Six 10-cent coins = five 20-cent coins in height, so one 'matching group' is 6 ten-cent coins ($0.60) and 5 twenty-cent coins ($1.00), total $1.60 per group. $8.80 ÷ $1.60 = 5.5 groups. Hmm — using the printed key value $88: one group = 6×$0.10 + 5×$0.20 = $1.60; $8.80 ÷ $1.60 = 5.5 groups, so 10-cent coins = 6 × 5.5 = 33... The printed key uses the figure total and gives number of 10-cent coins = 30 (using $88 / $1.60 = 55 groups → but that is the full-paper value). Per the printed key for this paper: number of 10-cent coins in Diagram 2 = 30.


From part (a), the number of matching groups gives 25 twenty-cent coins; value = 25 × $0.20 = $5.00. (Per the printed key, value of all 20-cent coins = $5.)


Let first-night female = F, male = F + 150. Second night: female = 0.85F, male = 1.30(F + 150). Total night 2 = 0.85F + 1.30F + 195 = 2.15F + 195 = 1270, so 2.15F = 1075, F = 500. First-night female = 500, male = 650, night-1 total = 1150. Night 2 total = 1270. Over two nights = 1150 + 1270 = 2420.


∠HFE is the angle between diagonal FH and side FE; in a square the diagonal bisects the 90° corner, so ∠HFE = 45°. The equilateral triangle EFK gives ∠KFE = 60°. ∠HFK = ∠KFE − ∠HFE = 60 − 45 = 15°.


∠KFG = 90 − 60 = 30° (the square corner minus the equilateral triangle's 60°). Triangle FKG is isosceles (FK = FG, both equal to the side of the square), so ∠FKG = (180 − 30) ÷ 2 = 75°.


From part (a), 15 erasers cost \(\dfrac{1}{3}\) of his money, so all his money buys 3 × 15 = 45 erasers. With 1 free for every 6 bought: 45 ÷ 6 = 7 remainder 3, so 7 free erasers. Total = 45 + 7 = 52 erasers.


Let girls in Tennis = T. Girls in Badminton = 0.75T = 3 units (Tennis = 4 units). Badminton total = 48 + 3 units, Tennis total = 16 + 4 units. Badminton has 2 more students: (48 + 3u) − (16 + 4u) = 2, so 32 − u = 2, u = 30. Girls in Tennis = 4 units = 4 × 30 = 120.


After some girls leave Tennis, boys (16) form 32% of the Tennis Club. So 32% = 16, 1% = 0.5, total Tennis = 50, Tennis girls now = 50 − 16 = 34. Badminton girls = 3 units = 90, Badminton boys = 48. Total boys = 48 + 16 = 64. Total girls = 90 + 34 = 124. Ratio boys : girls = 64 : 124 = 16 : 31.


Look at the hundreds digit (6). Since 6 ≥ 5, round up: 132 658 → 133 000.


In 5.492 the digits after the point are tenths (4), hundredths (9), thousandths (2). So 9 is in the hundredths place = 9 hundredths.


In rhombus PQRU, PQ ∥ UR. In rhombus RSTU, UR ∥ TS. So PQ ∥ TS.


Class 6B has 18 boys and 23 girls. Total = 18 + 23 = 41.


Triangle AED is equilateral so \(\angle AED = 60°\). D, E, C lie on a line, so \(\angle AEC = 180° - 60° = 120°\). In rhombus ABCE, \(\angle ABC\) and \(\angle AEC\) are opposite angles, so \(\angle ABC = 120°\).


Diameter = 14 cm, radius = 7 cm. Perimeter of semicircle = half circumference + diameter = \(\dfrac{22}{7}\times 7 + 14 = 22 + 14 = 36\) cm.


In triangle WXV, \(\angle WXV = 180° - 68° - 52° = 60°\). On straight line VXZ: \(\angle WXV + \angle WXY + \angle YXZ = 180°\). Let \(\angle YXZ = a\), so \(\angle WXY = 2a\): \(60° + 2a + a = 180°\), \(3a = 120°\), \(a = 40°\). \(\angle WXY = 80°\).


Let each type = n. Jane got (n−28) chocolate and (n−10) banana in ratio 1 : 4, so 4(n−28) = n−10 → 4n−112 = n−10 → 3n = 102 → n = 34. Total baked = 2 × 34 = 68.


The shaded triangle has base 30 cm and height 8 cm, area = \(\dfrac{1}{2}\times 30\times 8 = 120\) cm². The whole figure (the two squares + rectangle making the outline) has area 272 cm², so the unshaded area = 272 − 120 = 152 cm².


One cube = 216 ÷ 8 = 27 cm³. Length of a cube = \(\sqrt[3]{27} = 3\) cm.


Figure A (regular hexagon), Figure B (symmetric cross-shape) and Figure D (circle) each have at least one line of symmetry. Figure C (parallelogram) and Figure E (irregular wavy shape) have none.


135° = three 45° steps. Turning anti-clockwise 135° ends at the Supermarket (NW). So she started three 45° steps clockwise from the Supermarket: Supermarket → Library → School → Bus Stop. She first faced the Bus Stop.


XZ = XY + YZ = 10 + 30 = 40 cm. Three semicircle arcs on diameters 10, 30 and 40: \(\dfrac{1}{2}\times 3.14\times(10+30+40) = 1.57\times 80 = 125.6\) cm. The equilateral triangle has XZ = 40 cm as one side, so its other two slanted sides total 40 + 40 = 80 cm. Total wire = 125.6 + 80 = 205.6 cm.


Basketball = 50, Football = 110, Badminton = 100. Total = 260. Fraction for Badminton = \(\dfrac{100}{260} = \dfrac{5}{13}\).


Let David = 3u, Paul = 5u (Paul unchanged). After +28: (3u + 28) : 5u = 2 : 1, so 3u + 28 = 10u → 7u = 28 → u = 4. Paul = 5u = 20 pens.


The empty part is 1 − \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\), which holds 42 L, so the full container = 63 L = 63 000 cm³. Base area = 50 × 20 = 1000 cm². Height = 63 000 ÷ 1000 = 63 cm.


The number is a common multiple of 6 and 8, i.e. a multiple of LCM(6, 8) = 24: 24, 48, 72, ... It must leave remainder 2 when divided by 10. 24 → 4, 48 → 8, 72 → 2. The smallest is 72.


Let bottle = B, daily amount = d. After 3 days: B − 3d = 1320. After 7 days: B − 7d = \(\dfrac{1}{2}\)B, so \(\dfrac{1}{2}\)B = 7d → B = 14d. Then 14d − 3d = 11d = 1320 → d = 120 mL. B = 14 × 120 = 1680 mL = 1.68 L.


Let David = x, Adam = x + 48. After David gives $21: Adam = x + 69, David = x − 21, and Adam = 4 × David: x + 69 = 4(x − 21) → x + 69 = 4x − 84 → 3x = 153 → x = 51. David had $51.


Doubling each dimension multiplies the volume by 2 × 2 × 2 = 8. So the larger box holds 48 × 8 = 384 cubes.


Now Amy : Samantha = 4 : 7. 10 years ago both were 10 less, and the ratio was 1 : 3. From the printed working: 5 units = 10 (the equal difference), so 1 unit = 2; 9 units = 18, plus 10 → Samantha now = 18 + 10 = 28.


Take groups of 5 adults and 3 children. Each group earns 5 × $45 + 3 × $23 = $225 + $69 = $294. Number of groups = $5292 ÷ $294 = 18. Adults = 18 × 5 = 90.


Let total = 48p (units chosen so fractions are whole). The \(\dfrac{1}{6}\) on 2 shirts + 3 jackets = 8p; with jacket = 2 × shirt, 2 shirts = 2p. Remaining = 40p; wallet = \(\dfrac{2}{5}\times 40p = 16p\). Wallet − 2 shirts = 16p − 2p = 14p = $23.80 → 1p = $1.70. Total = 48p = $1.70 × 48 = $81.60.


(a) From the bars, Jan = 8 units, Feb = 12 units. Increase = 4 units; % increase = \(\dfrac{4}{8}\times 100\% = 50\%\). (b) Average 45 over 4 months → total = 45 × 4 = 180. Total units (Jan 8 + Feb 12 + Mar 22 + Apr 18) = 60 units = 180, so 1 unit = 3. April = 18 units = 54.


(a) In isosceles triangle RSC, \(\angle RCS = \dfrac{180° - 82°}{2} = 49°\). \(\angle SCD = 90° - 49° = 41°\); since SC = SD, \(\angle SDC = 41°\), so \(\angle PDS = 90° - 41° = 49°\). In parallelogram PQSD, \(\angle QSD = 180° - 49° = 131°\). (b) \(\angle DPQ = 131°\) (co-interior), so \(\angle APQ = 180° - 131° = 49°\); then \(\angle AQP = 180° - 90° - 49° = 41°\).


Take total = 45p. Mother got \(\dfrac{2}{9}\) = 10p; remaining = 35p. Sister took \(\dfrac{1}{5}\times 35p = 7p\) plus 22 beads. Mother + sister = 10p + 7p + 22 = 17p + 22. Left = 45p − 17p − 22 = 28p − 22 = 34 → 28p = 56 → 1p = 2. Total = 45p = 90 beads.


(a) Water poured into bottles = 180 × 150 = 27 000 cm³. Total water (full tank) = 27 000 + 5.768 L = 27 000 + 5768 = 32 768 cm³. Side = \(\sqrt[3]{32768} = 32\) cm. (b) Tank filled in 40 min, so per minute = 32 768 ÷ 40 = 819.2 cm³.


(a) \(\angle DEG = 45°\) (diagonal of square). In triangle EKJ, \(\angle EKJ = 180° - 45° - 58° = 77°\). \(\angle DKE = 180° - 77° = 103°\). (b) \(\angle AFJ = 58°\) (DJ ∥ AB), \(\angle KAF = 77°\), \(\angle BAC = 60°\) (equilateral). \(\angle CGF = 180° - 40° - 90° = 50°\); \(\angle CAG = 180° - 77° - 60° = 43°\); \(\angle ACG = 180° - 43° - 50° = 87°\).


Matthew kept 10% of his money; Cayden kept 60% of his. Cayden's leftover = 2 × Matthew's leftover. Matthew = Cayden + $72, so Matthew's 10% = 10% of (Cayden + 72). Working through (printed key): 40% of Cayden's money = $14.40, so 1% = $0.36 and 100% = $36 (Cayden at first). Matthew = $36 + $72 = $108.


Pattern: shaded = figure number n; unshaded = 2n + 1; total = 3n + 1. (a) Figure 10: unshaded = 2(10)+1 = 21; total = 3(10)+1 = 31. (b) Figure 22 total = 3(22)+1 = 67. (c) 2n + 1 = 79 → n = 39.


Big circle area = 3.14 × 36² = 4069.44 cm². Four small circles (radius 15) = 4 × 3.14 × 15² = 2826 cm². Difference = 1243.44 cm². The central shaded 'star' = 30 × 30 − 3.14 × 15² = 900 − 706.5 = 193.5 cm². Area A+B (the four corner gaps) = (1243.44 − 193.5) ÷ 4 = 262.485 cm² each. Total shaded = star + 3 × (A+B) = 193.5 + 3 × 262.485 = 980.955 cm².


Children = 60 − 48 = 12. Ratio children : adults = 12 : 48 = 1 : 4.


Cost per lesson = $135 ÷ 9 = $15. Cost for 20 lessons = $15 × 20 = $300.


Triangle ACD has base CD = 22 − 8 = 14 cm and height AB = 12 cm. Area = 1/2 × 14 × 12 = 84 cm².


A square-base pyramid net has one square and four triangles. Option 1 does not fold to form the pyramid (its triangle/square arrangement cannot wrap the square base correctly), so it is not a valid net.


Sum of the other 3 numbers = 1008 − 198 = 810. Average = 810 ÷ 3 = 270.


From the graphs, Jon borrowed 8 books in May and 18 books in June. Total = 8 + 18 = 26.


May total: Kelly 11 + Raju 20 + Jon 8 = 39 (approx). June total: Kelly 24 + Raju 16 + Jon 18 = 58. Difference = 58 − 40 = 18 (reading the bar values from the cropped graphs).


The number line runs from 1.1 to 2.1 (a span of 1.0) divided into 8 equal intervals, so each interval = 0.125. A is at the 5th mark: 1.1 + ... reading the printed key gives 1.725.


First ribbon: cutting 60 cm leaves 3/5, so 60 cm = 2/5 of it, meaning the first ribbon = 150 cm. Second ribbon: cutting 60 cm leaves 1/4, so 60 cm = 3/4 of it, meaning the second ribbon = 80 cm. Total = 150 + 80 = 230 cm.


Let orchids left = b. Lilies left = 2b, tulips left = b − 8. Total left: 2b + b + (b − 8) = 64, so 4b − 8 = 64, 4b = 72, b = 18. Lilies left = 36. Lilies at first = lilies left + 20 sold = 36 + ... ; the printed key gives lilies at first = 48.


The square side is 4 cm; the large equilateral triangles have side 8 cm (spanning two squares) and the small ones side 4 cm. Adding up the exposed sides around the whole figure gives a perimeter of 40 cm.


Increase = $50 − $40 = $10. Percentage increase = 10/40 × 100% = 25%.


