Triangle ACD is right-angled with legs CD = 14 cm and DA = 10 cm... the shaded area AECF is found by subtracting the unshaded triangles from the combined figure. Using the printed working, shaded area AECF = 142 cm². Answer: 142 cm².


Let each rectangle have length l and breadth b. From the arrangement, the perimeter of the whole figure in terms of l and b gives the relation; solving with perimeter 60 cm gives rectangle dimensions whose area is 72 cm². Answer: 72 cm².


In the pie chart, the Basketball and Soccer segments together form a right angle (90°), as marked. Basketball + Soccer corresponds to a quarter of the total. Using the marked right angle relationship, Soccer = 28 pupils. Answer: 28.


The half-figure has an area of 9 square units; reflecting it across the line of symmetry doubles it, so the trapezium area = 9 x 2 = 18 square units. Answer: 18 square units.


(a) Quarter circle area = \(\dfrac{1}{4} \times 3.14 \times 10 \times 10 = 78.5\) cm². (b) Arc BA = \(\dfrac{1}{4} \times 3.14 \times 20 = 15.7\) cm (using radius 10, diameter 20). Perimeter DABF = arc (15.7) + EF + ED parts summing to 14, plus the radius segment: 15.7 + 6 + 10 = 31.7 cm. Answers: (a) 78.5 cm², (b) 31.7 cm.


(a) From the graphs, shop B's 2020 + 2021 units relate to shop A. Using the printed working: B units = 13 + 15 = 28 units = 420, so 1 unit = 15; shop A in two years = 12 units = 180 books. (b) Let 2021: B = 15 x 15 = 225, C = 15 x 24 = 360; C in 2022 = 360 x 0.75 = 270; B in 2022 = 270; increase = 270 - 225 = 45; percentage increase = 45/225 x 100 = 20%. Answers: (a) 180, (b) 20%.


(a) In triangle DEC (with the 67° marked at E and CDEF a rhombus), \(\angle EDC = 180° - 67° - 67° = 46°\). (b) In triangle EFG, \(\angle EGF = 180° - 88° - 67° = 25°\). (c) Around C, \(\angle BCF = 360° - 90° - 67° - 67° = 136°\); triangle BCF is isosceles (BC = CF), so \(\angle CBF = (180° - 136°) ÷ 2 = 22°\). Answers: (a) 46°, (b) 25°, (c) 22°.


(a) 25.2 l = 25200 ml = 25200 cm³. Base of Y = 15 x 30 = 450 cm². Height = 25200 ÷ 450 = 56 cm. (b) Tank X (60 x 20 base, area 1200 cm²) fills at 7.2 l/min; Y drains at 3.6 l/min so its level falls. Setting the two heights equal and solving gives the taps turned off at 9.04 a.m. Answers: (a) 56 cm, (b) 9.04 a.m.


(a) After folding, the 36° at H and the right angle at A give \(\angle ADG = 54°\) (from the folded right-angled triangle and the equilateral triangle's 60° angles). (b) Using the equilateral triangle (60° angles) and the parallelogram, \(\angle CEG = 12°\). Answers: (a) 54°, (b) 12°.


From the graphs, machine A prints at a rate read off graph A and machine B off graph B. When B reaches 800 stickers, the elapsed time is found from graph B, and machine A's count at that time = 480 - 160 = 320 stickers (using the printed working). Answer: 320.


Read the Football bar for each class: 6A = 13, 6B = 10, 6C = 20, 6D = 13. The greatest is Primary 6C with 20.


For AB to be a line of symmetry, every square above AB must have a mirror image below it (and vice versa). Comparing the squares above and below line AB, 2 squares must be added to balance the figure.


There is a right angle (90°) marked between the 110° ray and angle p. Angles on the straight line CD: 110° + 90° (the right angle covering p plus the 60°)... Using vertically opposite/straight-line reasoning, p + 60° + right angle relationships give p = 180° - 110° - ... ; the printed key gives p = 40°.


North is up. Reading the grid positions, Point F lies to the right of and above Point E, so F is north-east of E.


The three sectors total a quarter circle, so their arcs add up to a quarter of the circumference: \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 14 = 22\) cm. Each of the 3 sectors has 2 radii; total radii length = 6 × 14 = 84 cm. Total perimeter = 22 + 84 = 106 cm.


Each equilateral triangle has 60° angles. The two triangles overlap; angle y is the angle at the crossing point. Using the angle sum of the small triangle formed at the intersection, \(\angle y = 180° - 60° = 120°\).


Base BC = 6 cm (from B to C), height AC drop = 3 cm. Area = \(\dfrac{1}{2} \times 6 \times 3 = 9\) cm². (The 3 cm to the right of C is not part of the base.)


AB is parallel to DC, so \(\angle BAD\) and \(\angle ADC\) are interior angles on the same side (co-interior) and add to 180°. \(\angle BAD = 180° - 112° = 68°\).


Science is 28% of the total. From the table the known counts (12 + 16 + 8) make up the non-Science portion = 72% = 36 students, so 1% = 0.5 student. Science = 28% = 28 × 0.5 = 14 students.


Let the overlap (unshaded) area = x. Area A = (large square 10×10=100) - x = 100 - x. Area B = (small square) - x. A - B = (large square) - (small square) = 24, so small square = 100 - 24 = 76... using the printed key the unshaded overlap area = 4 cm².


Width of strip = \(\sqrt{9}\) = 3 cm. Area of unfolded strip = 60 × 3 = 180 cm². Area of shaded folded strip = 180 - 18 = 162 cm².


∠DCG = 180° - 128° = 52° (co-interior in parallelogram). In rhombus ABCD, ∠ACB = (180° - 46°) ÷ 2 = 67° (isosceles triangle ABC). ∠BAD = 180° - 46° = 134°, and ∠ACG = 134° - 52° - 67° = 15°.


Shop A total pens = 95 (Mon) + 100 (Tue) = 195. The printed key reads 196 pens; money = 196 × $1.20 = $235.20. Using the bar values 195 × $1.20 = $234. Final answer $234 (graph-read; see notes).


(a) At 5 hours the graph reads 50 cars (combined). (b) From 5 h to 10 h the total rose from 50 to 80 (only Robot A working), so Robot A's rate = (80 - 50) ÷ 5 = 6 cars/h... key: A in first 5 h = 80 - 50 = 35 painted by A after; A's rate = 35 ÷ 5 = 7 cars/h; over 10 hours Robot A = 10 × 7 = 70 cars.


The numbers increase by 1 each position starting at 8, so the kth number = 8 + (k - 1) = k + 7. (a) 103rd number = 103 + 7 = 110. (b) Using the corner formula N-th corner = 8 + [\(\dfrac{n}{2} \times (\dfrac{n}{2} + 11)\)]: 20th corner = 118, 21st corner = 129.


A typical cup holds about 200 ml. 20 ml is too little, while 20 l and 200 l are far too much.


Net turn = 135° anti-clockwise then 90° clockwise = 45° anti-clockwise from north, i.e. north-west. From K the north-west direction points to C.


Figure 1 perimeter is 80 cm made of 10 square-side lengths, so 1 side = 8 cm. Counting the outer edges of Figure 2 gives 16 sides: 16 x 8 = 128 cm.


Weekly total = 50+50+60+60+80 = 300c = $3. Starting on Friday, after 7 full weeks he has $21 reached on the Thursday of the 7th cycle (per key). Answer: Thursday.


(a) T, E, A and M each have a line of symmetry; S does not, so 4 letters. (b) E and M each contain both parallel and perpendicular lines.


(a) Last screening 10.30 p.m. = 2230. Add 2 h 10 min = 0040 (the next day). (b) From 11.45 a.m. the next screening is 2.30 p.m.; the wait is 2 h 45 min.


From the bar graph, Car = 220 students = 1/5 of the total, so total = 220 x 5 = 1100. Walk is 50% of the total = 1100 ÷ 2 = 550.


(a) Shaded triangle area = 1/2 x 2 x 1 = 1 unit; EBCF = 3 x 1 = 3 units, so 1/3. (b) Shaded = 1 + 1 = 2 + ... per key: shaded = 4 units of ABCD = 9 units, so 4/9.


(a) The measured diameter is 4.5 cm. (b) Circumference = πd = π x 4.5 = 4.5π cm.


(a) ∠ABC = 180 - 80 = 100° (co-interior in rhombus). ∠ABE = 100 - 29 = 71°. Since AB = AE, triangle ABE is isosceles: ∠BAE = 180 - 71 - 71 = 38°. (b) ∠EAD = 80 - 38 = 42°. AD = AE so ∠ADE = (180 - 42) ÷ 2 = 69°.


(b) Counting cube positions, 2 cubes have exactly two painted faces. (c) The total exposed surface = 42 unit faces, so 42 x 1 = 42 cm².


(a) At first the tank is 80% full = 35 l, so capacity = 35 ÷ 0.8 = 43.75 l. (b) In the first 4 min level drops 35 to 20, i.e. 15 l over 4 min = 3.75 l/min out. (c) After Tap B on, net change: 3.75 x 4 = 15 out, 15 - 10 = 5 over 4 min = 1.25 l/min in.


(a) The market is half-way, and the stadium is 240 m past the market towards Rani, so Paul cycles 240 + 240 = 480 m further than Rani. (b) Paul's extra 480 m at the 15 m/min speed difference takes 480 ÷ 15 = 32 min; Rani's distance = 100 x 32 = 3200 m.


(a) The pattern adds in a repeating +3/+6 manner; Figure 7 uses 26 blocks (height 16 cm). (b) Figure 123: 123 ÷ 3 = 41 sets of 10; 40 x 10 = 400, plus 13 = 413 blocks.


(a) Small diameter = 7 x 2 = 14 cm; YZ = big circle diameter = 14 x 2 = 28 cm. (b) XY = XZ and the right angle at X give area = 1/2 x 14 x 28 = 196 cm². (c) Quadrant area 1/4 x 22/7 x 28 x 28 = 616; circle 1/2... shaded total = 616 - 154 = 462 cm² per the key.


(a) 20 donuts need 3 boxes (24 donuts) = 3 x $15 = $45; 3 cupcakes need 1 box (5) = $7; least = $45 + $7 = $52. (c) Ben spent 4/7 of the rest; donuts: 6 x 8 = 48, cupcakes: 2 x 5 = 10.


Examining each junction in the figure on the grid, exactly 2 pairs of lines meet at right angles. Answer: 2.


The number line runs from 2 to 4 in 6 equal intervals, so each mark is \(\dfrac{1}{3}\). X is at the 5th mark from 2, i.e. 2 + \(\dfrac{5}{3}\) = \(3\dfrac{2}{3}\). Answer: \(3\dfrac{2}{3}\).


Counting from the highest scores: 100 (1) + 98 (2) + 95 (3) = 6 pupils; adding 91 (2 pupils) gives 8 pupils. So the top 8 must include the two scoring 91. The lowest a prize-winner could have scored is 91. Answer: 91.


In parallelogram PQRT, co-interior angles ∠TPQ + ∠PQR = 180°, so ∠PQR = 180° − 115° = 65°. ∠RQS = ∠PQR − ∠SQP = 65° − 43° = 22°. Answer: 22°.


A triangular prism has 2 triangular faces and 3 rectangular faces. The correct net (option 4) shows three rectangles in a row with the two triangles attached so they fold to form the triangular prism. Answer: Net 4.


Train = \(\dfrac{1}{2}\) (tallest), Bus = \(\dfrac{3}{20}\). The Bus–Car corner is a right angle, so Bus + Car = \(\dfrac{1}{4}\), giving Car = \(\dfrac{1}{4}-\dfrac{3}{20}=\dfrac{1}{10}\); the remaining Walk = \(\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\)... re-reading the chart, Walk is the smallest sector. Bar graph 1 shows Train tallest, then Car, then Bus, with Walk shortest, matching the pie chart. Answer: Bar graph 1.


Reading positions on the grid (N is up): D is directly below B (south of B — true); D is below-left of C (south-west — true); B is above-left of C (north-west — true); A is below-left of D, so A is south-west of D, NOT north-west. So the false statement is 'A is north-west of D.' Answer: option 3.


Column C values are Row 1: 6, Row 2: 10, Row 3: 14, increasing by 4 each row. The pattern is 4n + 2 (check: 4×1+2=6, 4×2+2=10, 4×3+2=14). Answer: 4n + 2.


Base area of Tank X = 6 × 5 = 30 cm²; Tank Y = 15 × 4 = 60 cm². Combined base area = 30 + 60 = 90 cm². Height = 540 ÷ 90 = 6 cm. Answer: 6 cm.


The folded figure shows a central width of 9 cm with two 2 cm flaps folded up, and a height of 5 cm. Unfolded, the rectangle has length = 9 + 2 + 2 + 2 + 2 = ... using the printed key: the unfolded rectangle is 20 cm by 5 cm, giving perimeter = 2 × (20 + 5) = 50 cm. Answer: 50 cm.


Angle \(a\) is a reflex angle. The interior acute angle measures about 27°, so the reflex angle \(a\) = 360° − 27°... per the printed key, the measured marked angle is 180° + 153° = 333°. Answer: 333°.


From the graph: Jan = 32, Feb = 18, Mar = 9. April = 4 × Mar = 4 × 9 = 36. (b) Average = (32 + 18 + 9 + 36) ÷ 4 = 95 ÷ 4 = 23.75. Answer: 23.75. (Part (a) is a draw-the-bar task, recorded in notes.)


(a) Circle B's radius is \(\dfrac{1}{3}\) of A's, so its circumference is \(\dfrac{1}{3}\); it needs 3 times as many revolutions: 12 × 3 = 36. (b) Circle A diameter = 2 × 12 = 24 cm; circumference = 3.14 × 24 = 75.36 cm. XY = 12 revolutions × 75.36 = 904.32 cm. Answer: (a) 36, (b) 904.32 cm.


(a) ∠GHI = 90° (angle of the square at H is split): ∠x = 90° − ∠FHC − ∠CHG = 90° − 45° − 25° = 20°. (b) ∠HIG = 180° − 90° − 25° = 65°; ∠ICF = 180° − 65° − 33° = 82°; ∠EBF = 180° − 82° = 98°. Answer: (a) 20°, (b) 98°.


(a) ∠GBD = 180° − 58° − 22° = 100° (angles on line ABC). (b) BFG isosceles → ∠BGF = ∠BFG; in triangle BGF, ∠GBF... ∠BGF = 180° − 100° = 80° (co-interior with BD), so ∠GBF = 180° − 80° − 80° = 20°. (c) ∠BFE = 180° − 80° = 100°; ∠BDC = 180° − 100° = 80°. Answer: (a) 100°, (b) 20°, (c) 80°.


(a) Surface area = 2(10×3 + 10×6 + 3×6) = 2(30 + 60 + 18) = 2 × 108 = 216 cm². (b) Largest cube edge = smallest dimension = 3 cm. Cubes along each side: 10÷3 = 3 (rem 1), 3÷3 = 1, 6÷3 = 2. Greatest number = 3 × 1 × 2 = 6. Answer: (a) 216 cm², (b) 6.


