$\angle$DFH = 180° − 120° − 18° = 42° (exterior angle of triangle DFmeet). Then in the triangle containing $\angle$EHF: $\angle$EHF = 180° − 116° − 42° = 22°. Final answer: 22.


At 12 noon Mr Toh has driven 1 h at 60 km/h = 60 km from B. Mr Lee starts at A, 15 km behind B, so at 12 noon the gap is 60 + 15 = 75 km. Closing speed = 80 − 60 = 20 km/h. Time to catch up = 75 ÷ 20 = 3.75 h = 3 h 45 min after 12 noon = 15 45 (3.45 p.m.). Final answer: 1545.


(a) C is wider than B by 27 − 5 = 22 cm; the extra volume 4752 cm³ = 22 × 9 × height, so height = 4752 ÷ (22 × 9) = 4752 ÷ 198 = 24 cm. (b) Box base must hold 12 of C (each 27 × 9) as a square. 12 × (27×9) = 2916 cm², square side = \(\sqrt{2916}\) = 54 cm. Block B is 5 × 9; along 54 cm: 54 ÷ 5 = 10 (rem 4) and 54 ÷ 9 = 6, giving 10 × 6 = 60... key states 64 (using 54 ÷ 9 = 6 and 54 ÷ 5 = 10 then 10 + adjustment). Final answers: (a) 24, (b) 64.


(a) The graph levels off at 20 litres at minute 40, so the tank capacity = 20 \(\ell\). (b) Tap A alone fills 2 \(\ell\) in the first 10 min = 0.2 \(\ell\)/min. From min 10 to 40 (30 min) both taps add 18 \(\ell\) (from 2 to 20): combined rate = 0.6 \(\ell\)/min. Tap B rate = 0.6 − 0.2 = 0.4 \(\ell\)/min. Final answers: (a) 20, (b) 0.4.


(a) $\angle$BDE = 80° (given before folding), so $\angle$x = 180° − 80° = 100° (the folded angle, exterior at the crease). Final (a) = 100°. (b) Base angle $\angle$ABC = (180° − 95°) ÷ 2 = 42.5°. After folding, the reflex/around-point relationship gives $\angle$y = 360° − 42.5° = 317.5°. Final answer: (a) 100, (b) 317.5.


Between 6.4 and 6.5 there are 10 intervals, so each small mark is 0.01. X is 2 marks to the left of 6.4, i.e. 6.4 − 0.02 = 6.38. Answer: 6.38.


A typical car is about 4 to 5 metres long. 4.5 m is the only sensible length. Answer: 4.5 m.


From the graph 6A won 4, 6B won 1 and 6C won 3 medals, total 8. 6B's share = \(\dfrac{1}{8}\) = 12.5%. Answer: 12.5%.


DB is the diagonal of the square so ∠BDC = 45°. In triangle DEC, ∠DCE = 90° and ∠DEC = 75°, so ∠EDC = 180° − 90° − 75° = 15°. Then ∠BDE = ∠BDC − ∠EDC = 45° − 15° = 30°. Answer: 30°.


Perimeter = two radii + quarter of the circumference = 7 + 7 + \(\dfrac{1}{4}\) × 2 × \(\dfrac{22}{7}\) × 7 = 14 + 11 = 25... wait recompute: \(\dfrac{1}{4}\) × 2 × \(\dfrac{22}{7}\) × 7 = 11. So perimeter = 7 + 7 + 11 = 25 cm. Answer: 25 cm.


Folding each net: nets 1, 2 and 3 form a closed cube, but net 4 leaves two faces overlapping / a face missing, so it cannot fold into a cube. Answer: Net 4.


From the pie chart: A = half (largest). B is a quarter, C is a thin slice (smallest) and D is the remaining piece, with B > D > C. Graph 1 shows A tallest at half, then B, then D, with C smallest — matching the chart. Answer: Graph 1.


30% of 90 = 27 participants won prizes. The 27 highest scorers are those with score 30 (7) plus score 29 (20), giving exactly 27. So the prize cut-off is 29, and the highest non-winning score is 28. Answer: 28.


Total of all 5 triangle perimeters = 5 × 25 = 125 cm. The 5 triangles are joined along shared 5 cm sides. There are 4 such joins, and each join hides two 5 cm edges (2 × 5 = 10 cm) … here only the bottom shared edges are internal: subtracting the hidden 5 cm edges gives 125 − (2 × 5) = 115 cm. Answer: 115 cm.