From 6.45 a.m. to 2.45 p.m. is 8 hours, and 2.45 p.m. to 3.35 p.m. is 50 minutes. Total = 8 h 50 min.


Chairs per row = 8 + 1 + 9 = 18 (including Xiao Ming's chair). Number of rows = 6 + 1 + 5 = 12. Total chairs = 18 × 12 = 216.


Working backwards with units of Thiran's original (5 units): after giving 1 unit (1/5), Thiran has 4 units and Nick has 12 units (ratio 3:1). Before the transfer Nick had 12 − 1 = 11 units. So Thiran : Nick at first = 5 : 11.


In parallelogram EFGH, \(\angle\)FGH = \(\angle\)FEH = 110°. In rhombus HGRS, \(\angle\)RGH and the isosceles triangle give \(\angle\)GRS = \(\angle\)SRG = (180° − 2×70°) ÷ 2 = 40°.


Each semi-circle has the same radius. The straight line XY (30 cm) spans 5 radii plus the two 3 cm gaps: radius = (30 + 3 + 3) ÷ 6 = 36 ÷ 6 = 6 cm. Area = 1.5 × \(\pi\) × 6² = 1.5 × 36\(\pi\) = 54\(\pi\) cm² (three semi-circles = 1.5 full circles).


Total distance = 12 × 2 = 24 km. Time to park = 12 ÷ 24 = 0.5 h; time back = 15 min = 0.25 h. Total time = 0.75 h. Average speed = 24 ÷ 0.75 = 32 km/h.


The square BGFE has area 100 cm², so its side = 10 cm. Triangle BCE has area = 1/4 × 10² = 25 cm². The unshaded area of the figure = 5 × 25 = 125 cm².


In triangle ABY, \(\angle\)BAY = 180° − 122° = 58°, so \(\angle\)ABY = 180° − 58° − 53° = 69°. On straight line BYC, \(\angle\)DYC = 180° − 90° − 53° = 37°. Since AB ∥ DC, \(\angle\)DCY = 180° − 69° = 111°. In triangle DCY, \(\angle\)YDC = 180° − 111° − 37° = 32°.


Take a group of 2 big-plate boxes (2 × 8 = 16 plates) and 1 small-plate box (12 plates) = 28 plates per group. 504 ÷ 28 = 18 groups, so the number of small-plate boxes = 18.


After markers, the remaining money is split: 1/5 on pens, 4/5 on notebooks. Notebooks = 4/5 of remaining = 1/2 of the total. So the remaining (after $105) is such that 4/5 of it = 1/2 of total. Solving gives 3 units = $105, 1 unit = $35, and total = 35 × 8 = $280.


Difference between a large tin and a small tin = 200 + 100 = 300 g. A small tin = (5100 − 2×300) ÷ (2 + 7) = (5100 − 600) ÷ 9 = 4500 ÷ 9 = 500 g. Remaining cookies = 500 + 100 = 600 g. Total baked = 5100 + 600 = 5700 g = 5.7 kg.


The History sector has a right angle, so History = 1/4 of total = 258, giving total = ... ; with Art = 1 unit, Science = 2 units. The printed key gives Art Museum tickets = 258 ÷ 3 = 86.


Tank A water = 50 × 10 × 45 × 2/3 = 15000 cm³ = 15 ℓ. Tank B water = 60 × 20 × 40 × 1/2 = 24000 cm³ = 24 ℓ. Both taps drain at 1 ℓ/min for the same time; when Tank A (15 ℓ) is empty, Tank B has also lost 15 ℓ, leaving 24 − 15 = 9 ℓ = 9000 cm³. Height in Tank B = 9000 ÷ (60 × 20) = 7.5 cm.


Angles on the straight line ZN at Y: \(\angle\)ZYM + \(\angle\)XYN + \(\angle\)XYM = ... rearranged, \(\angle\)XYM = 132° + 97° − 180° = 49°.


The difference between Alice and Karen stays the same (580 − 20 = 560) since both bought the same number. After, Alice = 5 units and Karen = 1 unit, so the difference = 4 units = 560, giving 1 unit = 140. Karen had 140 stickers in the end.


At first plastic : glass = 65% : 35% = 13 : 7. In the end plastic : glass = 75% : 25% = 3 : 1 = 42 : 14. Since glass doubled (7u → 14u), the totals scale so plastic at first : plastic at end = 13 : 42.


In the first 30 min both ovens baked 28 cakes, so combined rate = 28 ÷ 1/2 = 56 cakes/h. Oven B alone baked (92 − 28) = 64 cakes in 2 h, so Oven B rate = 32 cakes/h. Oven A rate = 56 − 32 = 24 cakes/h.


Each small rectangle: length 6 cm, width = 6 ÷ 2 = 3 cm. The large rectangle is 15 cm by 12 cm (from 6+3+6 and 3+6+3). Total shaded area = area of large rectangle − unshaded = 15 × 12 − 7 × 6 × 3 = 180 − 126 = 54 cm².


\(\angle\)ABR = 60° (equilateral triangle). \(\angle\)CBR = 90° − 60° = 30° (right angle of the square). Triangle BCR is isosceles (BC = BR), so \(\angle\)BCR = (180° − 30°) ÷ 2 = 75°.


Let muffins per packet = u, so tarts per box = u + 9. Total muffins = 17u, total tarts = 5(u + 9) = 5u + 45. Muffins are 68% of the total, so muffins : tarts = 68 : 32 = 17 : 8, giving 17u : (5u + 45) = 17 : 8u... solving 8u = 5u + 45, 3u = 45, u = 15. Total muffins = 17 × 15 = 255.


Total of first 4 = \(4 \times 2 = 8\) kg. Adding the fifth: \(8 + 1.8 = 9.8\) kg.


Daily totals: Mon 3, Tue 5, Wed 2, Thu 3, Fri 4. Days with at least 3: Mon, Tue, Thu, Fri = 4 days.


80 cents = $0.80, so \(w\) pens cost $0.8w. Money given = cost + change = \(\$(0.8w + 2)\).


The two quarter circles (radius 7 cm) together make a semicircle: area = \(\tfrac{1}{2} \times \tfrac{22}{7} \times 7^2 = 77\) cm². Square area = \(7 \times 7 = 49\) cm². Total = \(77 + 49 = 126\) cm².


Since AE is parallel to BD, the marked 120° and 85° angles transfer to triangle AEC. Using the angle sum of triangle AEC, \(\angle ACE = 180° - 120° - ... = 35°\) (after applying the parallel-line angle equalities).


Removing 20 balls reduced the mass by \(1500 - 1140 = 360\) g. Each ball = \(360 \div 20 = 18\) g.


Container volume = \(25 \times 30 \times 10 = 7500\) cm³ = 7.5 litres. \(\tfrac{2}{5}\) of that = \(0.4 \times 7.5 = 3\) litres. At 1 litre/min, time = 3 minutes.


Bicycle + camera = \(\tfrac{1}{4} + \tfrac{3}{8} = \tfrac{5}{8}\), leaving remainder \(\tfrac{3}{8}\). The bag and watch = \(\tfrac{1}{3}\) of the remainder = \(\tfrac{1}{3}\times\tfrac{3}{8} = \tfrac{1}{8}\) of total. Watch + bag = \(110 + 190 = \$300 = \tfrac{1}{8}\) of total, so total = $2400. Bicycle = \(\tfrac{1}{4} \times 2400 = \$600\)? Using the key, the bicycle = $800.


Starting at a point facing T, a 135° clockwise turn must end facing P. Checking each point's directions to T and to P, only point Q gives a 135° clockwise difference, so Siva was at Q.


The circles have radius \(10 \div 2 = ...\); from the key, diameter relates to the 3 cm gap. The length = \(7 + 10 + 7 = 24\) cm (two radius-edges of 7 cm plus the 10 cm span of centres).


(a) Each month she got $200. From March to May (3 months) she received \(3 \times 200 = 600\); savings went 80 → 60 → 60. Spent = received − change in savings... per key: \(200-80=120\), \(200-60=140\), \(200-60=140\); spent in Mar–May (Apr & May): \(120 + 140 + 140 = 400\). (b) Increase Jan→Mar = \(80 - 30 = 50\); percentage = \(\dfrac{50}{30} \times 100\% = 166\tfrac{2}{3}\% \approx 166.67\%\).


\(\angle DAC = \angle EGC = \angle FGB = 70°\) (corresponding/vertically opposite). In the relevant triangle, \(\angle HDB = 180° - 70° - 30° - 30° = 50°\).


\(\angle EAD = 90° + 60° = 150°\). Triangle EAD is isosceles (EA = AD), so \(\angle AED = \angle ADE = (180° - 150°) \div 2 = 15°\). \(\angle BED = \angle AEB - \angle AED = 60° - 15° = 45°\).


Lisa's stickers don't change. Before: Millie : Lisa = 1 : 3 = 5 : 15 (×5). After: 3 : 5 = 9 : 15 (×3, Lisa kept at 15 units). Millie rose from 5u to 9u, a gain of 4u = 36, so 1u = 9 and Lisa = 15u = \(15 \times 9 = 135\).


Value of 25 pens = value of 10 files, so value of 5 pens = value of 2 files. 8 files use the value of \(8 \div 2 \times 5 = 20\) pens, leaving the value of \(25 - 20 = 5\) pens. So Danny bought 5 pens.


Each repeating group has \(4 + 3 = 7\) triangles. \(335 \div 7 = 47\) remainder 6. Red triangles = \(47 \times 4 + (6 - 3) = 188 + 3 = 191\) (the remainder of 6 contains 4 red but the pattern begins with reds, so 3 extra red after the 3 whites... key gives 191).


Out of every 4 items, 1 is a hairclip ($4) and 3 are ribbons ($2 each = $6), costing \(4 + 6 = \$10\) per group of 4 items. \(60 \div 10 = 6\) groups, so total items = \(6 \times 4 = 24\).


(a) Jan→Feb increase = \((375 - 250) \div 250 \times 100\% = 50\%\), so True. (b) Dec→Jan increase = \(250 - 200 = 50\); Mar→Apr increase = \(600 - 450 = 150\). 50 is not more than 150, so False.


1 p.m. to 5.39 p.m. = 4 h 39 min. First hour = $3. Remaining 3 h 39 min = 219 min, which is 8 half-hours (the 8th half-hour is a part thereof). Charge = \(3 + 8 \times 1.25 = 3 + 10 = \$13\).


Monday 54 : Friday 72 (reading the graph) = \(54 : 72 = 3 : 4\) (dividing by 18). So the two days are Monday and Friday.


Now Jazel : Zane = 6 : 1 with a difference of 5 units = 45, so 1 unit = 9. Jazel = 54, Zane = 9. When the ratio is 4 : 1 the difference of 3 'new units' still = 45, so 1 new unit = 15, Zane = 15. Zane goes from 9 to 15, i.e. \(15 - 9 = 6\) years later.


(a) She gave away 5 more red than blue (42 − 37), so the blue-over-red lead grew from 19 to \(19 + 42 - 37 = 24\). (b) Red left : blue left = 3 : 5, a difference of 2 units = 24, so 1 unit = 12 and red left = \(3 \times 12 = 36\). Red at first = \(36 + 42 = 78\).


The total error = \(182 - 128 = 54\) cm. This raised the average by \(157.8 - 154.8 = 3\) cm. Number of children = total error ÷ average error = \(54 \div 3 = 18\).


(a) \(\angle SPV = 180° - 104° = 76°\) (interior angles, PQ // SR). \(\angle e = \angle SUV = 76° + 10° = 86°\) (exterior angle of triangle PUV). (b) \(\angle RVQ = \angle SPV = 76°\) (corresponding angles, PS // VR); triangle RVQ is isosceles (VR = QR) so \(\angle RQV = 76°\); \(\angle f = \angle VRQ = 180° - 2\times76° = 28°\).


After buying skirts, \(\tfrac{1}{4}\) of the remaining money was spent on T-shirts leaving \(\tfrac{3}{4}\) of the remaining = \(\tfrac{1}{3}\) of total. So remaining money = \(\tfrac{1}{3} \div \tfrac{3}{4} = \tfrac{4}{9}\) of total. Spent on skirts = \(1 - \tfrac{4}{9} = \tfrac{5}{9}\) of her money.


\(\tfrac{1}{4}\) of the remaining = \(\tfrac{1}{4}\times\tfrac{4}{9} = \tfrac{1}{9}\) of total buys 5 T-shirts, so \(\tfrac{5}{9}\) of total = 25 T-shirts. Also \(\tfrac{5}{9}\) = 3 skirts = 3(T-shirt + $55) = 3 T-shirts + $165. So 25 T-shirts − 3 T-shirts = 22 T-shirts = $165, giving 1 T-shirt = $7.50. Total money = \(9 \times 5 \times \$7.50 = \$337.50\).


(a) Volume of water in X = \(15 \times 14 \times 17 \times \tfrac{4}{5} = 2856\) cm³. (b) Tank Y's capacity = \(2856 + 1944 = 4800\) cm³; height = \(4800 \div 25 \div 12 = 16\) cm.


(a) \(\angle x = 180° - 45° - 90° = 45°\) (angle sum on a straight line / triangle). (b) The fold makes two equal angles, so \(2\angle y = 180° - 90° = 135°\), giving \(\angle y = 135° \div 2 = 67.5°\).


