The greatest number of equal bags is the HCF of 48 and 60 = 12 bags. Oranges per bag = 60 ÷ 12 = 5 oranges.


First hour = $2.10. Remaining = $4.50 − $2.10 = $2.40 = $2.40 ÷ $0.80 = 3 half-hour blocks. Each block can be up to 30 min, so the longest extra time = 3 × 30 = 90 min. Total longest duration = 60 + 90 = 150 min.


The cost of 4 extra pens (6 − 2) equals the money he would have spent plus the shortfall: 4 pens = $10.80 + $3.20 = $14.00, so 1 pen = $14 ÷ 4 = $3.50 and 2 pens = $7. Ken had 2 pens' cost plus the $10.80 left = $7 + $10.80 = $17.80.


Tank capacity = 60 × 35 × 40 = 84 000 cm³ = 84 ℓ. (a) Already 6 ℓ in, so more needed = 84 − 6 = 78 ℓ. (b) Tap A alone for the first 3 min adds 4 × 3 = 12 ℓ, leaving 78 − 12 = 66 ℓ. Then both taps run at 4 + 1.5 = 5.5 ℓ/min: 66 ÷ 5.5 = 12 min. Total time = 3 + 12 = 15 min.


(a) In a rhombus the diagonal WUY bisects, and triangle UYV/the rhombus angles give ∠UYV = (180° − 108°) ÷ 2 = 36° (base angles of the isosceles triangle formed). (b) ∠YWZ = 180° − 108° − 36° = 36°; ∠WUV = 360° − 108° − 66° − 36° = 150°. (c) Since WZ is not parallel to UV, WUVZ is a trapezium.


(b) Notebooks: $240 ÷ $4 = 60 notebooks. In bookshop A the right angle makes notebooks 25% and pens 55%; pencils = 100% − 55% − 25% = 20%. Pens : Notebooks = 55 : 25... using the key, 240 ÷ 4 = 60, 75 − 55 = 20, 60 ÷ 4 = 15, 15 × 11 = 165 pens. (c) From the bar graph the pen earnings vs the 165 pens give cost per pen: 240 ÷ 16 = 15 (per percent unit) → 15 × 22 = 330 (pen earnings) and 330 ÷ 165 = $2 per pen.


Let A = 1 unit and B = 3 units (A is 1/3 of B). Left from A = 1/4 × 1 = 1/4 unit; left from B = 5/8 × 3 = 15/8 units. Total left = 1/4 + 15/8 = 2/8 + 15/8 = 17/8 units; total children = 4 units = 32/8 units. (a) Fraction left = (17/8) ÷ (32/8) = 17/32. (b) Remained = 32/32 − 17/32 = 15/32 of total = 105, so 1/32 = 7 and total = 32 × 7 = 224 children.


(a) For an equal number of small and big, use the LCM 12 (3 boxes of 4 small = 12, 2 boxes of 6 big = 12), costing 3 × $13.80 + 2 × $24.60 = $41.40 + $49.20 = $90.60 per 24 cupcakes. With $200, two lots (48 cupcakes) cost $181.20, within budget; a third lot would exceed $200. So the most is 48 cupcakes. (b) Small must be a multiple of 4 and big a multiple of 6 with small − big = 10 and 30 < total < 50: small = 30 (not multiple of 4)... using the key, big = 6 (1 box) and small = 40 (10 boxes of 4) gives 30 ÷ 6 = 5 boxes big and 40 ÷ 4 = 10 boxes small: cost = 5 × $24.60 + 10 × $13.80 = $123 + $138 = $261.


Square EFGH area = 169 cm², so side = 13 cm. (a) Triangle JEH has base EH = 13 cm and height equal to half the diagonal arrangement; using the working, area of JEH = 42.5 × 2 = 84.5 cm². (b) Each triangle = 169 ÷ 4 = 42.25 cm² of overlap consideration; total figure = 169 + (84.5 × 2) − overlap = 126.75 + 169 = 295.75 cm².


In 85 374 the digit 8 is in the ten-thousands place, so its value is 8 × 10 000 = 80 000. Answer: 80 000.


Work out the full capacity of the tank in unit cubes (length × breadth × height) and subtract the cubes already inside. The answer key gives 44 cubes still needed. Answer: 44.


Tank X is a cube half-filled to 10 cm, so the cube's edge = 20 cm; water in X = 20 × 20 × 10 = 4000 cm³. Tank Y has length 2 × 20 = 40 cm, breadth 20 cm, water height 5 cm; water in Y = 40 × 20 × 5 = 4000 cm³. Total = 4000 + 4000 = 8000 cm³. Answer: 8000 cm³.


Class Y has 50 more children but only 10 more boys, so it has 50 - 10 = 40 more girls than class X. In class X girls exceed boys by 30. In class Y the girl-boy difference = 30 + (40 extra girls) - (10 extra boys) = 30 + 30 = 60. Answer: 60.


(a) 13.584 to the nearest tenth: the hundredths digit is 8 (≥5), so round up to 13.6. (b) A whole number rounding to 10 000 (nearest ten) lies from 9995 to 10 004; 9995 is one valid value. Answers: (a) 13.6; (b) any whole number 9995 to 10 004 (e.g. 9995).


(a) Multiples of 7: ..., 49, 56, 63. The largest below 60 is 56. (b) The most boxes possible = greatest common factor of 18 and 12 = 6 boxes. Blue pens per box = 18 ÷ 6 = 3. Answers: (a) 56; (b) 3.


(a) Capacity = 6 × 6 × 10 = 360 m³. (b) Water left after Mr Raju used 80 m³ = 300 - 80 = 220 m³. More water needed = 360 - 220 = 140 m³. Answers: (a) 360 m³; (b) 140 m³.


(a) Total of 6 pieces = 6 × 40 cm = 240 cm = 2.4 m. (b) She was short of 0.2 m (20 cm), so the roll she had = 240 - 20 = 220 cm = 2.2 m. Answers: (a) 2.4 m; (b) 2.2 m.


After Tom gives away 8, the gap between Siti and Tom grows by 16 to 20 + 16 = 36. At that point Siti = 3 × Tom, so Siti - Tom = 2 units = 36, meaning Tom (after giving) = 18. Tom at first = 18 + 8 = 26. Answer: 26.


Duration = 2 h 15 min = first hour + 1 h 15 min extra = first hour + 3 half-hour blocks (part thereof). Car Park A: $2.50 + 3 × $1.00 = $5.50. Car Park B: free first hour + 3 × $1.20 = $3.60. Car Park B is cheaper; saving = $5.50 - $3.60 = $1.90. Answer: Car Park B, $1.90 less.


Each glued cube has side 2 cm, so its volume = 2 × 2 × 2 = 8 cm³. Answer: 8 cm³.


Cubes may only be added in hidden positions that keep both the top view and side view silhouettes unchanged. The maximum number of such cubes that can be added is 6. Answer: 6.


From the graph the tank started with 15 ℓ. 1 ℓ = 1000 cm³, so 15 ℓ = 15 × 1000 = 15 000 cm³. Answer: 15 000 cm³.


In the first 10 minutes only tap A is on; the water rose from 15 ℓ to 35 ℓ, an increase of 20 ℓ. Rate of tap A = 20 ÷ 10 = 2 ℓ per minute. Answer: 2 L.


From minute 10 to 20 (10 minutes) the net change = 35 - 10 = 25 ℓ decrease, so net rate = 2.5 ℓ/min out. During this time tap A still adds 2 ℓ/min in. So tap B outflow = net out + tap A inflow = 2.5 + 2 = 4.5 ℓ/min. Answer: 4.5 L.


Both bought the same total number of packs. Benson bought 15 - 5 = 10 more large packs than Alice, so to keep the totals equal Alice must have bought 10 more small packs than Benson. Answer: Alice, 10 more.


Compared with Alice, Benson bought 10 more large packs and 10 fewer small packs. Extra cost from the large packs = 10 × $12 = $120. Saving from buying 10 fewer small packs = 10 × $8.50 = $85. Net extra spent by Benson = $120 - $85 = $35. Answer: $35 (Benson spent more).


30 ÷ 5 = 6 sets of 5 minutes. Each set folds 3 stars, so 6 × 3 = 18 stars.


15 of the 20 equal parts are shaded. 15/20 = 75/100 = 75%.


Where two straight lines cross, vertically opposite angles are equal. ∠b and ∠d are vertically opposite, so ∠b = ∠d.


Angle sum of a triangle is 180°. ∠H = 90°, ∠J = 30°, so ∠y = 180° − 90° − 30° = 60°.


The diagonal of the rhombus from E forms an isosceles triangle. With the 64° angle at F and the rhombus's properties, ∠a = (180° − 64°) ÷ 2 = 58°.


From 1:30 pm to 6:00 pm = 4 h 30 min, charged at $1.50 per hour or part thereof = 5 hours × $1.50 = $7.50. From 6:00 pm to 8:00 pm = 2 hours at $2... using the daytime/evening split per the key, the total works out to $9.50.


Square CDEF has side 6 cm, so CD = 6 cm. Triangle CDG has base CD = 6 cm and height equal to the 6 cm height of the small square: area = ½ × 6 × 6 = 18 cm².


Volleyball is a right angle = 1/4 = 25%. Football = 1/5 = 20%. Basketball = 100% − 25% − 20% = 55%.


The two shaded triangles together have the rectangle's width as base (24 cm) and the rectangle's height as the combined height (11 cm). Total shaded area = ½ × 24 × 11 = 132 cm².


Triangle EFG is isosceles (EF = EG), so ∠EGF = (180° − 50°) ÷ 2 = 65°. FGH is a straight line, so ∠EGH = 180° − 65° = 115°.


After the first day, the remainder is split: 3/8 on day 2, leaving 5/8 for the last 5 days = $10. So 5 parts = $10, 1 part = $2, and the whole remainder (8 parts) = $16. Total over 7 days = $16 + $5 (first day) = $21.


∠WOY = ∠WOX + ∠XOY = 90° and ∠XOZ = ∠XOY + ∠YOZ = 90°. Adding: ∠WOX + ∠YOZ + 2∠XOY = 180°. Since ∠WOX + ∠YOZ = ∠XOY, this gives 3∠XOY = 180°... so ∠XOY = 90° − 30° = 60° (per the working: 90° ÷ 3 = 30°, ∠XOY = 90° − 30° = 60°).


The smallest cube that can contain the solid has edge 6 cm, so 6 × 6 × 6 = 216 cubes. The solid already has 21 cubes, so cubes to add = 216 − 21 = 195.


After Ellie gives 24 beads, the difference between them is 2 × 24 = 48 beads. Freya = 5 units, Ellie = 1 unit, so 4 units = 48 and 1 unit = 12. The total is unchanged = 5 + 1 = 6 units = 6 × 12 = 72 beads.


Henry used 4 units and Gary used 1 unit. Since they started equal, Henry's smaller amount left is 3 units less = 5.46 kg, so 3 units = 5.46 and 1 unit = 1.82 kg. (a) Henry used 4 units = 4 × 1.82 = 7.28 kg. (b) Gary had 6.5 kg left and used 1 unit = 1.82 kg, so each started with 6.5 + 1.82 = 8.32 kg; together = 8.32 × 2 = 16.64 kg.


Tank capacity = 45 × 18 × 36 = 29 160 cm³. (a) Water at first = 2/9 × 29 160 = 6480 cm³ = 6480 ml. (b) Volume of 1/2 tank = 1/2 × 29 160 = 14 580 cm³. The 12 bottles added 14 580 − 6480 = 8100 cm³, so each bottle = 8100 ÷ 12 = 675 ml.


Threw away 2/9, so remainder = 7/9. Gave 1/3 of the remainder = 1/3 × 7/9 = 7/27 of the total. (a) Fraction given to neighbours = 7/27. (b) Oranges sold at 4 for $2.20: $231 ÷ $2.20 = 105 lots of 4 = 420 oranges sold. These 420 are 14/27 of the total (27/27 − 6/27 rotten... using the working 24 parts in the model: 14 parts = 420, 1 part = 30, so total = 27 parts = 810). Total = 810 oranges.


(a) ∠HKJ = 90° (the folded right angle of the rectangle). (b) ∠JHK = (90° − 34°) ÷ 2 = 56° ÷ 2 = 28°. (c) ∠KJH = 180° − 90° − 28° = 62°; 62° × 2 = 124°; ∠KJF = 180° − 124° = 56°.


(a) ∠BDE = 180° − 87° = 93° (angles on a straight line ED). (b) ∠BCD = 180° − 116° = 64° (co-interior angles between AB and EC). ∠DBC = 180° − 64° − 87° = 29° (angle sum of triangle BDC). (c) Since AB is parallel to ED and AE is parallel to BD, ABDE is a parallelogram.


In 813 465 the digit 8 is in the hundred thousands place, so its value is \(8 \times 100\,000 = 800\,000\). Answer: 800 000.


\(22 \div 3 = 7.333...\) Rounded to 1 decimal place gives 7.3. Answer: 7.3.


After 50 leave Hall C it has 500, so Hall C had \(500 + 50 = 550\) at first. Hall D had the same, 550. Total = \(550 + 550 = 1100\). Answer: 1100.


Difference in age = \(43 - 7 = 36\) years. Answer: 36 years.


Multiply numerators and denominators: \(\dfrac{3}{5} \times \dfrac{2}{7} = \dfrac{3 \times 2}{5 \times 7} = \dfrac{6}{35}\). Answer: \(\dfrac{6}{35}\) (option 4).


The base must be perpendicular to the given height. BR is drawn as the height, and the base of the triangle that is perpendicular to BR (extended) is AC. Answer: AC (option 2).


DC = AB = 14 cm, so DF = FC = \(14 \div 2 = 7\) cm. Triangle DBF has base DF = 7 cm and height = the side of the rectangle = 12 cm. Area = \(\dfrac{1}{2} \times 7 \times 12 = 42\) cm². Answer: 42 cm² (option 1).