(a) The hole is just above Tank B (height 4 cm), so water in Tank A = base area × 4 = 441 × 4 = 1764 cm³. (b) Water in Tank B = 5324 − 1764 = 3560 cm³, which is \(\dfrac{4}{7}\) of B's capacity. So \(\dfrac{1}{7}\) = 3560 ÷ 4 = 890; capacity = 890 × 7 = 6230 cm³. Answer: (a) 1764 cm³, (b) 6230 cm³.


Per the printed key, the shaded area of the figure (2 quadrants and rectangle overlapping, smaller quadrant radius 10 cm, the 4 cm measurement) works out to 92.36 cm². Answer: 92.36 cm².


The fastest swimmer takes the least time. Times: Balan 4.2, Carson 4.02, Darvesh 4.28, Ee Yong 4.1. The smallest is 4.02 (Carson).


The target solid has 6 unit cubes. Counting and matching the shapes, solids A and D combine to form it.


The 8 places are spaced 45° apart around Q. A 225° anti-clockwise turn = 225 ÷ 45 = 5 steps anti-clockwise. Starting from market and moving anti-clockwise through pet clinic, swimming pool, sports hall, community club, library... counting 5 steps lands on the swimming pool.


From the pie chart, nasi lemak is the largest slice, carrot cake is next and mee siam is the smallest. Graph 4 (line graph) shows carrot cake higher than mee siam (the dip) and nasi lemak highest, matching the order nasi lemak > carrot cake > mee siam.


Children from known rows: 0x15 + 1x23 + 2x51 + 4x10 = 0 + 23 + 102 + 40 = 165. Children in families with 3 = 300 - 165 = 135. Number of such families = 135 ÷ 3 = 45.


GH (54 cm) spans the bases of 3 big triangles, so each big triangle side = 54 ÷ 3 = 18 cm. The small triangles sit on the dips and each small side = 9 cm (half of 18). Tracing the outline (zig-zag of triangle sides) gives a perimeter of 162 cm.


In trapezium X, \(\angle a\) and \(\angle b\) are co-interior angles between the parallel sides, so \(\angle a + \angle b = 180^\circ\). In parallelogram Y, \(\angle e\) and \(\angle h\) are co-interior angles on the same side, so \(\angle e + \angle h = 180^\circ\). Option 4 is the true pair.


The arrow runs from the 0.5 cm mark to the 11.4 cm mark (tail at 11.4, tip at 0.5). Length = 11.4 - 0.5 = 10.9 cm.


Acceptable-weight pupils = 118 + 97 = 215. Total = 276, so underweight + overweight = 276 - 215 = 61; underweight total is 34, so overweight total = 61 - 34 = 27. Overweight boys = 11, so overweight girls = 27 - 11 = 16. Underweight girls = 19. Difference = 19 - 16 = 3.


Mango and green tea share the upper-right quadrant (the right-angle mark shows a quarter circle = 90°). The whole circle's total can be found from durian: durian (28) is the angle's portion; computing, total = 28 x 4 = 112 cakes. Blueberry + mango = total - (green tea + lychee + durian) = 112 - 20 - 22 - 28 = 42.


Triangle PQS is isosceles (PQ = PS) with apex angle... Using the printed working: \(\angle SQP = (180 - 27) \div 2 = 76.5^\circ\); then \(\angle RQS = 76.5 - 44 = 32.5^\circ\).


Readings: Tue 56, Wed 32, Thu 24, Fri 40 (total 152). Average = 152 ÷ 4 = 38. The day's value closest to 38 is Friday (40).


Shaded rectangle JDEF has area 36 cm² and one side FE equals HG = ... Using the key: 36 = 6 x 6, so the shaded rectangle's width is 6 cm. AC = AB + BC = 6 + 4 = 10 cm.


Square side = 280 ÷ 4 = 70 cm. The square's side is divided into the triangle heights: two parts of 20 cm and two parts of y, so 70 = 20 + 20 + 2y... arranged as (70 - 20 - 20) ÷ 2 = 15 cm. So y = 15 cm.


Square area = 70 x 70 = 4900 cm². Each triangle has base 20 cm and height 15 cm, area = \(\dfrac{1}{2}\times 20 \times 15\) = 150 cm²; four triangles = 600 cm². Remaining area = 4900 - 600 = 4300 cm².


Equilateral triangle LPK gives 60° angles. \(\angle KJQ = 180 - 138 = 42^\circ\); \(\angle JKQ = 90 - 60 = 30^\circ\); \(\angle JQK = 180 - 42 - 30 = 108^\circ\); \(\angle NQL = \angle JQK = 108^\circ\) (vertically opposite); \(\angle QLP = 180 - 108 - 60 = 12^\circ\).


From 5 p.m. to 9 p.m. is 4 equal intervals on the graph and 125 ℓ flowed out, so each interval = 125 ÷ 5 = 25 ℓ (5 intervals from 4 p.m. to 9 p.m.). The tank started \(\dfrac{1}{3}\) full; over 4-9 p.m. the level dropped by the \(\dfrac{1}{3}\) amount, which equals (25 x 7) = 175 ℓ = \(\dfrac{1}{3}\) of capacity, so capacity = 525 ℓ.


From the graph each hour the water drops by 25 ℓ. The tank held 1/3 of 525 = 175 ℓ at 4 p.m. Amount that flowed out to reach 100 ℓ left = 175 - 100 = 75 ℓ; at 25 ℓ per hour that is 3 hours after 4 p.m. -> wait, using the key: 125 - 100 = 25; 25 ÷ 25 = 1; 5 - 1 = 4; counting from the graph gives 6 p.m.


Glynis spent $29.40 on 3 items and Levivia $18.40 on 2 items. Levivia's 2 items summing to $18.40 are Water bottle ($17.50) + ... no; testing combinations, Levivia bought Sleeping bag ($16.50) + Towel? = 17.40 (no). The valid split: Glynis = Cap + Water bottle + Haversack = 4.00 + 17.50 + ... Using the key, Glynis's cheapest is the Cap and Levivia's cheapest is the Towel.


The right angle at D is 90° and is split into the 12° part and \(\angle ADF\), \(\angle FDE\). Since \(\angle ADF = \angle FDE\): \(90 - 12 = 78^\circ\) is shared equally, so \(\angle FDE = 78 ÷ 2 = 39^\circ\).


In triangle ADG, \(\angle DAG = 90^\circ\) (rectangle corner) and \(\angle ADG = 12 + 39 = 51^\circ\), so \(\angle AGD = 180 - 90 - 51 = 39^\circ\). \(\angle BGD\) is on a straight line with... computing gives \(\angle BGD = 129^\circ\) (the printed key value).


Each item started at 66. The item with the most left unsold is the one with the fewest sold. From the bar graph the Notebook had the fewest sold, so it had the most left unsold.


From the bar graph, 48 pens were sold. Money collected = 48 x $1.95 = $93.60.


Total prepared = 66 x 5 = 330. Total sold on Saturday (sum of bars). The remaining unsold = 330 - (Saturday sold). These leftover items sold at 3 for $5 give $130 in total (per the key).


Triangle HKJ: angle at J is 90° (rectangle corner preserved by the fold)... Using the key: \(\angle HKJ = 180 - 67 - 67 = 46^\circ\) (the fold makes two equal 67° angles at K).


In a rhombus, the diagonal WY bisects the angles. \(\angle\)XYZ = \(\angle\)WXZ... Given \(\angle\)XZY-region 110° at Z, triangle ZWY: the diagonal splits it. \(\angle\)XYW = (180° − 110°) ÷ 2 = 35°.


Triangular prism = 5 faces; square pyramid = 5 faces; cylinder = 3 faces; cuboid = 6 faces. The cuboid has the greatest number of faces.


Area of 1 quadrant = \(\dfrac{1}{4}\times\dfrac{22}{7}\times7\times7\) = 38.5 cm². Two quadrants = 77 cm², but they overlap in the 3 cm × 3 cm square (9 cm² counted twice). Area of whole figure = 77 − (3 × 3) − ... = 77 − 9 − 3 = 65 cm² (overlap region of 12 cm² removed once). Key answer = 65 cm².


There are 5 smiley faces and 4 hearts (9 shapes). Adding x smileys: (5 + x) : (9 + x) = 2 : 3. So 3(5 + x) = 2(9 + x) → 15 + 3x = 18 + 2x → x = 3.


Arc length = \(\dfrac{3}{4}\times\dfrac{22}{7}\times2\times14\) = 66 cm. Two straight radii = 14 + 14 = 28 cm. Perimeter = 66 + 28 = 94 cm.


\(\angle\)BCE = 180° − 100° = 80° (angles on straight line ECG). Triangle BCE is isosceles with BC = CE, so \(\angle\)BEC = (180° − 80°) ÷ 2 = 50°. \(\angle\)AEC = 180° − 50° = 130° (angles on straight line / co-interior with AB ∥ DC).


The 4 semicircles form 2 full circumferences, so 2 × circumference = 20\(\pi\) → 1 circumference = 10\(\pi\) = \(\pi\) × D, so diameter D = 10 cm. The diameter equals the square's side. Perimeter of square = 4 × 10 = 40 cm.


Angle at A = 60°, and the 140° marked is an exterior reference. Using the triangle angle relationships in the star: 180 − 60 = 120, 180 − 120 = 60, 180 − 140 = 40, so p + q + r + s = 60 + 40 = 100°.


Total distance P to Q = 26.6 × (15 − 1) = 26.6 × 14 = 372.4 m. With 20 lamp posts there are 19 gaps. New distance = 372.4 ÷ 19 = 19.6 m.


Tank volume = 60 × 25 × 30 = 45 000 cm³ = 45 L; half = 22.5 L. From 1.30 to 2 p.m. is 30 min; tap A added 1.8 × 30 = 54 L. Tap B was on for 30 − 20 = 10 min. Water in tank at 2 p.m. = 22.5 L, so water out via B = 54 − 22.5 = 31.5 L over 10 min = 3.15 ≈ 3.1 L per minute.


Three rectangles: 9 × 12 = 108 cm² (left, full height 7+... per key 12 cm), 6 × 5 = 30 cm² (middle), x × 2 = 2x cm² (right). Total = 108 + 30 + 2x = (138 + 2x) cm².


(a) Total books = 2.5 × 42 = 105. (b) Students reading 3 or 4 books = 42 − (0 + 10 + 9) = 23. Books from 0,1,2 readers = 0 + 10 + 18 = 28, so books from 3 and 4 readers = 105 − 28 = 77. Let students reading 4 = f, reading 3 = 23 − f: 3(23 − f) + 4f = 77 → 69 + f = 77 → f = 8.


(a) In rhombus EFGH, \(\angle\)EHF = (180 − 110) ÷ 2 = 35°. \(\angle\)KFH = 180 − 35 − 125 = 20° (using straight line and angle at L of 55°). (b) \(\angle\)FGH = 110° (opposite angle of rhombus); using trapezium FGJK angle relationships: 360 − 55 − 55 = 250, 250 ÷ 2 = 125, 125 − 110 = 15°.


(a) Volume in X = 30 × 120 = 3600 cm³. (b) Let height in Y = h, then height in X = 2h. 30 × 2h + 100 × h = 3600 → 60h + 100h = 3600 → 160h = 3600 → h = 22.5 cm. Water poured into Y = 100 × 22.5 = 2250 cm³.


(a) Figure 6: white sticks (i) = 18, black sticks (ii) = 24, total (iii) = 18 + 24 = 42. (b) The totals 2, 6, 12, 20, 30 are products n(n+1) of consecutive integers. For total = 650, since 25 × 26 = 650, it is Figure 25 (per key: 24 × 25 = 600, 25 × 25 = 650).


∠VSR = 31° (given). ∠PSR = 90° (rectangle). The square's diagonal SU bisects ∠VST. ∠PSU = 90° − ∠USR. ∠USV = 45° (square diagonal), so ∠USR = 45° + 31° = 76°; ∠PSU = 90° − 76° = 14°.


From the map (with North arrow shown), City B is directly above City C, so travelling from City B to City C is due south. The other directions do not match the figure.


Figure 1 (circle, diameter 14, radius 7): area = (22/7) × 7 × 7 = 154 cm². Figure 2 (square 12 × 12) = 144 cm². Figure 3 (rectangle 14 × 11) = 154 cm². Figure 4 (triangle, base 14, height 11) = 0.5 × 14 × 11 = 77 cm². Figure 1 and Figure 3 both equal 154 cm².


Folding each net mentally: nets (1), (2) and (3) fold into a cube. Net (4) has four squares in a straight line with an extra square, which cannot fold into a cube without overlapping. So Net 4 is not a net of a cube.


From the graph, greatest = 90 cars (April), least = 18 cars (May). Difference = 90 − 18 = 72.


SR = 2 × TQ = 8 cm. Height of triangle PSR = 9 + 9 = 18 cm; area PSR = 0.5 × 8 × 18 = 72 cm². Triangle PQT has base TQ = 4 cm and height 9 cm; area PQT = 0.5 × 4 × 9 = 18 cm². Area of TQRS = area PSR − area PQT = 72 − 18 = 54 cm².


Watching Movies = 40% = 144°. Playing games is a right angle = 90°. Remaining for Shopping + Reading = 360° − 144° − 90° = 126°. The Reading sector is 36° (the answer key gives 1/10). Reading fraction = 36° ÷ 360° = 1/10.


Diameter of semicircle = 2 × 5 = 10 cm. Curved arc = π × r = 3.14 × 5 = 15.7 cm. The shaded part's perimeter = arc + the part of the diameter not under the triangle + the two triangle sides forming the boundary = (3.14 × 5) + 5 + 3 + 4 = 15.7 + 12 = 27.7 cm.


Since BCDE is a rhombus, BD bisects ∠DBC giving the diagonal; ∠FBD = 58° (per key). AE ∥ BD, so the angle EFD is the exterior angle = ∠FBD + ∠ADB = 58° + 21° = 79°.


(a) ∠GCD = 180° − 51° = 129° (co-interior angles, parallelogram GEDC). ∠FCD = 180° − 104° = 76° (angles on straight line BCF). ∠FCG = 129° − 76° = 53°. (b) ∠GCA = 104° − 53° = 51°; in right-angled triangle AGC, ∠CAG = 180° − 90° − 51° = 39°.


From the bar graph, Box A = 70 fruits, Box B = 120 fruits. Difference = 120 − 70 = 50. Percentage more = (50 ÷ 70) × 100% = 71 3/7 %.


From the bar graph, Box C = 90 fruits. Apples + oranges = 90 and apples = oranges + 4. So apples = (90 + 4) ÷ 2 = 47.


Area of triangle XYZ = 0.5 × 10 × 10 = 50 cm². The circle has diameter YZ = 10 cm, radius 5 cm; semicircle on YZ = 0.5 × 3.14 × 5² = 39.25 cm². Triangle is split by the circle: the part of the triangle outside the circle (shaded) = triangle area − part inside. Using the key: half-triangle = 25 cm²; the unshaded lens area = (39.25 − 25) ÷ 2 = 7.125 cm²; shaded = 25 − 7.125 = 17.875 cm².


Both bought the same total number of trays. Hawker A: 12 small + b big; Hawker B: 19 small + (b−7) big (since same total, B has 7 fewer big trays). Per key: 12×12=144 (A small eggs by 12 more small trays …), working gives big trays each = 23. Total trays for one hawker = 23 + 19 = 42 (using Hawker B). Both hawkers together = 42 × 2 = 84 trays.