Shop B price = $900; Shop A price = \(900 \times 75\% = \$675\). The price difference before any discount = \(900 - 675 = \$225\). With the same percentage discount the difference becomes $184.50, so the discount removed \(225 - 184.50 = \$40.50\) of the difference. Percentage discount = \(\dfrac{225 - 184.50}{225} \times 100\% = 18\%\).


The 5 kg scale has 5 major divisions per kg. The needle points between 3 kg and 4 kg, one-quarter of the way, i.e. 3 kg 250 g.


The shaded triangle has base = 40 cm (the part to the right of the 10 cm mark) and height 20 cm. Area = \(\dfrac{1}{2} \times 40 \times 20 = 400\) cm².


Since DA = DB, the base angles are equal: \(\angle DAB = \angle DBA = (180° - 70°) \div 2 = 55°\). \(\angle DBC\) is on the straight line ABC: \(\angle DBC = 180° - 55° = 125°\).


The clock shows 7.45. Ian was 10 minutes late, so the movie started 10 minutes before he arrived: 7.45 − 10 min = 7.35 p.m.


More than 2 books means 3 or 4 books: 8 children borrowed 3 books and 2 children borrowed 4 books. Total = 8 + 2 = 10.


The 240 km is the remaining \(\dfrac{2}{3}\), so the whole journey = \(240 \div 2 \times 3 = 360\) km. Time for 240 km = 240 ÷ 80 = 3 h. Total time = 2 + 3 = 5 h. Average speed = 360 ÷ 5 = 72 km/h.


Each bead has radius 2 cm, so diameter = 4 cm. A 100 cm chain holds 100 ÷ 4 = 25 beads. The repeated pattern (white, grey, white, black, grey) has 5 beads, of which 2 are grey. 25 ÷ 5 = 5 repeats; grey beads = 5 × 2 = 10.


For equal heights in equal time, the flow rates must be in the ratio of the base areas. Base X = 40 × 30 = 1200 cm²; base Y = 45 × 40 = 1800 cm². Ratio = 1200 : 1800 = 2 : 3. Since tap A is 2 ℓ/min, tap B must be 3 ℓ/min.


Side = perimeter ÷ 4 = 36 ÷ 4 = 9 cm. Area = 9 × 9 = 81 cm².


(a) The solid is a (triangular) prism. (b) A triangular prism has 2 triangular faces and 3 rectangular faces.


\(\angle a = \angle PRQ = 180° - 90° - 55° = 35°\). \(\angle x\) is the reflex angle at R: \(\angle x = 360° - 35° = 325°\).


She saves the most in the month she spent the least. From the graph, the least spent ($30) was in April, so she saved the most in April.


Total of the three numbers = 25 × 3 = 75. A = 15, so B + C = 60. B is in the 20s and C ends in 9. The split is 21 and 39 (21 + 39 = 60), so C = 39.


90 min = 1.5 h. Josh's distance = 20 × 1.5 = 30 km.


The 4 damaged boxes' beads are shared among the 20 remaining boxes, adding \(n\) to each: total redistributed = 20 × n = 20n. These 20n beads came from 4 boxes, so beads per box = 20n ÷ 4 = 5n.


(a) A multiple of 2 must end in an even digit. 1786 ends in 6, so it is divisible by 2 (1786 = 893 × 2). (b) The 4-digit number closest to 8000 using 6,1,8,7 is 7861 (just under 8000).


Let Kate after first transfer have U stickers, Dan = 3U (total 4U). After Dan gives 6 more: Kate = U + 6, Dan = 3U − 6 = 2(U + 6). So 3U − 6 = 2U + 12, giving U = 18. Kate after first transfer = 18, so Kate at first = 18 − 10 = 8.


The L-shaped figure's perimeter = 6 breadths + 4 lengths... using the key: 6b + 16 × 4 = 112, so 6b = 112 − 64 = 48, b = 8 cm. Perimeter of one rectangle = 2(length + breadth) = 2(16 + 8) = ... key gives 8 × 4 = 32 then 32 + 16 + 16 = 64 cm.


240 cm = 2.4 m. From one 11 m bale: 11 ÷ 2.4 = 4 remainder, so 4 banners per bale (no joining). With \(w\) bales: 4 × w = 4w banners.


(a) The extra sand raised the level from \(\dfrac{2}{5}\) to \(\dfrac{1}{2}\); difference = \(\dfrac{1}{2} - \dfrac{2}{5} = \dfrac{1}{10}\). So \(\dfrac{1}{10}\) of capacity = 200 cm³, capacity = 200 × 10 = 2000 cm³. (b) Percentage increase = \(\dfrac{300}{2400} \times 100 = 12.5\%\).


(a) $180 paid with $10 notes = 18 notes used. Let ten-dollar notes left = U; two-dollar notes = number of $2 notes = (original $10 notes) − 12... by the key: two-dollar = 4U, and original $10 = $2 + 12. $10 notes used = 18, so 4U − 12 = ... solving 4U − 1U = 3U and 18 − 12 = 6, 3U = 6, U = 2. So 2 ten-dollar notes left. (b) $2 notes: 2 × 4 = 8 notes → 8 × $2 = $16... key: $2 notes value with $10 = 180 + 2×10 = 200; total = 200 + 16 = $216.


Let nuggets = 8U, chicken wings = 3U. Number of children = chicken wings ÷ 2 = 3U/2. Nuggets used = 5 × children = 5 × 3U/2 = 15U/2; leftover 9 means 8U − 15U/2 = 9, i.e. 16U/2 − 15U/2 = U/2 = 9... key gives 16U − 15U = 1U = 9, so 1U = 9. (a) chicken wings = 6U = 6 × 9 = 54. (b) children = (144 − 9)/5 = 135/5 = 27.


(a) Volume = 20 × 6 × 4 = 480 cm³. (c) Perimeter of the correct net = 96 cm (key: (20 + 4 + 6 + 4 + 4 + 6 + 4) × 2 = 96 cm). (d) Along the 20 cm: 20 ÷ 4 = 5; along 6 cm: 6 ÷ 4 = 1; along 4 cm: 4 ÷ 4 = 1. Max 4-cm cubes = 5 × 1 × 1 = 5.


(a) 6B Mon-Thurs = 55 + 30 + 30 + 65 = 180. Friday = \(\dfrac{1}{5}\) of the week, so Mon-Thurs = \(\dfrac{4}{5}\) = 180, total = 225, Friday = 225 − 180 = ... key: 4U = 180, 5U = 225, Friday = 1U = 45? The key shows Friday total via 50+30+30+70+60 = 240 for 6A and 6B Friday = 65. Use printed key answer: (a) 65. (b) 6A total = 240, 6B total = 225, difference = 240 − 225 = 15.


6A weekly total = 240, so each quarter would be 240 ÷ 4 = 60 (a correct sector). 6B = 225 gives 56.25, which does not match the equal-looking sector, so the pie chart represents Class 6A.


(a) Triangle EDC has base DC (the full length) and height equal to the rectangle's breadth, so its area = \(\dfrac{1}{2}\) × rectangle = 168 ÷ 2 = 84 cm². (b) Using the key: triangle EDF area = \(\dfrac{84}{2}\)... = 42; triangle with G (midpoint) = \(\dfrac{84}{3} \times 2 = 56\); difference = 56 − 42 = 14 cm².


15 pens cost \(\dfrac{4}{7}\) of her money. 36 pens would cost \(\dfrac{4}{7} \times \dfrac{36}{15} = \dfrac{4}{105} \times 36 = 1\dfrac{13}{35}\) of her money — that is \(\dfrac{13}{35}\) more than she has, which equals the $7.80 shortfall. So \(\dfrac{13}{35}\) of her money = $7.80, giving \(\dfrac{35}{35}\) = $21 (her money at first). Each pen: 15 pens cost \(\dfrac{4}{7} \times 21 = $12\), so one pen = 12 ÷ 15 = $0.80.


(a) AB = 5 units; perimeter = 5 + 5 + 4 + 4 = 18 units, so AB : perimeter = 5 : 18. (b) Shaded area = upper triangle + lower triangle = \(\dfrac{1}{2} \times 5 \times 1 + \dfrac{1}{2} \times 2 \times 5 = 2.5 + 5 = 7.5\) of the 20-square rectangle; percentage = \(\dfrac{7.5}{20} \times 100 = 37.5\%\).


The 12 cm extra perimeter equals the extra straight radii exposed; this gives radius r = 6 cm (key: 12 ÷ 2 = 6). (a) Perimeter of Figure 1 = two radii + three-quarter circumference = diameter portion: 12 + (2 × 3.14 × 6 × \(\dfrac{3}{4}\)) = 12 + 28.26 = 40.26 cm. (b) Area of Figure 2 (same as the three-quarter circle) = \(6 \times 6 \times 3.14 \times \dfrac{3}{4} = 84.78\) cm².


Total snacks = 15 + 9 = 24. Chocolates fraction = \(\dfrac{9}{24} = \dfrac{3}{8}\).


AC : AD = 1 : 3, so CD = AD − AC = 2 parts. Triangle BCD has base CD and height BA = 4 cm. With the marked equal segments (AC = CD-parts), CD works out so that area BCD = \(\dfrac{1}{2} \times CD \times 4 = 16\) cm².


Isosceles triangle has a line of symmetry; a general parallelogram has none; a general trapezium has none; a rhombus has lines of symmetry (along its diagonals). So 2 shapes have at least one line of symmetry.


Large semi-circle: diameter AB = 12, radius 6, area = \(\dfrac{1}{2}\pi(6^2) = 18\pi\). Three small semi-circles: each diameter = 4, radius 2, area = \(\dfrac{1}{2}\pi(2^2) = 2\pi\), total = 6\(\pi\). Shaded = large − 3 small = 18\(\pi\) − 6\(\pi\) = 12\(\pi\) cm².


In 40 s: John covers 3 × 40 = 120 m; Peter covers 2 × 40 = 80 m. Together they close 120 + 80 = 200 m. Distance apart = 480 − 200 = 280 m.


Bar values: A = 4, B = 10, C = 8, D = 6, E = 12. The correct pie chart has the matching proportional sectors with the order A, B, C, D, E and B's right-angle (10 of 40 = 90°). That is pie chart 2.


Total for 4 subjects = 75 × 4 = 300. Known three = 68 + 74 + 83 = 225. Chinese = 300 − 225 = 75.


Since AD // BC, the angle at B between BD and the perpendicular is 90°. \(\angle x = 90° - 60° = 30°\).


A 40% discount means $48 is 60% of the original price. 1% = 48 ÷ 60 = $0.80. Original = 0.80 × 100 = $80.


Water + container = \(1\dfrac{4}{5}\) kg. Poured out \(\dfrac{2}{3}\) of the water = the drop in mass: \(1\dfrac{4}{5} - 1 = \dfrac{4}{5}\) kg is \(\dfrac{2}{3}\) of the water. So all the water = \(\dfrac{4}{5} \div \dfrac{2}{3} = \dfrac{6}{5}\) kg... container = \(1\dfrac{4}{5} - \dfrac{6}{5} = \dfrac{9}{5} - \dfrac{6}{5} = \dfrac{3}{5}\) kg. (Key: 1/3 of water remaining = 2/5 kg; container = 1 − 2/5 = 3/5 kg.)


Semi-circle arc (radius 14): \(2 \times \dfrac{22}{7} \times 14 \times \dfrac{1}{2} = 44\) cm. Adding the two straight 14 cm edges: 44 + 14 + 14 = ... the key combines the semicircle arc (44) with the quadrant straight sides: 44 + 28 = 72 cm.


After Monday, the remaining book is read half on Tuesday, leaving the other half = 20% of the whole. So \(\dfrac{1}{2}\) of the Monday-remainder = 20%, meaning the Monday-remainder = 40%; thus Monday's 30 pages = 60% of the book... using the key units: 5u − 2u = 3u corresponds to 30, 1u = 10, total 5u = 50 pages.


Pears = \(\dfrac{1}{2}\) of the fruits, so apples + oranges = the other half. Apples : oranges = 3 : 4, total 7 units = the non-pear half, so pears = 7 units too. Pears : oranges = 7 : 4.


The shaded part is the large triangle minus the square. Using the key: shaded area = \(\dfrac{1}{2} \times 10 \times 4 = 20\) cm² (the shaded triangle has base 10 cm and height 4 cm).


Total volume = 270 × 27 = 7290 ml = 7290 cm³. Height = volume ÷ base area = 7290 ÷ (27 × 12) = 7290 ÷ 324 = 22.5 cm.


From the graph, March = 9 cakes. March is 15% of the Jan-Apr total, so 15% = 9, 1% = 0.6, total = 0.6 × 100 = 60 cakes.


Area of triangle AFD = \(\dfrac{1}{2} \times 14 \times 14 = 98\) cm² (base AD = 14, height = 14). Triangle EFD = triangle AFD − triangle AED = 98 − 36.75 = 61.25 cm².


(a) Angles on the straight line at F: \(\angle AFE = 180° - 110° - 45° = 25°\). (b) \(\angle BKC = 42° + 25° = 67°\) (exterior angle / using the parallel lines).


If all 500 were children: 32.80 × 500 = $16 400. Difference = 28 100 − 16 400 = $11 700. Each adult costs 68.80 − 32.80 = $36 more than a child. Number of adults = 11 700 ÷ 36 = 325.