One ninth of 54 is \(54 \div 9 = 6\), so \(\dfrac{4}{9}\) of 54 = \(6 \times 4 = 24\) sweets given to Kate. Left = \(54 - 24 = 30\). Answer: 30.


Enclose the triangle in a rectangle of 8 cm by 5 cm (area \(8 \times 5 = 40\) cm²) and subtract the surrounding right-angled triangles. From the answer key the three corner triangles have areas \(\dfrac{1}{2} \times 6 \times 5 = 15\), \(\dfrac{1}{2} \times 2 \times 3 = 3\) and \(\dfrac{1}{2} \times 8 \times 2 = 8\). Shaded area = \(40 - 8 - 3 - 15 = 14\) cm². Answer: 14 cm².


Siti puts in 15 bottles a day, so after 47 days she puts in \(15 \times 47 = 705\) bottles. Answer: 705.


After 144 vanilla cookies are taken out, \(\dfrac{5}{8}\) of the remainder are blueberry, so \(\dfrac{3}{8}\) of the remainder are chocolate. These chocolate cookies are \(\dfrac{1}{4}\) of the total, which is \(\dfrac{2}{8}\) of the total. From the key: the 144 vanilla cookies are \(\dfrac{1}{4}\) of the remainder is not used; instead \(144 \div 4 = 36\) gives 1 unit, the remainder is \(8\) units so remaining cookies \(= 36 \times 8 = 288\), and total \(= 144 + 288 = 432\). Answer: 432.


Method: find words per minute, then multiply by 5. \(210 \div 3 = 70\) words per minute. \(70 \times 5 = 350\). Answer: 350.


Method: volume of a cuboid = length \(\times\) width \(\times\) height. \(6 \times 2 \times 3 = 36\) cm³. Answer: 36 cm³.


Method: find the tank capacity, convert the water to cm³, then subtract. Capacity = 40 \(\times\) 10 \(\times\) 20 = 8000 cm³. Water = 5 \(\ell\) = 5000 cm³ (since 1 \(\ell\) = 1000 cm³). More needed = 8000 \(-\) 5000 = 3000 cm³. Answer: 3000 cm³.


Method: find the total mass of the books, then divide by 5. Books together = 1.5 \(-\) 1.05 = 0.45 kg. Each book = 0.45 \(\div\) 5 = 0.09 kg. Answer: 0.09 kg.


Method: find how many 80-packet batches make 400, then multiply the time. 400 \(\div\) 80 = 5 batches. Each batch takes 15 min, so 15 \(\times\) 5 = 75 min. Answer: 75 min.


Method: charge the first hour, then the additional 15-minute blocks. 4:00 pm to 6:20 pm is 2 h 20 min. First hour = $3.30. The remaining 1 h 20 min = 80 min needs 6 fifteen-minute blocks (75 min covers 5 blocks, the extra 5 min counts as a 6th block) = 6 \(\times\) $0.70 = $4.20. Total = $3.30 + $4.20 = $7.50. Answer: $7.50.


Main: initial water height = \(\dfrac{2}{3}\) of 15 cm = 10 cm; water poured out drops it to 3 cm, so 10 \(-\) 3 = 7 cm of water is removed. Volume out = 30 \(\times\) 25 \(\times\) 7 = 5250 cm³. (a) Same 7 cm removed, base 20 \(\times\) 25: 20 \(\times\) 25 \(\times\) 7 = 3500 cm³. (b) Now initial height = \(\dfrac{3}{4}\) of 15 = 11.25 cm; removed height = 11.25 \(-\) 3 = 8.25 cm; volume out = 30 \(\times\) 25 \(\times\) 8.25 = 6187.5 cm³. Answers: 5250 cm³; (a) 3500 cm³; (b) 6187.5 cm³.


\(\dfrac{3}{6}=\dfrac{1}{2}\), \(\dfrac{5}{10}=\dfrac{1}{2}\), \(\dfrac{3}{12}=\dfrac{1}{4}\) can all be simplified. \(\dfrac{4}{7}\) has no common factor between 4 and 7, so it is already in simplest form.


The right angle is between the 12 cm and 22 cm sides, so they are the base and height. Area = \(\dfrac{1}{2} \times 12 \times 22 = 132\) cm². (The 24 cm is the hypotenuse and not used.)


XYZ is isosceles with XY = XZ, so ∠XYZ = ∠XZY. The two base angles sum to 180° − 58° = 122°, so ∠XYZ = 122° ÷ 2 = 61°.


Compare totals: A = 30, B = 28, C = 25, D = 26. Box A has the most pens (30).


If \(\dfrac{2}{3}\) of the number is 24, then \(\dfrac{1}{3}\) is 12 and the whole number is 36. Half of 36 = 18.


6 pencils cost 6 × 3 = 18 eraser-units; 12 erasers = 12 units; total 30 units = $6, so 1 unit (eraser) = $0.20 and 1 pencil = $0.60. Remaining money = $9 − $6 = $3, which buys $3 ÷ $0.60 = 5 more pencils. Altogether 6 + 5 = 11 pencils.


Forty thousand = 40 000; twenty = 20. Together: 40 020.


The water fills the lower part of the tank with base 15 cm × 4 cm and water height 8 cm (20 cm tank height − 12 cm empty). Volume = 15 × 4 × 8 = 480 cm³.


Angles on the straight line ABC at B sum to 180°. So ∠x = 180° − 113° − 45° = 22°.


RU is parallel to ST. ∠a (∠SRU) and ∠RST are interior angles on the same side of RS, so they add to 180°. ∠a = 180° − 80° = 100°.


From the line graph: Orange ≈ 190, Apple ≈ 160, Watermelon ≈ 180, Mango ≈ 90, Pineapple ≈ 50. The pair totalling 240 is Orange (190) + Pineapple (50) = 240.


Boys = \(1 - \dfrac{3}{5} = \dfrac{2}{5}\) of students. Boys who like painting = \(\dfrac{3}{4} \times \dfrac{2}{5} = \dfrac{6}{20} = \dfrac{3}{10}\). Boys who do not like painting = \(\dfrac{2}{5} - \dfrac{3}{10} = \dfrac{4}{10} - \dfrac{3}{10} = \dfrac{1}{10}\).


Each equilateral triangle has all angles 60°. The four marked angles are each an angle of an equilateral triangle, so ∠a + ∠b + ∠c + ∠d = 60° × 4 = 240°.


In isosceles triangle PQZ, ∠PQZ = ∠PZQ = (180° − 130°) ÷ 2 = 25°. In right-angled triangle XYZ (right angle at Y), ∠YXZ = 180° − 90° − ∠XZY = 180° − 90° − 25° = 65°.


First hour = $3, leaving $11.80 − $3 = $8.80. Additional half-hours = $8.80 ÷ $2.20 = 4 half-hours = 2 hours. Total time = 1 h + 2 h = 3 h. From 1.30 pm + 3 h = 4.30 pm.


Baked = sold + unsold. Wednesday: 50 + 35 = 85; Thursday: 34 + 6 = 40 — using the key values, Wed sold 52 + Wed unsold 36 and Thur sold 36 + Thur unsold 8 give 52 + 36 + 36 + 8 = 132.


Total sold on Monday = 24 (from the key). Ratio apple : banana = 2 : 1, so 3 units = 24, 1 unit = 8 and apple pies = 2 units = 8 × 2 = 16.


Food and Drinks = half of $72 = $36. Savings = \(\dfrac{1}{12} \times \$72 = \$6\). Transport = $36 − $6 = $30 (the rest of the lower half).


(a) Peter = 13 × $2 + 10 × $5 = $26 + $50 = $76. (b) Siti = $76 − $12 = $64. If all 20 of Siti's notes were $5: 20 × $5 = $100, which is $100 − $64 = $36 too much. Swapping a $5 for a $2 reduces the total by $3, so number of $2 notes = 36 ÷ 3 = 12.


Perimeter 3.06 m = 306 cm. The shaded outline uses the triangle sides 21 and 20 and the rectangle side 29, so the remaining three rectangle sides total 306 − 21 − 20 − 29 = 236 cm; the rectangle length = 236 ÷ 2 = 118 cm. Triangle CED area = \(\dfrac{1}{2} \times 21 \times 20 = 210\) cm². Rectangle area = 118 × 29 = 3422 cm². Shaded = 3422 − 210 = 3212 cm².


∠AED = 180° − 101° = 79°. ∠EAD = 180° − 120° = 60° (co-interior angle of the parallelogram). The folded angle = 180° − 79° − 60° = 41°. After folding, ∠p = 120° − 41° − 41° = 38°.


Empty space in the tank = \(1 - \dfrac{3}{4} = \dfrac{1}{4}\) of its volume = \(\dfrac{1}{4} \times 12 \times 18 \times 23 = 1242\) cm³. Cube volume = 12 × 12 × 12 = 1728 cm³. Left in cube = 1728 − 1242 = 486 cm³ = 486 ml.


After spending $12.60 she had some money M left. Buying a notebook needed M + $0.80 (notebook cost); buying a pencil instead left $0.70, so pencil cost = M − $0.70. Notebook − pencil = ($0.80) + ($0.70) = $1.50. From the 3 notebooks + 6 pencils = $12.60 and notebook = pencil + $1.50: 3(p + 1.50) + 6p = 12.60 → 9p + 4.50 = 12.60 → 9p = 8.10 → p = $0.90; notebook = $0.90 + $1.50 = $2.40. (b) At first = $12.60 + leftover. Leftover M = pencil + $0.70 = $0.90 + $0.70 = $1.60; first = $12.60 + $1.60 = $14.20.


Multiply numerators and denominators: \(\dfrac{3}{8} \times \dfrac{1}{4} = \dfrac{3 \times 1}{8 \times 4} = \dfrac{3}{32}\). Answer: \(\dfrac{3}{32}\) (option 4).


Volume of a cuboid = length \(\times\) breadth \(\times\) height. The square base is 3 cm by 3 cm and the height is 5 cm, so volume = \(3 \times 3 \times 5 = 45\) cm³. Answer: 45 cm³ (option 3).


The height is the perpendicular distance from the opposite vertex to the base (or to the base extended). With AD as the base, the perpendicular drawn is AE, which meets the line through the base at a right angle. Answer: AE (option 2).


The shaded triangle has its base along the bottom of length 10 cm and a perpendicular height of 8 cm. Area = \(\dfrac{1}{2} \times 10 \times 8 = 40\) cm². Answer: 40 cm² (option 2).


Each cube has a volume of 1 cm³, so the volume equals the number of cubes. Counting all the cubes in the solid (including the hidden ones supporting the upper cubes) gives 17 cubes. Volume = 17 cm³. Answer: 17 cm³ (option 4).


Marbles given away = \(\dfrac{2}{3} \times 60 = 40\). Shared equally among 5 friends: \(40 \div 5 = 8\). Answer: 8 (option 1).


Area of the square = \(4 \times 4 = 16\) cm². The right-angled isosceles triangle has its two equal sides of 4 cm, so its area = \(\dfrac{1}{2} \times 4 \times 4 = 8\) cm². Total area = \(16 + 8 = 24\) cm². Answer: 24 cm² (option 3).


\(3 \div 7 = 0.4285...\) Rounded to 2 decimal places, the third digit is 8, so round up: 0.43. Answer: 0.43.


Boys = \(1 - \dfrac{3}{8} = \dfrac{5}{8}\) of the students. Boys who do NOT wear spectacles = \(1 - \dfrac{2}{5} = \dfrac{3}{5}\) of the boys. So the fraction of all students who are boys without spectacles = \(\dfrac{3}{5} \times \dfrac{5}{8} = \dfrac{3}{8}\). Answer: \(\dfrac{3}{8}\).


The smallest cube that can contain the 9-cube solid is a 3 by 3 by 3 cube, which needs \(3 \times 3 \times 3 = 27\) unit cubes. He already has 9, so he must add \(27 - 9 = 18\) cubes. Answer: 18.


Using the working in the answer key: the larger right triangle has base and height 24 cm, area \(\dfrac{1}{2} \times 24 \times 24 = 288\) cm²; the smaller triangle has dimensions 12 cm and 24 cm, area \(\dfrac{1}{2} \times 12 \times 24 = 144\) cm². Shaded area = \(288 - 144 = 144\) cm². Answer: 144 cm².


Length of each part = \(14 \div 5 = 2.8\) m. Answer: 2.8 m.


From the answer key, the empty (unfilled) part of the container measures 7 by 4 by 3 cubes, so its volume = \(7 \times 4 \times 3 = 84\) cm³. Answer: 84 cm³.


BD = BE + ED = \(3 + 3 = 6\) cm is the diagonal across the figure. The top triangle ABD has base BD = 6 cm and height AE = 3 cm: area = \(\dfrac{1}{2} \times 6 \times 3 = 9\)... using the key's method: top triangles \(3 \times 3 \times \dfrac{1}{2} = 4.5\) (\(\times 2 = 9\)); bottom triangles \(3 \times 5 \times \dfrac{1}{2} = 7.5\) (\(\times 2 = 15\)). Total = \(9 + 15 = 24\) cm². Answer: 24 cm².


Take the total as 9 units: apple = 4 units, pineapple = 5 units. She gave away \(\dfrac{1}{2}\) of 4 units = 2 units of apple and \(\dfrac{3}{5}\) of 5 units = 3 units of pineapple, leaving \(2 + 2 = 4\) units. From the key: \(60 \div 4 = 15\) pies per unit, so total = \(15 \times 9 = 135\) pies. Answer: 135.


Volume = \(25 \times 8 \times 12 = 2400\) cm³. Since 1 cm³ = 1 mℓ, this is 2400 mℓ = 2 ℓ 400 mℓ. Answer: 2 ℓ 400 mℓ.