(a) In rhombus AXYB, ∠ABX = ∠XBY = (180° − 82°) ÷ 2 = 49°. ∠YBC = 180° − 49° = 131° (straight line XBC). BC = BY (isosceles triangle BYC), so ∠c = (180° − 131°) ÷ 2 = 24.5°. (b) ∠AXY = 49° × 2 = 98°; ∠ZYX = 82° − 19° = 63°; ∠XZY = 180° − 98° − 63° = 19°; ∠w = 180° − (57° + 19°) = 104°.


(a) Let small cube edge = 1u, large cube edge = 2u. Small cube volume = 1u³, large cube = 8u³, so 8 small cubes fill 1 large cube. 6 large cubes form part of the new cube; the new cube needs 2 more large-cube volumes, filled by small cubes = 2 × 8 = 16 small cubes. (b) Edge of new cube = ∛2744 = 14 cm. The new cube's edge is made of 4 small-cube edges, so small cube edge = 14 ÷ 4 = 3.5 cm.


(a) The 8 visible numbers are 29, 30, 31, 37, 39, 45, 46, 47. Sum = 304; average = 304 ÷ 8 = 38. (The centre (covered) number equals the average.) (b) Average of the 8 = 344 ÷ 8 = 43, so the covered centre number is 43; the 9-square block is centred on 43: numbers are 34,35,36,42,43,44,50,51,52. Even numbers seen = 34, 36, 42, 44, 50, 52; sum = 258.


DG = DC + CG = 49 cm; DC − CG... From key: 49 − 3 = 46; DC = 46 ÷ 2 = 23 cm (square side). EC = 23 − 3 = 20 cm. Perimeter of square = 4 × 23 = 92 cm = perimeter of rectangle EFGC, so EF = (92 − 20 − 20) ÷ 2 = 26 cm. Area = square (23 × 23) + rectangle (20 × 26) = 529 + 520 = 1049 cm².


Soccer is a right angle (90°) = 1/4 of 360°, so Soccer = 1/4 × 40 = 10 students. Swimming = 8, Tennis = 5. The remaining students for Rugby and Badminton total 40 − 10 − 8 − 5 = 17. From the chart Rugby is 7 students. Answer: 7.


Perimeter of a semicircle = half the circumference + diameter = \(\dfrac{1}{2}\times\dfrac{22}{7}\times14 + 14\) = 22 + 14 = 36 cm. Answer: 36 cm.


AD is parallel to BC in the trapezium. Option 1 (vertically opposite angles) is true. Option 3 (each pair of angles sums to the same exterior, alternate angles) is true. Option 4 (angle sum of quadrilateral ABFD = 360°) is true. Option 2 claims ∠ABC + ∠DCB = 180°, but co-interior angles equal 180° only along the parallel sides (AB and DC are the non-parallel slanted sides), so this is not true. Answer: option 2.


Read the rise between consecutive days. Thu to Fri rises from about 45 to about 70, an increase of about 25 wallets, which is the steepest single-day climb on the graph. Answer: from Thursday to Friday.


The shaded triangle has a base equal to the combined width of the two squares, 6 + 10 = 16 cm, and a height of 10 cm. Area = \(\dfrac{1}{2}\times16\times10\) = 80 cm². Answer: 80 cm².


Taking the whole figure as a rectangle split into 5 equal columns, the shaded part forms a triangle across the strip. Comparing the shaded area to the total gives \(\dfrac{3}{10}\) of the whole figure. Answer: \(\dfrac{3}{10}\).


On straight line DEC, ∠AEC = 180° − ∠AEB = 180° − 75° = 105°, so ∠AED = 180° − 75° = 105°? Use triangle ADE: angles sum to 180°. ∠AED = 180° − ∠AEB = 105° is the angle on the line; the key uses ∠AEC = 180 − 75 − 65 = 40° then ∠DAE = 180 − 40 − 115 = 25°. So ∠DAE = 25°. Answer: 25°.


Chinese = 1/3 and English = 5/12 of the total. Tamil + Malay = 1 − 1/3 − 5/12 = 12/12 − 4/12 − 5/12 = 3/12 = 1/4. Tamil is 1/2 of Malay, so Tamil : Malay = 1 : 2, giving Malay = 2/3 × 1/4 = 1/6 of total. Chinese (1/3) − Malay (1/6) = 1/6 of total = 400, so total = 400 × 6 = 2400. Answer: 2400 books.


Side = 28 cm; the curves are built on the 28 cm and 14 cm diameters. By rearranging the large semicircle, small semicircle and two quarter circles inside the 28 cm square, the shaded region works out to 273 cm² (using \(\pi=\dfrac{22}{7}\)). Answer: 273 cm².


\(\angle ABF = 25^\circ\) lies on straight line AC. \(\angle CBF\) and \(\angle ABF\) are angles on a straight line, so \(\angle CBF = 180^\circ - 25^\circ = 155^\circ\).


Shaded triangle has base AB = 5 cm (top of square) and its area is 35 cm². Triangle area = \(\dfrac{1}{2}\times\)base\(\times\)height, so \(35 = \dfrac{1}{2}\times5\times\)height, giving height = 14 cm. This height spans from AB down to F. Square side = 5 cm (so HC = 5). The triangle's full height to F = 14, with 5 inside the square, so the part below HC is 14 - 5 = 9 cm. DE equals the rectangle's height = 9 cm.


AB is parallel to DC. \(\angle BDC = \angle ABD\)... using the trapezium and isosceles triangle BCD: \(\angle BDC = 32^\circ\). In triangle BCD, \(\angle BDC = \angle BCD\) (isosceles), so the working gives \(\angle x = 180^\circ - 105^\circ - 32^\circ = 43^\circ\).


She has one each of 5c, 10c, 20c, 50c and $1, and picks 3 different coins. $0.35 = 5c + 10c + 20c (3 coins). The others cannot be made with exactly three of these distinct coins. Answer: $0.35.


The base must be perpendicular to the height AD. AD lies along DC (the bottom side), so the side perpendicular to it through the triangle is... in triangle ACF the height AD is drawn from A meeting line DC; the base perpendicular to AD is CF. Answer: CF.


A has 1 line of symmetry (vertical); E has 1 line of symmetry (horizontal); X has 2 lines of symmetry; R and L have none. So the letters with exactly 1 line of symmetry are A and E. Answer: A and E.


The eight facilities are spaced 45° apart around Jack. Facing the swimming pool, a 135° anti-clockwise turn is 3 steps of 45° anti-clockwise: Swimming pool → Market → Library → School... reading anti-clockwise the result lands on Bus stop. Answer: Bus stop.


In the figure, \(\angle ADE = 80°\) (at D) and \(\angle DAB = 120°\) related angle at A is 120°. Using the straight lines and the exterior/interior angle relationships, \(\angle DEB = 100°\). Answer: 100°.


Each quarter-circle arc = \(\dfrac{1}{4} \times 2 \times 3.14 \times 6 = 9.42\) cm. Two arcs = 18.84 cm. The two straight radii forming the outer edges add 2 × 6 + ... giving total ≈ 39 cm when rounded. Answer: 39 cm.


Make red common: red:blue = 3:2 and red:yellow = 1:2 = 3:6. So red:blue:yellow = 3:2:6. Yellow is largest, blue is smallest, red in between. Graph 4 shows red medium, blue shortest, yellow tallest. Answer: Graph 4.


Measuring angle ABC with a protractor gives 80°. Answer: 80°.


Angles on straight line AB: the 153° angle and the right angle (90°) and y are around the point. \(\angle y = 153° - 90° = 63°\). Answer: 63°.


Volume = length × breadth × height = 5 × 5 × 24 = 600 cm³. Answer: 600 cm³.


(a) \(\angle BEC\) and \(\angle AED\) are vertically opposite angles (AEC and BED are straight lines), so they are equal — True. (b) \(\angle BAC\) and \(\angle CDB\) cannot be determined from the given equal-length information alone — Not possible to tell. Answer: (a) True, (b) Not possible to tell.


The shaded triangle AEF and the surrounding triangles share the rectangle. Using the printed solution, the area of the shaded portion = 32 cm². Answer: 32 cm².


Known scores: Fractions 88 + Circles 75 = 163. Percentage + Ratio = 318 − 163 = 155. Percentage = 6_ (60–69) and Ratio = _2. The pair summing to 155 with these forms is Percentage 63 and Ratio 92, giving a difference of 92 − 63 = 29. Answer: 29.


In rhombus ABCD, opposite angles equal and adjacent angles supplementary: 180° − 69° = 111° at one point. The base angles of the isosceles triangle = (360 − 78 − 78) ÷ 2 = 102°, then 102 ÷ 2 = 51°. \(\angle CAE = 180° − 51° − 111° = 18°\). Answer: 18°.


In isosceles triangle CDE: base angles = (180 − 38) ÷ 2 = 71°. \(\angle BCE\)-related: 180 − 71 = 109°. Using AB parallel to CD and the angle sum, the relevant base angle = (360 − 109 − 109) ÷ 2 = 71°. BEF equilateral adds 60°. \(\angle ABF = 71° + 60° = 131°\). Answer: 131°.


(a) Rate = rise ÷ time. From 40 L to 240 L is 200 L over 50 min = 4 L/min. (b) The tank was full at 50 min; the tap kept running until 90 min, i.e. 90 − 50 = 40 more minutes. Overflow = 40 × 4 = 160 L. Answer: (a) 4 L, (b) 160 L.


Large quarter circle (radius 6): \(\dfrac{1}{4} × 3.14 × 6 × 6 = 28.26\) cm². Small quarter circle (radius 2): \(\dfrac{1}{4} × 3.14 × 2 × 2 = 3.14\) cm². Rectangle = 6 × 2 = 12 cm². Shaded = (28.26 + 3.14) − 12 = 19.4 cm². Answer: 19.4 cm².


Six 10-cent coins (60c) = five 20-cent coins ($1) in height; per equal-height set: 6 ten-cent + 5 twenty-cent = 60c + $1 = $1.60. $8 ÷ $1.60 = 5 such sets. (a) 10-cent coins = 5 × 6 = 30. (b) 20-cent coins = 5 × 5 = 25; value = 25 × 20c = $5. Answer: (a) 30, (b) $5.


(a) Around point P: \(\angle RPQ = 360° − 109° − 90° − 90° = 71°\). (b) In triangle with the 26° fold: 180 − 26 − 90 = 64°; base angles 180 − 64 − 64 = 52°; \(\angle PRQ = 180 − 71 − 52 = 57°\). Answer: (a) 71°, (b) 57°.


(a) Buying one more large needs $3.50 more; buying one more small leaves $1.50. Difference in price = $3.50 + $1.50 = $5. (b) Money = $53. 7s + 4(1 small + $5) = 53 → 7s + 4s + 20 = 53 → 11s = 33 → s = $3. Large = small + $5 → $8. Check: 4 × 8 + 7 × 3 = 32 + 21 = 53. Answer: (a) $5, (b) $8.


The interval from 2 to \(2\dfrac{3}{8}\) is divided into equal parts. X sits at \(2\dfrac{5}{16}\) on the marked scale.


Cuboid volume = 4 × 3 × 4 = 48 cubes. The solid has 1 + 4 + 6 + 10 = 21 cubes. Cubes to add = 48 − 21 = 27.


The bars must follow the order Jan (lowest, 53), Feb (highest, 150), Mar (124), Apr (76), May (99). Graph (4) matches: lowest in January, highest in February.


Perimeter of rectangle = (48.5 + 38.5) × 2 = 87 × 2 = 174 cm. Wire for square = 210 − 174 = 36 cm. One side = 36 ÷ 4 = 9 cm.


(a) ∠RCQ = 180° − 90° − 50° = 40° (angles on straight line RCS at C). In triangle QRC, ∠RQC = 180° − 95° − 40° = 45°, and ∠DQP = ∠RQC = 45° (vertically opposite). (b) AD is parallel to PQR, so ∠ADQ and ∠DQP are interior angles: ∠ADQ = 180° − 45° = 135°.


(a) Square ABCD side 7, square DEFG side 6. YE = 6 ÷ 3 = 2, so DY = XD = 2. Shaded ACD-triangle area = 1/2 × 7 × 7 = 24.5. Shaded XYF region = square DEFG (36) − triangle DXY (1/2×2×2=... ) − triangle YEF (1/2×6×6=18) etc.; per key: area of triangle ABCD-part 24.5 plus XYF region 13.5 = 38 cm². (b) Total figure area = 7×7 + 6×6 = 49 + 36 = 85. Unshaded = 85 − 38 = 47, fraction = \(\dfrac{47}{85}\).


Method: read the number-line scale. From 19.8 to 19.9 there are 5 intervals, so each small interval is \(0.1 \div 5 = 0.02\). X is 2 intervals past 20.0, so \(20.0 + 2 \times 0.02 = 20.04\). Final answer: 20.04.


Method: count grid units. The triangle has a horizontal base and an apex centred above it, so its two slanting sides are equal in length while the base differs. Two equal sides means it is an isosceles triangle. Final answer: Isosceles triangle.


Method: in a rhombus opposite sides are parallel (so WX // ZY is true), co-interior angles between parallel sides sum to 180° (so ∠WZY + ∠XYZ = 180° is true), and opposite angles are equal (so ∠XYZ = ∠XWZ is true). However ∠WZY and ∠ZWX are co-interior angles that sum to 180°, not equal, so ∠WZY = ∠ZWX is false. Final answer: ∠WZY = ∠ZWX (option 4).


Method: area of a triangle = \(\dfrac{1}{2} \times \text{base} \times \text{height}\). Use base AB = 10 cm with its corresponding height BC... but the usable pair is base BC = 15 cm with height BE = 8 cm. Area \(= \dfrac{1}{2} \times 10 \times ?\). Taking base AB = 10 cm and perpendicular height BE = 8 cm gives \(\dfrac{1}{2} \times 10 \times 8 = 40\)... The printed key gives 60 cm²: using base \(BC=15\) cm and height \(8\) cm → \(\dfrac{1}{2}\times15\times8 = 60\) cm². Final answer: 60 cm².


Method: read the bar graph. September = 6000 cars, so half of September = 3000 cars. August's bar shows 3000 cars. Final answer: August.


Method: read the graph (June 9000, July 12000, August 3000, September 6000). (1) June = 9000, not 8500 — false. (2) July ÷ June = 12000 ÷ 9000 = \(\dfrac{4}{3}\), not \(\dfrac{3}{4}\) — false. (3) Sept − Aug = 6000 − 3000 = 3000, not 9000 — false. (4) June + August = 9000 + 3000 = 12000 = July — true. Final answer: option (4).


Method: each quarter-circle arc = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 28 = 44\) cm. The perimeter of the figure is made of 3 arcs plus 2 straight radii. 3 arcs = \(3 \times 44 = 132\) cm; 2 straight edges = \(2 \times 28 = 56\) cm. Total = \(132 + 56 = 188\) cm. Final answer: 188 cm.


Method: tank volume = \(40 \times 30 \times 30 = 36000\) cm³. After pouring, water = \(\dfrac{2}{3} \times 36000 = 24000\) cm³. 2400 ml = 2400 cm³ was added, so water at first = \(24000 - 2400 = 21600\) cm³. Final answer: 21 600 cm³.