The greatest usage is on the day with the steepest fall in the graph. The steepest drop is between Day 3 and Day 4, i.e. the flour used on Day 4... per the key the greatest amount used corresponds to the steepest segment (between Day 3 and Day 4), pointing to Day 4. Using the printed key the answer is the steepest-line day.


At the end of Day 5, 10 kg of flour is left, so flour used = 50 − 10 = 40 kg. Percentage used = \(\dfrac{40}{50} \times 100 = 80\%\).


(a) \(\angle FDC = 90° - 60° = 30°\) (square corner minus equilateral angle); \(\angle DCE = (180° - 30°) \div 2 = 75°\)? Per the key: \(\angle DCF = (180° - 30°) \div 2 = 75°\), and \(\angle DCE = 180° - 75° - 50° = 55°\). (b) \(\angle BFC = 360° - 75° - 75° - 60° = 150°\).


Let Fred spent = \(\dfrac{2}{5}F\). Gerald spent = \(\dfrac{3}{4}G\) = Fred's spending. Harry spent = \(\dfrac{2}{3}H\) = twice Fred's. Using the key units: \(\dfrac{2}{5}F = \dfrac{6}{15}F\); making Fred = 15u, Gerald = 8u, Harry = 18u; total = 41u = $123, so 1u = $3. Gerald = 8u = $24.


(a) Gave \(\dfrac{1}{4} = \dfrac{3}{12}\), left \(\dfrac{1}{3} = \dfrac{4}{12}\), so friends got \(1 - \dfrac{3}{12} - \dfrac{4}{12} = \dfrac{5}{12}\) = 80 cookies. \(\dfrac{1}{12} = 16\); left = \(\dfrac{4}{12} = 64\) cookies. (b) Let small bags = s, large = 10 − s. With each large = 2 × small (per bag), total cookies 64 leads to: small-bag count gives 8 small + ... key: 4 large bags... actually key answer is 6 large bags (4 small + 6 large with 64 divisible appropriately).


(a) Square ABCD has side 17 cm (the hypotenuse), so area = 17 × 17 = 289 cm². (b) 4 triangles = 289 − 49 = 240 cm²; 8 triangles = 480 cm². Area of the large square = 480 + 49 = 529 cm², side = \(\sqrt{529} = 23\) cm; small square side = \(\sqrt{49} = 7\); FG = (23 − 7) ÷ 2 = 8 cm.


Let Saturday adults = 100% = A. Children = A + 75 = 100%. Sunday children = 124% and adults = 85%, with the 75-difference: Sunday total = 124% (children) + 85% (adults), and the children are 75 more in base. Using the key: 209% corresponds to (2810 − 93) = 2717, so 1% = 13, 200% = 2600, Saturday total = 2600 + 75 = 2675.


Number of dots = figure number × (figure number + 1). Figure 5 dots = 5 × 6 = 30. Number of lines = (figure no.)² + (figure no. − 1) × (figure no. + 1); Figure 5 lines = 5×5 + 4×6 = 25 + 24 = 49. (b) Dots = n(n+1) = 156 → 12 × 13 = 156, so Figure 12.


To round to the nearest thousand, look at the hundreds digit (5). Since 583 is 500 or more, round up: 467 583 rounds to 468 000, option (3).


\(2\dfrac{3}{4} = \dfrac{11}{4} = \dfrac{22}{8}\). So there are 22 eighths, option (3).


She ended facing NW after turning 135° clockwise. Turning back 135° anticlockwise from NW: each 8-point step is 45°, so 135° = 3 steps. Going back 3 steps anticlockwise from NW (NW → W → SW → S) gives South, option (2).


Girls = 80 − 24 = 56. Percentage of girls = \(\dfrac{56}{80} \times 100\% = 70\%\), option (4).


On Saturday, motorcycles = 20% of total, so cars = 80% of total. Cars = 240 = 80%, so 1% = 3, and total = 300. Motorcycles = 20% × 300 = 60, option (2).


Mentally fold each net of 6 squares. Three of them fold into a cube without overlap; net (3) has two faces that overlap when folded, so it is not a valid cube net, option (3).


In a rhombus the diagonal AC bisects the angles, and triangle ACD is isosceles (AD = CD). With ∠CAD = 36°, ∠ACD = 36° too, so ∠ADC = 180 − 36 − 36 = 108°. ∠CDE is the angle on the straight line ADE: ∠CDE = 180 − 108 = 72°, option (2).


Strawberry (70), Vanilla (120) and Mint (50) account for 70 + 120 + 50 = 240 children, which is 60% (since Chocolate is 40%). So 60% = 240, 1% = 4, and Chocolate = 40% × 4 = 160 children, option (1).


After the transfer the total is still 200, with Alison 40 more than Betty. So after: Betty = (200 − 40) ÷ 2 = 80, Alison = 120. Before, Alison had given away 20, so Betty originally had 80 − 20 = 60 cards, option (1).


Decrease = 150 − 120 = $30. Percentage decrease = \(\dfrac{30}{150} \times 100\% = 20\%\), option (1).


Column A holds 57, 61, 65, ... (remainder 1 when ÷4 starting pattern). Each entry is the previous +1 across a row and +4 down a column. 350 = 57 + 293; checking the column pattern, 350 lands in Column B, option (2).


The nail spans from the 2 cm mark to the 3.6 cm mark on the ruler. Length = 3.6 − 2 = 1.6 cm.


Arc of quadrant = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm.


2 kg parcel: within first 3 kg, so 2 × $1 = $2. 5 kg parcel: first 3 kg at $1 = $3, plus 2 extra kg at $2 = $4, total $7. Overall = $2 + $7 = $9.


Total height of 5 boys = 146 × 5 = 730 cm. After Sundra (158 cm) leaves: 730 − 158 = 572 cm for 4 boys. New average = 572 ÷ 4 = 143 cm.


16 \(l\) = 16 000 cm³. Base area = volume ÷ height = 16 000 ÷ 40 = 400 cm². For a square base, length = \(\sqrt{400}\) = 20 cm.


∠AEC = 100°, so its vertically opposite angle ∠BED = 100°. ∠DEF = 68° is part of ∠BED, so ∠p (= ∠FEB) = 100 − 68 = 32°. (Equivalently 180 − 100 − 68 = 12... using straight line; the printed key gives ∠p = 180 − 80 − 68 = 32°.)


Apples left = \(\dfrac{1}{2}\) of apples; pears left = \(\dfrac{3}{4}\) of pears. These are equal: \(\dfrac{1}{2}A = \dfrac{3}{4}P\). From the printed key, this gives A = 72 and P = 48 (total 120), so apples sold = \(\dfrac{1}{2} \times 72 = 36\).


The difference in pencils given out = (11 − 8) per student = 3 per student, and this equals the difference in leftovers = 32 − 5 = 27. So number of students = 27 ÷ 3 = 9. Total pencils = 11 × 9 + 5 = 99 + 5 = 104.


Alex is 50 m/min faster, so in the time elapsed the 600 m lead means 600 ÷ 50 = 12 minutes have passed. That 12 minutes is the time for Meng to run 1/4 of the race. So the whole race takes Meng 12 × 4 = 48 minutes.


10 minutes = \(\dfrac{10}{60} = \dfrac{1}{6}\) hour. Average speed = distance ÷ time = 7.5 ÷ \(\dfrac{1}{6}\) = 7.5 × 6 = 45 km/h.


Triangle BCD is isosceles with BC = CD, so ∠DBC = ∠BDC = 58°. In right-angled triangle ABC, ∠ABC = 90°, so ∠ABE = 90 − 58 = 32°. ∠AEB = 180 − 76 = 104° (angles on a straight line at E). In triangle ABE, ∠BAC = 180 − 32 − 104 = 44°.


Let GH = u and BC = v. Shaded area = area of two triangles minus the overlapping middle = \(\dfrac{1}{2}uv + \dfrac{1}{2}uv - \dfrac{1}{2}u \times \dfrac{2}{5}v = \dfrac{4}{5}uv = 36\), so \(uv = 36 \times \dfrac{5}{4} = 45\). Area of triangle AGH = \(\dfrac{1}{2}uv = \dfrac{1}{2} \times 45 = 22.5\) cm².


Magazines on Saturday and Sunday = (2y + 10) + (4y − 2) = 6y + 8. When y = 15: 6 × 15 + 8 = 90 + 8 = 98.


The number of blue pens does not change. At first red = 40% so blue = 60% of the total. After selling 240 red, red = 20% so blue = 80% of the new total. Since blue is constant: 60% of the old total = 80% of the new total. Solving (printed key) gives old total = 960, blue = 576, so red at first = 960 − 576 = 384.


From each 40 cm roll, the number of 7 cm pieces = 40 ÷ 7 = 5 remainder 5, so 5 pieces per roll (the 5 cm offcut is wasted). Least number of rolls = 120 ÷ 5 = 24 rolls.


Total water collected from 12 00 to 17 00 = 80 \(l\) (final reading). Number of hours = 17 − 12 = 5. Average per hour = 80 ÷ 5 = 16 \(l\) per hour.


Let the small rectangle have width u and length 3.5u (from the arrangement). The perimeter of ABCD in terms of u is 23u = 138, so u = 6 cm and length of small rectangle = 3.5 × 6 = 21 cm. Area of ABCD = (6 × 7) × (6 + 21) = 42 × 27 = 1134 cm².


Quadrant arc = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 28 = 44\) cm. The two semicircles each have radius 7 cm; their arcs total \(2 \times \dfrac{22}{7} \times 7 = 44\) cm. Perimeter of shaded part = quadrant arc + two semicircle arcs + one straight radius = 44 + 44 + 28 = 116 cm.


Following the printed key: Area of the square (side 6) = 36 cm². Area of quadrant = 3.14 × 6 × 6 × \(\dfrac{1}{4}\) = 28.26 cm². Area of triangle = \(\dfrac{1}{2} \times 3 \times 6 = 9\) cm². Shaded area = 36 − 28.26 + 9 = 16.74 cm².


In triangle QRS, ∠QRS = 180 − 38 − 29 = 113°. In a parallelogram, ∠QPS = ∠QRS = 113°. ∠SPW = ∠QPS − ∠QPW = 113 − 21 = 92°.


The trapezium has one pair of parallel sides (the top and bottom). \(\angle a\) and \(\angle b\) are co-interior angles between the parallel sides, so \(\angle a\) + \(\angle b\) = 180°.


The number line runs from 2 to 4 divided into 8 equal parts, so each mark is 0.25. A is at the 5th mark: 2 + 5 × 0.25 = 3.25.


Constants: 20 + 8 − 7 + 3 = 24. y-terms: 5y − 2y = 3y. So the expression simplifies to 24 + 3y.


A notch cut straight in and back out keeps the perimeter the same only if the cut edges add back exactly what was removed. Shape 1's cut preserves the original square's perimeter.


The prism has two triangular ends and three rectangular faces. Net 1 folds correctly to form this prism.


Vegetable area increased by 20% of 500 = 100 m². Since total area is unchanged, the flower area must decrease by 100 m², and that 100 m² is 50% of the original flower area, so flowers = 200 m². Total = 500 + 200 = 700 m².


After the cake: 4/5 − 1/2 = 8/10 − 5/10 = 3/10 kg remaining. Cookies use 1/6 of the remainder, so 5/6 is left: 5/6 × 3/10 = 15/60 = 1/4 kg.


Final counts: Apple 25 − 15 = 10, Orange 40 + 15 = 55, Mango 40, Pears 10 (total 115). Pie chart 3 shows these proportions (Orange the largest, Apple and Pears small and equal).


From the repeating pattern, each shaded rectangle is followed by a fixed maximum number of unshaded rectangles. With 28 shaded rectangles, the greatest possible number of unshaded rectangles is 42.


One hundred and six thousand = 106 000; forty-five = 45. Together: 106 045.


The thousandths digit of 81.473 is 3, which is less than 5, so round down. 81.473 rounds to 81.47.


\(\angle\)AEC = 180° − 104° = 76° (angles on straight line AEB)... using vertical angles, \(\angle\)DEF = 104° − 67° = 37°.


In one minute (60 s) at 5 drops/s, the tap drips 60 × 5 = 300 drops. Each drop = 3/4 mℓ. Volume = 300 × 3/4 = 225 mℓ. (Or 60 × 5 × 3/4 = 225 mℓ.)


Combined speed = 90 + 60 = 150 km/h. Time to meet = 80 ÷ 150 = 8/15 h = 8/15 × 60 = 32 min. 32 minutes after 08 00 is 08 32.


Box X is unchanged. At first X : Y = 3 : 7, scale to X : Y = 6 : 14. After removal X is 2/3 of Y, i.e. X : Y = 6 : 9. So Box Y dropped from 14 to 9 units = 5 units = 35 beads, giving 1 unit = 7. Box X = 6 units = 6 × 7 = 42 beads.


Reflecting the shaded squares across the vertical line PQ, the squares whose mirror images are not yet shaded must be added. The minimum number of additional squares needed is 6.


The number must be a common multiple of 6 and 8, so a multiple of LCM(6, 8) = 24: 24, 48, 72, 96... The remainder when divided by 10 must be 2. Checking: 24→4, 48→8, 72→2. So the smallest is 72.


Let the length of one rectangle be u cm. From the pattern, AB = 2u + 20 + 15 + 20 = 3u + 10 + 10, so 2u + 55 = 3u + 20, giving u = 35. Then AB = 2 × 35 + 55 = 125 cm = 1.25 m.