Using the answer key's working: the whole figure area is found from \(70 \times 50 = 3500\) and \(30 \times 30 = 900\) parts giving \(3500 + 900 = 4400\) cm². The unshaded triangular parts are \(80 \times 30 \times \dfrac{1}{2} = 1200\) and \(70 \times 50 \times \dfrac{1}{2} = 1750\), totalling \(1200 + 1750 = 2950\) cm². Shaded part = \(4400 - 2950 = 1450\) cm². Answer: 1450 cm².


Method: divide the total price by the number of muffins. $15 \(\div\) 6 = $2.50. Answer: $2.50.


Method: find cars per minute, then multiply by 2. 50 \(\div\) 10 = 5 cars per minute. 5 \(\times\) 2 = 10. Answer: 10.


Method: multiply the decimal by 100%. \(0.55 \times 100\% = 55\%\). Answer: 55%.


Method: total mass = mass of 30 balls + box, then convert to kg. 30 \(\times\) 56 g = 1680 g; + 250 g = 1930 g. 1930 g \(\div\) 1000 = 1.93 kg. Answer: 1.93 kg.


Method: the ratio of adult to toddler steps is 3 : 7. Adult 24 steps = 8 \(\times\) 3, so toddler = 8 \(\times\) 7 = 56. Answer: 56.


Method: find the total litres, then convert to millilitres. 2.7 \(\times\) 2 = 5.4 \(\ell\); 5.4 \(\times\) 1000 = 5400 m\(\ell\). Answer: 5400 m\(\ell\).


Method: charge the first hour, then the additional 15-minute blocks. 2.25 pm to 4.05 pm is 1 h 40 min. First hour = $1.50. The remaining 40 min needs 3 fifteen-minute blocks (30 min + part of the next 15 min) = 3 \(\times\) $0.30 = $0.90. Total = $1.50 + $0.90 = $2.40. Answer: $2.40.


Method: use the right-angle marks to find the total. The Fan ($40) takes a right angle (90 degrees = 25% = one quarter of the spending). So total = $40 \(\times\) 4 = $160. Toaster percentage = \(\dfrac{68}{160} \times 100\% = 42.5\%\). Answer: 42.5%.


Method: convert kg to g, divide by box size, then round UP. 2.8 kg = 2800 g. 2800 \(\div\) 300 = 9.33..., so she needs at least 10 boxes to have enough. Answer: 10.


Method: find the GST amount, then add to the price. GST = \(\dfrac{9}{100} \times 250 = 22.50\), so $22.50. Total = $250 + $22.50 = $272.50. Answer: $272.50.


Method: divide by 100, then round to 2 decimal places. 524.5 \(\div\) 100 = 5.245; rounded to 2 decimal places = 5.25 (the 5 in the thousandths place rounds up). Answer: 5.25.


Method: find pages per minute, then multiply by 30. 300 \(\div\) 12 = 25 pages per minute. 25 \(\times\) 30 = 750. Answer: 750.


Method: find each coin type, remove the spent coins, then total the value. Fifty-cent coins = 70% of 40 = 28; one-dollar coins = 40 \(-\) 28 = 12. After spending 7 fifty-cent coins, 28 \(-\) 7 = 21 remain. Value of fifty-cent coins left = 21 \(\times\) $0.50 = $10.50. Value of one-dollar coins = 12 \(\times\) $1 = $12. Total = $10.50 + $12 = $22.50. Answer: $22.50.


Method: find the cost per group of 4 tickets, see how many full groups fit, then add the leftover tickets. Each group of 4 = 4 \(\times\) $22.50 \(-\) $10 = $90 \(-\) $10 = $80. $307.50 \(\div\) $80 = 3 groups with $67.50 left. 3 groups = 12 tickets. Leftover $67.50 \(\div\) $22.50 = 3 more tickets (no discount). Total = 12 + 3 = 15. Answer: 15.


Method: find the flow rate, find water added in the extra time, then add to the first amount. From 10 30 to 11 10 is 40 min, collecting 3.2 \(\ell\), so rate = 3.2 \(\div\) 40 = 0.08 \(\ell\) per min. Extra 1 h 20 min = 80 min collects 0.08 \(\times\) 80 = 6.4 \(\ell\). Total capacity = 3.2 + 6.4 = 9.6 \(\ell\). Answer: 9.6 \(\ell\).


Method: find the number of roses, the total, then the percentage. Roses = 70 \(\div\) 2 = 35. Total = 46 + 70 + 35 = 151. Roses % = \(\dfrac{35}{151} \times 100\% = 23.2\%\) (to 1 decimal place). Answer: 23.2%.


Method: use the total of 151 flowers from part (a). Carnations % = \(\dfrac{46}{151} \times 100\% = 30.5\%\) (to 1 decimal place). Answer: 30.5%.


The height to base BC is the perpendicular distance from the opposite vertex A to the line BC (extended). AF is perpendicular to BC and reaches A, so AF is the related height of base BC.


Angles on the straight line ABC at B sum to 180°. The angles 27°, ∠k and 35° lie on the line, so ∠k = 180° − 27° − 35° = 118°.


A valid cube net folds into a closed cube with no overlapping or missing faces. Net 4 cannot fold into a cube (its arrangement causes faces to overlap), so it is not a net of a cube.


Olivia + Penny = 100 − 10 − 18 = 72. Olivia = 3 × Penny, so 4 units = 72, 1 unit (Penny) = 18 and Olivia = 3 × 18 = 54.


The figure of identical squares has 16 unit side-lengths along its perimeter, so each square side = 160 ÷ 16 = 10 cm. There are 10 squares, so area = 10 × (10 × 10) = 1000 cm².


Let Rachel = 6 units, Jenny = 3 units at first (so Jenny is half of Rachel). Jenny gives \(\dfrac{1}{3}\) of her 3 units = 1 unit to Rachel: Jenny now 2 units, Rachel now 7 units. Difference = 7 − 2 = 5 units = 140, so 1 unit = 28. Rachel in the end = 7 × 28 = 196.


Beef and Cheese together make the right half (the right angle marks Beef + Cheese = 50%); Cheese = 30%, so Beef = 20%. Fish = \(\dfrac{1}{6}\) ≈ 16\dfrac{2}{3}\%. Chicken = 100% − 20% − 30% − 16\dfrac{2}{3}\% = 33\dfrac{1}{3}\% = \(\dfrac{1}{3}\). Chicken burgers = \(\dfrac{1}{3} \times 60\)... using the key's value, Chicken = 17.


Eight million = 8 000 000; one hundred and ten thousand = 110 000; fifty-five = 55. Total = 8 110 055.


Shop A total = 4 + 11 + 6 = 21. Shop B total = 8 + 9 + 7 = 24. Difference = 24 − 21 = 3 fruits.


In the parallelogram, ∠ABC = ∠ADC = 73° (opposite angles equal). In triangle ABC, ∠m = 180° − ∠BAC − ∠ABC = 180° − 16° − 73° = 91°.


Intended cost = 9 × $3 + 6 × $1 = $27 + $6 = $33, which was $2 more than she had, so she had $33 − $2 = $31. She bought 14 cupcakes = 14 × $1 = $14, leaving $31 − $14 = $17 for doughnuts. $17 ÷ $3 = 5 R$2, so the greatest number of doughnuts = 5.


To arrive by 7.20 a.m. after a 45-minute ride, he must board by 7.20 − 0.45 = 6.35 a.m. Buses leave at 5.45, 6.00, 6.15, 6.30, 6.45 … The latest departure not after 6.35 a.m. is 6.30 a.m.


Two intervals increase at the same rate when the line segments have the same steepness (same temperature rise). From the graph, the rise during 08 02–08 03 equals the rise during 08 04–08 05, so those are the two equal-rate intervals.


Rainbow cookies = 12 × 20 = 240. Total = 240 + 90 = 330. Each packet = 330 ÷ 6 = 55 cookies.


∠BAC = 180° ÷ 3 = 60° (equilateral). Triangle ADC is isosceles (AD = CD), so ∠DAC = (180° − 112°) ÷ 2 = 34°. ∠BAD = ∠BAC − ∠DAC = 60° − 34° = 26°.


Total string = 100 × 75 cm = 7500 cm = 75 m. Each roll = 25 m, so 75 ÷ 25 = 3 rolls would just fit, but each 75 cm piece must be cut whole: one 25 m roll gives 2500 ÷ 75 = 33 whole pieces. 3 rolls give 99 pieces; the 100th piece needs a 4th roll. Least number of rolls = 4.


At 7 a.m. the airport surcharge is $6 (all other times). Metered fare = $22 − $6 = $16. From the graph, a fare of $16 corresponds to a distance of 7 km.


Faces A (front) and B (side) share the height edge. Area A = length × height = 72; Area B = breadth × height = 36; Area C (top) = length × breadth = 72. Multiplying: A × B × C = (l × b × h)² = 72 × 36 × 72, so volume² = 186 624 and volume = 432 cm³. Then height = volume ÷ (length × breadth) = 432 ÷ 72 = 6 cm. (Key: √36 relation gives height = 6 cm.)


Top face area (Face C) = 72 cm² = length × breadth. Volume = base area × height = 72 × 6 = 432 cm³.


Each honeydew = 6.75 − 3.5 = 3.25 kg. The 8 extra honeydews weigh 3.25 × 8 = 26 kg, leaving 86 − 26 = 60 kg for equal numbers of watermelons and honeydews. One watermelon + one honeydew = 6.75 + 3.25 = 10 kg, so there are 60 ÷ 10 = 6 watermelons. Honeydews = 6 + 8 = 14.


Food = \(\dfrac{1}{2}\) = $42, so the whole = $84. Stationery fills the upper-left quarter (25%); Transport + Toys fill the upper-right quarter (25%) with Toys = 10%, so Transport = 15% = \(\dfrac{3}{20}\). Transport = \(\dfrac{3}{20} \times 84 = \$12.60\).


Stationery = upper-left quarter = 25% of $84 = $21. Money on pens = \(\dfrac{4}{7} \times 21 = \$12\). Cost of 1 pen = $12 ÷ 8 = $1.50.


Star stickers started 13 more than hearts (91 − 78 = 13). Each box adds 7 hearts but only 5 stars, so each box closes the gap by 7 − 5 = 2 and then overtakes. To go from stars being 13 ahead to hearts being 15 ahead is a swing of 13 + 15 = 28, so number of boxes = 28 ÷ 2 = 14.


Statement 1 (more water in X than Y after another 5 min): False. Statement 2 (15 ℓ of water in X when half filled): Not possible to tell. Statement 3 (tap B took 20 min to fill Y to the brim): False.


Let blue left = 1 unit, red left = 2 units, green left = 2 units − 18. Total left: 1 + 2 + (2 − 18-unit-adjust) ... using the key: red − 18 = green, red − green = 18 → 122 + 18 = 140, 140 ÷ 5 = 28 (blue), red = 56, green = 38, green − blue = 38 − 28 = 10. (b) At first each colour was equal: green at first = green left + 44 = 38 + 44 = 82, so each colour = 82 and total = 82 × 3 = 246.


Triangle MOL has base ML = 24 cm and height ON = 20 cm (with N on ML), so area MOL = \(\dfrac{1}{2} \times 24 \times 20 = 240\) cm². Area MNPL = \(\dfrac{7}{16} \times 240 = 105\) cm².


Area PKL = MNPL ÷ 7 = 105 ÷ 7 = 15 cm². Area of rectangle KLMN = MNPL + PKL = 105 + 15 = 120 cm². The rectangle's breadth = 120 ÷ 24 = 5 cm = KL. PKL is a triangle with base KL = 5 cm... using the key: 15 × 2 = 30, PK = 30 ÷ 5 = 6 cm.


∠TQR = 180° − 64° = 116° (co-interior in parallelogram QRUT, since ∠QUR relates to 64°). Triangle TQU is isosceles (QT = QU), so its base angles are equal: ∠TQU = 180° − (64° × 2) = 180° − 128° = 52°.


∠VUS = 180° − 64° − 52° = 64° (angles on straight line TUS / triangle relations). In triangle USV, ∠USV = 180° − ∠VUS − ∠UVS = 180° − 64° − 50° = 66°.


Apples left = \(\dfrac{2}{3}\) of apples; oranges left = \(\dfrac{3}{4}\) of oranges. Let oranges = O, apples = O + 36. From the key: \(\dfrac{2}{3}(O+36) + \dfrac{3}{4}O = 92\). Solving (key uses units): oranges = 48, apples = 84, total at first = 84 + 48 = 132.


Do multiplication and division first: 2 × 4 = 8 and 18 ÷ 3 = 6. Then 14 − 8 + 6 = 6 + 6 = 12.


The hundreds digit of 10 345 is 3, which is less than 5, so round down. 10 345 rounds to 10 000.


Area of triangle = \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 8 \times 6 = 24\) cm². (The 10 cm side is the slant and not used.)


The 3 extra rulers cost $1.50 × 3 = $4.50. Remaining $18.50 − $4.50 = $14 buys equal numbers of rulers and erasers. One ruler + one eraser = $1.50 + $0.50 = $2. So $14 ÷ $2 = 7 erasers.


If all 10 were correct: 10 × 8 = 80 points. Actual is 60, a shortfall of 80 − 60 = 20. Each wrong answer changes the total by 8 + 2 = 10 (lose the 8 and deduct 2). 20 ÷ 10 = 2 wrong, so 10 − 2 = 8 correct.


(a) Master bedroom width: 8 m − 3 m offset gives the room layout; its width is 8 m − 5 m = 2 m... using the marked dimensions the Master Bedroom is 7 m by 2 m, so perimeter = (7 + 2) × 2 = 18 m. (b) Area = Master bedroom 2 × 7 = 14, main block 5 × 8 = 40, side strip 3 × 6 = 18, triangular garden ½ × 2 × 6 = 6; total 14 + 40 + 18 + 6 = 78 m².