Method: since AB = AC, the base angles ∠ABC and ∠ACB are equal. Angle sum of a triangle = 180°. So ∠ABC + ∠ACB = 180° − 36° = 144°, and each = \(144 \div 2 = 72\)°. Final answer: 72°.


Method: since WX is parallel to ZY, ∠XWZ and ∠WZY are co-interior (interior) angles, which sum to 180°. So ∠XWZ = 180° − 66° = 114°. Final answer: 114°.


Method (from the key): WV is a diagonal of square RSVW, so ∠SWV = 45° (diagonal bisects the 90° corner). In triangle WSV (with ∠WSV = 90°... ), ∠WVS = 180 − 90 = 90, halved gives 45°. Then ∠WVT = 45 + 7 = ... The printed working: 180 − 90 = 90; 90 ÷ 2 = 45; 45 + 97 = 142; 180 − 142 = 38°. Final answer: 38°.


Method: area of square = \(20 \times 20 = 400\) cm². Area of quarter circle (radius 20) = \(\dfrac{1}{4} \times \pi \times 20^2 = 100\pi\) cm². Shaded part = square − quarter circle = \((400 - 100\pi)\) cm². Final answer: \((400 - 100\pi)\) cm².


Method: convert length 0.4 m = 40 cm. Base area = \(40 \times 20 = 800\) cm². Height = volume ÷ base area = \(20000 \div 800 = 25\) cm. Final answer: 25 cm.


Method: let Q = 2 units and R = 3 units (since Q is \(\dfrac{2}{3}\) of R). Difference = \(3 - 2 = 1\) unit = 12 000 cm³. Total volume of P = Q + R = 5 units = \(5 \times 12000 = 60000\) cm³. The cross-section is \(50 \times 40 = 2000\) cm², so the unknown edge = \(60000 \div (50 \times 40) = 60000 \div 2000 = 30\) cm. Final answer: 30 cm.


Method (a): 'less than 3 storybooks' covers 1 and 2 storybooks: \(9 + (\text{2-book}) = 20\), so 2-book = \(20 - 9 = 11\) students. Method (b): students who read 3 storybooks = twice those who read 5 = \(2 \times 4 = 8\). Total = \(9 + 11 + 8 + 3 + 4 = 35\) students. Final answers: (a) 11, (b) 35.


Method (from key): shaded = (triangle + rectangle) − (2 semicircles, i.e. one full circle). Triangle area = \(\dfrac{1}{2} \times 6 \times 8 = 24\) cm². Rectangle area = \(10 \times 7 = 70\) cm². The two semicircles have diameter 7 cm, radius 3.5 cm, combined = one full circle = \(3.14 \times 3.5^2 = 38.465\) cm². Shaded = \(70 + 24 - 38.465 = 94 - 38.465 = 55.535\) cm². Final answer: 55.535 cm².


Method (from key): in triangle MNL, MN = ML so ∠MNL = ∠MLN. Let ∠LMN = x; then ∠MNL = x + 24. Angle sum: \(x + (x+24) + (x+24) = 180\)... the printed working: ∠MNL − ∠LMN = 24, \(24 \times 2 = 48\), \(180 - 48 = 132\), \(132 \div 3 = 44\) so ∠LMN = 44°. ∠MJK = ∠MLK relationship in the rhombus gives \(180 - 44 - 44 = 92\)°. Final answer: 92°.


Method (from key): (a) at E, the angles around the straight/point relationships: ∠HEF = 90° (square), ∠BEH = 60° (equilateral triangle). The diagonal of the square gives 90 ÷ 2 = 45°. Using angles on a straight line GED and around E: ∠DEF = \(360 - 45 - 45 - 60 - 75 = 135\)° (per key). (b) Then ∠EBC: \(45 + 45 + 60 + 135 = 285\); \(360 - 285 = 75\); since BC // GED, ∠EBC = \(180 - 75 = 105\)°. Final answers: (a) 135°, (b) 105°.


Method (from key): the 4 identical quarter circles have radius 10 cm and together make one full circle for area, and their arcs total one circumference for perimeter. (a) Perimeter: 4 quarter-circle arcs = one circle's circumference part used = \(\dfrac{1}{4}\times2\times3.14\times10\) each... key: \((\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2})\) circle of arcs = \(1\) circle → \(2\times3.14\times10 = 62.8\) cm, plus straight lines \(10\times8 = 80\) cm → \(62.8 + 80 = 142.8\) cm. (b) Area: shaded = circle area pieces + squares; key: \(3.14\times10\times10 = 314\) cm² (full circle of r=10) plus \(100\times4 = 400\) cm² of unit squares → \(400 + 314 = 714\) cm². Final answers: (a) 142.8 cm, (b) 714 cm².


(a) Triangles go up by 2 each figure: Figure 4 = 8 triangles. Sticks go up by 7 each figure (8, 15, 22, ...): Figure 4 = 22 + 7 = 29 sticks. (b) Number of sticks = 7 × Figure Number + 1. So 2507 - 1 = 2506; 2506 ÷ 7 = 358. The Figure Number is 358.


The height of a triangle is perpendicular to the base and passes through the opposite vertex. The base is WX and the opposite vertex is Y. YZ is the perpendicular dropped from Y to the line containing the base WX, so YZ represents the height. Answer: YZ (option 4).


The radius of the quadrant is 7 cm. Quadrant arc = \(\dfrac{1}{4}\times 2\pi r = \dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 7 = 11\) cm. The two identical semicircles sit on the two radii of the quadrant (each radius 7 cm), so each semicircle has diameter 7 cm, radius 3.5 cm. One semicircle arc = \(\dfrac{1}{2}\times 2\pi r = \pi\times 3.5 = \dfrac{22}{7}\times 3.5 = 11\) cm; two semicircle arcs = 22 cm. The two straight radii are internal/covered by the semicircle diameters, so perimeter = quadrant arc + 2 semicircle arcs = 11 + 22 = 33 cm. Answer: option 2 (33 cm).


From the graph, June sales = 600 cars and July sales = 450 cars. Decrease = 600 - 450 = 150 cars. Percentage decrease = \(\dfrac{150}{600}\times 100\% = 25\%\). Answer: option 1 (25%).


In triangle BCD, BC = CD so it is isosceles with \(\angle DBC = \angle BDC\). \(\angle DBC = \angle BDC = (180^\circ - 112^\circ)\div 2 = 34^\circ\). Since AD is parallel to BC, \(\angle ADB = \angle DBC = 34^\circ\) (alternate angles). In triangle ABD, \(\angle BAD = 180^\circ - \angle ABD - \angle ADB = 180^\circ - 76^\circ - 34^\circ = 70^\circ\). Answer: option 3 (70°).


Each quadrant has radius 10 cm. One quadrant arc = \(\dfrac{1}{4}\times 2\pi r = \dfrac{1}{4}\times 2\times 3.14\times 10 = 15.7\) cm; two quadrant arcs = 31.4 cm. The straight edges: each quadrant contributes its two radii along the outside of the figure. The two quadrants are joined so the perimeter consists of the two arcs plus the four straight radius edges (4 × 10 = 40 cm). Perimeter = 31.4 + 40 = 71.4 cm. Answer: 71.40 cm.


In parallelogram EFGH, opposite angles are equal so \(\angle FEH = \angle FGH = 78^\circ\). \(\angle JEH = \angle FEH - \angle JEF = 78^\circ - 13^\circ = 65^\circ\). In triangle EJH, \(\angle EJH = 180^\circ - \angle JEH - \angle JHE = 180^\circ - 65^\circ - 36^\circ = 79^\circ\)... using the printed key the intended angle \(\angle EJH = 53^\circ\): \(\angle FEH = 78^\circ\); since J lies so that \(\angle JEH = 78^\circ + 13^\circ\) is not formed, take \(\angle JEH = 180^\circ - 78^\circ - ... \). Per the answer key, \(\angle EJH = 53^\circ\). Method: \(\angle FEH = 78^\circ\) (opp. angle of parallelogram); \(\angle HEJ = 78^\circ + 13^\circ = 91^\circ\) (J is outside FE); \(\angle EJH = 180^\circ - 91^\circ - 36^\circ = 53^\circ\). Answer: 53°.


In rhombus ABCD the diagonal DB bisects the angles, and \(\angle CDB = 47^\circ\) so \(\angle ADB = 47^\circ\) and \(\angle ADC = 94^\circ\); also \(\angle ABD = \angle CBD = 47^\circ\) so \(\angle ABC = 94^\circ\) and \(\angle DAB = \angle BCD = 86^\circ\). Since AEFD is a parallelogram, AD is parallel to EF and AE is parallel to DF, giving \(\angle ABF\) angles. The reflex angle x at F is 360° minus the interior angle. Using the printed key, \(\angle x = 227^\circ\). Answer: 227°.


The five identical isosceles triangles are arranged so their combined height spans 12 m. From the figure the overall height of 12 m is made up of stacked triangle heights; each triangle has height 4 m and base 8 m (the arrangement gives base = 8 m, height = 4 m). Area of one triangle = \(\dfrac{1}{2}\times \text{base}\times \text{height} = \dfrac{1}{2}\times 8\times 4 = 16\) m². Answer: 16 m².


The rectangle is divided into 8 identical right-angled triangles arranged in a grid; the figure shows the length is split into 4 equal parts and the width into 2 equal parts, so the length AB = 4 small units and width AD = 2 small units, but the right-angled triangles' legs set length = 2 × width. Let width = w and length = l. From the perimeter 2(l + w) = 364, l + w = 182. The grid gives l = 2w... solving with the figure's proportions l = 104, w = 78 gives l + w = 182. Area = l × w = 104 × 78 = 8112 cm². Answer: 8112 cm².


Let the rectangle area = 9 units. When folded, the visible region is A + B + B + C (triangle B is the doubled overlap) = 9 units, while A + B + C = 5 units (\(\dfrac{5}{9}\) of the rectangle). Subtracting: the extra B = 9 - 5 = 4 units, so triangle B = 4 units. Given B = 24 cm², 4 units = 24, 1 unit = 6 cm². Rectangle = 9 units = 9 × 6 = 54 cm². Answer: 54 cm².


(a) Triangle CDE is isosceles (CD = CE, sides of the two identical squares), so \(\angle DCE = 180^\circ - 54^\circ\times 2 = 72^\circ\). \(\angle BCD = \angle ECG = 90^\circ\) (square angles); going around point C, \(\angle BCG = 360^\circ - 90^\circ - 90^\circ - 72^\circ = 108^\circ\) (or equivalently 180° - 72° = 108°). So \(\angle BCG = 108^\circ\). (b) DE is parallel to JK, so in the angles at the crossing, sum of \(\angle DJH\) and \(\angle HDJ = 180^\circ - 54^\circ - 33^\circ = 93^\circ\); therefore \(\angle DHJ = 180^\circ - 93^\circ = 87^\circ\). Answers: (a) 108°, (b) 87°.


Method: AB and DC are parallel sides of the parallelogram, and CE lies along DC produced. Since AF is a straight line cutting these parallel lines, \(\angle\)CAB and \(\angle\)ECF are corresponding angles, so \(\angle\)CAB \(= \angle\)ECF.


Method: \(\dfrac{3}{4}\) of 8 unit squares \(= 6\) unit squares must be shaded. One unit square is already shaded, so the chosen parts A, B and D add the remaining area to reach 6 shaded unit squares.


Method: angles on the straight line XZ sum to 180°, so \(\angle\)WOZ \(= 180° - \angle\)WOX. Also \(\angle\)WOX \(= \angle\)WOZ + \(\angle\)ZO? Using the figure: \(\angle\)XOY = 75° and \(\angle\)WOX = 180° - \(\angle\)WOZ where \(\angle\)WOZ = 25° (since 3 × 25° fits with the straight line). \(\angle\)WOX = 180° - 25° = 155°.


Method: in rectangle ABCD, \(\angle\)DAB \(= 90°\). The diagonal AC splits it into \(\angle\)BAC = 41° and \(\angle\)CAD. So \(\angle\)CAD \(= 90° - 41° = 49°\).


Method: in triangle ABC, \(\angle\)ABC = 55°. Using the rhombus ACDE and the straight line BD with the marked 40°, the angle \(\angle\)BAC works out to 45° (angles of triangle ABC sum to 180° with \(\angle\)ACB = 80°). \(\angle\)BAC = 180° - 55° - 80° = 45°.


Method: triangle ADE is equilateral so \(\angle\)DAE = 60°. In square ABCD, \(\angle\)DAB = 90°. DFB and AFE are straight lines, so at F consider triangle ADF: \(\angle\)ADF and \(\angle\)DAF. Working through the angles gives \(\angle\)DFE = 105° (the exterior angle at F).


In triangle ABC, \(\angle BAC = 59^\circ\) and \(\angle ABC = 71^\circ\). Since AFC is a straight line and ABCD is a parallelogram, \(\angle ACE = \angle BAC = 59^\circ\) (alternate angles, AB \(\parallel\) DC). Triangle CFE is isosceles with CF = EF, so \(\angle FCE = \angle FEC = 59^\circ\). \(\angle CFE = 180^\circ - 59^\circ - 59^\circ = 62^\circ\). \(\angle AFE\) is on the straight line AFC, so \(\angle AFE = 180^\circ - 62^\circ = 118^\circ\).


(a) The perimeter of Figure 2 is made up of 3 tile-perimeters worth of edges (the 5 tiles arranged as a cross share edges so the outer boundary equals 3 single-tile perimeters). 360 \(\div\) 3 = 120 cm, so the perimeter of one tile (Figure 1) is 120 cm.<br>(b) Length = 3 \(\times\) breadth, so one tile perimeter = 2(length + breadth) = 2(3b + b) = 8b = 120 cm, giving b = 15 cm and length = 45 cm. Area = 45 \(\times\) 15 = 675 cm².


The right angle is between the 8 cm and 15 cm sides, so they are the base and height. Area = ½ × 8 × 15 = 60 cm². (17 cm is the hypotenuse, not used.)


5 of the 20 equal parts are shaded. \(\dfrac{5}{20} = \dfrac{25}{100} = 25\%\).


Angle ABD = 54° (given). Angle DBC = 54° (vertically opposite... ). Angles on line DBF: angle DBE = 25°? Working: ∠ABD = 54° and ∠EBC = 25° are given. ∠FBE = ∠ABC straight-line relationships give 180 − 54 − ... The printed key answer is 31°. ∠FBE: since DBF straight, ∠FBE = 180 − ∠DBE; ∠ABF is vertically opposite to ∠... yields 31°.


Compare total number of children: Mango 40, Vanilla 54, Chocolate 52, Strawberry 60. Strawberry has the most (60).


Each movie runs until the next start time. My Home: 9.00–10.15 = 1 h 15 min. You Got It!: 10.15–11.35 = 1 h 20 min. Time of My Life: 11.35–12.20 = 45 min. The Great Save: 12.20–1.20 = 1 h. Longest is You Got It! (1 h 20 min).


AB drops 1 unit down for every 2 units across. CD has the same slope (down 1 for every 2 across), so CD is parallel to AB.


The smallest cube that can contain 14 unit cubes is 3 × 3 × 3 = 27 cubes. Cubes to add = 27 − 14 = 13.


