November 2020 Paper 3 Foundation

There are 100 cm in 1 m, so divide by 100.

$300 \div 100 = 3$ metres

One third means divide by 3.

$24 \div 3 = 8$

Per cent means out of 100.

$40\% = \frac{40}{100} = \frac{2}{5}$

Squaring means multiplying the number by itself.

$2.5 \times 2.5 = 6.25$

The negative number furthest from zero is smallest.

$-6, -4, -3, 0, 1, 2, 7$

(a) Bars above the 1.4 line: years 4, 5, 6, 7 and 8, so 5 years.

(b) Years 5 and 6 both reach exactly 1.5 million.

Count the squares: 16 in total, 4 unshaded, so 12 shaded.

$\frac{12}{16} = \frac{3}{4}$

Boxes needed: $200 \div 25 = 8$

Cost: $8 \times £9.75 = £78$

(a) $2 \times 5 = 10$, and $300 \div 10 = 30$

(b) $3.5 \times 12 = 42$

(c) The reciprocal of 20 is $\frac{1}{20} = 0.05$

(a) Up from 2 minutes to the curve, across: 80 °C.

(b) Across from 50 °C to the curve, down: 8 minutes.

(c) From 100 to about 73 is a drop of about 27. From about 47 to 40 is a drop of about 7. So yes, Pam is right.

Angles in a quadrilateral add to $360°$:

$360 - 130 - 95 - 65 = 70°$

Angles on a straight line add to $180°$:

$y = 180 - 70 = 110$

(a)(i) Subtract 5 each time: 20, 15

(ii) Keep going: 10, 5, 0, −5 are terms 8 to 11, so $-5$ is the 11th term.

(b) $4 \times 9 + 3 = 39$

Missing sides: $10 - 4 = 6$ cm and $7 - 5 = 2$ cm.

Perimeter: $4 + 5 + 6 + 2 + 10 + 7 = 34$ cm

(a) $3x + 2x = 5x$ and $5y - 4y = y$, giving $5x + y$

(b) $5p = 22 - 7 = 15$, so $p = 15 \div 5 = 3$

(a) Shop A: $30 \div 4 = 7.5$, so 8 packs: $8 \times £1.60 = £12.80$

Shop B: $30 \div 6 = 5$ packs: $5 \times £2.70 = £13.50$

Shop A is cheaper.

(b) Shop A: 7 packs + 2 singles $= £11.20 + £0.80 = £12$. Shop B is still £13.50, so no change: shop A.

(a) Total cards: $5 + 2 + 4 = 11$, so $P(\text{blue}) = \frac{5}{11}$

(b) $1 - 0.3 = 0.7$

Set compasses to 6 cm. Draw an arc from each end of the 8 cm line.

Join both ends to the point where the arcs cross.

The graph shows £50 = 65 dollars.

$345 \div 65 = 5.307...$

$5.307... \times 50 \approx £265$

(a) $40\% = 56$, so $1\% = 1.4$ and $100\% = 140$

(b) $\frac{18}{56} \times 100 = 32.14... \approx 32\%$

Total for all three: $3 \times 80 = £240$

Carl has $240 - 65 - 100 = £75$

Known notes: $3 \times 5 + 20 = £35$

$75 - 35 = £40$, so $40 \div 10 = 4$ notes

(a) $P(\text{heads}) = \frac{1}{2}$, so $\frac{1}{2} \times 50 = 25$

(b) Simon: more trials give a more reliable estimate of the probability.

From above you see the square top of the cube, 6 cm by 6 cm.

The pyramid's sloping edges meet at the centre, so draw both diagonals of the square.

(a) Add the powers: $n^{3+5} = n^8$

(b) Subtract the powers: $c^{3-2}d^{4-1} = cd^3$

(c) $5x > 14$, so $x > \frac{14}{5} = 2.8$

Cycling: $30 \div 24 = 1.25$ h

Running: $12 \div 8 = 1.5$ h

Total: $2.75$ h $= 2$ hours $45$ minutes

Half of 0.1 is 0.05.

Lower bound $9.4 - 0.05 = 9.35$, upper bound $9.4 + 0.05 = 9.45$

$9.35 \le m < 9.45$

(a) 3 kg covers $5 \times 9 = 45$ m², so 1 kg covers $15$ m².

New lawn: $10 \times 14 = 140$ m² needs $140 \div 15 = 9.33$ kg.

She has $5 \times 2 = 10$ kg, so yes.

(b) Now she has $4 \times 2 + 1 = 9$ kg, less than 9.33 kg, so she does not have enough.

(a) Each spinner has 3 equal sections, one of which is 2: $P(2) = \frac{1}{3}$, $P(\text{not } 2) = \frac{2}{3}$

(b) $\frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$

(a) The lines cross at $(-2, 4)$, so $x = -2$, $y = 4$.

Check: $5(-2) - 9(4) = -46$ ✓

(b) Read where the curve meets the $x$-axis: $x \approx 0.6$ and $x \approx 3.4$

Triangle area: $\frac{1}{2} \times 4 \times 6 = 12$ cm²

Volume: $12 \times 10 = 120$ cm³

Mass: $120 \times 0.8 = 96$ g