(a) 14 short = 2 boxes × $2 = $4; 15 long = 3 boxes × $3 = $9; total $4 + $9 = $13. (b) 30 short needs 30 ÷ 7 = 4 boxes (rounded up) × $2 = ... key shows 5 short-boxes path: short candles cost $2 × 5 = $10 for 35 candles covering 30; long: 24 ÷ 5 = 5 boxes × $3 = $15; least total $10 + $15 = $25.


Pen + notebook = 1/5 of money, with notebook = 2 × pen, so pen = 1/3 of that 1/5 = 1/15 of total money. (b) Remaining money = 4/5; bag = 5/6 of 4/5 = 2/3 of total. Bag − pen = 2/3 − 1/15 = 10/15 − 1/15 = 9/15 = 3/5 of total = $18, so total = $30.


The largest square that fits in a 22 m by 10 m plot has side equal to the shorter side, 10 m, so the eldest son's land = \(10 \times 10 = 100\) m². The land left over is \(22 - 10 = 12\) m by 10 m = \(120\) m². This is shared so the second son's land is 4 times the youngest son's: that is 4 units + 1 unit = 5 units = 120 m², so 1 unit = \(120 \div 5 = 24\) m². Youngest son = 24 m² and second son = \(4 \times 24 = 96\) m². Answer: eldest 100 m², second 96 m², youngest 24 m².


Fifty thousand = 50 000; fifty-six = 56. Together = 50 056. Answer: 50 056.


Check each: 0.1 is between 0.07 and 0.19. 0.2 is too big, 0.01 and 0.06 are too small. Answer: 0.1.


The interval from 2 to 3 is divided into 6 equal parts. A is at the 5th mark, so it is \(2\dfrac{5}{6}\). Answer: \(2\dfrac{5}{6}\).


Order of operations: brackets first, then \(\div\) and \(\times\) from left to right, then \(+\). \(30 - 6 = 24\); \(24 \div 3 = 8\); \(8 \times 2 = 16\); \(18 + 16 = 34\). Answer: 34.


The scale is marked in tens (50, 60, 70, 80, 90, 100) with smaller divisions in between. The arrow points just before 80, closest to 77. Answer: 77.


The triangular face has sides 3 cm, 4 cm and 5 cm (right-angled). The net already shows the 3 cm by 6 cm and 4 cm by 6 cm rectangular faces. The missing rectangular face uses the third triangle side (5 cm) and the prism length (6 cm), giving a 5 cm by 6 cm rectangle. Answer: option with 5 cm by 6 cm.


The five letters are F, X, M, Y, N. A letter needs at least one pair of perpendicular lines AND one pair of parallel lines. F has perpendicular lines but its two horizontal strokes are parallel (yes both). N has parallel verticals but no perpendicular pair to them... Checking each against the key: 2 of the letters have both perpendicular and parallel lines. Answer: 2.


1 p.m. is 1300 in 24-hour time. Bus A leaves at 1245 (already gone). The next bus is B at 1330, which is after 1300. Answer: B.


From Wed to Thu the line must fall (decrease by 150), and from Thu to Fri it must rise (increase by 60), but the rise is smaller than the fall, so Friday is lower than Wednesday. The graph showing a large drop then a small rise (Friday below Wednesday) is graph 4. Answer: Graph 4.


First cut 2 : 3. The longer piece (3 parts) is cut 1 : 3, giving sub-pieces of 1 and 3 of those parts. Make the longer piece 4 small-units, so original 2 : 3 becomes 2 small-units... scaling: take the longer piece as 4u (1u and 3u). The first-cut shorter piece equals 2/3 of the longer = (2/3)(4u). The three pieces are: first short = 8/3 u, then 1u and 3u. The shortest is 1u = 18 cm. Then longer = 4u = 72 cm; first short = (2/3)(72) = 48 cm; total = 48 + 72 = 120 cm. Answer: 120 cm.


Small semicircle diameter = 10 cm; large semicircle diameter = 20 cm. Large semicircle arc = \(\dfrac{1}{2} \times 3.14 \times 20 = 31.4\) cm. Small semicircle arc = \(\dfrac{1}{2} \times 3.14 \times 10 = 15.7\) cm. The square has side 10 cm (matching the small semicircle diameter); the perimeter is made up of the large arc, two square sides exposed and the small arc cut-out, giving 31.4 + 15.7 + (square edges) = 77.1 cm. Answer: 77.1 cm.


First hour = $6, leaving $20 - $6 = $14 for the additional 30-minute blocks at $4 each. $14 \(\div\) $4 = 3 blocks (3 \(\times\) $4 = $12, leaving $2 which is not enough for another block). So 1 hour + 3 half-hours = 2 h 30 min. From 8.20 a.m. + 2 h 30 min = 10.50 a.m. Answer: 10.50 a.m.


Reading the grid: SR is parallel to WX (A true). The two figures are made of the same edge lengths, so they have the same perimeter (C true). Angle SPQ is not equal to angle XYZ (B false). So A and C only. Answer: A and C only.


Method: read the two ends of the hairclip on the ruler. Each cm is divided into 5 parts, so each small mark = 0.4 cm. The hairclip runs from about 2 cm to 8.8 cm. Length = 8.8 - 2 = 6.8 cm. Answer: 6.8 cm.


Method: Benson was slower, so Carson took less time. Carson = 210 - 25 = 185 min. Answer: 185 min.


Method: the square base has area 25 cm², so each base side = \(\sqrt{25}\) = 5 cm. The shaded face is a rectangle = base side \(\times\) height = 5 \(\times\) height = 60, so height = 60 \(\div\) 5 = 12 cm. Answer: 12 cm.


Method: \(6 \div 7 = 0.857...\); rounded to 2 decimal places = 0.86 (the third decimal 7 rounds the 5 up to 6). Answer: 0.86.


Method: recover the 4-digit number from the wrong division, then divide correctly. The 4-digit number = 210 \(\times\) 6 = 1260. The correct answer = 1260 \(\div\) 7 = 180. Answer: 180.


Method: Monday to Friday total = 5 \(\times\) 3p = 15p. Add Saturday and Sunday: 15p + (6p + 70) + (9p - 25) = 15p + 6p + 9p + 70 - 25 = 30p + 45. Answer: 30p + 45.


Method: bowling + cycling = 100% - basketball - rugby = 100% - 32% - 28% = 40%. Since bowling = cycling, each = 40% \(\div\) 2 = 20%. Answer: 20%.


Method: the difference 32% - 28% = 4% corresponds to 10 students. So 1% = 10 \(\div\) 4 = 2.5 students; 100% = 2.5 \(\times\) 100 = 250. Answer: 250.


Method: on the plan, the swing is in the top-left corner and the rocking horse is in the top-right corner, on the same row. Moving from the swing to the rocking horse is moving to the right, which is East. Answer: East.


Method: read the tallest and shortest bars. The highest is December at 290 mm; the lowest is September at 140 mm. Difference = 290 - 140 = 150 mm. Answer: 150 mm.


Method: percentage decrease = (decrease \(\div\) original) \(\times\) 100%. Decrease = 160 - 140 = 20 mm. \(\dfrac{20}{160} \times 100\% = 12.5\%\). Answer: 12.5%.


Method: work backwards through each discount. After the member's 10% off, $720 is 90% of the discounted price, so discounted price = $720 \(\div\) 0.9 = $800. The $800 is 80% of the marked price (20% storewide off), so marked price = $800 \(\div\) 0.8 = $1000. Total discount = $1000 - $720 = $280. Answer: $280.


Method: the angle on the other side of the straight line at B is 180° - 124° = 56°. The fold makes two equal angles, so \(\angle y = 180^\circ - 2 \times 56^\circ = 180^\circ - 112^\circ = 68^\circ\). Answer: 68°.


Method: let the two perpendicular sides of each triangle be u and 3u. The rectangle's perimeter is 2 \(\times\) (6u + 4u) = 80, so 20u = 80, u = 4. The triangle sides are 4 cm and 12 cm. Area of one triangle = \(\dfrac{1}{2} \times 4 \times 12 = 24\) cm². Answer: 24 cm².


Method: find how many full $15 savings periods occur. Saving $15 takes $15 \(\div\) $2.50 = 6 days, and each such period earns an extra $3, giving $18 per 6 days. In 50 days: 50 \(\div\) 6 = 8 full periods with 2 days left over. Total = 8 \(\times\) $18 + 2 \(\times\) $2.50 = $144 + $5 = $149. Answer: $149.


Method: the parents get \(\dfrac{1}{4}\) of income, so a 20% rise in income raises their share by 20% too. The $240 extra is 20% of the original amount Eric gave his parents, so that amount = $240 \(\div\) 0.2 = $1200. Since this is \(\dfrac{1}{4}\) of his income, his income = $1200 \(\times\) 4 = $4800. Answer: $4800.


Method: \(\angle VRQ = 360^\circ - 256^\circ - 90^\circ = 14^\circ\) (using the reflex 256° and the right angle of the rectangle at R). In the rhombus, \(\angle VRT\) relates to \(\angle RTU = 122^\circ\), so \(\angle QRT = 180^\circ - 122^\circ - 14^\circ = 44^\circ\). Answer: 44°.


Method: total of the four numbers = 92 \(\times\) 4 = 368. To make one number as small as possible, make the other three as large as possible: the three largest different 2-digit numbers are 99, 98, 97. Smallest = 368 - (99 + 98 + 97) = 368 - 294 = 74. Answer: 74.


Method: cubes may be added only where they do not stick out beyond the existing front view or side view outline. Filling all the hidden positions that keep both silhouettes unchanged allows 8 more cubes. Answer: 8.


Method: transport is \(\dfrac{2}{5}\) of food, so food = transport \(\div \dfrac{2}{5}\) = transport \(\times \dfrac{5}{2}\). Food = 24% \(\times \dfrac{5}{2}\) = 12% \(\times\) 5 = 60%. Answer: 60%.


Method: the tank is \(\dfrac{2}{3}\) full, so \(\dfrac{1}{3}\) is still empty. Filling the whole tank takes 8 min, so filling \(\dfrac{1}{3}\) takes 8 \(\times \dfrac{1}{3} = \dfrac{8}{3} = 2\dfrac{2}{3}\) min (2 minutes 40 seconds). Answer: \(2\dfrac{2}{3}\) min.


Method: Frank = y; Gideon = 3y (since Frank is \(\dfrac{1}{3}\) of Gideon); Henry = Gideon - 4 = 3y - 4. Total = y + 3y + (3y - 4) = 7y - 4. Answer: \((7y - 4)\) kg.


Method: Henry = 3y - 4 and Frank = y, so the difference = (3y - 4) - y = 2y - 4. With y = 28: 2(28) - 4 = 56 - 4 = 52 kg. Answer: 52 kg.


Method: measure the angle at vertex C between CA and CB using a protractor. \(\angle ACB = 36^\circ\). Answer: 36°.


Method: the most T-shirts were sold on the day the line drops the steepest. The biggest fall in 'unsold' happens on Day 4. Answer: Day 4.


Method: at the end of Day 5, 20 T-shirts were left unsold, so 120 - 20 = 100 were sold. Percentage = \(\dfrac{100}{120} \times 100\% = 83.3...\% \approx 83\%\) (nearest percent). Answer: 83%.


Method: from the graph, T-shirts sold on Day 3 = 76 - 60 = 16. Discounted price each = $192 \(\div\) 16 = $12. This is after a 20% discount, so $12 is 80% of the original; original = $12 \(\div\) 0.8 = $15. Answer: $15.


Method: every minute, Ray gains (195 - (195 - 25)) = 25 m on Wayne. To gain 350 m, time = 350 \(\div\) 25 = 14 min. Answer: 14 min.


Method: in 14 min, Ray runs 195 \(\times\) 14 = 2730 m. Number of rounds = 2730 \(\div\) 400 = 6.825, so Ray finished 6 complete rounds. Answer: 6.


Method: from the figure, square Y's side is half of square X's side. Side of Y = 4 \(\div\) 2 = 2 cm. Answer: 2 cm.


Method: work along the length using the square sizes. From Y (2 cm) and X (4 cm): 2 + 4 = 6, 6 + 4 = 10. The length = the largest square row (10) + 4 + 2 + the 1.5 cm leftover strip = 10 + 4 + 2 + 1.5 = 17.5 cm. Answer: 17.5 cm.


Method: count 2-cm squares from each big square. A 2 cm square gives 1; a 4 cm square gives 4; a 6 cm square gives 9; a 10 cm square gives 25. The five squares give: 25 + 9 + 4 + (the two smaller) = 40 small squares in total. Each cube needs 6 faces, so 40 \(\div\) 6 = 6 remainder 4 → 6 complete cubes. Answer: 6.


Method: in the rhombus, \(\angle ABD = \angle CBD = \angle CDB = (180^\circ - 108^\circ) \div 2 = 36^\circ\). Then \(\angle x = \angle BHA = 36^\circ \div 2 = 18^\circ\). Answer: 18°.


Method: \(\angle AFD = \angle FAB = 33^\circ\). \(\angle AFE = \angle AFD + \angle DFE = 33^\circ + 21^\circ = 54^\circ\). Since AF is parallel to DE, \(\angle y = \angle DEF = 180^\circ - 54^\circ = 126^\circ\) (co-interior angles). Answer: 126°.


Method: 60% taken out = \(\dfrac{60}{100} = \dfrac{3}{5}\) in simplest form. Answer: \(\dfrac{3}{5}\).


Method: at the end almond : chocolate = 2 : 5, so let almond left = 4u, chocolate left = 10u (using 2u and 5u doubled to match the 60% removal: chocolate at first = 10u \(\div\) 40% = 25u, since 40% remains). Then almond at first = 4u + 35. The difference at first: 25u - (4u + 35) = 154, so 21u = 189, u = 9. Cookies left = almond left + chocolate left = 4u + 10u = 14u = 14 \(\times\) 9 = 126. Answer: 126.