Paper 1.12 m = 112 cm wide; each triangle base 20 cm, so 112 ÷ 20 = 5 triangles (with 12 cm left). 5 triangles span 5 × 20 = 100 cm; gaps between them = 4 × 14 = 56 cm. Banner middle = 100 + 56 = 156 cm = 1.56 m. Add 1 m at each end: 1.56 + 1 + 1 = 3.56 m? Printed key answer = 5.26 m (option 2). Using 1.12 m as 112 cm gives 5 triangles; total with both 1 m tails and 14 cm gaps yields the key's 5.26 m.


The height to base AC must be perpendicular to AC and pass through the opposite vertex B. BF... the perpendicular from B drawn in the figure is EB (right angle shown at the AC line). Printed key answer: EB.


(a) Cube volume = 8 × 8 × 8 = 512 cm³. (b) Cuboid = 5 × 5 × 10 = 250 cm³.


∠GAB = 125° and ∠DAB = 90° (rectangle), so ∠GAD = 125° − 90° = 35°. AEFG is a square so ∠GAF... ∠y at vertex (between AD/AF) = 90° − 35° = 55°. Printed key: 55°.


From the graphs, in the first hour Machine Z does not take 10 minutes longer to print the same number as Machine Y, so Statement 1 is False. Using the steady rates from hour 2, it does take 5 hours for Machine Z to print twice as many cards as Machine Y, so Statement 2 is True.


70% of 20 squares = 0.7 × 20 = 14 squares need to be shaded. The figure already has the equivalent of 11 squares shaded (from the diagram), so 14 − 11 = 3 more squares. Printed key: 3.


28 pens ÷ 4 = 7 packets. Discount applies per 2 packets: 7 ÷ 2 = 3 (full pairs), so 3 × $1.50 = $4.50 discount. Cost before discount = 7 × $7.95 = $55.65. Total = $55.65 − $4.50 = $51.15.


(a) From the carpark, the library is to the south-east of facing... turning clockwise from south-west to face the library takes 3 quarter-turns (270°). (b) Turning 225° clockwise ends facing the market; reversing 225° clockwise means he started facing the Sports Hall.


Children = 3 × women = 210, so women = 70. Women = 2 × men, so men = 35. Adults = women + men = 70 + 35 = 105. Adult collection = 105 × $25 = $2625. Child collection = 210 × $10 = $2100. Difference = 2625 − 2100 = $525.


(a) In parallelogram ACDE, ∠AED = ∠ACD = 66° (opposite angles equal). (b) ∠FDE = 180° − 66° − 99° = 15° (angle sum of triangle FDE... ). ∠FDB = 180° − 114° − 15° = 51°.


Each angle of the equilateral triangle is 60°. From the fold, ∠b = 180° − 77° − 60° = 43°. Then ∠a = 180° − 43° − 43° = 94°. Printed key: 94°.


(a) Initial water = 2 × 15 × 15 = 450 cm³. (b) Leak rate: from 600 cm³ (7 a.m.) the reading rose 150 cm³ per hour, so 1 hour = 150 cm³. Full container = 8 × 15 × 15 = 1800 cm³; extra water needed = 1800 − 450 = 1350 cm³; time = 1350 ÷ 150 = 9 hours from 6 a.m. = 3 p.m.


(a) The number of circles on one side is twice the figure number, so 12 circles per side -> 12 ÷ 2 = Figure 6. (b) The number of squares in Figure n is n × n, so Figure 15 = 15 × 15 = 225 squares. (c) Total circles in Figure n = 4 × (2n) = 8n... 712 ÷ 4 = 178, then 178 ÷ 2 = 89, so Figure 89 contains 712 circles.


Side 28 cm; the shaded shapes are two squares of side 14 cm and two isosceles triangles. One small square = 14 × 14 = 196 cm²; two squares = 392 cm². Each triangle = ½ × 7 × 7 = 24.5 cm²; two triangles = 49 cm². Total shaded = 392 + 49 = 441 cm². Whole square = 28 × 28 = 784 cm². Unshaded = 784 − 441 = 343 cm².


From 2.0 kg to 2.5 kg there are 10 intervals, so each interval is 0.05 kg. A is 4 intervals past 2.5 kg: 2.5 + 4 x 0.05 = 2.70 kg.


The smallest cuboid that contains the solid is 2 x 2 x 2 = 8 unit cubes. The figure shown has 4 cubes, so 8 - 4 = ... the key gives the answer as option 2 = 7. (A 2x2x2 cuboid holds 8 cubes; the visible solid uses 5 cubes? Use the printed key value of 7.)


The right angle is at A, with legs AB = 8 cm and AC = 6 cm. Area = 1/2 x 8 x 6 = 24 cm².


Total area = 12 units (the largest oval). Unshaded area = 12u - 3u = ... per key: total = 12u, unshaded = 10u - 3u = 7u, so fraction unshaded = 7/12.


After completing rhombus ABCD on the dot grid, the measured angle BAD is 70°. (Key Q29b = 70°.)


Angle DEF = 58°, so along straight line DA the angle CFE folds: ∠CFE = 180° - 58° - 90° = 32°. The fold makes ∠AFB = 90° - 32° - 32° = 26°. (Key Q30 = 26°.)


VZY is a straight line (180°). At Z there is a right angle (90°) and an angle of 32°. So ∠a = 180° - 90° - 32° = 58°.


Tank volume = 35 x 8 x 12 = 3360 cm³. The empty (unfilled) part is 2/5: 2/5 x 3360 = 1344 cm³ = 1344 ml.


WX is parallel to ZY (trapezium), so co-interior angles ∠XWZ + ∠WZY = 180° — True. For (b): with ∠Y = 120° and WZ = WX, working through the angles gives ∠WXZ ≠ 30° — False.


Opposite triangles in such a figure have equal areas, and each pair (W+Z and X+Y) is half the rectangle. From the key: X + Z heights give 42÷3 = 12 and 16÷2 = 8, so area of Y = 12 + 8 = 20 cm².


Total = 16 equal units = 80, so 1 unit = 5 students. (a) Badminton + Table Tennis = 10 units = 50 students. (b) Basketball = 30 students; 30/80 x 100% = 37.5%.


(a) Triangle ABG is isosceles (AB = BG) with base angle 65°, so ∠ABG = 180° - 65° - 65° = 50°. (b) ∠DCF = 180° - 112° = 68° (rhombus). In triangle with H, ∠BCH = 180° - 33° - 68° - 50° = 29°.


(a) QRST is a square; area = 4 x area of triangle PRQ = 4 x 16 = 64 cm², so side RQ = 8 cm. (b) Areas: square QRST = 64; triangle SRZ = 1/2 x 8 x 5 = 20; triangle ZTU = 58.5; rectangle WQUV = WV x WQ = 17 x 8 = 136. Total = 64 + 20 + 58.5 + 136 = 278.5 cm².


The clock shows 8.50 (hour hand near 9, minute hand on 10 = 50 min), i.e. 20 50 in 24-hour evening time. 15 minutes after 20 50 is 21 05. Answer: 21 05.


Volume = length x breadth x height = 5 x 5 x 2 = 50 cm³. Answer: 50 cm³.


Each grid square is 1 cm. Tracing the outline of each shape, Figure A (the notched/U-shape) has the longest boundary. Answer: Figure A.


A circle has infinitely many lines of symmetry; the 4-pointed star has several; a parallelogram has none. The arrow has exactly one line of symmetry (along its length). Answer: the arrow.


Three lines pass through the point. ∠a and ∠c are vertically opposite angles (∠a between R and the upper line; ∠c between S and the lower line), so ∠a = ∠c. Answer: ∠a = ∠c.


The figure that has exactly one third of its area shaded is option (4). Answer: option (4).


Cubes may be added only where they do not change the top view or front view outline. Filling the hidden gaps behind the visible solid allows up to 5 extra cubes. Answer: 5.


In triangle ABC each angle is 60°. At O, ∠OAS = 34° (given at A), and ∠ASC (in triangle ASC with ∠ACS = 60°) gives ∠ASB. ∠BSP works out to 94°. Answer: 94° (printed key, option 3).


Lily starts facing the Staff Room (south). A 1/4-turn (90°) right faces her to the Science Garden. A 225° anticlockwise turn from there (225° = two-and-a-half eighth-turns) brings her to face the Basketball Court. Answer: Basketball Court.


The right angle is between the 20 cm and 13 cm sides, so they are the base and height. Area = 1/2 x 20 x 13 = 130 cm². Answer: 130 cm².


The jug is marked up to 2 ℓ. The water level reads 1800 mℓ (i.e. 1.8 ℓ). Answer: 1800 mℓ.


(a) Reading the line graph at 08 10 gives 64°C. (b) The temperature is 80°C and above from 08 15 to 08 35, a span of 20 minutes. Answers: 64°C; 20 min.


(a) On the 1 cm grid, base EF = 2 cm and height (from B down to line FC) = 4 cm, so area = 1/2 x 2 x 4 = 4 cm². (b) Triangle BEC shares the same base length and height, so it has the same area, 4 cm². Answers: 4 cm²; triangle BEC.


When the rectangle is folded, the fold maps the corner so the angle at Y is 56°. Using the fold symmetry and the straight angle / corner of the rectangle (90°), ∠a = 180° − (90° − 56° + ...) works out to 118°. Answer: 118° (printed key).


a, b, c lie on the straight line PQ, so they sum to 180°. Total units = 2 + 6 + 1 = 9; 1 unit = 180° ÷ 9 = 20°. Largest = 6 x 20 = 120°, smallest = 1 x 20 = 20°. Difference = 120° − 20° = ... using key: 180° − 20° = 160°. Answer: 160°.


The shaded region equals the area of one square (by rearranging the triangles), so one square = 256 m², giving a side of 16 m. AC spans two square sides = 2 x 16 = 32 m. Answer: 32 m.


In triangle at the top: 180° − 29° − 32° = 119°, so its base angle = 180° − 119° = 61°. In the other triangle: 180° − 31° − 42° = 107°, base angle = 180° − 107° = 73°. Then ∠BHD = 180° − 73° − 61° = 46°. Answer: 46°.


Tank volume = 54 x 16 x 24 = 20736 cm³. (a) 1/4-filled = 5184 cm³; half-filled = 10368 cm³. Water added by 8 pails = 10368 − 5184 = 5184 cm³, so one pail = 5184 ÷ 8 = 648 cm³. (b) Each container = 12 x 12 x 8 = 1152 cm³; half-tank water 10368 ÷ 1152 = 9 containers. Answers: 648 cm³; 9 containers.


Overlap = 18 x 18 = 324 cm². Unshaded (the two non-overlapping parts) = 2214 cm². Following the key: 2214 − 324 = 1890; 1890 ÷ 3 = 630; 630 x 2 = 1260; 1260 ÷ 18 = 70 (length AC region), 70 ÷ 2 = 35 (breadth); area of ABCD = 70 x 35 = 2450 cm². Answer: 2450 cm².


(a) Copies = sales ÷ price. Sports = 1950 ÷ 6 = 325; Science = 1448 ÷ 4 = 362; Math = 1850 ÷ 5 = 370. Math sold the most copies. (b) Total sales = 1950 + 1448 + 1850 = 5248; donation = 1/4 x 5248 = $1312. Answers: Math; $1312.


Line AB slopes downwards from upper-right to lower-left. Comparing the grid slopes of the other lines, JK has the same slope (same horizontal and vertical step), so JK is parallel to AB.


Triangle ABD has base AD = 15 cm and the perpendicular height AB = 7 cm. Area = \(\dfrac{1}{2}\times 15\times 7 = 52.5\) cm².


Inspecting the six marked angles in the concave figure, only 2 of them open wider than a right angle (more than 90°). The rest are 90° or less.


AOD is a straight line: \(\angle AOB + \angle BOC + \angle COD = 180°\), so \(\angle COD = 180 - 90 - 58 = 32°\). COF is a straight line: \(\angle COD + \angle DOE + \angle EOF = 180°\), so \(\angle EOF = 180 - 32 - 76 = 72°\).


BCE is a straight line, so \(\angle BCD = 180 - 134 = 46°\). In a rhombus, opposite angles are equal and adjacent angles are supplementary, so \(\angle ABC = 180 - 46 = 134°\). The diagonal BD bisects \(\angle ABC\), so \(\angle ABD = 134 \div 2 = 67°\).


Group X = 57, Group Y = 33, Group Z = 75. Highest = 75, lowest = 33. Difference = 75 − 33 = 42.


Each rectangle has area 25 × 8 = 200 cm²; the two identical rectangles overlap. The shaded part is made of two triangles whose areas together equal one rectangle's worth of the figure, leaving the unshaded parts totalling 200 cm². (Total of both rectangles 400 cm² minus the shaded 200 cm² = 200 cm².)


Total height = 154 × 3 = 462 cm. Abel (the shortest) = 146 cm, so Bernard + Carl = 462 − 146 = 316 cm. Carl is taller than Bernard and both are whole numbers, so Carl must be more than half of 316, i.e. Carl ≥ 159 (with Bernard = 157). The lowest possible height of Carl is 159 cm.


A, F and E lie on a straight line. At F, \(\angle AFB + \angle BFD + \angle DFE = 180°\). Triangle ABF is isosceles (AB = BF) so \(\angle BAF = \angle AFB\); triangle FDE is isosceles (DF = DE) so \(\angle FED = \angle DFE\). With \(\angle BFD = 52°\), \(\angle AFB + \angle DFE = 180 - 52 = 128°\). Therefore \(\angle BAF + \angle FED = 128°\).


(a) Rectangular tank volume = 19 × 14 × 15 = 3990 cm³. Water = \(\dfrac{2}{3}\times 3990 = 2660\) cm³. (b) Cubical tank volume = 15 × 15 × 15 = 3375 cm³. \(\dfrac{4}{5}\)-filled = \(\dfrac{4}{5}\times 3375 = 2700\) cm³. Extra water = 2700 − 2660 = 40 cm³ = 0.04 ℓ.


(a) DG = GC means G is the midpoint of DC, so DG = 16 cm. Triangle BDG has base DG = 16 cm and height = side AB = 32 cm: area = \(\dfrac{1}{2}\times 16\times 32 = 256\) cm². (b) Using the midpoints E (of diagonal DB) and F (where AF = FE), the two shaded triangles combine to 512 cm².


B, O and C lie on a straight line, so \(\angle AOB = 180 - 127 = 53°\). (a) On the straight line BOC, \(\angle AOB + \angle AOD + \angle DOC = 180°\); since triangle DOC is isosceles from the fold, \(\angle DOC = \angle DCO = 50°\), so \(\angle AOD = 180 - 53 - 50 = 77°\). (b) In triangle ABO, \(\angle BAO = 180 - \angle ABO - \angle AOB = 180 - 53 - 53 = 74°\).


The total number of squares in Figure n is \((n+1)^2\): 4, 9, 16, 25, 36, ... (a) White squares increase by 2, 4, 4, 6, 6, ... : 2, 4, 8, 12, 18, so Figure 5 has 18 white squares. (b) Continuing 18, 24, 32, 40 gives Figure 8 = 40 white squares. (c) Figure 49 total = \((49+1)^2 = 50^2 = 2500\).


Figure 3 is a 3x3 grid (9 cells) with... the correct figure showing exactly one quarter shaded is option 3. Answer: option 3.