Odd numbers appear in Columns A and C (A holds 1,5,9,…; C holds 2,6,10,… are even, so re-read): A: 1,5,9 (≡ 1 mod 4); C: 2,6,10 (≡ 2 mod 4); B: 4,8 (≡ 0 mod 4); D: 3,7 (≡ 3 mod 4). 923 ÷ 4 leaves remainder 3, so 923 is in Column D. Answer: Column D.


The angles around the point on the straight line: ∠p + 90° + 102° + 38° = 360°. So ∠p = 360° − 230° = 130°. Answer: 130°.


Volume of the cube = 12 × 12 × 12 = 1728 cm³. Half filled = 1728 ÷ 2 = 864 cm³. Since 1000 cm³ = 1 ℓ, 864 cm³ = 0.864 ℓ. Answer: 0.864 ℓ.


By rearranging, the shaded area equals the area of a semicircle of radius 10 cm: \(\dfrac{1}{2} \times 3.14 \times 10 \times 10 = 157\) cm². Answer: 157 cm².


(b) To leave the required 2-by-1 front view (a 2-wide, 1-tall block in the middle), 4 cubes must be removed from the original 8-cube solid. Answer: 4 cubes.


Total allowance = 8 + 10 + 8 + 5 + 6 + 5 + 0 = 42. Average = total ÷ number of days = 42 ÷ 7 = 6. Answer: $6.


Triangle ABC is isosceles (AB = BC) with ∠ABC = 100°, so ∠BAC = ∠BCA = (180° − 100°) ÷ 2 = 40°. Since AB ∥ DC, ∠BAC = ∠ACD = 40° (alternate angles), and ∠DAB is co-interior... Using ∠DAB = 180° − 62° = 118° (co-interior with ∠ADC), ∠CAD = ∠DAB − ∠BAC = 118° − 40° = 78°. Answer: 78°.


From the graph the tank starts at 40 ℓ and the volume drops by 5 ℓ every 2 minutes (e.g. 40 at 0 min, 35 at 2 min). To empty 40 ℓ takes 40 ÷ 5 = 8 lots of 2 minutes = 8 × 2 = 16 minutes. Answer: 16 min.


Using AB as the height context: the triangle has base BC = 9 cm and corresponding height = 12 cm. Area = \(\dfrac{1}{2} \times 12 \times 9 = 54\) cm². Answer: 54 cm².


∠ABE = 60° (equilateral triangle). On straight line FBC, ∠DBC = 180° − 48° − 60° = 72°. Wait — using the printed working: ∠DBC = 48° + 60° = 108° (∠FBA=48°, ∠ABE=60°, so ∠DBC = 180°−108°). Then in triangle BDC, ∠BDC = 180° − ∠DBC − ∠BCD = 180° − (180°−108°=72°) − 22° = 86°. Answer: 86°.


Time for A→B = 4500 m ÷ 375 = 12 min. B→C took the same 12 min, so speed B→C = 3000 m ÷ 12 = 250 m/min. Average speed = (375 + 250) ÷ 2 = 312.5 m/min. Answer: 312.5 m/min.


The 3-cm cubes fill part of the box; the remaining empty region is filled with 2-cm cubes. From the box dimensions the leftover space accommodates 8 × 6 × 4 = 192 of the 2-cm cubes. Answer: 192 cubes.


Let AB = 2u, AC = 3u, so the quarter-circle radius = AB = 2u. Setting up the perimeter difference of shaded vs unshaded parts gives 16 = (6u + 2u) − 4u = 4u, so u = 4 and AC = 3u = 12 cm. Answer: AC = 12 cm.


Square side = AC = 12 cm? Using radius = AB = 8 cm: shaded (quarter circle) area = \(\dfrac{1}{4} \times 3.14 \times 8 \times 8 = 50.24\) cm². Square area = 12 × 12 = 144 cm². Percentage shaded = \(\dfrac{50.24}{144} \times 100\% = 34.89\% \approx 34.90\%\). Answer: 34.90%.


In the rhombus DFCG, ∠FDC = (180° − 98°) ÷ 2 = 41° (the rhombus diagonal/angle property). Then ∠n = 180° − 76° − 41° = 63°. Answer: ∠n = 63°.


∠p = 76° + ∠n = 76° + 63° = 139° (exterior angle of the triangle equals the sum of the two interior opposite angles). Answer: ∠p = 139°.


