June 2024 Paper 2 Foundation

Negative numbers are smaller than positive ones, and $-3$ is further below zero than $-1$.

So the order is $-3,\ -1,\ 2,\ 4,\ 7$.

There are $1000$ ml in a litre.

$5000 \div 1000 = 5$ litres.

$\frac{31}{100}$ means 31 hundredths.

$\frac{31}{100} = 0.31$

Multiples of 7: $28, 35, 42, \dots$

The only one between 30 and 40 is $35$.

$2m$ means $2 \times m$.

So the missing number is $2$.

(a) Key: one square = 4 houses. January = $4 + 4 + 2 = 10$ (half a square is 2).

(b) $11 = 4 + 4 + 3$, so draw 2 full squares and $\frac{3}{4}$ of a square.

(c) February $= 4 + 4 + 1 = 9$, March $= 12$, April $= 11$.

May $= 60 - (10 + 9 + 12 + 11) = 60 - 42 = 18$

(a) Line up the ruler's zero with one end of the line and read the other end: about $8.7$ cm.

(b) Put the protractor's centre on the vertex and a zero line along one arm. The angle is acute, so read the smaller scale: about $67^\circ$.

(c) A hexagon is any shape with 6 straight sides.

(a) A is 2 across and 1 up: $(2, 1)$.

(b) B is $(-2, -5)$. Midpoint $= \left(\frac{2 + (-2)}{2}, \frac{1 + (-5)}{2}\right) = (0, -2)$

(c) Go 4 left from $O$, then 2 up, and mark a cross labelled C.

Miles driven $= 47879 - 47241 = 638$

Mileage pay $= 638 \times 0.47 = £299.86$

Total $= 299.86 + 80 = £379.86$

Work systematically, fixing the first throw then the second:

HHH, HHT, HTH, HTT

THH, THT, TTH, TTT

There are $2 \times 2 \times 2 = 8$ outcomes.

The three sections are not equal: section 2 is a right angle ($90^\circ$) and the others are larger.

So the outcomes are not equally likely.

Majid is not correct: $P(2) = \frac{90}{360} = \frac{1}{4}$, not $\frac{1}{3}$.

White bars $= \frac{2}{3} \times 24 = 16$

Milk bars $= 24 - 16 = 8$

Weight $= 8 \times 35 = 280$ g

(a) $2c \times 3d = 2 \times 3 \times c \times d = 6cd$

(b) $T = 3 + 2 \times (-4) = 3 - 8 = -5$

(a) Speed $= 36 \div 3 = 12$ km/h

(b) Tuesday time $= 36 \div 16 = 2.25$ hours $= 2$ h $15$ min

Total $= 3$ h $+ 2$ h $15$ min $= 5$ h $15$ min

$5$ h $15$ min $< 5$ h $20$ min, so Lizzie is correct.

Interest per year $= 2.5\%$ of $3500 = 0.025 \times 3500 = £87.50$

Total interest $= 87.50 \times 6 = £525$

(a) Go up from 6 ounces to the line and across: about $170$ g.

(b) $1$ kg $= 1000$ g, which is off the graph. Read $200$ g $\approx 7$ ounces.

$1000 = 5 \times 200$, so $1000$ g $\approx 5 \times 7 = 35$ ounces.

1. Set compasses to 4 cm and draw an arc centred at $A$. Points inside it are less than 4 cm from $A$.

2. Construct the perpendicular bisector of $CB$. Points on the $C$ side are closer to $C$ than to $B$.

3. Shade the part of the triangle that is inside the arc and on the $C$ side of the bisector.

Children pay $100\% - 56\% = 44\%$

$0.44 \times 1800 = £792$

Each child pays $792 \div 66 = £12$

The expression is $\dfrac{\sqrt{35.2 + 1.7^3}}{4.6^2 - 8.91}$ (the square root covers the whole numerator).

Numerator: $\sqrt{35.2 + 4.913} = \sqrt{40.113} = 6.33348\ldots$

Denominator: $21.16 - 8.91 = 12.25$

(a) $6.33348\ldots \div 12.25 = 0.5170189759\ldots$

(b) $0.52$ (2 s.f.)

19 cm is the hypotenuse.

$CB^2 = 19^2 - 10^2 = 361 - 100 = 261$

$CB = \sqrt{261} = 16.155\ldots \approx 16.2$ cm

(a) $90 = 2 \times 45 = 2 \times 3 \times 15 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$

(b) $A = 2^2 \times 3 = 12$, $B = 2 \times 3^2 = 18$.

LCM uses the highest power of each prime: $2^2 \times 3^2 = 36$

The formula is $H = \dfrac{72}{n}$.

Monday: $72 \div 6 = 12$ hours

Tuesday: $72 \div 9 = 8$ hours

Difference $= 12 - 8 = 4$ hours

Total discs $= 24 \div 0.16 = 150$

Red and blue $= 150 - 24 = 126$

One part $= 126 \div (5 + 4) = 14$

Red $= 5 \times 14 = 70$

(a) $x = -1$: $1 + 1 = 2$. $x = 1$: $1 - 1 = 0$. $x = 3$: $9 - 3 = 6$.

(b) Plot $(-2,6), (-1,2), (0,0), (1,0), (2,2), (3,6)$ and join with a smooth curve.

(c) Draw $y = 4$ and read where it crosses the curve: $x \approx -1.6$ and $x \approx 2.6$.

Take the shares as $1 : 6 : 14$ parts (21 parts).

Tina gives Andy $\frac{3}{7} \times 14 = 6$. Now Andy $= 7$, Tina $= 8$.

Tina gives Luke $12.5\%$ of $8 = 1$. Now Luke $= 7$, Tina $= 7$.

Final: Andy 7, Luke 7, Tina 7 parts, so Tina is correct.

Angles in a quadrilateral add to $360^\circ$:

$(2x+15) + (4x+15) + (4x+8) + (3x-3) = 360$

$13x + 35 = 360 \Rightarrow x = 25$

$A = 65^\circ$, $B = 115^\circ$, so $A + B = 180^\circ$.

Co-interior angles add to $180^\circ$, so $AD$ is parallel to $BC$ and $ABCD$ is a trapezium.

The bases are $BC = 6$ and $EF = 1.5$, so the scale factor is $1.5 \div 6 = \frac{1}{4}$.

$DE = 8 \times \frac{1}{4} = 2$ cm

(a) The median is about the 60th/61st value. $34 < 60 \le 34 + 29 = 63$, so it is in $100 < w \le 150$.

(b) Midpoints: 75, 125, 175, 225, 275.

$\sum fx = 2550 + 3625 + 4725 + 4275 + 3025 = 18200$

Mean $\approx 18200 \div 120 = 151.67 \approx 152$ g