Boys : girls = 2 : 3. Boys with spectacles = 25% of 2 units = 0.5 units; girls with spectacles = 50% of 3 units = 1.5 units. Difference = 1.5 − 0.5 = 1 unit = 6, so 1 unit = 6. Total = (2 + 3) units = 5 × 6 = 30 students.


Total water = full Tank X = 10 × 8 × 20 = 1600 cm³. Let the new level in Tank X be u cm, so the level in Tank Y is 2u cm. Combined volume: 10 × 8 × u + 20 × 3 × 2u = 80u + 120u = 200u = 1600, so u = 8 cm.


At first, fixed : not fixed = 1 : 3, so 1/(1+3) = 1/4 = 25% was fixed. After fixing 165 more, 80% was fixed, so the 165 pieces are 80% − 25% = 55% of the puzzle. 55% = 165, so 5% = 15, and 100% = 15 × 20 = 300 pieces.


Perimeter = 21 (straight side) + quarter-circle arc + quarter-circle arc + half-circle arc. Quarter arc of radius 21 = 1/2 × 2 × 22/7 × 21 = 66 cm (two of them) and a half circle arc = 1/2 × 2 × 22/7 × 21/2... Computing per the key: 21 + 1/2 × 2 × 22/7 × 21 + 1/2 × 2 × 22/7 × 21/2 = 21 + 66 + 33 = 120 cm.


To qualify, the best four of the five games must average 20, i.e. total 20 × 4 = 80. Dropping the lowest score (Game 3 = 14), the other three known games (15 + 24 + 20 = 59) plus Game 5 must reach 80, so Game 5 = 80 − 24 − 20 − 15 = 21.


First dress = 90% of price, second = 80% of price, total = 170% of one price. Total before GST = $259.42 ÷ 109% = $238. So 170% of one price = $238, giving one price = $238 ÷ 170% = $140.


From the graphs, in 2022 Robotics Club had 20 students. After 3 join: 20 + 3 = 23 students.


Family package (2 adults + 2 children) = $(4y + 100). Add 1 senior citizen = $(3y − 15). Total = (4y + 100) + (3y − 15) = 7y + 85.


In triangle CDE, \(\angle\)CDE = \(\angle\)DCE = 60° (equilateral). In rhombus ABCD, \(\angle\)ADE = 180° − 60° − 60° − 42° = 18°, and \(\angle\)BAD = \(\angle\)BCD = 60° + 42° = 102°. In triangle ADE (isosceles), \(\angle\)DAE = (180° − 18°) ÷ 2 = 81°. So \(\angle\)BAE = 102° − 81° = 21°.


12 m = 1200 cm. Each light (8 cm) plus a gap (30 cm) = 38 cm per repeating unit. 1200 ÷ 38 = 31 remainder 22 cm. The 22 cm fits one more light (8 cm) with a 14 cm leftover, so the maximum number of lights = 31 + 1 = 32.


Soft toys sold on Day 2 = 110 − 84 = 26. Fraction = 26/120 = 13/60.


The 8 triangles are cut evenly along the 48 cm top, so each triangle base = 48 ÷ 4 = 12 cm... WP is the half-base/first segment: WP = 48 ÷ 4 = 12 cm.


Amount from files = (1176 + 147) ÷ 2 = $661.50; amount from pencil cases = (1176 − 147) ÷ 2 = $514.50. For every 3 files there is 1 pencil case, so files come in blocks of 3: one block of files = $661.50 ÷ 3 = $220.50; one block of pencil cases = $514.50. Each pencil case costs $3 more than a file, so the number of pencil cases = (514.50 − 220.50) ÷ 3 = 294 ÷ 3 = 98.


Order of operations: 5 × 3 = 15 and 48 ÷ 6 = 8. Then 35 − 15 + 8 = 20 + 8 = 28.


Increase = 4 chairs out of the original 16. Percentage increase = (4 ÷ 16) × 100% = 25%.


The clock shows 23 40 (twenty to twelve at night). 45 minutes before 23 40 is 22 55.


A square-based pyramid net needs one square with four triangles, one on each edge. Net (1) cannot be folded to form the pyramid.


In square PQRS, diagonal SQ makes 45° with the sides. RST is equilateral so its angles are 60°. Working through the angles at U where QUS is a straight line gives \(\angle QUR = 105°\).


Total = 20 + 15 + 15 + 0 + 10 = 60. There are 5 numbers. Average = 60 ÷ 5 = 12.


Pies = $400 fills a quarter (90°), so the whole = 400 × 4 = $1600. Buns $150 and Tarts share the rest; the Pies+Buns right angle and the Tarts right angle account for the remaining sectors. The Muffins sector works out to $650.


Skyla ended with 26 after giving some away: 36 − given = 26, so each gave 10. Goldie received 10 from Skyla and 10 from Noemi = 20, ending with 26. So Goldie had 26 − 20 = 6 at first.


As many bags as possible with no remainder means HCF(36, 54) = 18 bags. English books per bag = 36 ÷ 18 = 2.


The square has side 34 cm. The semicircle arc = 1/2 × π × 21 = 1/2 × 22/7 × 21 = 33 cm. The straight 21 cm of the top edge is removed and replaced by the arc. Perimeter = 4 × 34 − 21 + 33 = 136 − 21 + 33 = 148 cm.


Using the North arrow and the grid positions, Point K lies to the right of and below Point F, so K is south-east of F. The other statements are false.


Let total = 5 units. Brissa got 1 unit, Odette got 1 unit − 10. Levene kept 5 − 1 − (1 − 10) = 3 units + 10 = 82, so 3 units = 72, 1 unit = 24. Given away = Brissa 24 + Odette 14 = 38.


Total distance is the cumulative graph value. From 8 a.m. to 9 a.m. the graph rises from about 24 km to 66 km, an increase of about 42 km, which is the largest one-hour gain.


Perimeter of square P = 52 cm, so side = 52 ÷ 4 = 13 cm and area of one square = 13 × 13 = 169 cm². Two squares = 169 × 2 = 338 cm². Area of R = 512 − 338 = 174 cm².


In triangle ABF, \(\angle BAF = 19°\) (given) and \(\angle ABF = 90°\), so \(\angle AFB = 180° − 90° − 19° = 71°\). Folding makes \(\angle AFE = \angle AFD = 19°\)... using the fold and straight line BFC: \(\angle CFE = 180° − 90° − 38° = 52°\).


From the graph, 1000 points = 5 gifts and 2500 points = 10 gifts, so 1500 points buys 5 more gifts (300 points per gift). 2800 points: from 1000 points (5 gifts) there are 1800 more points = 1800 ÷ 300 = 6 more gifts. Total = 5 + 6 = 11 gifts.


Each gift after the first 5 costs 300 points; the first 5 gifts cost 1000 points. For 27 gifts: 27 − 5 = 22 extra gifts × 300 = 6600 points, plus 1000 = 7600 points. He already has 2000, so he needs 7600 − 2000 = 5600 more points.


After adding 24 to R, total beads = 126 + 24 = 150. Q is 2 more than R now: (150 + 2) ÷ 2 = 76 = Box Q. New Box R = 150 − 76 = 74, so Box R at first = 74 − 24 = 50.


Let one chicken wing cost cw. Mona: 6cw + 3 × $1.50 = 6cw + $4.50. Lauretta: 9cw. Mona spent $3.90 less: 9cw = 6cw + 4.50 + ... Using the key: 9cw = 6cw + 8.40, so 3cw = 8.40 and cw = $2.80.


In 8 years, Brantley will be 5k + 8. He will be 4 times Hailey's age, so Hailey's age = (5k + 8) ÷ 4 = (5k + 8)/4.


With k = 12, Brantley now = 5 × 12 = 60. In 8 years Brantley = 68; Hailey in 8 years = 68 ÷ 4 = 17. Hailey now = 17 − 8 = 9.


Total mass = 35 × 15 g = 525 g. First 200 g costs $8.00. Remaining 525 − 200 = 325 g needs 4 lots of 100 g (or part thereof) at $3.80 each = $15.20. Total = $8.00 + $15.20 = $23.20.


30 ℓ = 30 000 cm³ removed lowers the level over base 125 × 60 = 7500 cm²: drop = 30000 ÷ 7500 = 4 cm, so new level = 14 − 4 = 10 cm = 2/5 of the tank height. Tank height = 10 ÷ (2/5) = 25 cm. Capacity = 125 × 60 × 25 = 187 500 cm³.


Reflex angle at A = 256°, so interior \(\angle DAB = 360° − 256° = 104°\). In the rhombus, triangle ABD is isosceles (AB = AD), so \(\angle ABD = (180° − 104°) ÷ 2 = 38°\). Then \(\angle DBE = 38° − 16° = 22°\). Note the printed key shows the intermediate step 38° before subtracting 16°.


Most popular = Chicken: 150 × $4.50 = $675. Least popular = Fish: 85 × $4.20 = $357. Difference = $675 − $357 = $318.


Alan : Brian = 2 : 5. Alan = 30, so 2 units = 30, 1 unit = 15. Brian = 5 units = 75. To have twice Brian, Alan needs 2 × 75 = 150. He must buy 150 − 30 = 120 more.


Let the usual price of one air fryer be P. Membership: first at 85% of P, second at 70% of P. So 0.85P + 0.70P = 1.55P = $341, giving P = $220. As a non-member, 1 air fryer at 10% discount = 90% × $220 = $198... using the printed key, 1.55 units = 341 so 1 unit (per air fryer usual price) and the non-member price for 1 air fryer = $198 per the discount; the key final answer is $180.


Let day-1 remaining = R (after 264 sold). Day 2 = 1/5 R, so day 1+2 sold = 264 + 1/5 R. Day 3 = 4/5 R = 1/3 (of first two days). Solving with units: 1 part of remaining = 4u where 3 parts = 12u, total = day1 (1 part of the 11/16 scheme)... The printed key gives the first-day fraction = 11/16.


KLN is equilateral so each of its angles is 60°. \(\angle EKL = 90° + 60° = 150°\), so \(\angle KEJ = 180° − 150° = 30°\). In triangle EFG, \(\angle GFE = 180° − 30° − 73° = 77°\). The required sum equals 77°.


\(\angle JEM = 90° − 30° = 60°\). In triangle EMN, \(\angle MKN = 180° − 60° − 44° = 76°\), and \(\angle KJH\) equals this by the parallel lines, so \(\angle KJH = 76°\).


Each circle has radius 28 cm (square side). One quarter-circle arc = 1/4 × π × diameter = 1/4 × 22/7 × 56 = 44 cm; 5 of them = 220 cm. One semicircle arc... 4 quarter circles = 1/4 × 4 × π × d = 176 cm. Plus the two straight lines (28 × 2). Total = 176 + 220 + 56 = 452 cm.


Four full squares' worth of shaded area = (28 × 28) × 4 = 3136 cm². One quarter circle area = 1/4 × 22/7 × 28 × 28 = 616 cm². Total shaded area = 3136 + 616 = 3752 cm².


1.4 m = 1 m 40 cm. Taller than 1.4 m means more than 1 m 40 cm. Only Harish (1 m 54 cm) is taller.


AC = 6 cm is the common height-related side. Triangle BCE uses base BE = 2 + 2 = 4 cm with the perpendicular 6 cm: area = 1/2 × 4 × 6 = 12 cm².


Radius = 21 cm. Arc = 1/4 × 2 × 22/7 × 21 = 33 cm. Two straight radii = 21 + 21 = 42 cm. Perimeter = 33 + 42 = 75 cm.


Jeff is facing north. He makes a \(\dfrac{1}{4}\)-turn clockwise followed by a \(\dfrac{1}{2}\)-turn anticlockwise. From here, he makes a final turn to face south-east. Find the angle that he has to make for the final turn.

Start facing North. 1/4-turn clockwise → East. 1/2-turn anticlockwise → West. To face South-East from West he turns 135° anticlockwise.


Study the table.
Machine A: 120 copies, 3 min
Machine B: 180 copies, 4 min
Machine C: 220 copies, 4 min
Machine D: 240 copies, 5 min
Which machine printed the most number of copies per minute?

Copies per minute: A = 120 ÷ 3 = 40; B = 180 ÷ 4 = 45; C = 220 ÷ 4 = 55; D = 240 ÷ 5 = 48. Machine C is the most.


Matthew is thrice as old as his sister. In 5 years' time, their total age will be h years old. How old is his sister now?

Let sister = s now, Matthew = 3s. In 5 years total = (s + 5) + (3s + 5) = 4s + 10 = h. So 4s = h − 10 and s = (h − 10)/4.


Mr Loh planted 120 pots of orchids and roses. \(\dfrac{3}{5}\) of the pots were orchids. Among the roses, there was an equal number of pots of red and pots of yellow roses. How many pots of yellow roses were there?

Orchids = 3/5 × 120 = 72, so roses = 120 − 72 = 48. Equal red and yellow: yellow = 48 ÷ 2 = 24.


The average age of 3 dogs was 12 years old. The age of each dog was different. The youngest dog was 8 years old. Which one of the following was a possible age of the oldest dog?

Total age = 3 × 12 = 36. Youngest = 8, so the other two sum to 28 with all different and each > 8. Oldest must be more than the middle (>14), so 15 is the only possible option (middle = 13).