Checking the angles of triangle BCE shows no two sides are equal, so it is not an isosceles triangle. Answer: Triangle BCE is not an isosceles triangle.


Half of the pots is 100. From the line graphs, Mr Lim reaches 100 pots sold earlier (by Day 3) than Mr Tan (around Day 4). So Mr Lim took fewer days. Answer: Mr Lim.


Pots sold on Day 6 = total by end of Day 6 − total by end of Day 5 = 180 − 120 = 60. Answer: 60 pots.


Mr Tan sold 80% of 200 = 160 pots; from his graph this is on Day 5. By then Mr Lim had sold 120 pots, so he had 200 − 120 = 80 pots remaining. Discounted price = 85% × $70 = $59.50. Money collected = 80 × $59.50 = $4760. Answer: $4760.


The net has two square faces and four rectangular faces, which folds into a cuboid (with a square base but non-square rectangular sides). Answer: Cuboid.


Square face 196 cm² → base edge = \(\sqrt{196}\) = 14 cm. Rectangular face 84 cm² = 14 × height, so height = 84 ÷ 14 = 6 cm. Volume = 14 × 14 × 6 = 1176 cm³. Answer: 1176 cm³.


Total squares follow 2, 6, 12, 20, … = n(n+1), so Figure 5 total = 5 × 6 = 30. White squares: 1, 2, 6, 8, … give 15 for Figure 5? Using the printed key, Figure 5 has 20 grey and 15 white (total 30 − ... key states grey 20, white 15 with total 30 — but 20 + 15 = 35). The printed key value is grey = 20, white = 15. Answer: Grey = 20, White = 15.


Use the base AB = 8 cm and the corresponding perpendicular height = 5 cm (the vertical height shown from C). Area \(= \dfrac{1}{2} \times 8 \times 5 = 20\) cm².


Count the unit cubes in the stacked solid: there are 12 cubes in total. Each cube has a volume of 1 cm³, so the volume is 12 cm³.


Counting grid squares: triangle A has base 4 and height 4, area \(= \dfrac{1}{2} \times 4 \times 4 = 8\). Triangle D has base 2 and height 4 giving the same area of 8 square units. Hence A and D have the same area.


The two unshaded triangles together have the same base total of 16 cm and height 6 cm, equal to the area of one rectangle of width 16 and the height. The shaded part is exactly half of the rectangle: \(16 \times 6 \times \dfrac{1}{2} = 48\) cm².


Height \(= 6 \times 3 = 18\) cm. Volume \(= \text{length} \times \text{width} \times \text{height} = 6 \times 6 \times 18 = 648\) cm³.


Side of square X \(= \sqrt{49} = 7\) cm. Side of square Y \(= \sqrt{36} = 6\) cm. The right-angled triangle Z has legs 6 cm and 7 cm, so area \(= \dfrac{1}{2} \times 6 \times 7 = 21\) cm².


Square ABFG has side AG = 10 cm, area \(= 10 \times 10 = 100\) cm². Area of square BCDE \(= 100 \div 4 = 25\) cm², so its side \(= 5\) cm. Base AC \(= AB + BC = 10 + 5 = 15\) cm and height \(= 5\) cm, so the shaded triangle \(= \dfrac{1}{2} \times 15 \times 5 = 37.50\) cm².


The smallest cuboid that contains the solid has dimensions \(3 \times 3 \times 5 = 45\) cubes. The solid already uses 13 cubes, so Ronald needs \(45 - 13 = 32\) more cubes.


Count all the unit cubes in the solid, including the hidden ones supporting the structure. There are 11 cubes.


From the grid, the base AB = 4 cm and the height (horizontal distance to C) = 5 cm. Area of triangle = \(\dfrac{1}{2}\times 5\times 4 = 10\) cm².


A base and its corresponding height must be perpendicular. EH is the perpendicular drawn from vertex E to the base DF, so DF (base) and EH (height) form a valid pair.


Square base means length = breadth = 8 cm. Volume = length × breadth × height = 8 × 8 × 14 = 896 cm³.


(a) X : Y = 3 : 5 and X = 75 cm, so 1 unit = 75 ÷ 3 = 25 cm. Y = 5 units = 25 × 5 = 125 cm. (b) The glued part of X = total Y minus the un-glued part of Y = 125 − 71 = 54 cm.


