}{500} = \frac{1}{5}$

Multiplication comes before subtraction (BIDMAS).

So $5 - 3 \times 4 = 5 - 12 = -7$, not $2 \times 4 = 8$.

(a) $11 + 5 = 16$, then $16 \div 2 = 8$

(b) Work backwards from the output: $28 \div 4 = 7$

The first box turns 10 into 7, so it is $-3$ (or $\times 0.7$).

(a) Base area $= 10 \times 4 = 40$ cm², so height $= 200 \div 40 = 5$ cm

(b) Faces: $6 \times 5 = 30$, $6 \times 3 = 18$, $5 \times 3 = 15$

Each appears twice: $2(30 + 18 + 15) = 126$ cm²

(a) The highest common factor of 6 and 15 is 3: $3(2a + 5)$

(b) $4(3y + 1) = 28$

$3y + 1 = 7$

$3y = 6$, so $y = 2$

The difference is 2 teas: $8.50 - 4.50 = £4$, so 1 tea costs $£2$.

1 coffee $= 4.50 - 2 = £2.50$

$4 \times 2 + 3 \times 2.50 = 8 + 7.50 = £15.50$

Total parts $= 4 + 3 + 1 = 8$, so one part $= 24 \div 8 = 3$ litres.

Blue $= 12$, yellow $= 9$, white $= 3$.

She has only 7 litres of yellow, so no, she does not have enough.

Multiply both numbers by 10: $8184 \div 12$

$81 \div 12 = 6$ r 9, $98 \div 12 = 8$ r 2, $24 \div 12 = 2$

So the answer is $682$.

(a) $0.10 + 0.30 + 0.05 + 0.25 = 0.70$, so $P(1) + P(2) = 0.30$

$P(2) = 0.30 \div 2 = 0.15$

Estimate $= 0.15 \times 500 = 75$

(b) P(2) is really more than 0.15, so the number of 2s would be more than 75.

(a) $3\frac{1}{2} - 1\frac{1}{6} = 3\frac{3}{6} - 1\frac{1}{6} = 2\frac{2}{6} = 2\frac{1}{3}$

(b) $5\frac{1}{4} = \frac{21}{4}$ and $2\frac{1}{3} = \frac{7}{3}$

$\frac{21}{4} \div \frac{7}{3} = \frac{21}{4} \times \frac{3}{7} = \frac{63}{28} = \frac{9}{4} = 2\frac{1}{4}$

AB is parallel to DC, so angle $ACD = e$ (alternate angles are equal).

Angles in triangle $ACD$ add up to $180°$:

angle $CAD = 180 - 3e - e = 180 - 4e$

(a) Round: $4.96 \approx 5$ miles, $30.4 \approx 30$ mph

Time $= 5 \div 30 = \frac{1}{6}$ hour $= 10$ minutes

(b) The distance was rounded up and the speed rounded down, so the estimate is too big: an overestimate.

Angle sum $= (5 - 2) \times 180 = 540°$

$c + 155 + c + 3c + 2c = 540$

$7c = 385$, so $c = 55$

Angle $a = 55°$

The gradient is change in volume ÷ change in time.

Here $-20 \div 50 = -0.4$, so water leaves the tank at 0.4 litres per second: the rate of change of volume.

$y = 3w - 10$

Add 10: $y + 10 = 3w$

Divide by 3: $w = \frac{y + 10}{3}$

Find two numbers that multiply to $-15$ and add to $-2$: $-5$ and $3$.

$(x - 5)(x + 3) = 0$

$x = 5$ or $x = -3$