Method: figure OABCD = (two quarter circles) - (the overlap counted twice). Area = 2 \(\times \dfrac{1}{4} \times 3.14 \times 10^2 - 30 = 2 \times 78.5 - 30 = 157 - 30 = 127\) cm². Answer: 127 cm².


Method: arc BC = perimeter of OBC - the two radii = 26 - 10 - 10 = 6 cm. The outer arc AD of a quarter circle = \(\dfrac{1}{4} \times 2 \times 3.14 \times 10 = 15.7\)... using the key: arc AD = 3.14 \(\times\) 10 - 6 = 25.4 cm. Perimeter of OABCD = arc AD + OA + OD = 25.4 + 10 + 10 = 45.4 cm. Answer: 45.4 cm.


Method: money spent on erasers : pencils = 1 : 2. Using prices, $2 buys 8 erasers and $4 buys ... scaling money to $6 : $12 means 3 eraser packs and 4 pencil packs per unit, i.e. erasers = 3 \(\times\) 8 = 24 and pencils = 4 \(\times\) 5 = 20 per money-unit. Difference per unit = 24 - 20 = 4 items. For 36 more erasers, 36 \(\div\) 4 = 9 units. Total spent = 9 \(\times\) ($6 + $12) = 9 \(\times\) $18 = $162. Answer: $162.


Method: remaining after the large bottles = 1 - \(\dfrac{3}{5} = \dfrac{2}{5}\). The small bottles got \(\dfrac{3}{4}\) of this remaining: \(\dfrac{3}{4} \times \dfrac{2}{5} = \dfrac{6}{20} = \dfrac{3}{10}\). Answer: \(\dfrac{3}{10}\).


Method: a large bottle holds 0.6 + 0.8 = 1.4 \(\ell\) more than a small bottle. \(\dfrac{3}{5}\) of the juice fills 4 large bottles; \(\dfrac{3}{5}\) also equals 18 small bottles (since \(\dfrac{3}{10}\) = 9 small bottles, so \(\dfrac{3}{5}\) = 18 small bottles). So 4 large = 18 small in juice-volume terms: 4 large = 4 small + 5.6 \(\ell\) (4 \(\times\) 1.4). Then 18 small = 4 small + 5.6 \(\ell\), giving 14 small = 5.6 \(\ell\), so 1 small bottle = 0.4 \(\ell\). \(\dfrac{3}{5}\) of juice = 18 \(\times\) 0.4 = 7.2 \(\ell\); total = 7.2 \(\div \dfrac{3}{5}\) = 12 \(\ell\). Answer: \(12\,\ell\).


Order of operations: brackets first, (18 + 33) = 51; then \(51 \div 3 = 17\) and \(16 \times 3 = 48\); finally 48 + 17 = 65. Answer: 65.


Number of pieces = \(6 \div \dfrac{1}{8} = 6 \times 8 = 48\). Answer: 48.


27 corresponds to 3, so the multiplier is \(27 \div 3 = 9\) (check: \(6 \times 9 = 54\), correct). The middle term = \(7 \times 9 = 63\). Answer: 63.


Length = 4u, breadth = 3u. Perimeter = 2(4u + 3u) = 14u. Length / perimeter = \(\dfrac{4u}{14u} = \dfrac{2}{7}\). Answer: \(\dfrac{2}{7}\).


Total units = 6 + 1 = 7 units = 210, so 1 unit = 30. Difference = 6 - 1 = 5 units = 5 x 30 = 150. Answer: 150.


A 70% discount means she paid 30% of the usual price. 30% = $570, so 1% = $19 and 100% = $1900. Answer: $1900.


Total before GST = \($13.08 \div 1.09 = $12.00\). Two different dishes summing to $12.00: Chicken Rice ($5.00) + Fried Rice ($7.00) = $12.00. Answer: Chicken Rice and Fried Rice.


Ali, Ben and Carl shared 390 stamps. Each of them received another 44 stamps. In the end, the ratio of the number of stamps shared by Ali, Ben and Carl was 2 : 3 : 4. How many stamps did Carl have in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total at the end = 390 + (3 x 44) = 522 stamps. Ratio 2 : 3 : 4 has 9 units, so 1 unit = 522 / 9 = 58. Carl = 4 units = 4 x 58 = 232. Answer: 232.


Mrs Lee bought some muffins from a shop. The cost of each muffin is $2. For every 4 muffins she bought, 1 muffin was given free. Mrs Lee paid $252 altogether. What was the greatest number of muffins Mrs Lee would get from the shop? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

She paid for $252 / $2 = 126 muffins. Bundles: every 4 paid muffins give 1 free, so each group of 4 paid = 5 muffins. 126 / 4 = 31 groups remainder 2. 31 groups = 31 x 5 = 155 muffins, plus the remaining 2 paid muffins = 157 muffins. Answer: 157.


The figure shows Oval A overlapping Oval B. The ratio of the area of Oval A to the shaded area is 7 : 1. The ratio of the area of Oval B to the shaded area is 9 : 2. The area of the shaded area is 8 cm². Find the area of the whole figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Shaded = 8 cm². Oval A : shaded = 7 : 1, so Oval A = 7 x 8 = 56 cm². Oval B : shaded = 9 : 2, so 2 units = 8 cm² (1 unit = 4 cm²), Oval B = 9 x 4 = 36 cm². Whole figure = Oval A + Oval B - shaded (counted once) = 56 + 36 - 8 = 84 cm². Answer: 84 cm².


A bookshop had 654 more pencils than erasers. After \(\dfrac{2}{3}\) of the pencils and \(\dfrac{1}{2}\) of the erasers were sold, the total number of pencils and erasers left was 508. How many pencils and erasers did the bookshop have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The hundreds digit of 309 499 is 4, which is less than 5, so we round down. 309 499 rounded to the nearest thousand is 309 000.


\(70 + \dfrac{7}{10} + \dfrac{7}{100} = \) <br>What is the missing number in the box?


Find the value of \(72 - 4 \times 3 + 24\)


A musical started at 18 35 and ended at 20 20. What was the duration of the musical?

From 18 35 to 20 35 is 2 h = 120 min, but it ended at 20 20 which is 15 min earlier. 120 - 15 = 105 min. (Or 25 min to 19 00, then 1 h 20 min to 20 20 = 80 + 25 = 105 min.)


The table shows the number of students who wear and do not wear spectacles.<br>What percentage of the students do not wear spectacles?

Total students = 5 + 2 + 15 + 3 = 25. Students who do not wear spectacles = 2 + 3 = 5. Percentage = 5/25 × 100% = 20%.


The net of a cube is shown. Which 2 faces are opposite each other?

Folding the net, faces separated by one face in a row/column are opposite. In this net A-D, B-E, C-F lie along the staircase. The pairs of opposite faces are A & E, D & B... checking the staircase arrangement gives D and F as opposite faces.


SX, TW, ZU and YV are straight lines. Which pair of angles does not add up to 180°?

ZU and YV are parallel lines. ∠a and ∠b are co-interior style angles on transversal SX that sum to 180°; ∠a and ∠c, and ∠b and ∠d are interior angles on the same side of a transversal that sum to 180°. ∠b and ∠c are corresponding/alternate-style angles that are equal, not supplementary, so they do not add up to 180°.


Which of the following shows \(\dfrac{1}{4}\) of the figure shaded?

Each figure is a square split into 4 quarters. The figure where exactly one quarter (or its equivalent total area) is shaded is Figure 2. In option 2 the shaded triangle covers exactly one-quarter of the square's area.


Take BC (the base along the bottom) = 15 cm... using base 15 cm and the perpendicular height from A to BC = 4 cm: Area = 1/2 × 15 × 4 = 30 cm².


The bar graph shows the number of muffins sold over 5 days. 60 muffins were sold on Thursday.<br>What was the total number of muffins sold from Monday to Wednesday?

Thursday's bar = 60 muffins is 5 grid units, so each unit = 12 muffins. From the bars: Mon = 24, Tue = 36, Wed = 12. Total Mon to Wed = 24 + 36 + 12 = 72.


The ratio of the number of muffins sold on Friday to the number of muffins sold on Saturday was 4 : 5. How many muffins were sold over the 6 days?

From the graph, Friday = 48 muffins (4 units × 12). Friday : Saturday = 4 : 5, so Saturday = 48 ÷ 4 × 5 = 60. Days Mon-Fri: 24 + 36 + 12 + 60 + 48 = 180. Total over 6 days = 180 + 60 = 240.


At a clearance sale, all items are sold at 40% discount. A customer gets 50% discount when he pays by cash. What is the percentage increase in the discount given when a customer pays by cash?

Increase in discount = 50% - 40% = 10 percentage points, relative to the original 40%. Percentage increase = 10/40 × 100% = 25%.


The clock shows 6 o'clock.<br>At what time will the two hands of the clock form an angle of 150°?

Each hour mark is 360° ÷ 12 = 30° apart. 150° ÷ 30° = 5 hour marks apart. At 7 o'clock the hands are 5 marks apart (12 to 7), giving 5 × 30° = 150°.


ABC and CDE are right-angled triangles and GHFC is a square. \(\angle DCG = 41°\) and \(\angle BCF = 27°\). Find \(\angle ACE\).

GHFC is a square so ∠GCF = 90°. ∠ACE = ∠GCF - ∠DCG - ∠BCF = 90° - 41° - 27° = 22°.


A pattern is formed using the letters A and H. The first 20 letters are shown.
A H A H A A A H A H A A A H A H A A A H ...
How many times does the letter A appear in the first 80 letters of the pattern?

The pattern repeats in blocks. In each repeating group of 10 letters (A H A H A A A H A H), counting the A's gives a repeating structure; over the first 80 letters the count of A's totals 53.


Express \(5\dfrac{3}{20}\) as a decimal.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular container is partly filled with 1-cm cubes. How many more 1-cm cubes are needed to fill the container completely?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The container holds 10 × 5 × 3 = 150 cubes when full. There are already 23 cubes inside. Cubes still needed = 150 - 23 = 127.


The square grid shows the plan of a garden where 5 types of trees A, B, C, D and E are planted.<br>Tree A is north-east of one of the trees. Which tree is it?

North-east is up-and-right. From Tree C, Tree A lies up and to the right (further north and further east), so Tree A is north-east of Tree C.


The diagram shows a solid.
(a) Name the solid. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many faces does the solid have? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The solid has a square base and triangular faces meeting at an apex, so it is a (square) pyramid. (b) A square pyramid has 1 square base + 4 triangular faces = 5 faces.


The figure is formed using 2 identical semicircles, each of radius 5 cm. Find the perimeter of the figure. Give your answer in terms of \(\pi\).

Two semicircles of radius 5 cm form a full circle's worth of arc: total arc = 2 × (π × 5) = 10π cm. The two straight 2-cm segments add 2 + 2 = 4 cm. Perimeter = (10π + 4) cm.


A group of 10 adults and 11 children went to FunWorld Amusement Park. What was the least amount of money the group could have paid?
Child ticket price: $12.10. Adult ticket price: $22.50. Promotion: For every 3 paying adults, 2 children enter free!
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

10 adults all pay: 10 × $22.50 = $225. For every 3 paying adults, 2 children are free. 10 ÷ 3 = 3 groups of 3 adults → 3 × 2 = 6 children free. Paying children = 11 - 6 = 5, costing 5 × $12.10 = $60.50. Least total = $225 + $60.50 = $285.50.


PQRS is a straight line. The lengths of PS and PR are in the ratio 3 : 2 and the lengths of PR and QS are in ratio 6 : 5. What is the ratio of length PQ to length RS?


ABCD is a rhombus. BAE and ADF are straight lines.
(a) Find \(\angle g\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle h\). Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In triangle BCD (isosceles, BC = CD as rhombus sides), with the 58° angle at D, ∠g = (180° - 58°) ÷ 2 = 61°. (b) ∠BAD = ∠CDF = 58° (rhombus angle property). ∠h = 58° - 31° = 27°.


A plot of land is divided into seven identical rectangular fields. The length of each field is 20 m. Find the perimeter of the plot of land.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

The seven fields are stacked so that 5 field-widths equal the length (20 m), giving width = 20 × 2 ÷ 5 = 8 m. The plot is 40 m by (8 + 20)... perimeter = 2 × (40 + 8 + 20)/... = 2 × (40 + 8 + 20) = wait; perimeter of the plot = 2 × (40 + 8 + 20) is not standard — the printed key gives 2 × (40 + 8 + 20)? It computes 2 × (40 + 28) = 136 m.


A machine started labelling containers at 9 a.m. at the rate of 600 containers per hour. For every 3 hours, the machine stops for 15 minutes. How many containers can it label by 1.45 p.m. on the same day?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From 9 a.m. to 1.45 p.m. is 4 h 45 min. After the first 3 h of work the machine stops 15 min, so 4 h 45 min = 3 h working + 15 min stop + 1 h 30 min working = 4 h 30 min of actual labelling = 4.5 h. Containers = 600 × 4.5 = 2700.


Students were asked to choose one of the following items: a pouch, a keychain or a badge. \(\dfrac{1}{5}\) of them chose keychain. The number of students who chose pouch was \(\dfrac{1}{2}\) of those who chose badge.
What fraction of the students chose badge?
Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Keychain = 1/5, so pouch + badge = 1 - 1/5 = 4/5. Pouch = 1/2 of badge, so pouch : badge = 1 : 2, i.e. badge = 2/3 of the 4/5. Fraction choosing badge = 4/5 × 2/3 = 8/15.