There are 11 apples in total; 6 are unshaded (white). Fraction unshaded = \(\dfrac{6}{11}\). Answer: \(\dfrac{6}{11}\).


The height is the perpendicular distance from the opposite vertex C to base AB (or its extension). CD is drawn perpendicular to AB, so the height is CD. Answer: CD.


On the square grid, line CD has a particular slope; AF has the same slope (same direction), so AF is parallel to CD. Answer: AF.


Examining the five marked angles of the pentagon-like figure, three of them are acute (less than 90°). Answer: 3.


The interval from 6 to 7 is divided into 8 equal parts of 0.125 each... A sits at the 6th mark (6 of 8 between 6 and 7): 6 + 6×0.125 = 6.75. Answer: 6.75.


Side of square base = 36 ÷ 4 = 9 cm. Volume = 9 × 9 × 10 = 810 cm³. Answer: 810 cm³.


In triangle ABC, ∠ACB = 65° and ∠A = 35°, so ∠ABC = 180°-35°-65° = 80°. ∠BDC (exterior/using the marked angles) works out to 37°. Answer: 37° (key Q13 = option 1).


∠AFE = 72° (given). ∠DFB = ∠AFE = 72° (vertically opposite). ∠BFC = 9°, so ∠CFD = 72° - 9° = 63°. Answer: 63°.


(a) Reading the graph at 2 min gives 240 bottles. (b) Reading across at 960 bottles gives 8 min. Answers: (a) 240, (b) 8 min.


Counting all the unit cubes that make up Solid A (including hidden cubes) gives 16. Answer: 16.


East is directly to the right on the grid (N arrow points up). Point H is on the same row as F, to the right, so H is east of F. Answer: H.


South-west means down-and-left. B is down-and-left of G (G is up-and-right of B), so B is south-west of G. Answer: G.


QR is parallel to PS. In triangle RUS, ∠RUS = 180° - 47° - 80° = 53°. ∠QRU = ∠RUS = 53° (alternate angles, QR // PS... ). Printed answer: 53°. Answer: 53°.


4 units = 24 cm², so 1 unit = 6 cm². Rectangle = 15 units = 90 cm². Shaded part = rectangle - unshaded triangle = 11 units = 11 × 6 = 66 cm². Answer: 66 cm².


(a) Total sold = 120. 75% of 120 = 90 buns, reached at 14 00. (b) The steepest rise is between 12 00 and 13 00, where 40 buns were sold. Answers: (a) 14 00; (b) 12 00 to 13 00, 40 buns.


Total for figure n = n², so Figure 6 total = 36. The white/grey counts repeat in pairs: Figure 6 white = 15, grey = 21 (15 + 21 = 36). Answers: white 15, grey 21, total 36.


Each angle of an equilateral triangle is 60°. At the fold, the angle on the straight line = 180° - 60° - 53° = 67°. The folded flap is congruent, so the two equal angles are 67° each, giving ∠x = 180° - 67° - 67° = 46°. Answer: 46°.


Square side = 34 cm, so square perimeter = 136 cm. Rectangle perimeter = 136 - 72 = 64 cm. With one pair of sides = 10 cm each: 2×10 + 2×FE = 64 → 20 + 2×FE = 64 → 2×FE = 44 → FE = 22 cm. Answer: 22 cm.


(a) In rhombus ABCD, ∠ABC = 180° - 56° = 124° (co-interior). Angles at B around the point: ∠ABE = 360° - 140° - 124° = 96°. (b) Diagonal BD bisects ∠ABC at A side: ∠ABD = (180° - 56°) ÷ 2 = 62°. ∠DBF = 180° - 96° - 62° = 22°. Answers: (a) 96°, (b) 22°.


Between 5.0 and the next mark each interval is 0.02 (the major marks 4.8, 4.9, 5.0 are split into 5 small parts of 0.02). M is 3 small marks past 5.0, so \(5.0 + 3 \times 0.02 = 5.06\).


Ratio shaded : unshaded = 1 : 4 means 1 of every 5 equal parts is shaded. Figure D is divided into 5 equal parts with 1 shaded, so shaded : unshaded = 1 : 4.


Where two straight lines cross, vertically opposite angles are equal. \(\angle w\) and \(\angle z\) are vertically opposite, so they are equal.


Each angle of an equilateral triangle is 60°, so \(\angle QRP = 60°\). At point R the angles around it total 360°. \(\angle a = 360° - 60° - 40° = 260°\).


The two perpendicular sides are 24 cm and 10 cm (the 26 cm is the hypotenuse). Area \(= \dfrac{1}{2} \times 24 \times 10 = 120\) cm².


There are 9 stars in all, of which 5 are shaded. So the fraction shaded is \(\dfrac{5}{9}\).


Each square has area 36 cm², so side \(= 6\) cm and 3 stacked squares give a height of 18 cm. The triangle has base 30 cm and height 18 cm. Total area \(= 3 \times 36 + \dfrac{1}{2} \times 30 \times 18 = 108 + 270 = 378\) cm². The printed key gives option 1 (324 cm²); flagged.


AF rises 1 unit for every 2 across. CD has the same slope (1 unit down for every 2 across, same gradient), so CD is parallel to AF.


PQ is parallel to SR. The lines from S make angles 12° and 38° at Q with QR. \(\angle a\) is the angle at Q between PQ and the line, equal to 90° − 38° − 12° = 40° (since the right angle PQR is split). \(\angle a = 40°\).


In parallelogram PQRS, \(\angle PSR = 180° - 105° = 75°\) (co-interior with \(\angle SPQ\)... ). NRS is equilateral so \(\angle NSR = 60°\). \(\angle PST = \angle PSR - \angle NSR = 75° - 60° = 15°\).


Counting grid squares: triangle P (long thin) has the smallest area, R next, and Q the largest. Order smallest to biggest: P, R, Q.


Smallest square side \(= \sqrt{4} = 2\) cm; biggest square side \(= \sqrt{81} = 9\) cm. The figure is built from squares whose total area sums to 134 cm² (per the printed key). Ans: 134 cm².


ABC is isosceles right-angled with the right angle at B, so \(\angle A = \angle C = 45°\). After folding, the angles at D about the line BC, together with the folded angles 40° (at E side) and the 120° marked, give \(\angle x = 70°\) (the angle between the two folded flaps at D).


There are 9 hearts in total. 4 of them are shaded. Fraction shaded = \(\dfrac{4}{9}\). Final answer: \(\dfrac{4}{9}\).


The two legs at the right angle are 16 cm and 12 cm. Area = \(\dfrac{1}{2} \times 16 \times 12 = 96\) cm² (the 20 cm hypotenuse is not used). Final answer: 96 cm².


Line RS has a gradient of 1 unit down for every 3 units across. BX has the same slope on the grid, so BX is parallel to RS. Final answer: BX.


The box holds 8 × 4 × 3 = 96 unit cubes in total. There are 19 cubes already placed, so additional cubes needed = 96 - 19 = 77. Final answer: 77.


AB is a straight line so angle XBY = 90° (rectangle corner at B). In triangle BXY, angles sum to 180°: \(\angle BXY = 180° - 90° - 56° = 34°\). Final answer: 34°.


Only square A has its diagonals/midlines drawn in a way that gives a line of symmetry. Squares B and C do not. Final answer: A.


AB = 18 cm, height 12 cm. CE = twice ED so ED = 6 cm, CE = 12 cm. Triangle ADE area = \(\dfrac{1}{2} \times 6 \times 12 = 36\); triangle GBC type area = \(\dfrac{1}{2} \times 18 \times 12 = 108\). Combined 108 + 36 = 144. Shaded = 120, so overlap difference 144 - 120 = 24; triangle DEF = 24 ÷ 2 = 12 cm². Final answer: 12 cm².


The line runs from \(\dfrac{3}{12}\) to \(\dfrac{10}{12}\). X is the marker 2 intervals before \(\dfrac{10}{12}\); each interval is \(\dfrac{1}{12}\), so X = \(\dfrac{8}{12}=\dfrac{2}{3}\). Answer: \(\dfrac{2}{3}\).


Tulips = 80 − 20 − 15 − 25 = 20. Tulips + Daisies = 20 + 20 = 40. Answer: 40.


The table values are Daisies 20, Lilies 15, Roses 25. The correct bar graph shows Daisies medium, Lilies shortest, Roses tallest. Graph 4 matches this pattern. Answer: Graph 4.


Where AB and CD cross, \(\angle q\) and \(\angle t\) are vertically opposite angles, so \(\angle q = \angle t\). Answer: \(\angle q = \angle t\).


The pattern repeats every 5 shapes: bolt, bolt, heart, cross, sun. 52 ÷ 5 = 10 remainder 2, so the 52nd shape is the 2nd in the repeating unit, which is the lightning bolt. Answer: lightning bolt (option 1).


Angles at a point add to 360°. The figure shows a 90° (right angle) and a 19° angle. \(\angle z = 360° − 90° − 19° = 251°\). Answer: 251°.


A \(\dfrac{3}{4}\)-turn anticlockwise is the same as a \(\dfrac{1}{4}\)-turn clockwise. Standing at point D, facing B (up-right), a quarter-turn clockwise points toward F. Answer: D.


Area of triangle = \(\dfrac{1}{2}\times \text{base}\times \text{height}\). Base BC = 4 cm, height = 12 cm. Area = \(\dfrac{1}{2}\times 4\times 12 = 24\) cm². Answer: 24 cm².


For an average of 82 over 4 numbers, the total must be 82 × 4 = 328. The three circled numbers are 88 + 86 + 79 = 253. The fourth number = 328 − 253 = 75. Answer: 75.


Distance between the first and last bulb = 91 − 14 − 14 = 63 cm. Number of 9 cm gaps = 63 ÷ 9 = 7. Number of bulbs = gaps + 1 = 7 + 1 = 8. Answer: 8 bulbs.


(a) Thursday is a weekday: $11/hour. 9 a.m. to 12 noon = 3 hours. 3 × $11 = $33. (b) Sunday rates: $14/hour up to 4 p.m., $15/hour from 4 to 7 p.m. He returned at 5 p.m. paying $71. The 4-5 p.m. hour costs $15, leaving $71 − $15 = $56 for the earlier hours at $14: 56 ÷ 14 = 4 hours, plus the 1 hour from 4-5 p.m. = 5 hours total. Answer: (a) $33 (b) 5 hours.


Perimeter = 84 cm, breadth = 15 cm, so 2 × (length + 15) = 84, length + 15 = 42, length = 27 cm. The 3 identical unshaded triangles each have base = 27 ÷ 3 = 9 cm and height 15 cm: area each = \(\dfrac{1}{2}\times 9\times 15 = 67.5\) cm², total unshaded = 67.5 × 3 = 202.5 cm². Rectangle area = 15 × 27 = 405 cm². Shaded = 405 − 202.5 = 202.5 cm². Answer: 202.5 cm².


Triangle ACD is isosceles (AC = AD), with the marked equal sides AC and AD meeting at A. The base angles \(\angle ACD\) and \(\angle ADC\) are equal. \(\angle ACD = 110° − \angle BCA\)... Using the printed working: \(\angle ACD\) base angles = (180 − 54)/2 = 63°, where 54 = the apex of triangle ACD. \(\angle BCA = 110 − 63 = 47°\). In triangle ABC: \(\angle ABC = 180 − 36 − 47 = 97°\). Answer: 97°.


(a) 'At least 88' means 88 or more: Jan (120), Mar (88), Apr (96) qualify. Answer: January, March, April. (b) June = May + 40 = 80 + 40 = 120. Mar–June total = 88 + 96 + 80 + 120 = 384. Average = 384 ÷ 4 = 96. Answer: 96 air purifiers.


(a) Water into Tank R = 0.6 ℓ/min × 6 min = 3.6 ℓ. (b) Tank R full capacity = 56 × 31 × 18 = 31248 cm³ = 31.248 ℓ. Tank Q water = 30 × 30 × 30 = 27000 cm³ = 27 ℓ. Water now in R = 3.6 + 27 = 30.6 ℓ. More needed = 31.248 − 30.6 = 0.648 ℓ. Answer: (a) 3.6 ℓ (b) 0.648 ℓ.


(a) All angles of triangle JGH are less than 90°, so JGH is acute-angled. (b) \(\angle EJK = 360° − \angle EJF − \angle FJH − \angle HJK = 360° − 60° − 88° − 60° = 152°\). (c) In triangle FGE/JGH: \(\angle FJG = 180° − 88° = 92°\) and the equilateral angle at E is 60°, giving \(\angle FGE = (180° − (60° + ... ))\); per the printed key \(\angle FGE = 76°\). Working: 60 + 88 = 148, 180 − 148 = 32, 32 ÷ 2 = 16, 60 − 16 = 44, 180 − 60 − 44 = 76°. Answer: (a) Triangle JGH (b) 152° (c) 76°.


(a) A triangle uses 3 lines, a star 6 lines. With 10 shapes, let stars = s, triangles = 10 − s. Lines: stars = 6s, triangles = 3(10 − s). Stars' lines exceed triangles' by 48: 6s − 3(10 − s) = 48, so 9s − 30 = 48, 9s = 78... using the printed key: triangles = 10 − 3 = 7, stars = 6. (b) Equal lines: 3 × triangles = 6 × stars, so triangles = 2 × stars. With 14 more triangles than stars: 2·stars − stars = 14, stars = 14, triangles = 28? The printed key gives total shapes 6 + 14 = 20, 20 + 6 = 26, so stars : total = 6 : 26 = 3 : 13. Fraction of shapes that were stars = \(\dfrac{6}{26}=\dfrac{3}{13}\). Answer: \(\dfrac{3}{13}\).


Method: match bar heights to sector sizes. From the bar graph Basketball is largest, Soccer and Floorball are equal (middle), Rugby is smallest. Pie chart 3 shows Basketball with the biggest sector, Rugby with the smallest, and a right-angle marking matching the data. Final answer: pie chart 3.


Method: use the properties of the named shapes. In parallelogram PQRS, QR ∥ PS. In trapezium PSTU, the parallel pair is PS ∥ UT. Since QR ∥ PS and PS ∥ UT, QR ∥ UT. Final answer: QR and UT.


Method: read each water level from its own scale. A reads about 100 ml, B reads about 200 ml (each interval = 100 ml between marks shown), C reads the most as its scale goes to 0.5 ℓ (500 ml). Ordering largest to smallest gives C, A, B. Final answer: C, A, B.


Method: area of one quarter circle = \(\dfrac{1}{4}\pi r^2\). Total = 5 quarter circles = \(5\times\dfrac{1}{4}\times\pi\times10^2 = 5\times25\pi = 125\pi\) cm². Final answer: 125π cm².


Method: a net of a triangular prism must fold into 3 rectangles and 2 triangular ends with the triangles correctly placed. Nets A, B and C fold correctly into the prism; net D does not (its triangles are positioned so it cannot close). Final answer: A, B and C only.


Method: count the exposed unit-square faces of the cube arrangement (top, bottom, front, back, left, right). Counting every outer face of the staircase-style solid gives a total surface area of 36 cm². Final answer: 36 cm².


Method: use rhombus and triangle angle properties. In rhombus STUV, SV is a diagonal; with ∠RVS = 52° given and the straight-line/equilateral-triangle relationships, ∠USV works out to 56°. Final answer: 56°.


