November 2023 Paper 3 Foundation

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

$35\% = 35 \div 100 = 0.35$

8061 lies between 8000 and 8100.

The tens digit is 6, which is 5 or more, so round up: 8100.

On a number line, numbers less than $-5$ are further to the left.

So $-6$, $-7$ or $-10$ all work.

The grid has $3 \times 7 = 21$ squares and 6 of them are shaded.

Fraction shaded $= \frac{6}{21} = \frac{2}{7}$

In 27.963 the first digit after the decimal point is the tenths digit.

So the 9 is worth $\frac{9}{10} = 0.9$.

(a) Each circle is 12 cakes, so $24 \div 12 = 2$ circles for banana.

(b) Chocolate $5 \times 12 = 60$; vanilla $3\frac{1}{4} \times 12 = 39$; lemon $1\frac{1}{2} \times 12 = 18$; banana 24.

Year 7 total $= 60 + 39 + 18 + 24 = 141$, which is less than 150, so Year 8 sold more.

Total lengths needed: $8050 \div 25 = 322$

Lengths still to swim: $322 - 178 = 144$

(a) $97 - 91 = 6$, so each term is 6 less than the one before.

(b) 5th term $= 73$, 6th $= 67$, 7th $= 61$. Difference $= 73 - 61 = 12$

(c) Every term is odd (odd minus 6 stays odd). 52 is even, so it cannot appear. The terms go 55, 49, missing 52.

$12\text{ kg} = 12000\text{ g}$

Food per day $= 3 \times 105 = 315$ g; per week $= 315 \times 7 = 2205$ g

$12000 \div 2205 = 5.44...$, so the bag lasts 5 complete weeks.

(a) It has 5 sides, so it is a pentagon.

(b) A pentagonal prism has $5 + 5 + 5 = 15$ edges (two pentagons plus 5 joining edges).

Total length $= 15 \times 7.5 = 112.5$ cm

Total parts $= 2 + 16 + 7 = 25$

Green is 7 of these parts, so the fraction is $\frac{7}{25}$.

$3\frac{1}{4}$ hours $= 3$ hours 15 minutes.

14 10 back 3 hours is 11 10; back 15 more minutes is 10 55.

(a) The terms are all $h^3$: $8 + 14 - 2 = 20$, so $20h^3$.

(b) $9y + 12y = 21y$, and $21y \div 3 = 7y$.

Change each to a decimal:

$\frac{7}{12} = 0.583...$, $57\% = 0.57$, $\frac{6}{11} = 0.545...$

Smallest first: $\frac{6}{11}$ (0.545), 0.558, 0.56, 57% (0.57), $\frac{7}{12}$ (0.583)

(a) Couples $= 100 - 64 = 36$. Couples abroad $= 11$, so couples UK $= 36 - 11 = 25$.

Families abroad $= 18 - 11 = 7$, so families UK $= 64 - 7 = 57$.

(b) Not abroad means UK: $\frac{57}{64}$

$\frac{x}{7} + 9 = 4$

Subtract 9: $\frac{x}{7} = -5$

Multiply by 7: $x = -35$

UK hourly pay: $473.28 \div 32 = £14.79$

Australian hourly pay in pounds: $26.40 \div 1.796 = £14.699...$

$£14.70 < £14.79$, so no, Blake is not correct.

The shape is symmetrical, so the triangle's base $= 14 - 3.8 - 3.8 = 6.4$ cm.

Triangle area $= \frac{1}{2} \times 6.4 \times 6 = 19.2$ cm²

Rectangle area $= 3.5 \times 19.2 = 67.2$ cm²

$14w = 67.2$, so $w = 67.2 \div 14 = 4.8$

From the elevation the cuboid is 8 cm long and 4 cm high.

Depth $= 224 \div (8 \times 4) = 7$ cm

The plan (view from above) is length × depth: draw a rectangle 8 cm by 7 cm.

(a) Put the decimal point after the first digit: 4.68. The point moves 5 places, so $4.68 \times 10^5$.

(b) A power of $-4$ means move the point 4 places left: $0.0005037$

Expected number $=$ probability × number of spins

$0.4 \times 200 = 80$

(a) Midpoints 12.5, 17.5, 22.5, 27.5, 32.5.

$\sum fx = 25 + 140 + 292.5 + 577.5 + 195 = 1230$

Mean $\approx 1230 \div 50 = 24.6$ °C

(b) No. The median is about the 25.5th value. $2 + 8 + 13 = 23$, so it is in the $25 < T \le 30$ group, not the middle group.

(a) $\le 4$ includes 4, so the circle at 4 should be solid. The line should start with an open circle at $-3$, but Jenna started at $-2$.

(b) $5y - 7 < 16$ gives $5y < 23$, so $y < 4.6$.

The greatest integer is 4.

Multiples of 30: 30, 60, 90, 120

Multiples of 24: 24, 48, 72, 96, 120

LCM $= 120$, so $120 \div 30 = 4$ packs and $120 \div 24 = 5$ boxes.

Total work $= 30 \times 4 = 120$ machine-hours.

In 6 hours: $120 \div 6 = 20$ machines.

Car time $= 143 \div 55 = 2.6$ hours $= 2$ hours 36 minutes ($0.6 \times 60 = 36$).

Total $= 2$ h 36 min $+ 5$ h 20 min $= 7$ hours 56 minutes

Area of face on the table $= \text{force} \div \text{pressure} = 504 \div 3.5 = 144$ cm²

A cube has 6 equal faces: $6 \times 144 = 864$ cm²

$864 < 900$, so the surface area is less than 900 cm².

The line crosses the $y$-axis at 3, so $c = 3$.

From (0, 3) to (1, 1): down 2 for every 1 across, so gradient $= -2$.

Equation: $y = -2x + 3$

The diagonal of the square is a diameter of the circle.

$d = \sqrt{3.5^2 + 3.5^2} = \sqrt{24.5} = 4.949...$ cm

Circumference $= \pi d = \pi \times 4.949... = 15.55... \approx 15.6$ cm