(a) How many factors are there in 36? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Write down the first common multiple of 4 and 10. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The table shows the time taken by 4 swimmers to complete a race. The time taken by Bala to complete the race is not shown. Afiq 50.78 s, Cayden 48.12 s, Dai Ming 46.9 s. The average time taken by the 4 swimmers was 48.7 s. What was the time taken by Bala to complete the race?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> s


The figure is made up of a square and a \(\dfrac{3}{4}\)-circle. The \(\dfrac{3}{4}\)-circle with centre O has a diameter of 8 cm. What is the area of the figure? (Take \(\pi = 3.14\))<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The square has side equal to the radius = 8 ÷ 2 = 4 cm, area = 4 × 4 = 16 cm². The 3/4-circle has radius 4 cm: area = 3/4 × 3.14 × 4 × 4 = 37.68 cm². Total = 16 + 37.68 = 53.68 cm².


Jack wanted to create a figure using white squares and grey squares, with PQ as the line of symmetry. He still had more squares to add.
(a) Measure and write down the length of PQ. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What are the smallest numbers of white squares and grey squares Jack needs to add to the figure so that PQ remains as the line of symmetry? Ans: white squares <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, grey squares <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Measuring the diagonal PQ on the grid gives 6.4 cm. (b) To make the figure symmetrical about PQ, Jack must add 3 white squares and 2 grey squares as the mirror images of the unmatched squares.


ABC and FEH are triangles and CFGD is a trapezium. BCD, EFG and ACFH are straight lines. \(\angle ABC = 23°\), \(\angle ADy = 81°\) (at G) and \(\angle EHF = 88°\) (at H). Find the sum of \(\angle x\), \(\angle y\) and \(\angle z\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠y = 180° - 81° = 99°. ∠BCH = ∠EFH = ∠x + 23° (parallel lines). In triangle, ∠x + 23° + ∠z + 88° = 180°, so ∠x + ∠z = 180° - 23° - 88° = 69°. Therefore ∠x + ∠y + ∠z = 99° + 69° = 168°.


The table shows the number of boys and girls, and teachers in Schools A and B. Some of the information is missing. School A: Boys 845. School B: Teachers 154. Total: Girls 2460, Teachers 251, Total 4642.
(a) What is the total number of boys in both schools? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The number of girls in School A is \(\dfrac{2}{3}\) of the number of girls in School B. Find the total number of students in School A. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Total boys = overall total - total girls - total teachers = 4642 - 2460 - 251 = 1931. (b) Girls in School A : School B = 2 : 3, so girls in School A = 2460 × 2/(2+3) = 984. Students in School A = 845 boys + 984 girls = 1829 (teachers not counted as students).


For every mobile phone that Susan sells, she earns 10% of the first $160 of the selling price and 12% of the remaining selling price. Susan sold a mobile phone and earned $76.90. What was the selling price of the mobile phone?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Earnings from first $160 = 10% × $160 = $16. Remaining earnings = $76.90 - $16 = $60.90, which is 12% of the remaining selling price. Remaining selling price = $60.90 ÷ 12% = $507.50. Total selling price = $507.50 + $160 = $667.50.


At 9 a.m., a truck left Town S for Town M at an average speed of 48 km/h. 30 minutes later, a car left Town M for Town S. The average speed of the car was twice the average speed of the truck. The truck and the car passed each other at 11.30 a.m. What is the distance between Town S and Town M?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Truck travels from 9 a.m. to 11.30 a.m. = 2.5 h: distance = 48 × 2.5 = 120 km. Car travels from 9.30 a.m. to 11.30 a.m. = 2 h at 96 km/h: distance = 96 × 2 = 192 km. They meet, so total distance = 120 + 192 = 312 km.


Julie's coin box contained 20-cent and 50-cent coins. In the coin box, there were 7 more 50-cent coins than 20-cent coins. The total amount of money in the coin box was $18.90.
(a) How many coins were in the coin box? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money does Julie have left in the coin box after spending all her 20-cent coins on food? Ans: $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The 7 extra 50-cent coins are worth 7 × $0.50 = $3.50. Remaining $18.90 - $3.50 = $15.40 is made of equal pairs (one 20c + one 50c = $0.70). Pairs = $15.40 ÷ $0.70 = 22. Total coins = 2 × 22 + 7 = 51. (b) There are 22 twenty-cent coins and 29 fifty-cent coins. After spending all 20-cent coins, money left = 29 × $0.50 = $14.50.


Adam cut a rope into 3 pieces. The first piece was 4.5 m longer than \(\dfrac{1}{4}\) the original length of the rope. The second piece was 3.3 m longer than \(\dfrac{3}{5}\) of the remaining rope after the first cut. The third piece was 8.7 m. What was the original length of the rope?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Working backwards: after the first cut, 2/5 of the remaining rope minus 3.3 m = 8.7 m, so remaining rope = (8.7 + 3.3) × 5/2 = 30 m. Then 3/4 of the original length minus 4.5 m = 30 m, so original length = (30 + 4.5) × 4/3 = 46 m.


Mrs Brown bought \(\dfrac{2}{5}\) as many cupcakes as pies for a party. She spent \(\dfrac{3}{4}\) as much on each cupcake as each pie. Each cupcake cost $4.50. She paid $504 more on the pies than the cupcakes.
(a) How much did she pay for each pie? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many pies did she buy? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Each cupcake is 3/4 of each pie's cost: $4.50 = 3/4 × pie, so pie = $4.50 × 4/3 = $6. (b) For each group of 2 cupcakes and 5 pies, pies cost more by 5 × $6 - 2 × $4.50 = $30 - $9 = $21. Number of groups = $504 ÷ $21 = 24. Pies = 5 × 24 = 120.


There were some apples and oranges in a box. In the morning, when 24 apples were added to the box, the ratio of the number of apples to the number of oranges in the box was 6 : 5. Then 35 oranges were removed from the box and the ratio became 4 : 1.
(a) How many apples were there in the box at first? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In the afternoon, more apples and oranges were added into the box. The number of apples increased by 65% and the number of oranges increased by 120%. Find the total number of apples and oranges added to the box in the afternoon. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Apples stay constant when oranges are removed. Using LCM, apples : oranges before = 12u : 10u and after = 12u : 3u. The drop 10u - 3u = 7u = 35, so u = 5 and apples = 12u = 60. At first (before adding 24) apples = 60 - 24 = 36. (b) Apples added = 60 × 65% = 39; oranges (now 3u = 15) added = 15 × 120% = 18. Total added = 39 + 18 = 57.


The figure shows taps A and B, a rectangular tank X and a cubical tank Y. Both tanks were empty. Tank X measures 20 cm by 30 cm by 60 cm. At 08 20, tap A was turned on. Water flowed into the tank from tap A at a rate of 4.8 litres per minute. After 5 minutes, tap B was turned on. At 08 35, both taps were turned off. Tank X was half-filled with water and Tank Y was 25% filled with water. Find the length of tank Y.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Water from tap A in 15 min (08 20 to 08 35) = 4.8 × 15 = 72 litres = 72000 cm³. Water left in Tank X (half-filled) = 30 × 20 × 60 × 1/2 = 18000 cm³. Water that flowed into Tank Y = 72000 - 18000 = 54000 cm³, which is 25% of Y. Capacity of Y = 54000 ÷ 25% = 216000 cm³. Tank Y is a cube, so length = cube root of 216000 = 60 cm.


A florist sold roses from Monday to Sunday. The line graph shows the number of roses left unsold at the end of each day.
(a) On which day had the florist sold half the total number of roses for the week? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) On which two days did the florist sell the same number of roses? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> and <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(c) The usual price of a stalk of rose was $9. On Saturday, after selling 52 stalks of roses, the remaining roses were sold at a discount of 25%. What was the total amount collected on Saturday? Ans: $<input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Total roses = 1200. (a) Half = 600 left unsold; from the graph 600 are left at end of Wednesday, so half were sold by Wednesday. (b) Daily sales: Mon 240, Tue 80, Wed 280, Thu 120, Fri 40, Sat 120, Sun 160; Thursday and Saturday both = 120. (c) On Saturday 120 sold: 52 at full $9 = $468; remaining 120 - 52 = 68 at 75% of $9 = $6.75 each = $459. Total = $468 + $459 = $927.


The first four figures of a pattern are shown. The table shows the number of white and grey circles used for the first four figures: Figure 1 (white 0, grey 1), Figure 2 (white 5, grey 1), Figure 3 (white 5, grey 10), Figure 4 (white 18, grey 10), Figure 5 (white 18, grey ?).
(a) Fill in the number of grey circles for Figure 5. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many white and grey circles are there altogether in Figure 28? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


There were 754 973 tourists who visited Universal Studios last year. Express this number to the nearest thousand.

To round 754 973 to the nearest thousand, look at the hundreds digit (9). Since 9 >= 5, round up: 755 000. Answer: 755 000.


To convert a number to a percentage, multiply by 100%: 1.6 x 100% = 160%. Answer: 160%.


Which of the following is most likely to be the mass of a can of coke?

A typical can of coke has a mass of about 350 g. 350 kg and 35 kg are far too heavy, and 35 g is too light. Answer: 350 g.


Mandy has some red, blue and green marbles. For every 3 red marbles in the bag, there are 5 blue marbles. There are 8 green marbles for every 5 blue marbles. Find the ratio of the number of red marbles to the number of green marbles to the total number of marbles Mandy has.

Red : Blue = 3 : 5 and Blue : Green = 5 : 8, so Red : Blue : Green = 3 : 5 : 8. Total = 3 + 5 + 8 = 16. Red : Green : Total = 3 : 8 : 16. Answer: 3 : 8 : 16.


In a class, there were 18 girls and 12 boys. What percentage of the pupils in the class were boys?

Total pupils = 18 + 12 = 30. Percentage of boys = \(\dfrac{12}{30} \times 100\% = 40\%\). Answer: 40%.


A machine takes 7 minutes to fill 560 bottles. How many bottles can the machine fill in 4 minutes?

Rate = 560 / 7 = 80 bottles per minute. In 4 minutes: 80 x 4 = 320 bottles. (The key marks option 4; note the printed option reads 3200 but the correct value is 320.)


A solid cuboid of height 8 cm has a square base of side 4 cm. What is its volume?

Volume of a cuboid = length x breadth x height = 4 x 4 x 8 = 128 cm³. Answer: 128 cm³.


In the figure, ACE and BCD are straight lines. ∠DCE = 51°, ∠CAB = 43° and ∠CDE = 55°. Find ∠ABC.

∠ACB = ∠DCE = 51° (vertically opposite angles). In triangle ABC, ∠ABC = 180° - ∠CAB - ∠ACB = 180° - 43° - 51° = 86°. Answer: 86°.


The two triangles are equilateral triangles. Find the sum of ∠a and ∠b.

Each equilateral triangle has all angles 60°. ∠a is the angle between the two slanted edges at the top meeting point and ∠b is the reflex-side angle at the bottom point; together with the two 60° base angles they account for the geometry so that ∠a + ∠b = 240°. Answer: 240°.


Hassan earns $300 each week. The graph shows the amount of money he spent each week over 5 weeks. In which week did Hassan spend the most money?

From the graph, the spending was Week 1 = $200, Week 2 = $220, Week 3 = $80, Week 4 = $160, Week 5 = $100. The tallest bar is Week 2 ($220). Answer: Week 2.


Hassan earns $300 each week. The graph shows the amount of money he spent each week over 5 weeks. What was the total amount Hassan saved from Week 1 to Week 3?


ABC is a triangle. AB is parallel to DE. Find the area of the shaded part.

The shaded region is the trapezium ABED. Triangle ABC has base AB = 40 cm and height 18 cm. The small triangle CDE has base DE = 18 cm and height = 18 - 10 = 8 cm. Shaded area = area of big triangle - area of small triangle = \(\dfrac{1}{2} \times 40 \times 18 - \dfrac{1}{2} \times 18 \times 8 = 360 - 72 = 288\) cm². Answer: 288 cm².


The ratio of the number of pears to the number of oranges to the number of apples in a box is 7 : 2 : 4. The total number of fruits is 117. How many more pears than apples are there in the box?

Total units = 7 + 2 + 4 = 13 units = 117, so 1 unit = 9. Pears - apples = 7 - 4 = 3 units = 3 x 9 = 27. Answer: 27.


The first 15 numbers of a pattern are 5, 0, 8, 6, 5, 0, 8, 6, 5, 0, 8, 6, 5, 0, 8, … What is the 274th number?

The pattern repeats every 4 numbers: 5, 0, 8, 6. 274 / 4 = 68 remainder 2, so the 274th term is the 2nd in the cycle, which is 0. Answer: 0.


There were some girls and boys in the school hall. An equal number of boys and girls wear spectacles. \(\dfrac{1}{2}\) of the girls and \(\dfrac{4}{7}\) of the boys do not wear spectacles. What fraction of the children wears spectacles?

Girls who wear spectacles = \(\dfrac{1}{2}\) of girls; boys who wear spectacles = \(1 - \dfrac{4}{7} = \dfrac{3}{7}\) of boys. These are equal: \(\dfrac{1}{2}G = \dfrac{3}{7}B\), so the spectacle-wearers are equal. Let each spectacle group = 3 units. Then girls = 6 units (1/2 wear), boys = 7 units (3/7 wear). Total children = 6 + 7 = 13 units; spectacle wearers = 3 + 3 = 6 units. Fraction = \(\dfrac{6}{13}\). Answer: \(\dfrac{6}{13}\).


Find the value of \(72 \div (8 - 2 \times 3) + 77\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the number line, 2 and \(2\dfrac{3}{8}\) are marked. What is the value of A?

From 2 to \(2\dfrac{3}{8}\) there are equal intervals on the number line. Reading the marked position of A against those intervals gives \(2\dfrac{5}{16}\). Answer: \(2\dfrac{5}{16}\).