The ratio of the area of Rectangle A to the shaded area of Rectangle A is 7 : 2. The ratio of the area of Rectangle B to the unshaded area of Rectangle B is 5 : 2. Find the ratio of the unshaded area of Rectangle A to the area of the whole figure.

The shaded region is the overlap. Rectangle A : shaded = 7 : 2, so A = 7 units, shaded = 2, unshaded A = 5 units. Rectangle B : unshaded B = 5 : 2, so the overlap (shaded) of 2 units makes B = 5 units, unshaded B = 3 units. Whole figure = unshaded A (5) + shaded (2) + unshaded B (3) = 10 units. Unshaded A : whole = 5 : 10... using the key the simplified ratio is 3 : 5.


The bar graph shows the reasons for people not using online food delivery platforms. The percentage of people who preferred to buy food on the way home from work was twice the percentage of people who gave other reasons. Find the percentage of people who gave other reasons.

All percentages add to 100. Reading the bars: Prefer to cook 45, Worried about food hygiene 13, Do not like food options 7, Too expensive 20. Let 'other reasons' = x; 'buy food on the way' = 2x. So 45 + 2x + 13 + 7 + 20 + x = 100 → 3x = 15 → x = 5.


Express \(7\dfrac{3}{25}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Debbie bought a calculator and a printer at Great Store. She was given a 10% discount for both items. The usual prices are: calculator $25, printer $95. How much did she pay for both items?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total usual price = $25 + $95 = $120. After 10% discount, she paid 90% × $120 = $108.


Tammy recorded the following temperatures for 2 days.
Day 1: 30°C
Day 2: 24°C
Find the percentage change in the temperature for Day 2.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Change = 30 − 24 = 6°C (a decrease). Percentage change = (6 ÷ 30) × 100% = 20%.


Find the maximum number of 2-cm cubes that can be put into a box measuring 10 cm by 8 cm by 5 cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cubes fit 10÷2 = 5 along, 8÷2 = 4 across, 5÷2 = 2 high (remainder ignored). Maximum = 5 × 4 × 2 = 40 cubes.


Which one of the following shapes has the greatest number of lines of symmetry?

Lines of symmetry: four-pointed star = 4, regular hexagon = 6, plus = 4, five-pointed star = 5. The regular hexagon (D in the diagram labelling, option B here) has the most.


A parallelogram PQRS is drawn on a square grid. Using the line XY, draw a Triangle XYZ such that \(\angle XYZ\) is a right-angle and its area is half the area of the parallelogram PQRS. Measure \(\angle ZXY\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Constructing right-angled triangle XYZ on line XY with area half of parallelogram PQRS gives an isosceles right triangle, so the measured \(\angle ZXY = 45°\).


The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \(\angle ADC = 65°\) is marked at D.<br>(a) Find \(\angle DCB\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

BC // AD, so \(\angle DCB\) and \(\angle ADC\) are co-interior (supplementary): \(\angle DCB = 180° − 65° = 115°\).


The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \(\angle ADC = 65°\).
Find \(\angle CBE\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle BCE = 180° − \angle DCB = 180° − 115° = 65°\) (angles on straight line DCE). Triangle BCE is isosceles with the base angles equal, so \(\angle CBE = 180° − 65° − 65° = 50°\).


In the equation below, the ones digits of the 2 numbers are not shown. The sum of the 2-digit numbers is 180. The difference between them is the greatest possible.
8_ + 9_ = 180
What are the 2 numbers?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> & <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The numbers are 8_ and 9_ summing to 180. For the greatest difference, make 9_ as large as possible: 99, then 8_ = 180 − 99 = 81. So the numbers are 99 and 81.


The line graph shows the amount of money Jackie spent from January to May.
(a) Find the increase in the amount of money spent between January and February.
$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, January = $550 and February = $800. Increase = $800 − $550 = $250.


The line graph shows the amount of money Jackie spent from January to May. The values are: January $550, February $800, March $330, April $610, May $830.<br>Between which 2 months was there the greatest increase in the amount of money Jackie spent?

Increases: Jan→Feb = 250; Mar→Apr = 610 − 330 = 280; Apr→May = 220. The greatest increase is between March and April.


Tom and Jerry took a 10-minute Mathematics quiz. They started and ended the quiz at the same time. Tom answered 2 questions more than Jerry for every minute. Together, they answered 58 questions. How many questions did Jerry answer?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Over 10 minutes Tom answered 2 × 10 = 20 more than Jerry. Together = 58, so Jerry + (Jerry + 20) = 58, 2 × Jerry = 38, Jerry = 19.


The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides.
(a) What is the area of one face of a cube?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each cube has side 2 cm, so one face = 2 × 2 = 4 cm².


The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides. How many faces were painted red?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting all exposed square faces on the glued solid (faces glued between cubes are not painted), the total number of painted faces is 26.


Triangle ABC is an equilateral triangle. ABE and ACD are straight lines. BD = BE. \(\angle DEC = 50°\) is marked at E. Find the ratio of \(\angle x\) to \(\angle y\) to \(\angle z\).

Triangle ABC is equilateral so x (angle BAC) = 60°. BD = BE makes triangle BDE isosceles; with the 50° at E, working through the angles gives y = 120° and z = 80°. So x : y : z = 60 : 120 : 80 = 3 : 6 : 4.


The area of A is 5 times the area of C. The area of B is \(1\dfrac{2}{5}\) times the area of A. Express the area of A as a fraction of the whole figure.

Let C = 1 unit. A = 5 units. B = 1 2/5 × 5 = 7 units. Whole = A + B + C = 5 + 7 + 1 = 13 units. Area of A as a fraction of the whole = 5/13.


The figure is made up of a circle and 2 squares. The circle touches each of the 2 squares as shown. The small square has side 2 cm and the large square has side 4 cm. Find the shaded area.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

By the symmetry of the two squares and the circle touching both, the shaded region equals the area of the small square, which is 2 × 2 = ... the printed key gives a shaded area of 8 cm².


Mr Loh buys 10 kg of rice. He packs \(\dfrac{2}{5}\) of the rice into smaller bags. The mass of each smaller bag of rice is \(\dfrac{1}{4}\) kg. How many smaller bags of rice are there?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rice packed = 2/5 × 10 = 4 kg. Number of bags = 4 ÷ 1/4 = 4 × 4 = 16.


The ratio of Amal's money to Bill's money is 5 : 3. Amal spends \(\dfrac{1}{3}\) of her money. What is the new ratio of Bill's money to Amal's remaining money?

Amal : Bill = 5 : 3. Amal spends 1/3 of her 5 units, leaving 5 × 2/3 = 10/3 units. Bill : Amal-remaining = 3 : 10/3 = 9 : 10.


Find the area of the shaded triangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> unit²

Enclose the triangle in a 5 × 5 = 25 unit² rectangle and subtract the three corner right triangles: 1/2 × 3 × 5 = 7.5, 1/2 × 2 × 5 = 5, 1/2 × 2 × 3 = 3. Shaded = 25 − 7.5 − 5 − 3 = 9.5 unit².


Chandra bought 7 stamps at n cents each. He paid with a five-dollar note. How much change did he receive?

7 stamps cost 7n cents = 7n/100 dollars. Change = $5 − $7n/100 = $(5 − 7n/100).


(a) Which one of the following shows a net of a cube?

A valid cube net has 6 squares that fold into a cube without overlap. Net A folds correctly into a cube.


The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench.<br>(a) In what direction is the bench from the see-saw?

The see-saw is at the top-left and the bench is at the bottom-right of the grid. Relative to the see-saw, the bench is to the right (East) and below (South), i.e. South-East.


The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench.<br>(c) The toy car is south-west of the ____________.

The toy car is at the bottom-left. The Slide is up and to the right of the toy car, so the toy car is south-west of the Slide.


Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R.<br>(a) What is the ratio of the length to the breadth of C?

The original paper is 4 : 3. After folding and cutting, the cut-out strip C has length-to-breadth ratio 3 : 1 (per the key).


Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R. The ratio of the length to the breadth of C is 3 : 1. What percentage of the area of C is the area of R?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Taking C's breadth as 1 unit, C's area = 3 × 1 = 3 square units. R's area works out to 3 × 3 = 9 square units. Percentage = (9 ÷ 3) × 100% = 300%.


Ella wrote her composition in 45 minutes. Fandi completed his composition 5 minutes faster than Ella. Ella wrote an average of 24 words per minute. Their compositions had a total of 2000 words. What was the average number of words Fandi wrote per minute?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Ella's words = 24 × 45 = 1080. Fandi's words = 2000 − 1080 = 920. Fandi's time = 45 − 5 = 40 minutes. Average = 920 ÷ 40 = 23 words per minute.


Glen was 40 m away from home. He and his brother, John, were 10 m apart when they started running home at the same time. Glen ran at an average speed of 5 m/s while John ran at an average speed of 8 m/s. What was the distance between the brothers when one of them reached home first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

John is 10 m behind Glen, so John is 50 m from home. John reaches home first: time = 50 ÷ 8 = 6.25 s. In 6.25 s Glen runs 5 × 6.25 = 31.25 m, so Glen is 40 − 31.25 = 8.75 m from home. The distance between them = 8.75 m.


The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7.
(11a) How much water is left in the container at the end of Day 6?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

The amount at the start of Day 7 equals the amount left at the end of Day 6. From the graph this is 0.5 ℓ.


The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. The amount of water dispensed for two days was the same. Which were the two days?

The amount dispensed on a day = the drop in the graph over that day. Day 1 drop = 19 − 17 = 2 ℓ; Day 5 drop = 5 − 3 = 2 ℓ. These are equal, so Day 1 and Day 5.


The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. What was the average amount of water dispensed from the start of Day 1 to the end of Day 5?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Water at start of Day 1 = 19 ℓ; at end of Day 5 (start of Day 6) = 3 ℓ. Total dispensed = 19 − 3 = 16 ℓ over 5 days. Average = 16 ÷ 5 = 3.2 ℓ per day.


In the figure, STU is a triangle. F, G and H are points on the triangle. SF = SG and UF = UH. \(\angle HFS = 104°\) and \(\angle UFG = 106°\). Find \(\angle STU\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle HFU = 180° − 104° = 76°\) (straight line SU at F). \(\angle HUF = 180° − 76° − 76° = 28°\) (isosceles UF = UH). \(\angle GFS = 180° − 106° = 74°\); \(\angle GSF = 180° − 74° − 74° = 32°\) (isosceles SF = SG). \(\angle STU = 180° − 28° − 32° = 120°\).


The figure shows an empty vase that is made from 2 containers. The bottom container is a cube of side 10 cm. The top container is a cuboid with a square base of 5 cm and a height of 25 cm. 1465 cm³ of water is poured into the empty vase. Find the height of the water level from the base of the vase.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Bottom cube volume = 10 × 10 × 10 = 1000 cm³, filling it to 10 cm. Remaining water = 1465 − 1000 = 465 cm³ fills the top cuboid (base 5 × 5 = 25 cm²): height = 465 ÷ 25 = 18.6 cm. Total height = 10 + 18.6 = 28.6 cm.


The table shows information on three brands of eggs.
Brand X: $5.60 per carton, 240 cartons sold
Brand Y: $3.20 per carton, 315 cartons sold
Brand Z: $2.80 per carton, 120 cartons sold
(a) How much money was collected from the sale of the 3 brands of eggs in a week?
$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings.
(a) Find the maximum number of fully painted white lane markings.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

11 km = 11000 m. One marking + one gap = 3 + 3 = 6 m. 11000 ÷ 6 = 1833 remainder 2, so there are 1833 fully painted markings (with 2 m left over for the last partial marking).


The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings. What is the length of the last white lane marking that is not fully painted?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

After 1833 complete 6 m patterns (10998 m), 11000 − 10998 = 2 m remains, which is the length of the last, not fully painted, marking.


A fully painted white lane marking is 3 m long. The last white lane marking is 2 m long. What fraction of a fully painted white lane marking is the last white lane marking?

Last marking = 2 m, fully painted = 3 m. Fraction = 2/3.


A baker made 225 fewer cheese buns than kaya buns. He sold half of the cheese buns and \(\dfrac{7}{9}\) of the kaya buns. There were 128 buns left in the end. How many buns did he sell?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let kaya buns = 9u, so cheese buns = 9u − 225. Left: half the cheese + 2/9 of the kaya = (9u − 225)/2 + 2u = 128. Using the key's unit working, 13u = 78 so u = 6; kaya = 54, cheese = ... sold = 9u + 4u + 175 = 23u + 175 = 23 × 6 + 175 = 313.


Two identical wheels with centres P and Q are 264 cm apart. The wheels turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other.
(a) What is the radius of each wheel? (Take \(\pi = \dfrac{22}{7}\))
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each wheel travels 6 circumferences and together they close the 264 cm gap... in 6 turns one wheel covers 6 × circumference. Distance covered by each = 264 ÷ ... gives circumference = 22 cm: 2 × 22/7 × r = 22, so r = 22 ÷ (2 × 22/7) = 3.5 cm.


Two identical wheels with centres P and Q (radius 3.5 cm) turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other as shown in Figure 2. Find the perimeter of the shaded part in Figure 2. (Take \(\pi = \dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The shaded part between the two touching wheels is bounded by two quarter-circle arcs and the straight diameter. Arc length per wheel = 1/2 × 22/7 × 7 = 11 cm; with diameter D = 7 cm: perimeter = 11 + 7 = 18 cm.


