Thirty-one million, seventy-two thousand

1. (a) Measuring with a protractor at vertex A, between ray AC and ray AB: x ≈ 46°
2. (b) Measuring with a ruler from A to B: AB ≈ 84 mm
3. (c) Measure the full length of AB and mark the point exactly halfway between A and B (about 42 mm from each end) as M.
4. (d) Using a protractor or set square at M, draw a line at 90° to AB through M.

The reciprocal of 0.4 is 1÷0.4 = 2.5

Converting each to a decimal: &frac67; = 0.857, 8.6×10⁻¹ = 0.86, &frac{11}{13} = 0.846, 86.5% = 0.865

11/13 < 6/7 < 8.6×10⁻¹ < 86.5%

1. (a) A square has 4 lines of symmetry: the two diagonals, plus the two lines connecting the midpoints of opposite sides (one horizontal, one vertical).
2. (b) A shape with rotational symmetry of order 2 (but not necessarily order 4, so not required to be a square) is a parallelogram (a rectangle or rhombus would also be accepted, as both have order-2 rotational symmetry).

25 − (−4) = 25+4 = 29°C

6.55 × 4 = 26.20. Adding the fixed charge: 26.20 + 15.50 = $41.70

Travel spending = ¼ × 750 = $187.50. Remaining after travel = 750−187.50 = $562.50.

After also spending on food, he has $437.50 left, so food spending = 562.50−437.50 = $125.

As a fraction of the original $750: 125÷750 = 1/6

(a) Sorting each group of leaves in order:

Key: 4|8 represents 48 pizzas

(b) With 12 values, the sorted list is: 22, 28, 36, 39, 41, 44, 48, 49, 52, 54, 57, 57. The median is the mean of the 6th and 7th values: (44+48)÷2 = 46

a−b = (5−6, −7−(−7)) = (−1, 0)

1. (a) The common difference is −6 (17−23=−6, etc.). Next terms: 5−6=−1, then −1−6=−7.
   −1, −7
2. (b) nth term = 23 + (−6)(n−1) = 23−6n+6 = 29 − 6n

The first two significant figures are 4 and 6; the next digit (2) rounds down: 0.046

A∪B means “in A OR in B (or both)”. Shade the entire area covered by both circles — circle A, circle B, and their overlap — leaving only the region outside both circles unshaded.

The highest common factor of 20x and 90x² is 10x: 10x(2 − 9x)

Faster runners take less time, so as one variable increases the other decreases: negative correlation

1. (a) Area = πr² = 36π → r²=36 → r=6. Circumference = 2πr = 2π(6) = 12π cm
2. (b) Volume = base area × height: 36π × h = 540π → h = 540÷36 = 15

1.74 × 10⁵

Since 8.5 cm is adjacent to angle x and 14 cm is the hypotenuse: cos(x) = 8.5÷14 = 0.6071

x = cos⁻¹(0.6071) = 52.7° (1 d.p.)

Convert to improper fractions: 2¼ = &frac94;, 1&frac78; = &frac{15}{8}

Dividing is the same as multiplying by the reciprocal: &frac94; ÷ &frac{15}{8} = &frac94; × &frac{8}{15} = &frac{72}{60} = &frac65;

As a mixed number: 1 1/5

(x−4)(x−7) = x²−7x−4x+28 = x² − 11x + 28

Using the law 5^7−x=5³: 7−x = 3 → x = 4

Rounding to the nearest 10 cm (0.1 m) means l could be up to half of 0.1 m (0.05 m) above or below 4.5:

4.45 ≤ l < 4.55

The universal set is {1,2,3,4,5,6,7,8,9,10,11}.

S∩T (in both) = {1, 7}. S only (not T) = {4, 10}. T only (not S) = {3, 5, 9, 11}. Outside both = {2, 6, 8}.

Since 10 (in S only) and 2, 8 (outside both) are already placed, the remaining numbers to add are:

S only: add 4   S∩T (overlap): 1, 7   T only: 3, 5, 9, 11   Outside both: add 6

The sample proportion is 13/30. Applying this to the whole school: 570 × 13÷30 = 19 × 13 = 247

1. (a) Triangle ABD is right-angled at B, with hypotenuse AD=21.2 and base AB=16.5. By Pythagoras: BD = √(21.2²−16.5²) = √(449.44−272.25) = √177.19 = 13.3 m (1 d.p., as required)
2. (b) Triangle BCD is also right-angled at B (since ABC is a straight line and angle ABD=90°, so BD is perpendicular to the line, making angle DBC=90° too). Using the unrounded height BD=13.3113 m and angle BCD=48°: sin(48°) = BD/CD → CD = 13.3113÷sin48° = 13.3113÷0.74314 = 17.9 m (3 s.f.)