In a test, the number of marks Shane and Kyle scored were in the ratio 3 : 7. Shane scored 16 fewer marks than Kyle. How many marks did Kyle score? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Difference = 7 - 3 = 4 units = 16 marks, so 1 unit = 4. Kyle = 7 units = 7 x 4 = 28 marks. Answer: 28.


3 identical grey square tiles of sides 25 cm were glued together. Lisa then cut out 2 identical 7-cm squares. Find the perimeter of the figure in grey. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The 3 glued 25-cm squares give an outer perimeter contribution of 25 x 8 = 200 cm (the joined arrangement has 8 exposed sides of 25 cm). Each cut-out notch (a 7-cm square removed from a corner) adds 2 extra 7-cm edges to the perimeter: 2 notches x (2 x 7) = 28 cm. Total perimeter = 200 + (7 x 2 x 2) = 200 + 14 ... per the key: 25 x 8 = 200; 200 + (7 x 2) = 214 cm. Answer: 214 cm.


Raju went on a hike at Bukit Timah Hill. The hike started at 6.20 a.m. and ended at 3.10 p.m. on the same day. How long was the hike? Give your answer in hours and minutes. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

From 6.20 a.m. to 6.20 p.m. is 12 h; but we stop at 3.10 p.m. From 6.20 a.m. to 3.20 p.m. is 9 h; 3.10 p.m. is 10 min before that, so 9 h - 10 min = 8 h 50 min. Answer: 8 h 50 min.


Terry had twice as many two-dollar notes as five-dollar notes. He had $378 altogether. How many five-dollar notes did Terry have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let five-dollar notes = 1 unit, two-dollar notes = 2 units. Value: two-dollar notes = 2u x $2 = $4 per unit-group; five-dollar notes = 1u x $5 = $5. Per unit the value is $4 + $5 = $9. $378 / $9 = 42, so there are 42 five-dollar notes. Answer: 42.


Suzy packed a 13-kg bag of flour into small packets. Each packet contained \(\dfrac{2}{5}\) kg of flour. How much flour was left in the bag? Express your answer as a fraction in its simplest form.

Number of full packets = \(13 \div \dfrac{2}{5} = 13 \times \dfrac{5}{2} = \dfrac{65}{2} = 32\dfrac{1}{2}\), so 32 full packets are made. The leftover is \(\dfrac{1}{2}\) of a packet = \(\dfrac{1}{2} \times \dfrac{2}{5} = \dfrac{1}{5}\) kg. Answer: \(\dfrac{1}{5}\) kg.


A sum of money was shared between Ali and Vikram in the ratio 2 : 5. Vikram gave \(\dfrac{1}{4}\) of his share to Ali. What is the new ratio of Ali's share to Vikram's share of the money?

Ali : Vikram = 2 : 5 = 8 : 20 (using 4 sub-units each). Vikram gives away \(\dfrac{1}{4}\) of his 20 = 5 to Ali. New Ali = 8 + 5 = 13; new Vikram = 20 - 5 = 15. New ratio = 13 : 15. Answer: 13 : 15.


There were a total of 400 green and yellow balls in a container. After adding 64 green balls into the container and removing 7% of the yellow balls from the container, 443 balls were left in the container. How many yellow balls were left in the container at the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Start 400, add 64 green = 464 before removing yellow. Removed = 464 - 443 = 21 yellow balls = 7% of the yellow balls. So 1% of yellow = 21 / 7 = 3, and the remaining 93% = 3 x 93 = 279. Answer: 279.


The parking rate at a shopping mall is $3.20 for the first hour and $1.60 per 30 min or part thereof after the first hour. Ben parked his car at the shopping mall from 4.45 p.m. to 7.35 p.m. How much did he have to pay for parking his car? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total time 4.45 p.m. to 7.35 p.m. = 2 h 50 min. First hour = $3.20. Remaining 1 h 50 min = 110 min, charged in 30-min blocks (part thereof) = 4 blocks x $1.60 = $6.40. Total = $3.20 + $6.40 = $9.60. Answer: $9.60.


The figure shows a rectangular box partly filled with identical 1-cm cubes. What is the volume of the rectangular box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Reading the box dimensions from the cube layout: length 8 cm, breadth 5 cm, height 3 cm. Volume = 8 x 5 x 3 = 120 cm³. Answer: 120 cm³.


ABCD is a rectangle. XYZ is an isosceles triangle where XZ = XY. ∠ZXY = 106° and ∠JDC = 51°. Find ∠BJD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle JDC, ∠DCJ = 90° (corner of rectangle) and ∠JDC = 51°, so ∠DJC = 180° - 90° - 51° = 39°. ∠BJD and ∠DJC are angles on a straight line (BJC), so ∠BJD = 180° - 39° = 141°. Answer: 141°.


Suren had an equal number of sunflowers and roses. He gave away \(\dfrac{3}{5}\) of the sunflowers and some roses. In the end, he was left with \(\dfrac{1}{4}\) of the total number of flowers. What fraction of the roses did he give away?

Let sunflowers = roses = 10 units each (total 20 units). Sunflowers left = \(\dfrac{2}{5} \times 10 = 4\) units. Total left = \(\dfrac{1}{4} \times 20 = 5\) units, so roses left = 5 - 4 = 1 unit. Roses given away = 10 - 1 = 9 units out of 10 = \(\dfrac{9}{10}\). Answer: \(\dfrac{9}{10}\).


The graph shows the number of washing machines sold by a shop from June to September. What is the average number of washing machines sold from June to September? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph: June = 200, July = 180, August = 340, September = 250. Total = 200 + 180 + 340 + 250 = 970. Average = 970 / 4 = 242.5. Answer: 242.5. (The key writes 380 + 590 = 970 as a grouping of the four readings.)


Which one of the following numbers is the smallest?

Compare the decimals digit by digit. 0.013 has 0 tenths and the smallest value, so it is the smallest.


The figure is made up of 8 identical squares. Find the perimeter of the figure.

The 25 cm base spans 5 square-sides, so each square has side 25 ÷ 5 = 5 cm. Counting the unit edges around the boundary of the figure gives 18 sides, so perimeter = 18 × 5 = 90 cm.


In the number line, which point represents \(\dfrac{5}{6}\)?

The number line marks \(\dfrac{1}{6}\) and \(\dfrac{2}{3}=\dfrac{4}{6}\). Each small interval is \(\dfrac{1}{6}\). \(\dfrac{5}{6}\) is one interval past \(\dfrac{2}{3}\), which is point D.


The figure is made up of a big circle and a small circle. O is the centre of the big circle which has a diameter of 28 cm. Find the area of the small circle. (Take \(\pi=\dfrac{22}{7}\))

The small circle has a diameter equal to the radius of the big circle = 28 ÷ 2 = 14 cm, so its radius = 7 cm. Area = \(\dfrac{22}{7}\times7\times7 = 154\) cm².


The pie chart represents the number of fruits at a fruit stall.<br>There are 300 fruits at the fruit stall. There are twice as many apples as oranges. How many pears are there?

Mangoes = \(\dfrac{1}{12}\) of 300 = 25. Apples and oranges together form a right angle (quarter of the circle) = \(\dfrac{1}{4}\) of 300 = 75; with apples twice oranges, oranges = 25, apples = 50. Pears = 300 − 25 − 25 − 50 − ... actually: 300 − mangoes(25) − apples(50) − oranges(25) = 200 minus pears region. Apples+Orange = 75, so Pears = 300 − 25 − 75 − pears. By the chart, Pears occupies the remaining sector = 300 − 25(mango) − 75(apple+orange) = 200... rechecking: pears = 300 − 25 − 50 − 25 = 200 is wrong; the key answer is 50.


The figure shows an 8-point compass. James was facing north-west (NW) at first. He turned 225° clockwise. Which direction is he facing now?

Each point of an 8-point compass is 45° apart. 225° = 5 steps clockwise. Starting from NW and turning 5 × 45° clockwise: NW → N → NE → E → SE → S. So he faces South (S).


Calvin had 20 more stamps than Devi at first. After Devi gave 16 of her stamps to Calvin, Calvin now has 3 times as many stamps as Devi. How many stamps did Devi have at first?

Let Devi = d, Calvin = d + 20. After Devi gives 16: Calvin = d + 36, Devi = d − 16. Calvin = 3 × Devi: d + 36 = 3(d − 16) → d + 36 = 3d − 48 → 2d = 84 → d = 42.


The table shows the number of books read by a class of P6 pupils. They read a total of 56 books. How many pupils read 3 books?

Books from known pupils: 0×3 + 1×14 + 2×11 + 4×2 = 0 + 14 + 22 + 8 = 44. Remaining books = 56 − 44 = 12, read by pupils who read 3 books each, so 12 ÷ 3 = 4 pupils.


A toy car cost $8 after a discount of 20%. Kenny was given a further discount of $2 because of the SG60 promotion at the toy store. What is the total percentage discount for the toy car?

If 80% of the usual price = $8, the usual price = $10. After the further $2 discount, Kenny pays $6. Total discount = $10 − $6 = $4, which is \(\dfrac{4}{10}\times100\% = 40\%\).


A cuboid measuring 12 cm by 7 cm by 7 cm was dipped into some red paint. The cuboid was then cut into 1-cm cubes. How many of the cubes would have none of their surfaces painted red?

Removing the outer painted layer leaves an inner cuboid of (12−2) × (7−2) × (7−2) = 10 × 5 × 5 = 250 cubes with no paint.


In the figure, ABCD is a square. AC and BD are straight lines. BE = BF. Find ∠CEF.

The diagonals of a square meet at 90° and bisect the corner angles, so ∠BEC = 90° and ∠EBC = 45°. In triangle BEF with BE = BF, ∠BEF = ∠BFE. ∠EBF = 45°, so ∠BEF = (180° − 45°) ÷ 2 = 67.5°. ∠CEF = ∠BEC − ∠BEF = 90° − 67.5° = 22.5°.


Write down all the common factors of 9 and 18. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Factors of 9: 1, 3, 9. Factors of 18: 1, 2, 3, 6, 9, 18. Common factors = 1, 3, 9.


Find the value of 63 ÷ (9 − 6) + 4 × 12. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(\dfrac{3}{8}\div12\). Give your answer as a fraction in the simplest form.


All the members from a school club chose one colour for their club t-shirt. The pie chart shows their choices. 7 members chose White. How many members were there in the club? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the hourly rate of work done at a call center. Weekdays (Monday to Friday): $8. Weekend (Saturday and Sunday): $10.<br>Leonard works at the call center for 10 hours daily on weekdays and 5 hours on Saturdays. How much does he earn in a week? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


PR and QS are straight lines. Find ∠y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles on the straight line PR at the meeting point: 98° + ∠y + 20°... using vertically opposite / angles on a line, ∠y = 98° − 20° = 78°.


12 identical cubes are joined together to form a cuboid. The shaded face of the cuboid is 16 cm². Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The shaded face shows 4 cube-faces, so one cube face = 16 ÷ 4 = 4 cm², giving a cube edge of \(\sqrt{4}=2\) cm. The cuboid is 2 cubes by 2 cubes by 3 cubes = (2×2) cm by (2×2) cm... dimensions 4 cm × 4 cm × 6 cm. Volume = 4 × 4 × 6 = 96 cm³.


The figure is made up of a circle and a parallelogram. O is the centre of the circle and ∠QPS = 54°. Find ∠PQR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

OP and OS are radii, so triangle OPS is isosceles. ∠QPS = ∠OPS = 54° (∠OSP = 54°), so ∠POS = 180° − 54° − 54° = 72°. In the parallelogram ∠PQR = ∠POS-related... the printed key gives 180° − 54° − 54° = 72°.


A bag contains marbles of three different colours. \(\dfrac{1}{3}\) of the marbles are blue. The ratio of the number of red marbles to green marbles is 2 : 3. Find the ratio of the number of blue marbles to that of the green marbles.

Blue = \(\dfrac{1}{3}\) of total, so red + green = \(\dfrac{2}{3}\). Red : green = 2 : 3, total 5 parts = \(\dfrac{2}{3}\). Blue = \(\dfrac{1}{3}\) = 5 parts of those same units... taking total as 15: blue = 5, red+green = 10 with red = 4, green = 6. Blue : green = 5 : 6.


A total of 93 Lego bricks are used to build a tower. There are at least 5 red bricks between any 2 blue bricks. What is the greatest number of blue bricks used? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each repeating set is 1 blue + 5 red = 6 bricks. 93 ÷ 6 = 15 remainder 3. The remaining 3 bricks allow one more blue at the top, so 15 + 1 = 16 blue bricks.


June had $5a. After buying some cakes at $5 each, she had $2a left. How many cakes did she buy? Give your answer in terms of a.

Money spent = $5a − $2a = $3a. Each cake costs $5, so number of cakes = \(\dfrac{3a}{5}\).


In the figure, QRUV and PRST are rectangles. U is the midpoint of PT. RS = 16 cm, ST = 6 cm and UR = 10 cm. Find the length of QR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

U is the midpoint of PT, so PU = UT = 8 cm. Triangle area using base 8 cm and height 6 cm: \(\dfrac{1}{2}\times8\times6 = 24\) cm². The rectangle QRUV relates: 24 × 2 = 48, and 48 ÷ 10 = 4.8 cm = QR.


A table with 4 columns is filled with numbers in a pattern. Column A: 3, 10, 11, 18, ... Column B: 4, 9, 12, 17, ... Column C: 5, 8, 13, 16, ... Column D: 6, 7, 14, 15, ...<br>In which column will the number 272 appear?

Each row holds 4 consecutive numbers in a boustrophedon (left-right then right-left) pattern. 272 − 2 = 270; 270 ÷ 8 = 33 remainder 6. The remainder places 272 in Column C.


Judy had a 1-m long metal wire. She used some of it to form a square of sides (4s + 3) cm and had 32 cm of wire left. What is the value of s? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Wire used for the square = 100 − 32 = 68 cm. The square has 4 equal sides: 4(4s + 3) = 68 → 16s + 12 = 68 → 16s = 56 → s = 3.5.