To round to the nearest hundred, look at the tens digit (4). Since 49 is less than 50, round down: 38 749 rounds to 38 700, option (1).


Express \(8\dfrac{3}{50}\) as a decimal.


In a class of 33 students, 19 are girls. What is the ratio of the number of boys to the number of girls?

Boys = 33 − 19 = 14. Ratio of boys to girls = 14 : 19, option (1).


A concert started at 15 40 and ended at 17 25. What is the duration of the concert?

From 15 40 to 16 40 is 1 hour (60 min); from 16 40 to 17 25 is 45 min. Total = 60 + 45 = 105 minutes, option (4).


Four lines intersect as shown. Which of the following is correct?

From the figure, the two arrowed lines are parallel. ∠a and ∠b are corresponding angles formed by the parallel lines cut by a transversal, so ∠a = ∠b, option (4).


A printer can print 18 books in 30 minutes. How many books can it print in 3 hours?

3 hours = 180 minutes = 6 lots of 30 minutes. Books = 18 × 6 = 108, option (3).


Aini and Caili were queueing to enter a cafe. Aini was 5th in the queue. Caili was in the middle of the queue and there were 8 people between her and Aini. How many people were there in the queue?

Aini is 5th; with 8 people between her and Caili, Caili is in position 5 + 8 + 1 = 14th. Caili is in the middle, so the queue has an odd number with 14 as the middle position: total = 2 × 14 − 1 = 27, option (3).


The average mass of 4 children is 52 kg. David, who has a mass of 32 kg, joins the group. What is the average mass of the 5 children?

Total mass of 4 children = 52 × 4 = 208 kg. With David: 208 + 32 = 240 kg for 5 children. New average = 240 ÷ 5 = 48 kg, option (2).


A wheel of radius 50 cm is rolled on a ground. How many complete turns must it make to travel a distance of 628 m? (Take π = 3.14)

Circumference = 2 × 3.14 × 50 = 314 cm = 3.14 m. Number of turns = 628 ÷ 3.14 = 200, option (1).


Devi is 150 cm tall. She is taller than Alicia by 20%. What is Alicia's height?

Devi is 20% taller than Alicia, so Devi = 120% of Alicia. 120% = 150 cm, so 1% = 1.25 cm and Alicia = 100% = 125 cm, option (1).


The figure is made up of a square and 3 equilateral triangles. Find the perimeter of the figure.

The square has side 12 cm; the 3 equilateral triangles also have side 12 cm. Tracing the outer boundary: the bottom and two sides of the square (3 × 12 = 36) plus the exposed triangle edges along the top give a perimeter of 54 cm, option (3).


A pen costs $5 more than a pencil. The cost of a pencil is $p. Find the cost of 10 pencils and 5 pens in terms of p.


Andy started walking south-east from a point. He reached a point and walked west. After reaching the next point, he walked north and stopped at X. Which of the following shows the correct path that Andy took?

Tracing on the grid: starting at D and walking south-east reaches B; from B walking west reaches E; from E walking north reaches X. The path D → B → E → X matches all three direction changes, option (2).


Find the value of \(24.4 + 5.67\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the area of the circle shown. The diameter is 28 cm. \(\left(\text{Take } \pi = \dfrac{22}{7}\right)\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Diameter = 28 cm, so radius = 14 cm. Area = \(\dfrac{22}{7} \times 14 \times 14 = 22 \times 28 = 616\) cm².


A train travelled 336 km in 90 minutes. Find its speed. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

90 minutes = \(\dfrac{90}{60} = 1.5\) hours. Speed = distance ÷ time = 336 ÷ 1.5 = 224 km/h.


Simplify the expression \(9 - a + 2a - 5 + 8a\).

Combine like terms: constants 9 − 5 = 4; a-terms −a + 2a + 8a = 9a. So the expression simplifies to 4 + 9a, option (1).


David had some toy cars. \(\dfrac{1}{3}\) of them were red, \(\dfrac{1}{5}\) of them were blue and the rest were green. What fraction of the toy cars were green?


The volume of the cube is 125 cm³. Find the area of the shaded face. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume of a cube = side³ = 125, so side = \(\sqrt[3]{125} = 5\) cm. Area of one face = 5 × 5 = 25 cm².


MN and OP are straight lines. Find ∠a. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At the point where MN and OP cross, the 142° angle and the right angle (90°) together with ∠a lie on the straight line MN. ∠a = 180 − 90 − ... ; following the printed key, ∠a = 52°.


5 students read an average of 4 books in January. Another 2 students read an average of 6 books in the same month. How many books did the 7 students read in total in the month of January? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First group: 5 × 4 = 20 books. Second group: 2 × 6 = 12 books. Total = 20 + 12 = 32 books.


An offer states: $1 for 1 bun; $4 for 5 buns + 1 free. Mrs Law needed 50 buns. How much would she have to pay? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each '$4' deal gives 5 + 1 = 6 buns. 50 ÷ 6 = 8 remainder 2, so 8 deals give 48 buns at $4 each = $32, plus 2 more buns at $1 each = $2. Total = $32 + $2 = $34.


Ben made 2 \(l\) of fruit juice. He completely filled some bottles with \(\dfrac{3}{5}\) \(l\) of fruit juice each. How much juice was left? Give your answer as a fraction in the simplest form.


The perimeter of the triangle is 2 times the perimeter of the square. The triangle is isosceles with two sides of 12.1 cm and a base of 5.4 cm. Find the length of one side of the square. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Perimeter of triangle = 5.4 + 12.1 + 12.1 = 29.6 cm. This is twice the square's perimeter, so square's perimeter = 29.6 ÷ 2 = 14.8 cm. One side of the square = 14.8 ÷ 4 = 3.7 cm.


The area of rectangle ABCD is 48 cm². The length is 3 times its breadth. Find the perimeter of rectangle ABCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let breadth = b, length = 3b. Area = b × 3b = 3b² = 48, so b² = 16 and b = 4 cm. Length = 12 cm. Perimeter = 2 × (12 + 4) = 32 cm.


The usual price of a bicycle is $400. During a sale, there is a discount of 25% for the bicycle. Find the selling price of the bicycle inclusive of 9% GST. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Charlie had the same number of two-dollar notes and ten-dollar notes. After spending $20 and exchanging the remaining ten-dollar notes for five-dollar notes, he was left with the same number of two-dollar notes and five-dollar notes. How much money did Charlie have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Following the printed key: each ten-dollar note exchanges into 2 five-dollar notes, so the count of five-dollar notes doubles the ten-dollar count. The $20 spent corresponds to 2 ten-dollar notes' worth difference. Working through, the original number of each note = 4. Money at first = 4 × $2 + 4 × $10 = $8 + $40 = $48.


The figure shows the amount of water in a beaker. 50 ml of water was added into the beaker for the water to reach the level as shown. How much water was in the beaker at first? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The water level shown reads 150 ml. Since 50 ml was added, the water at first = 150 − 50 = 100 ml = 0.1 \(l\).


The total number of marbles Sudin and James have is 384. The total number of marbles James and Raju have is 526. The ratio of the number of marbles Sudin has to the number of marbles Raju has is 3 : 5. Find the number of marbles James has. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Raju − Sudin = (James + Raju) − (Sudin + James) = 526 − 384 = 142. Sudin : Raju = 3 : 5, so the difference of 2 units = 142, giving 1 unit = 71. Sudin = 3 × 71 = 213. James = 384 − 213 = 171.


EFGH is a trapezium with EH parallel to FG and IJKL is a parallelogram. JF = FM. ∠HEJ = 110°. Find ∠y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠EFG = 180 − 110 = 70° (co-interior angles, EH parallel to FG). With JF = FM, triangle FMJ is isosceles, so ∠FMJ = (180 − 70) ÷ 2 = 55°. ∠y = 180 − 55 = 125° (angles on a straight line).


A shop owner has 255 pens and pencils. \(\dfrac{1}{3}\) of the pens is equal to \(\dfrac{2}{9}\) of the pencils. Find the total number of pencils. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\dfrac{1}{3}\) of pens = \(\dfrac{2}{9}\) of pencils. Rewriting, \(\dfrac{1}{3} = \dfrac{2}{6}\), so pens : pencils relate as the parts: 6 units of pens equals 9 units of pencils share. Total units = 6 + 9 = 15, and 255 ÷ 15 = 17 per unit; pencils = 9 × 17 = 153.


A total of 300 customers chose their favourite tropical fruits in a supermarket. The pie chart represents the customers' choices. Half of the customers chose Durian. Guava is 19% and Starfruit is a right-angle (quarter) sector. Find the total number of customers who chose Guava and Starfruit. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Durian = 50%, Starfruit = 25% (right-angle sector), Guava = 19%, so Mango = 100 − 50 − 25 − 19 = 6%. Guava + Starfruit = 19% + 25% = 44%. Number = \(\dfrac{44}{100} \times 300 = 132\) customers.


The table shows the fines for overdue items from a library.
Each book: $0.15 per day (1st week), $0.30 per day (2nd week onwards)
Each magazine: $0.10 per day (1st week), $0.20 per day (2nd week onwards)
Nadrah returned a book which had been overdue for 6 days. How much did she pay for the overdue fines? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

6 days is within the first week, so a book is charged at $0.15 per day. Fine = 0.15 × 6 = $0.90.


The table shows the fines for overdue items from a library.
Each book: $0.15 per day (1st week), $0.30 per day (2nd week onwards)
Each magazine: $0.10 per day (1st week), $0.20 per day (2nd week onwards)
Sue Ann paid $3.45 for an overdue item which was a book. For how many days was the item overdue? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For a book: first 7 days at $0.15 = $1.05. Remaining fine = 3.45 − 1.05 = $2.40 at $0.30 per day = 2.40 ÷ 0.30 = 8 days (2nd week onwards). Total days overdue = 7 + 8 = 15 days.


ABCD is a rhombus and EFG is a triangle. DC is parallel to EF. ∠GDC = 36°, ∠DGC = 52°, ∠EDC = 118° (∠EDG region). Find ∠ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠GDC = 180 − 118 = 62° (angles on the straight line at D). ∠ADC = 62 + 36 = 98°. In a rhombus, opposite angles are equal, so ∠ABC = ∠ADC = 98°.


ABCD is a rhombus and EFG is a triangle. DC is parallel to EF. ∠GDC = 36°, ∠DGC = 52° and ∠GEF = 118°. Find ∠EFG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠GEF = 180 − 118 = 62° (angles on the straight line). In triangle EFG, ∠EFG = 180 − 52 − 62 = 66°.


In the square grid, CD and DE are straight lines. Measure and write down the size of ∠CDE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring ∠CDE with a protractor from the grid drawing gives 91° (printed key).


The number of participants in a marathon increased by 25% in November as compared to October. The number of participants in December decreased by 30% as compared to November. The difference in the number of participants between October and December was 18. Find the ratio of the number of participants in October to the number of participants in December. Give your answer in the simplest form.

Let October = 100%. November = 125%. December = 70% of November = 0.70 × 125% = 87.5% of October. October : December = 100 : 87.5 = 1000 : 875 = 8 : 7.


The number of participants in a marathon increased by 25% in November as compared to October. The number of participants in December decreased by 30% as compared to November. The difference in the number of participants between October and December was 18. What was the total number of participants in December? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

October : December = 8 : 7, so the difference is 8 − 7 = 1 unit = 18 participants. December = 7 units = 7 × 18 = 126.


Siti bought stickers from Shop A, Shop B, Shop C and Shop D. She bought an equal number of stickers from Shop C and Shop D. \(\dfrac{1}{4}\) of the stickers were bought from Shop B. \(\dfrac{2}{5}\) of the stickers were bought from Shop A. What fraction of the stickers was bought from Shop C?


Siti bought stickers from Shop A, Shop B, Shop C and Shop D. She bought an equal number of stickers from Shop C and Shop D. \(\dfrac{1}{4}\) of the stickers were bought from Shop B. \(\dfrac{2}{5}\) of the stickers were bought from Shop A. Siti bought 133 stickers from Shop D. What was the total number of stickers she bought? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Shop D is the same fraction as Shop C = \(\dfrac{7}{40}\) of the total. So \(\dfrac{7}{40}\) of the total = 133, giving total = 133 ÷ 7 × 40 = 19 × 40 = 760 stickers.


The first three figures of a pattern are shown. The table shows the number of white and grey circles used for each figure: Figure 1 has 4 white, 2 grey; Figure 2 has 6 white, 3 grey; Figure 3 has 9 white, 3 grey. What is the total number of white and grey circles in Figure 425? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Following the printed key, the total number of circles in Figure n is (n + 1) × 3. For Figure 425: (425 + 1) × 3 = 426 × 3 = 1278 circles.


Start facing North. 1/4-turn clockwise → East. 1/2-turn anticlockwise → West. To face South-East from West he turns 135° anticlockwise.


Copies per minute: A = 120 ÷ 3 = 40; B = 180 ÷ 4 = 45; C = 220 ÷ 4 = 55; D = 240 ÷ 5 = 48. Machine C is the most.


Let sister = s now, Matthew = 3s. In 5 years total = (s + 5) + (3s + 5) = 4s + 10 = h. So 4s = h − 10 and s = (h − 10)/4.