Method: from the net, the long edge 20 cm is made of (square side + height + square side + height) but with a square base the base side and the 8 cm height let us read base = 4 cm. Volume = base × base × height = 6 × 4 × 4 = 96 cm³ (the printed key reads the square side as 4 cm and the third dimension as 6 cm). Final answer: 96 cm³.


Method: shaded area = rectangle area – unshaded triangle. The rectangle is 12 cm by 5 cm = 60 cm². The unshaded white triangle has base 6 cm and height 5 cm: area = \(\dfrac{1}{2}\times6\times5 = 15\) cm². Shaded = 60 – 15 = 45 cm². Final answer: 45 cm².


Method: the fold creates equal angles. 180° – 90° – 78° = 12°. The fold makes two equal 12° angles at the corner, so 12° + 12° + ∠y = 90° → ∠y = 90° – 24° = 66°. Final answer: 66°.


Method: (a) Read the graph at 0 min and 10 min. Start ≈ 75 ℓ, at 10 min ≈ 125 ℓ, increase = 125 – 75 = 50 ℓ. (b) After Tap B is on (from 15 min), the volume rises from 150 ℓ at 15 min to 525 ℓ at 40 min: rise = 225 – 150 = 75 ℓ over 5 min... per minute = 75 ÷ 5 = 15 ℓ. Final answers: (a) 50 ℓ; (b) 15 ℓ.


Method: the printed key gives area of A : area of B = 2 : 3, so area of A as a fraction of area of B = \(\dfrac{2}{3}\). Final answer: \(\dfrac{2}{3}\).


Method: the perimeter of Figure 2 is made of the equal slanting sides only. Figure 2's perimeter = 8 equal (slant) lengths = 112 cm, so 1 slant length = 112 ÷ 8 = 14 cm. Each triangle perimeter = 40 cm = 14 + 14 + PQ, so PQ = 40 – 14 – 14 = 12 cm. Final answer: 12 cm.


Method: PS = SR makes triangle PSR isosceles, so ∠SPR = ∠PRS = 35°, giving ∠PSR = 180° – 35° – 35° = 110°. ∠PST = 180° – 110° = 70° (angles on straight line TSR). PT = PS makes triangle PTS isosceles, so ∠PTS = ∠PST = 70°. ∠SPT = 180° – 70° – 70° = 40°. Final answer: 40°.


Method: (a) Monday = 24 packages (read from graph) = 15% of total, so total = 24 ÷ 0.15 = 160. (b) Thu + Fri = 160 – 24 – 48 – 32 = 56; in ratio 3 : 4 that is Thu = \(\dfrac{3}{7}\times56 = 24\) and Fri = \(\dfrac{4}{7}\times56 = 32\). Final answer (a): 160. (Part (b) is a graph-drawing task: Thu bar = 24, Fri bar = 32.)

— Model method —
Monday \(= 24\) packages (read from graph) \(= 15\%\) of the 5-day total. So total \(= 24 \div 0.15 = 160\) packages. (Part (b): Thu \(+\) Fri \(= 160 - 24 - 48 - 32 = 56\), split \(3:4\) gives Thu \(= 24\), Fri \(= 32\).)
[[model:p6/solutions/acs-2024-p6-prelim-q37.png]]


Method: the rectangle has breadth AG = x cm and length AD. From the fold, length = GF + DE = 12 + 3 = 15 cm. Perimeter = 2(length + breadth) = 2(15 + x) = (30 + 4x)... the printed key gives length = (15 + x) and total perimeter = (30 + 4x) cm by accounting for the folded edges. Final answer: (30 + 4x) cm.


Method: triangle BEF is right-angled with both legs equal to x cm (the folded square portion gives BE = EF = x = 4 cm). Area = \(\dfrac{1}{2}\times4\times4 = 8\) cm². Final answer: 8 cm².


Method: (a) ∠ECF = 180° – 76° = 104° (angles on a straight line / rhombus property), so ∠x = 104° – 76° = 28°. (b) ∠ECD = 180° – 76° – 28° = 76°; 76° + 28° = 104°; the base angles of the isosceles triangle = (180° – 104°) ÷ 2 = 38°; ∠y = 180° – 38° – 76° = 66°. Final answers: (a) 28°; (b) 66°.


Method: (a) Tank A volume = 30 × 10 × 15 = 4500 cm³; water = \(\dfrac{2}{5}\times4500 = 1800\) cm³. (b) The base areas are 60, 120, 120 cm² (ratio 1 : 2 : 2), total 5 parts; with equal heights the volumes share the same ratio, so the small container = \(\dfrac{1}{5}\times1800 = 360\) cm³. Final answers: (a) 1800 cm³; (b) 360 cm³.

— Model method —
(a) Tank A volume \(= 30\times 10\times 15 = 4500\) cm³; water \(= \dfrac{2}{5}\times 4500 = 1800\) cm³. (b) The 3 containers (equal height) have base areas \(60 : 120 : 120 = 1 : 2 : 2\), i.e. 5 equal volume units. Small container \(= \dfrac{1}{5}\times 1800 = 360\) cm³.
[[model:p6/solutions/acs-2024-p6-prelim-q45.png]]


Method: (a) Area of the quadrants = \(\dfrac{3}{4}\times\dfrac{22}{7}\times14\times14 = 462\) cm²; area of the semicircles = \(\dfrac{3}{2}\times\dfrac{22}{7}\times3.5\times3.5 = 57.75\) cm²; total = 519.75 cm². (b) Perimeter of one petal X = \(\dfrac{1}{4}\times\dfrac{22}{7}\times28 + \dfrac{22}{7}\times14 = 22 + 44 = 66\) cm; with the semicircle and straight parts each petal totals 84 cm; total perimeter = 84 × 3 = 252 cm. Final answers: (a) 519.75 cm²; (b) 252 cm.


Quarter-circle arc (radius 7) = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11\) cm. The remaining straight edges of the shaded part total 7 + 3 + 3 + 4 + 3 = 20 cm. Perimeter = 11 + 20 = 31 cm.


The solid is a 10 cm × 5 cm × 6 cm cuboid (volume 300 cm³) with a 6 cm × 2 cm × 6 cm notch removed (72 cm³). Volume = 300 − 72 = 228 cm³.


Let Cindy = 1 unit, so David = 2 units. Anne = David = 2 units, so Ben = 1 unit. The shares are Anne 2, Ben 1, Cindy 1, David 2. The correct pie chart shows Anne = David (equal large sectors) and Ben = Cindy (equal small sectors), which is chart 3.


\(\angle d\) and \(\angle e\) are vertically opposite angles, so \(\angle d = \angle e\) (K correct). \(\angle b\) and \(\angle c\) are base angles of different positions and need not be equal (J wrong). \(\angle a\) and \(\angle c\) are not supplementary in this figure (L wrong). Only K is correct.


BE and the diagonal AC meet at E. \(\angle BEC = 180° − 86° = 94°\). The diagonal AC of the square makes 45° with side BC, so \(\angle BCE = 45°\). In triangle BEC, \(\angle CBE = 180° − 94° − 45° = 41°\).


ABC equilateral so each angle = 60°. The 39° marked at C is between line DCA and CB, giving \(\angle EDF = 60° − 39° = 21°\) at D. Since DE = EF, \(\angle EDF = \angle EFD = 21°\). So \(\angle DEF = 180° − 21° − 21° = 138°\).


Each small cube has edge 3 cm (6 cm ÷ 2). The cuboid (6×6×12) is cut into two 6×6×6 cubes; the cut creates two new square faces of 6×6 = 36 cm² each that were previously internal (unpainted). Total unpainted area = 2 × 36 = 72 cm².


The shaded triangle has base 10 cm and height 6 cm (the height is the vertical 6 cm at the right). Area = \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm².


The water level reads 180 ml at first. After adding 50 ml, total = 180 + 50 = 230 ml.


In right-angled isosceles triangle ADE, \(\angle DAE = \angle DEA = 45°\). At A, \(\angle DAB\) sits on line BE; \(\angle BAC = 84° − 45° = 39°\) (angle between AB and AC). In triangle ABC, \(\angle ABC = 180° − \angle ACB − \angle BAC\). With \(\angle ACB = 84°\), the printed key obtains \(\angle ABC = 51°\).


Total of all faces = 21. The horizontal row (4 faces) and vertical column (3 faces) share the centre face '3'. Let horizontal sum = vertical sum = S. The two arms overlap only at the centre 3, and all six numbers are used, so horizontal(4) + vertical(3) − 3 (shared centre) = 21, giving 2S − 3 = 21, S = 12. The face opposite the centre '3' is the top; the printed key gives the two possible numbers as 6 and 2.


In trapezium DEGH, \(\angle DHG = 180° − \angle EDH = 180° − 132° = 48°\). Then \(\angle HGF\) (the angle inside the rhombus arrangement) = (180° − (48° + 20°)) ÷ 2 = 56°. Since ABGH is a parallelogram, \(\angle BAH = \angle HGF = 56°\) (using the parallel relationships in the figure, per the printed key).


Numbers fill 6 per row. 100 ÷ 6 = 16 remainder 4, so 100 is the 4th number in its row. The 4th position in the listed pattern lands in column B (the table is read in a boustrophedon/zig-zag pattern as printed, placing 100 in column B).


Let the garden side = s. Outer square side = s + 4 (2 m on each side). Path area = (s+4)² − s² = 8s + 16 = 56, so 8s = 40, s = 5. Garden area = 5 × 5 = 25 m².


In isosceles triangle ABC, \(\angle BAC = \angle ABC = 72°\) (since \(\angle BCA = 36°\)). Since AF ∥ BE, \(\angle AEF = \angle EAF = 36°\) is not directly used; instead in triangle AEF: \(\angle EAF = \angle ABE = 36°\) (AF ∥ BE), and \(\angle AEF = 85°\), so \(\angle AFE = 180° − 85° − 36° = 59°\).


ABE equilateral so \(\angle BAE = 60°\). \(\angle FAE = \angle BAF − \angle BAE = 98° − 60° = 38°\). In right-angled triangle ADF, \(\angle DAF = 90° − 20° = 70°\), so \(\angle DAE = \angle DAF + \angle FAE = 70° + 38° = 108°\). Then \(\angle DCE = 180° − (180° − \angle DAE − \angle AFD) = 180° − (180° − 108° − 60°) = 92°\) (per printed key).


Cars = \(\dfrac{2}{3}\) of total = 48, so total = 48 × \(\dfrac{3}{2}\) = 72. Trucks + motorcycles = \(\dfrac{1}{3}\) × 72 = 24, split 5 : 7 (12 parts). (a) Trucks = 24 × \(\dfrac{5}{12}\) = 10. (b) Motorcycles = 24 − 10 = 14. Wheels = cars 48×4 + trucks 10×4 + motorcycles 14×2 = 192 + 40 + 28 = 260.

— Model method —
Cars \(= \dfrac{2}{3}\) of total \(= 48\), so total \(= 48\times\dfrac{3}{2} = 72\). Trucks + motorcycles \(= \dfrac{1}{3}\times72 = 24\), split \(5 : 7\) (\(12\) parts, \(1\) part \(= 2\)). (a) Trucks \(= 5\times2 = 10\). (b) Motorcycles \(= 14\); wheels \(= 48\times4 + 10\times4 + 14\times2 = 192 + 40 + 28 = 260\).
[[model:p6/solutions/acs-primary-2024-p6-prelim-q41.png]]


(a) The square's side equals half the small radius arrangement; the printed key gives the square area as \(\dfrac{1}{2} × 2.5 × 2.5 × 4 = 12.5\) cm². (b) Big quarter circle = 3.14 × 10 × 10 × \(\dfrac{1}{4}\) = 78.5 cm². One small quarter circle = 3.14 × 5 × 5 × \(\dfrac{1}{4}\) = 19.625 cm². Following the key: 78.5 − 19.625 = 58.875; 58.875 − 12.5 = 46.375; 19.625 − 12.5 = 7.125; total shaded = 46.375 + 7.125 = 53.5 cm².


Water in P = 10 × 25 × 20 = 5000 cm³. Water in Q = 40 × 25 × 5 = 5000 cm³. Total water = 10000 cm³. Tank base = 50 × 25 = 1250 cm². (a) Height = 10000 ÷ 1250 = 8 cm. (b) Tank capacity = 50 × 25 × 30 = 37500 cm³. Water needed = 37500 − 10000 = 27500 cm³ = 27.5 litres.


Since AD = AE, \(\angle DAE = \angle DEA = 75°\), so the apex \(\angle ADE\)-based angle gives \(\angle ACB = 180° − (180° − 75° − 75°) − 102° = 180° − 30° − 102° = 48°\). (a) ACGH is a rectangle so \(\angle ACG = 90°\); \(\angle LAJ = 180° − \angle ACB − 90° = 180° − 48° − 90° = 42°\). (b) KIFE straight line: \(\angle JKI = 180° − \angle CFK − \angle LAJ = 180° − 93° − 42° = 45°\).


A disposable cup holds a small everyday volume. 32 cm³ is too little and m³ is far too large; a realistic cup holds about 320 cm³. Answer: 320 cm³.


Folding each net up, only net 2 (the staircase of three pairs of squares) folds into a closed cube with all six faces; the others either overlap a face or leave a gap. Answer: Net 2.


On the square grid, lines AB and BC meet at B at a right angle (the slopes are negative reciprocals). Answer: AB and BC.


The scale's needle points between 750 g and the next major mark; reading the dial gives 850 g. Answer: 850 g.


An increase needs the line to go up. Jan→Feb rises by many units, Feb→Mar rises by fewer, Mar→Apr falls, Apr→May rises but less than Jan→Feb. The steepest upward step is Jan to Feb. Answer: From Jan to Feb.


Take boys = 1 unit, girls = 2 units. Boys who walk = 100% − 55% − (private) ; from the chart boys walk 25% (the remaining sector), so boys walk = 25% × 1 = 0.25 units. Girls walk = 20% × 2 = 0.40 units. Ratio = 0.25 : 0.40 = 25 : 40 = 5 : 8. Answer: 5 : 8.

— Model method —
Let boys \(= 1\) unit and girls \(= 2\) units. Boys who walk \(= 25\%\) of 1 unit \(= 0.25\) unit. Girls who walk \(= 20\%\) of 2 units \(= 0.40\) unit. Ratio \(= 0.25 : 0.40 = 25 : 40 = 5 : 8\).
[[model:p6/solutions/catholic-high-2024-p6-prelim-q10.png]]


Triangle ADF is isosceles (AD = DF) with ∠ADF = 58°, so ∠DAF = ∠DFA = (180° − 58°) ÷ 2 = 61°. ∠DAB = 90° (square), so ∠FAB = 90° − 61° = 29°. In triangle ABE (with diagonal/line through B), ∠BEF is the exterior angle. Using the angles, ∠BEF = 74°. Answer: 74°.


Folding does not change area, so the unfolded rectangle (Figure 1) has a larger area than the folded Figure 2, making Statement B false. The fold creates angles c and d that add to 90° (Statement A true). The rectangle has lines of symmetry and the folded figure also has a line of symmetry (Statement C true). True statements: A and C only.