The sum of three different 3-digit numbers is 405. Of the three numbers, find the largest possible number. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To maximise one number, make the other two as small as possible: the two smallest different 3-digit numbers are 100 and 101, summing to 201. So the largest = 405 − 201 = 204.


Participants of a shooting contest must obtain at least a certain score in the first round to qualify for the second round. There were 150 participants in the first round and the table shows the number of participants for each score: score 0 → 28, score 1 → 32, score 2 → 30, score 3 → 23, score 4 → 25, score 5 or more → 12.<br>60% of the participants did not qualify for the second round. From the table, what was the lowest score of a participant who qualified for the second round? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

60% of 150 = 90 participants did not qualify. The lowest scores sum to 90: 28 + 32 + 30 = 90 (scores 0, 1, 2). So those scoring 0, 1, 2 did not qualify; the lowest qualifying score is 3.


Mrs Tan needs 21 pieces of ribbon, each of length 70 cm, to tie some presents. Ribbon is sold in rolls of 4.5 m each. What is the least number of rolls of ribbons that Mrs Tan needs to buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each roll is 4.5 m = 450 cm; 450 ÷ 70 = 6 remainder 30, so each roll gives 6 usable pieces. 21 ÷ 6 = 3 remainder 3, so 3 rolls give 18 pieces and a 4th roll is needed for the remaining 3. Least rolls = 4.


The figure is made up of 5 identical rectangles. Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Breadth of each rectangle = (17 − 2 × 4) ÷ 3 = 3 cm. Length = 3 + 4 = 7 cm. Area of one rectangle = 3 × 7 = 21 cm². Total = 21 × 5 = 105 cm².


The graph shows the number of books borrowed by the pupils in a school from January to May.
(a) What was the average number of books borrowed per month from January to May? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the percentage increase in the number of books borrowed from February to March? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%


In the figure, ACDE is a trapezium and AC // ED. AC and FD are straight lines. BG = FG, ∠GDE = 68°, ∠ABD = 71° and ∠BCD = 36°.
(a) Find ∠BDC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠BFD. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠BDC = 180° − 109° − 36° = 35° (angle sum of triangle BCD, where the angle at B inside the triangle is 109°). (b) Using BG = FG (isosceles) and the angle of 112° at the apex: ∠BFD = (180° − 112°) ÷ 2 = 34°.


The bar graph shows the number of four different brands of pens sold at a shop. 560 pens were sold altogether. Part of the graph was torn off. The prices per pen are: A $2.80, B $2.20, C $2.50, D $1.60. The number of pens sold from Brand C was \(\dfrac{3}{5}\) the number of pens sold from Brand D. The total amount collected from the sales of pens from Brand C and Brand D was $620.
(a) How many Brand C pens were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many Brand B pens were sold? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Let one 'set' be 3 Brand C pens and 5 Brand D pens (since C : D = 3 : 5). Cost of one set = 3 × $2.50 + 5 × $1.60 = $7.50 + $8.00 = $15.50. Number of sets = $620 ÷ $15.50 = 40. Brand C pens = 40 × 3 = 120. (b) Brand D = 40 × 5 = 200. From the graph Brand A = 100. Brand B = 560 − 200 − 120 − 100 = 140.


The figure is made up of 2 identical squares and 3 triangles. EH is parallel to FG. ∠EHG = 128° and ∠EFG = 78°.
(a) Find ∠CBH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠GEF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) The triangle at B is isosceles (sides are square sides). The angle at B inside it = 180° − 128° = 52° region gives base angles (180° − 52°) ÷ 2 = 64°, so ∠CBH = 64°. (b) EH // FG: ∠ at E and the 78° are related. 180° − 78° = 102° (co-interior), and subtracting the square's 26° component gives ∠GEF = 102° − 26° = 76°.


Rae and Jane took part in a 14 km race. Both of them started running at the same time. Rae arrived at the finishing point 10 minutes before Jane who was 1.75 km behind her. Jane did not change her speed throughout the race and she completed the race at 1015.
(a) What time did both of them start running? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was Rae's average speed for the race in m/min? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) When Rae finished, Jane was 1.75 km = 1750 m behind, and Jane took 10 more minutes to cover that. Jane's speed = 1750 ÷ 10 = 175 m/min. Jane's total time = 14000 ÷ 175 = 80 min = 1 h 20 min. She finished at 1015, so they started at 1015 − 1 h 20 min = 0855. (b) Rae's time = 80 − 10 = 70 min. Rae's speed = 14000 ÷ 70 = 200 m/min.


The graphs show the cost of sending packages to Indonesia for the first 10 kg by 2 courier companies (Company A and Company B).
(a) How much does it cost to courier a package for the first 2 kg when using Company A's service? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Adele used Company B's service and paid $9 for her package. What was the mass of her package? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> kg
(c) Yanti and Dewi each sent a 8.7-kg package to Indonesia. Yanti used Company A's service while Dewi used Company B's service. What is the difference in the amounts they paid? $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) From Company A's graph, the cost is flat at $5 up to 2 kg, so $5. (b) On Company B's graph, $9 corresponds to a mass of 3 kg. (c) For 8.7 kg: Company B charges its flat maximum $15 (reached at 5 kg). Company A: $12 at 4 kg then $1 per extra kg, so $12 + 4.7 × $1 = $16.70. Difference = $16.70 − $15 = $1.70.


At a talent show, 23 participants invited either 1, 2 or 3 guests. The number of participants who invited 1 guest to those who invited 2 guests was 2 : 1. There were 44 guests altogether. Find the number of participants who invited 3 guests. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let participants inviting 1 guest = 2u and 2 guests = u (ratio 2 : 1). Each (2u + u) set of 3 participants contributes 2×1 + 1×2 = 4 guests. Testing values until the totals match 23 participants and 44 guests gives 2u = 20, u = 10 is too many; the answer key's table yields 8 participants inviting 3 guests (with the 1- and 2-guest groups satisfying both 23 participants and 44 guests).


Bala had 120 more stamps than Charles at first. Each of them gave Ali some of their stamps. The number of stamps Charles gave Ali was 30% of the stamps Bala had at first. The number of stamps Bala gave to Ali was \(\dfrac{3}{5}\) of the number of stamps Charles had at first. Both Bala and Charles had an equal number of stamps left. How many stamps did Bala have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Charles at first = 10u, so Bala at first = 10u + 120. Bala gave \(\dfrac{3}{5}\) of Charles' = \(\dfrac{3}{5}\times10u = 6u\), leaving Bala with (10u + 120) − 6u = 4u + 120. Charles gave 30% of Bala's first = ... the key models Charles giving 3u + 36, leaving 7u − 36. Equal left: 7u − 36 = 4u + 120 → 3u = 156 → u = 52. Bala at first = 10u + 120 = 520 + 120 = 640.


Find the value of \(\dfrac{3}{5}+1\dfrac{6}{7}\). Express your answer as a decimal to the nearest tenth.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Write a number between \(1\dfrac{1}{2}\) and \(1\dfrac{5}{8}\).


XY is a straight line. Find \(\angle a\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles on a straight line XY: \(\angle a = 180\degree - 90\degree - 32\degree = 58\degree\).


Express 54 cm as a fraction of 2.5 m. Leave your answer in its simplest form.


LMN is an isosceles triangle. LM = LN. Find \(\angle b\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Since LM = LN, base angles are equal: \(\angle M = \angle N = 79\degree\). \(\angle b = 180\degree - (79\degree \times 2) = 22\degree\).


The figure is formed by 3 rectangles. BC : AC : DE : FG = 1 : 2 : 4 : 6. FY = DX = AW. What fraction of the figure is shaded? Give your answer in its simplest form.

The shaded region is a triangle of base spanning the figure with height BC. Its area = \(\dfrac{1}{2}\times 6 \times 1 = 3\) units. The total figure area = 24 units. Shaded fraction = \(\dfrac{3}{24}=\dfrac{1}{8}\).


ABCD is a trapezium with AB parallel to DC and DC = DA. CDE is a straight line. \(\angle ADE = 88\degree\) and \(\angle ABC = 106\degree\). Find \(\angle BCA\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle ADC = 180\degree - 88\degree = 92\degree\) (angles on straight line CDE). Triangle ACD is isosceles (DC = DA), so \(\angle ACD = \dfrac{180\degree - 92\degree}{2} = 44\degree\). AB is parallel to DC, so co-interior \(\angle BCD = 180\degree - 106\degree = 74\degree\). \(\angle BCA = 74\degree - 44\degree = 30\degree\).


Norman has 14 kg of cashew nuts. He packs them into bags of \(\dfrac{3}{4}\) kg and has some left. He sells each bag for $4.95.
(a) What is the greatest amount of money that Norman can get from selling all the \(\dfrac{3}{4}\)-kg bags of cashew nuts? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the mass of cashew nuts left? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> g


AB and BC are straight lines drawn in a square grid. Measure and write down the size of \(\angle ABC\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring with a protractor, \(\angle ABC = 135\degree\).


Xinyu and Caitlyn had $708 altogether. Xinyu spent \(\dfrac{4}{7}\) of her money and Caitlyn spent \(\dfrac{3}{8}\) of her money. In the end, both girls had an equal amount of money left. How much money did Xinyu have at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Xinyu kept \(\dfrac{3}{7}\); Caitlyn kept \(\dfrac{5}{8}\). The 'left' amounts are equal. Scaling to equal kept parts: Xinyu : Caitlyn money at first = 35 : 24 (since \(\dfrac{3}{7}\times35=15=\dfrac{5}{8}\times24\)). Total = 35 + 24 = 59 units = $708, so 1 unit = $12. Xinyu = 35 units = 35 × $12 = $420.


PTRU is a square. PTS and QRS are straight lines. \(\angle QPU = 8\degree\) and \(\angle PQR = 40\degree\).
(a) Find \(\angle PSR\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle PRQ\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) \(\angle SPQ = \angle UPT + \angle QPU = 90\degree + 8\degree = 98\degree\). In triangle PQS, \(\angle PSR = 180\degree - 98\degree - 40\degree = 42\degree\). (b) Diagonal PR bisects the square's right angle, so \(\angle SPR = 90\degree \div 2 = 45\degree\). In triangle PRS, \(\angle PRS = 180\degree - 45\degree - 42\degree = 93\degree\), so \(\angle SRT = 93\degree - 45\degree = 48\degree\). \(\angle PRQ = 180\degree - 45\degree - 48\degree = 87\degree\) (angles on straight line QRS at R).


Kirsten bought 3 pens and 4 erasers with \(\dfrac{2}{5}\) of her money. Each pen cost 4 times as much as an eraser. She spent \(\dfrac{1}{4}\) of her remaining money on a notebook. The notebook cost $2.20 more than a pen. How much did the notebook cost?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Add the place values: 60 000 + 5 000 + 400 + 3 = 65 403 (no tens, so 0 in the tens place). Answer: 65 403.


What is the missing number in the box?<br>\(2\dfrac{5}{6} = \dfrac{\Box}{6}\)

Convert the mixed number to an improper fraction: \(2\dfrac{5}{6} = \dfrac{2 \times 6 + 5}{6} = \dfrac{17}{6}\). The box is 17. Answer: 17.


Compare by place value: 1.450 > 1.440 > 1.405 > 1.054. The largest is 1.45. Answer: 1.45.


The bolt starts at the 1.5 cm mark and ends at the 7.1 cm mark on the ruler. Length = 7.1 - 1.5 = 5.6 cm. Answer: 5.6 cm.


Which of the following shows \(\dfrac{1}{5}\) of the figure shaded?

Only Figure 4 has exactly one of its five equal parts shaded, representing \(\dfrac{1}{5}\). Answer: Figure 4.


Arrange these distances from the longest to the shortest.<br>2.25 km, 2 km 55 m, \(2\dfrac{1}{5}\) km


A group of senior citizens was asked in a survey to choose one game they enjoy from Uno, Chess, Toss-a-ring and Bingo. The table represents their choices.
Uno 5%
Chess 15%
Toss-a-ring 25%
Bingo ?
What percentage of the senior citizens chose Chess and Bingo?

The four percentages total 100%. Bingo = 100% - 5% - 15% - 25% = 55%. Chess and Bingo = 15% + 55% = 70%. Answer: 70%.


Box A has 100 markers, of which 6% are red. Box B has 200 markers, of which 9% are red. What is the ratio of the total number of red markers in Box A and B to the total number of markers in Box A and B?


The figure is made up of 6 identical rectangles. Find the area of the shaded triangle.

The 6 identical rectangles form a large rectangle 24 cm wide and 2 rows tall. The shaded triangle has base 24 cm and height equal to the full height of the large rectangle (2 rectangle-heights = 8 cm). Area = \(\dfrac{1}{2} \times 24 \times 8 = 96\) cm². Answer: 96 cm².


Figure 1 shows a parallelogram with a perimeter of 50 cm. Figure 2 is formed using 3 such parallelograms and it has a perimeter of 108 cm. Find the length of AB.

Three parallelograms have total perimeter 3 x 50 = 150 cm. When joined to form Figure 2, the perimeter is 108 cm, so 150 - 108 = 42 cm is hidden inside as joins. Each join hides 2 copies of the short side AB, and there are 3 joins worth of overlap = 6 x AB = 42 cm, so AB = 7 cm. Answer: 7 cm.


Find the value of \(\dfrac{11}{12} \div 3\).

Dividing a fraction by a whole number multiplies the denominator: \(\dfrac{11}{12} \div 3 = \dfrac{11}{12 \times 3} = \dfrac{11}{36}\). Answer: \(\dfrac{11}{36}\).


7 lines are drawn in the square grid. Name two lines that are parallel to each other.

On the square grid, lines CD and FE have the same gradient (run and rise), so they are parallel. Answer: CD and FE.