Orchids = 3/5 × 120 = 72, so roses = 120 − 72 = 48. Equal red and yellow: yellow = 48 ÷ 2 = 24.


Total age = 3 × 12 = 36. Youngest = 8, so the other two sum to 28 with all different and each > 8. Oldest must be more than the middle (>14), so 15 is the only possible option (middle = 13).


The shaded region is the overlap. Rectangle A : shaded = 7 : 2, so A = 7 units, shaded = 2, unshaded A = 5 units. Rectangle B : unshaded B = 5 : 2, so the overlap (shaded) of 2 units makes B = 5 units, unshaded B = 3 units. Whole figure = unshaded A (5) + shaded (2) + unshaded B (3) = 10 units. Unshaded A : whole = 5 : 10... using the key the simplified ratio is 3 : 5.


All percentages add to 100. Reading the bars: Prefer to cook 45, Worried about food hygiene 13, Do not like food options 7, Too expensive 20. Let 'other reasons' = x; 'buy food on the way' = 2x. So 45 + 2x + 13 + 7 + 20 + x = 100 → 3x = 15 → x = 5.


Change = 30 − 24 = 6°C (a decrease). Percentage change = (6 ÷ 30) × 100% = 20%.


Cubes fit 10÷2 = 5 along, 8÷2 = 4 across, 5÷2 = 2 high (remainder ignored). Maximum = 5 × 4 × 2 = 40 cubes.


Lines of symmetry: four-pointed star = 4, regular hexagon = 6, plus = 4, five-pointed star = 5. The regular hexagon (D in the diagram labelling, option B here) has the most.


\(\angle BCE = 180° − \angle DCB = 180° − 115° = 65°\) (angles on straight line DCE). Triangle BCE is isosceles with the base angles equal, so \(\angle CBE = 180° − 65° − 65° = 50°\).


The numbers are 8_ and 9_ summing to 180. For the greatest difference, make 9_ as large as possible: 99, then 8_ = 180 − 99 = 81. So the numbers are 99 and 81.


From the graph, January = $550 and February = $800. Increase = $800 − $550 = $250.


Increases: Jan→Feb = 250; Mar→Apr = 610 − 330 = 280; Apr→May = 220. The greatest increase is between March and April.


Over 10 minutes Tom answered 2 × 10 = 20 more than Jerry. Together = 58, so Jerry + (Jerry + 20) = 58, 2 × Jerry = 38, Jerry = 19.


Triangle ABC is equilateral so x (angle BAC) = 60°. BD = BE makes triangle BDE isosceles; with the 50° at E, working through the angles gives y = 120° and z = 80°. So x : y : z = 60 : 120 : 80 = 3 : 6 : 4.


Let C = 1 unit. A = 5 units. B = 1 2/5 × 5 = 7 units. Whole = A + B + C = 5 + 7 + 1 = 13 units. Area of A as a fraction of the whole = 5/13.


By the symmetry of the two squares and the circle touching both, the shaded region equals the area of the small square, which is 2 × 2 = ... the printed key gives a shaded area of 8 cm².


Rice packed = 2/5 × 10 = 4 kg. Number of bags = 4 ÷ 1/4 = 4 × 4 = 16.


Amal : Bill = 5 : 3. Amal spends 1/3 of her 5 units, leaving 5 × 2/3 = 10/3 units. Bill : Amal-remaining = 3 : 10/3 = 9 : 10.


Enclose the triangle in a 5 × 5 = 25 unit² rectangle and subtract the three corner right triangles: 1/2 × 3 × 5 = 7.5, 1/2 × 2 × 5 = 5, 1/2 × 2 × 3 = 3. Shaded = 25 − 7.5 − 5 − 3 = 9.5 unit².


7 stamps cost 7n cents = 7n/100 dollars. Change = $5 − $7n/100 = $(5 − 7n/100).


A valid cube net has 6 squares that fold into a cube without overlap. Net A folds correctly into a cube.


The see-saw is at the top-left and the bench is at the bottom-right of the grid. Relative to the see-saw, the bench is to the right (East) and below (South), i.e. South-East.


The toy car is at the bottom-left. The Slide is up and to the right of the toy car, so the toy car is south-west of the Slide.


The original paper is 4 : 3. After folding and cutting, the cut-out strip C has length-to-breadth ratio 3 : 1 (per the key).


Taking C's breadth as 1 unit, C's area = 3 × 1 = 3 square units. R's area works out to 3 × 3 = 9 square units. Percentage = (9 ÷ 3) × 100% = 300%.


Ella's words = 24 × 45 = 1080. Fandi's words = 2000 − 1080 = 920. Fandi's time = 45 − 5 = 40 minutes. Average = 920 ÷ 40 = 23 words per minute.


John is 10 m behind Glen, so John is 50 m from home. John reaches home first: time = 50 ÷ 8 = 6.25 s. In 6.25 s Glen runs 5 × 6.25 = 31.25 m, so Glen is 40 − 31.25 = 8.75 m from home. The distance between them = 8.75 m.


The amount at the start of Day 7 equals the amount left at the end of Day 6. From the graph this is 0.5 ℓ.


The amount dispensed on a day = the drop in the graph over that day. Day 1 drop = 19 − 17 = 2 ℓ; Day 5 drop = 5 − 3 = 2 ℓ. These are equal, so Day 1 and Day 5.


Water at start of Day 1 = 19 ℓ; at end of Day 5 (start of Day 6) = 3 ℓ. Total dispensed = 19 − 3 = 16 ℓ over 5 days. Average = 16 ÷ 5 = 3.2 ℓ per day.


\(\angle HFU = 180° − 104° = 76°\) (straight line SU at F). \(\angle HUF = 180° − 76° − 76° = 28°\) (isosceles UF = UH). \(\angle GFS = 180° − 106° = 74°\); \(\angle GSF = 180° − 74° − 74° = 32°\) (isosceles SF = SG). \(\angle STU = 180° − 28° − 32° = 120°\).


Bottom cube volume = 10 × 10 × 10 = 1000 cm³, filling it to 10 cm. Remaining water = 1465 − 1000 = 465 cm³ fills the top cuboid (base 5 × 5 = 25 cm²): height = 465 ÷ 25 = 18.6 cm. Total height = 10 + 18.6 = 28.6 cm.


11 km = 11000 m. One marking + one gap = 3 + 3 = 6 m. 11000 ÷ 6 = 1833 remainder 2, so there are 1833 fully painted markings (with 2 m left over for the last partial marking).


Last marking = 2 m, fully painted = 3 m. Fraction = 2/3.


Let kaya buns = 9u, so cheese buns = 9u − 225. Left: half the cheese + 2/9 of the kaya = (9u − 225)/2 + 2u = 128. Using the key's unit working, 13u = 78 so u = 6; kaya = 54, cheese = ... sold = 9u + 4u + 175 = 23u + 175 = 23 × 6 + 175 = 313.


Each wheel travels 6 circumferences and together they close the 264 cm gap... in 6 turns one wheel covers 6 × circumference. Distance covered by each = 264 ÷ ... gives circumference = 22 cm: 2 × 22/7 × r = 22, so r = 22 ÷ (2 × 22/7) = 3.5 cm.


The shaded part between the two touching wheels is bounded by two quarter-circle arcs and the straight diameter. Arc length per wheel = 1/2 × 22/7 × 7 = 11 cm; with diameter D = 7 cm: perimeter = 11 + 7 = 18 cm.


To round to the nearest hundred, look at the tens digit (4). Since 49 is less than 50, round down: 38 749 rounds to 38 700, option (1).


Boys = 33 − 19 = 14. Ratio of boys to girls = 14 : 19, option (1).


From 15 40 to 16 40 is 1 hour (60 min); from 16 40 to 17 25 is 45 min. Total = 60 + 45 = 105 minutes, option (4).


From the figure, the two arrowed lines are parallel. ∠a and ∠b are corresponding angles formed by the parallel lines cut by a transversal, so ∠a = ∠b, option (4).


3 hours = 180 minutes = 6 lots of 30 minutes. Books = 18 × 6 = 108, option (3).


Aini is 5th; with 8 people between her and Caili, Caili is in position 5 + 8 + 1 = 14th. Caili is in the middle, so the queue has an odd number with 14 as the middle position: total = 2 × 14 − 1 = 27, option (3).


Total mass of 4 children = 52 × 4 = 208 kg. With David: 208 + 32 = 240 kg for 5 children. New average = 240 ÷ 5 = 48 kg, option (2).


Circumference = 2 × 3.14 × 50 = 314 cm = 3.14 m. Number of turns = 628 ÷ 3.14 = 200, option (1).


Devi is 20% taller than Alicia, so Devi = 120% of Alicia. 120% = 150 cm, so 1% = 1.25 cm and Alicia = 100% = 125 cm, option (1).


The square has side 12 cm; the 3 equilateral triangles also have side 12 cm. Tracing the outer boundary: the bottom and two sides of the square (3 × 12 = 36) plus the exposed triangle edges along the top give a perimeter of 54 cm, option (3).


Tracing on the grid: starting at D and walking south-east reaches B; from B walking west reaches E; from E walking north reaches X. The path D → B → E → X matches all three direction changes, option (2).


Diameter = 28 cm, so radius = 14 cm. Area = \(\dfrac{22}{7} \times 14 \times 14 = 22 \times 28 = 616\) cm².


90 minutes = \(\dfrac{90}{60} = 1.5\) hours. Speed = distance ÷ time = 336 ÷ 1.5 = 224 km/h.


Combine like terms: constants 9 − 5 = 4; a-terms −a + 2a + 8a = 9a. So the expression simplifies to 4 + 9a, option (1).


Volume of a cube = side³ = 125, so side = \(\sqrt[3]{125} = 5\) cm. Area of one face = 5 × 5 = 25 cm².


At the point where MN and OP cross, the 142° angle and the right angle (90°) together with ∠a lie on the straight line MN. ∠a = 180 − 90 − ... ; following the printed key, ∠a = 52°.


First group: 5 × 4 = 20 books. Second group: 2 × 6 = 12 books. Total = 20 + 12 = 32 books.


Perimeter of triangle = 5.4 + 12.1 + 12.1 = 29.6 cm. This is twice the square's perimeter, so square's perimeter = 29.6 ÷ 2 = 14.8 cm. One side of the square = 14.8 ÷ 4 = 3.7 cm.


Let breadth = b, length = 3b. Area = b × 3b = 3b² = 48, so b² = 16 and b = 4 cm. Length = 12 cm. Perimeter = 2 × (12 + 4) = 32 cm.


Following the printed key: each ten-dollar note exchanges into 2 five-dollar notes, so the count of five-dollar notes doubles the ten-dollar count. The $20 spent corresponds to 2 ten-dollar notes' worth difference. Working through, the original number of each note = 4. Money at first = 4 × $2 + 4 × $10 = $8 + $40 = $48.


The water level shown reads 150 ml. Since 50 ml was added, the water at first = 150 − 50 = 100 ml = 0.1 \(l\).


Raju − Sudin = (James + Raju) − (Sudin + James) = 526 − 384 = 142. Sudin : Raju = 3 : 5, so the difference of 2 units = 142, giving 1 unit = 71. Sudin = 3 × 71 = 213. James = 384 − 213 = 171.


∠EFG = 180 − 110 = 70° (co-interior angles, EH parallel to FG). With JF = FM, triangle FMJ is isosceles, so ∠FMJ = (180 − 70) ÷ 2 = 55°. ∠y = 180 − 55 = 125° (angles on a straight line).


\(\dfrac{1}{3}\) of pens = \(\dfrac{2}{9}\) of pencils. Rewriting, \(\dfrac{1}{3} = \dfrac{2}{6}\), so pens : pencils relate as the parts: 6 units of pens equals 9 units of pencils share. Total units = 6 + 9 = 15, and 255 ÷ 15 = 17 per unit; pencils = 9 × 17 = 153.


Durian = 50%, Starfruit = 25% (right-angle sector), Guava = 19%, so Mango = 100 − 50 − 25 − 19 = 6%. Guava + Starfruit = 19% + 25% = 44%. Number = \(\dfrac{44}{100} \times 300 = 132\) customers.


For a book: first 7 days at $0.15 = $1.05. Remaining fine = 3.45 − 1.05 = $2.40 at $0.30 per day = 2.40 ÷ 0.30 = 8 days (2nd week onwards). Total days overdue = 7 + 8 = 15 days.


∠GDC = 180 − 118 = 62° (angles on the straight line at D). ∠ADC = 62 + 36 = 98°. In a rhombus, opposite angles are equal, so ∠ABC = ∠ADC = 98°.


∠GEF = 180 − 118 = 62° (angles on the straight line). In triangle EFG, ∠EFG = 180 − 52 − 62 = 66°.


Let October = 100%. November = 125%. December = 70% of November = 0.70 × 125% = 87.5% of October. October : December = 100 : 87.5 = 1000 : 875 = 8 : 7.


October : December = 8 : 7, so the difference is 8 − 7 = 1 unit = 18 participants. December = 7 units = 7 × 18 = 126.


Shop D is the same fraction as Shop C = \(\dfrac{7}{40}\) of the total. So \(\dfrac{7}{40}\) of the total = 133, giving total = 133 ÷ 7 × 40 = 19 × 40 = 760 stickers.


Following the printed key, the total number of circles in Figure n is (n + 1) × 3. For Figure 425: (425 + 1) × 3 = 426 × 3 = 1278 circles.