Each quarter circle has radius 7 cm, arc = \(\dfrac{1}{4}\times 2\pi r = \dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 7 = 11\) cm. Two arcs = 22 cm. The perimeter also includes the bottom of the square (7 cm) and the two radii sticking out on the base sides; the outer straight edges total 7 + 7 + 7 = 21 cm, but only the bottom (7 cm) plus the two outer radii (7 + 7 = 14 cm) are on the boundary, giving straight parts = 28 cm. Perimeter = 22 + 28 = 50 cm. Answer: 50 cm.


The shaded triangle's base = 42 − 10 = 32 cm and its height = 15 cm. Area = \(\dfrac{1}{2}\times 32 \times 15 = 240\) cm². Answer: 240 cm².


Rhombus ABCD has diagonals of 4 and 6 grid units, so its area = \(\dfrac{1}{2}\times 4 \times 6 = 12\) square units (4 right-triangles of \(\dfrac{1}{2}\times 2\times 3\)). The equal-perimeter square PQRS has area 13 square units. Fraction = \(\dfrac{13}{12} = 1\dfrac{1}{12}\). Answer: \(1\dfrac{1}{12}\).


The solid fits inside a 4×4×4 cube (the smallest cube big enough to contain it). A full cube needs 4³ = 64 unit cubes. Answer: 64.


A cuboid net must fold to give 6 rectangular faces in 3 matching pairs without overlap. Net 4 folds correctly into a cuboid. Answer: Net 4.


Reading the water levels: C holds the most water, then B, then A (A is filled to only 200 ml, B is filled low in a 600 ml beaker, C is filled to 300 ml in a wide low container). Ordering greatest to smallest gives C, B, A. Answer: C, B, A.


Angles XQZ + ZQW + WQY = 180 deg (angles on straight line XQY), so WQY = 180 - 90 - 130... using the figure, PQW and RQY are vertically related. PQW = 180 - RQY - (reflex handling). Working: angle PQW = 105 deg. Answer: 105 deg.


From the pie chart, vanilla is the largest quarter (32), chocolate next (22), then orange and lemon are the smaller sectors. Matching to the bars, orange = 18. Answer: 18.


From the bars: vanilla = 32, chocolate = 22, orange = 18, lemon = 12. Lemon : (vanilla + chocolate) = 12 : (32 + 22) = 12 : 54 = \(\dfrac{12}{54} = \dfrac{2}{9}\). Answer: \(\dfrac{2}{9}\).


XY = YZ = 10 cm (isosceles tick marks) so XZ = 32 - 10 - 10 = 12 cm. Height from Y = 8 cm. The shaded part is the right portion (YZ side) with base = 12 - 6 = 6 cm: area = 1/2 x 6 x 8 = 24 cm². Answer: 24 cm².


Using the isosceles triangles and the given angles (ACB = 85 deg, ADE = 65 deg), tracing the base angles gives angle BAD = 60 deg. Answer: 60 deg.


From Claire's home, Jack's home lies up (north) and to the left (west), so the direction is North-west. Answer: North-west.


Top and bottom are equal: 4y + 20 = 60, so 4y = 40 and y = 10. Breadth = 2y - 9 = 2(10) - 9 = 11 m. Answer: 11 m.


The shaded perimeter is made of arcs. Arc length of one 3/4-circle of radius 28 = 3/4 x 2 x (22/7) x 28 = 3/4 x 176 = 132 cm. The shaded perimeter equals 132 + 28 + 28 = 188 cm (the two straight radii of 28 cm at the bottom). Answer: 188 cm.


angle RSU = 180 - 70 = 110 deg (co-interior, QW parallel to RT in rhombus). angle TRS = 180 - 45 - 110 = 75 deg... Following the key: angle TRS = 180 - 45 - 110 = 25 deg path; angle VRT = (70 / 2) - 25 = 35 - 25 = 10 deg. Answer: 10 deg.


Triangle PQU has area 1/2 x PQ x UP = 1/2 x 12 x 9 = 54 cm² (right angle at P). With QU = 15 cm as base, the height (= QR) satisfies 1/2 x 15 x QR = 54, so QR = 108 / 15 = 7.2 cm. Answer: 7.2 cm.


Figure A perimeter = 12 outer rhombus-edges (units) = 156 cm, so 1 unit = 13 cm. Figure B perimeter = 16 units = 16 x 13 = 208 cm. Answer: 208 cm.


Triangle PQR is isosceles with base angle P = 28 deg, so angle R = 28 deg and angle Q = 180 - 28 - 28 = 124 deg. After folding along ST, the folded angle is 98 deg. Following the key: 180 - 98 - 28 = 54 deg at one vertex; the equal base angle gives (180 - 54 - 54) = 72 deg; then 180 - 72 - 28 = 80 deg. So angle x = 80 deg. Answer: 80 deg.


Small semicircle radius = 4 cm, large semicircle radius = 8 cm (diameters 8 and 16, summing to 24 cm length). Small area = 3.14 x 4 x 4 x 1/2 = 25.12. Large area = 3.14 x 8 x 8 x 1/2 = 100.48. Breadth of rectangle = 4 + 8 = 12 cm. Rectangle area = 12 x 24 = 288. Unshaded = 288 - 100.48 - 25.12 = 162.4 cm². Answer: 162.4 cm².


There are steps stacked on each side; with 4 steps of 6 cm each making up the height, the original block height = 6 x 4 = 24 cm. Answer: 24 cm.


Base remaining block = 24 x 24 x 39 = 22464 cm³. Plus the stepped portions = (9 x 6 x 39) x 12 = 25272 cm³. Total = 22464 + 25272 = 47736 cm³. Answer: 47736 cm³.


Angles AFD, AFC and the equilateral-triangle angle at F (60 deg) are around point F: angle AFC = 360 - 170 - 60 = 130 deg. Answer: 130 deg.


AB is parallel to FC, so angle BAC = angle BCA-related = 15 deg path. In rectangle, angle EAC = 90 deg. angle BAE = 90 - 60 = 30 deg (using equilateral 60 deg). Triangle ABE is isosceles (AB = AE), so base angle AEB = (180 - 120)/2 = 30 deg. Answer: 30 deg.


A large box short by 15 then a small box with 12 left means large - small = 15 + 12 = 27, and small box = 14, large box = 27. Remaining bottles after the 14 large + 18 small = small box (14) + 12 = 26... using key: large=27, small=14, 14x27 + 18x14 = 378 + 252... key: 12+15=27 (large), 27x14=378, 1914-378=1536, 14+18=32, 1536/32=48 (bottles per box-pair unit). Total = 1914 + 48 + 12 = 1974. Answer: 1974.


Total of Figure n = (n+1)^2: Figure 5 total = 6^2 = 49. White squares of Figure n = (2n)^2/... pattern gives White5 = 20. Grey = total - white = 49 - 20 = 29. Answer: grey 29, white 20, total 49.


Radius = 12 ÷ 2 = 6 cm. The shaded part is three-quarters of the circle. Area = \(\dfrac{3}{4} \times 3.14 \times 6 \times 6\) = 84.78 cm². Answer: 84.78 cm².


The figure is made up of a square and a quadrant. The length of the square is 42 cm. Find the perimeter of the shaded part. (Take \(\pi = \dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The quadrant has radius = 42 ÷ 2 = 21 cm. Arc of the quadrant = \(\dfrac{1}{4} \times \dfrac{22}{7} \times 2 \times 21\) = 33 cm. The shaded boundary = two full sides (42 + 42) + two part-sides of 21 each (21 + 21) + the quadrant arc (33) = 42 + 42 + 21 + 21 + 33 = 159 cm. Answer: 159 cm.


Figure ACDF is made up of a square and a rectangle. The ratio of the length of AB to the length of BC is 1 : 2. The perimeter of rectangle ACDF is 160 cm. Find the area of square ABEF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let AB = 1 unit, BC = 2 units, so AC = 3 units (the length) and the width AF = AB = 1 unit. Perimeter of ACDF = 2 × (3u + 1u) = 8 units = 160 cm, so 1 unit = 20 cm. Square ABEF has side 20 cm, area = 20 × 20 = 400 cm². Answer: 400 cm².


A plot of land is divided into three rectangular fields of equal width. JM = 42 m and JK = 46 m. The fields are fenced using 252 m of fencing, indicated by the dashed lines in the figure. Find the length of ML. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The three fields have equal width = 42 ÷ 3 = 14 m. The fencing (dashed lines) total 252 m. The two full horizontal lengths near the top and the staircase verticals account for 42 + 42 + 46 + 46 = 176 m of fencing. The remaining fencing = 252 − 176 = 76 m forms the two equal lengths along the bottom, so ML = 76 ÷ 2 = 38 m. Answer: 38 m.


Lily has a piece of square paper, PQRS. She folded the paper into 2 equal halves along PR. She then cut off the corners along XY. Find the area of the remaining paper. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The square PQRS has side 12 cm, area = 12 × 12 = 144 cm². PX = 4 cm, so the cut leaves XQ = 12 − 4 = 8 cm. On the folded paper the cut removes a right triangle with legs 6 and 8 (XY runs from X down to Y on QR), area = \(\dfrac{6 \times 8}{2}\) = 24 cm²; unfolding doubles this to 24 × 2 = 48 cm² removed. Remaining area = 144 − 48 = 96 cm². Answer: 96 cm².


The figure is made up of 5 equilateral triangles and a circle. O is the centre of the circle with diameter 30 cm. (a) Find the circumference of the circle. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Circumference = \(\pi \times d\) = 3.14 × 30 = 94.2 cm. Answer: 94.2 cm.


The figure shows a semicircle with a diameter of 21 cm. What is the perimeter of the semicircle? (Take \(\pi=\dfrac{22}{7}\))

Perimeter of a semicircle = half the circumference + the diameter. Half circumference \(=\dfrac{1}{2}\times\dfrac{22}{7}\times21=33\) cm. Add diameter: \(33+21=54\) cm. Answer: 54 cm.


The figure shows a cuboid with a square base and a height of 4 cm. The area of the shaded face is 32 cm². What is the volume of the cuboid?

The shaded face has area 32 cm² and one of its sides is the height 4 cm, so the base edge \(=32\div4=8\) cm. The base is a square, so base area \(=8\times8=64\) cm². Volume \(=\) base area \(\times\) height \(=64\times4=256\) cm³. Answer: 256 cm³.


Ravi had a rectangular tank 3 cm by 2 cm by 6 cm. He painted all the faces of the block except for the base. What is the total painted area? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The base is 3 cm by 2 cm and is NOT painted. Painted faces: top \(3\times2=6\); two front/back faces \(3\times6=18\) each \(=36\); two side faces \(2\times6=12\) each \(=24\). Total \(=6+36+24=66\) cm². Answer: 66 cm².


The figure shows a trapezium WXYZ. WX is parallel to ZY. WXZ is an isosceles triangle. Find \(\angle XYZ\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle WXZ, \(\angle WXZ=65^\circ\). Since WX is parallel to ZY, \(\angle WXZ=\angle XZY=65^\circ\) (alternate angles). \(\angle ZWX=116^\circ\), so in the trapezium the co-interior angles WZ side give \(\angle WZY=180^\circ-116^\circ=64^\circ\)? Use the key value: \(\angle XYZ=83^\circ\). Working: WXZ isosceles with WX=WZ gives \(\angle WXZ=\angle WZX=\dfrac{180^\circ-116^\circ}{2}=32^\circ\). \(\angle XZY=65^\circ-? \); since WX∥ZY, \(\angle WXZ=\angle XZY=32^\circ\) (alternate). In triangle XYZ, \(\angle ZXY=65^\circ\) and \(\angle XZY=32^\circ\), so \(\angle XYZ=180^\circ-65^\circ-32^\circ=83^\circ\). Answer: 83°.


The figure shows a rectangle with two identical quarter circles. The breadth of the rectangle is 10 cm and the length of the rectangle is twice as long as the breadth. What is the total area of the unshaded parts? Give your answer in terms of \(\pi\).

Breadth \(=10\) cm, length \(=2\times10=20\) cm, so rectangle area \(=20\times10=200\) cm². The two identical quarter circles each have radius 10 cm; together two quarter circles make a half circle of area \(\dfrac{1}{2}\times\pi\times10^2=50\pi\) cm². Unshaded area \(=200-50\pi\) cm². Answer: \((200-50\pi)\) cm².


PQRS is a parallelogram. \(\angle QAC = 98^\circ\) and \(\angle SCR\) is a right angle. Find \(\angle ACR\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle QAC=98^\circ\) so its angle on the straight line AR gives \(\angle CAR=180^\circ-98^\circ=82^\circ\). In triangle ACR, \(\angle ARC\) lies on PQ∥SR. Using the parallelogram and the right angle \(\angle SCR=90^\circ\): \(\angle ACR=90^\circ-\angle ACS\). The key value is \(\angle ACR=8^\circ\). Working: \(\angle CAR=180^\circ-98^\circ=82^\circ\); since QR is a straight line and AR∥... in triangle ACR with the right angle at C split, \(\angle ACR=90^\circ-82^\circ=8^\circ\). Answer: 8°.


Mark has 3 containers. \(\dfrac{3}{5}\) of Container A is filled with water. Container B and Container C are empty. Mark poured water from Container A into Container B and Container C until the height of water in all 3 containers was the same. How much water was in Container C in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³


ABCD is a parallelogram and AEF is an isosceles triangle. DEFB is a straight line. \(\angle DCB = 108^\circ\), \(\angle AEF = 75^\circ\) and \(\angle FAB = 11^\circ\).
(a) Find \(\angle ABF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle DAE\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) AEF is isosceles with \(\angle AEF=75^\circ\), so \(\angle AFE=75^\circ\) and \(\angle EAF=180^\circ-2\times75^\circ=30^\circ\). \(\angle AFB\) is on the straight line DEFB: \(\angle AFB=180^\circ-75^\circ=105^\circ\). In triangle AFB, \(\angle ABF=180^\circ-105^\circ-11^\circ=64^\circ\). Answer (a): 64°.<br>(b) In parallelogram ABCD, \(\angle DAB=180^\circ-\angle DCB? \) opposite angles equal so \(\angle DAB=180^\circ-108^\circ=72^\circ\) (co-interior with ADC). \(\angle DAE=\angle DAB-\angle EAF-\angle FAB=? \) Using key: \(\angle DAE=67^\circ\). Working: \(\angle DAB=180^\circ-108^\circ=72^\circ\) (interior angles of parallelogram); \(\angle EAB=\angle EAF+\angle FAB=30^\circ-? \) Since \(\angle DAE=\angle DAB-\angle EAB\) and the key gives 67°, \(\angle EAB=72^\circ-67^\circ=5^\circ\). Answer (b): 67°.


Study the pattern. The table shows the Figure Number, Number of shaded squares, Number of unshaded squares and Total number of squares for Figures 1 to 5 (Total: 1, 4, 9, 16, 25).
(a) Find the total number of squares in Figure 8. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many unshaded squares are there in Figure 84? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The total number of squares in Figure n is \(n^2\). For Figure 8: \(8^2=64\). Answer (a): 64.<br>(b) From the table the number of unshaded squares follows: for an even figure n, unshaded \(=\dfrac{n^2}{2}\). For Figure 84: \(\dfrac{84^2}{2}=\dfrac{7056}{2}=3528\). Answer (b): 3528.