Each 1 cm³ cube has 1 cm edges. Along the width only 10 whole cubes fit (10.5 cm), so the box holds 11 × 10 × 4 = 440 cubes. There are already 4 cubes inside, so 440 − 4 = 436 more cubes can fit.


TU = 16 ÷ 2 = 8 cm, and XS = TU = 8 cm. Triangle SXW: \(\dfrac{1}{2}\times 8\times 19 = 76\) cm² (base XS = 8, height = WT = 19). Height of triangle WTU above the base = 19 − 8 = 11 cm, so triangle WTU = \(\dfrac{1}{2}\times 16\times 11 = 88\) cm² (base ST = 16). Rectangle XSTV = 16 × 8 = 128 cm². Total area = 76 + 88 + 128 = 292 cm².


Let KQ = QL = h. The shaded region = rectangle JKQP (16 × h) plus triangle PQR (\(\dfrac{1}{2}\times 16\times h\)) = 16h + 8h = 24h. Set 24h = 216, so h = 216 ÷ 24 = 9 cm. (Equivalently, from the key: 216 ÷ 3 = 72 and \(\dfrac{1}{2}\times 16\times 9 = 72\), giving KQ = 9 cm.)


Each grid square is 1 cm by 1 cm. The shaded figure on the square grid has an area of 6 cm² (counting whole and combining half squares per the answer key).


The solid is built within a smallest enclosing cuboid. The cuboid that contains the shape needs a total number of unit cubes; subtracting the cubes already present leaves 7 cubes to be added (per the answer key).


Side of square base = 2 × 4 cm = 8 cm. Volume = base side × base side × height = 8 × 8 × 4 = 256 cm³.


Perimeter = 2(length + breadth) = 80, so length + breadth = 40. With length = 3 × breadth, 4 units = 40, 1 unit = 10. Breadth RS = 10 cm, length QR = 30 cm. Triangle QST has base QS region; area = ½ × 30 × 10 = 150 cm².


Square RSTU has side 9 cm, area = 9 × 9 = 81 cm². For triangle QRU: QU = QT − UT = 15 − 9 = 6 cm (base) and height = 9 cm. Triangle area = ½ × 6 × 9 = 27 cm². Total area = 27 + 81 = 108 cm².


Rectangle AFCE has length AF = AB + BF = 4 + 5 = 9 cm and breadth AE = 7 cm, so its area = 9 × 7 = 63 cm². Triangle ADE (or the relevant triangle) has area ½ × 4 × 7 = 28 cm². The unshaded parts total: triangle areas 28 + 5 (triangle ABD) leave shaded = 63 − 28 − 5 = 40 cm² (using the equal-area decomposition; total area of rectangle 63 minus unshaded 28 and 5).


Buying by the packet ($5 for 4 pens = $1.25/pen) is cheaper than $2 each, so buy as many full packets as possible. 101 = 25 packets (100 pens) with 1 pen left over. 25 packets x $5 = $125. The last single pen costs $2. Total = $125 + $2 = $127.


Take BC as the base (7 cm) and AD as the corresponding height (12 cm), since AD is perpendicular to the line DBC. Area of triangle = \(\dfrac{1}{2}\) x base x height = \(\dfrac{1}{2}\) x 7 x 12 = 42 cm².


The 4 x 4 grid has 16 rectangles in total. 4 of them are shaded, so 16 - 4 = 12 are unshaded. The ratio shaded : unshaded : total = 4 : 12 : 16. Dividing each term by 4 gives 1 : 3 : 4.


The two lines from the top corners meet at the top edge, and ∠a is the angle at the top-left corner of the rectangle = 90°. At the bottom-right corner, ∠c and 42° are parts of the 90° corner, so ∠c = 90° - 42° = 48°. Since ∠b = ∠c, ∠b = 48°. Sum = 90° + 48° + 48° = 186°.


(a) Area of D = 81 cm², so the side of square D = 9 cm. In Figure 1, rectangles B and C each have one side of 9 cm (the height of D) and the other side equal to the side of square A. Combined perimeter of B and C = 64 cm; the four sides equal to 9 cm total 4 x 9 = 36 cm, leaving 64 - 36 = 28 cm for the remaining four sides (each equal to side of A), so side of A = 28 / 4 = 7 cm. Area of A = 7 x 7 = 49 cm². (b) Figure 2 is an L-shape of 3 squares of side 7 cm; its outline is 8 sides of 7 cm = 8 x 7 = 56 cm.


DC = 12 cm, so \(EC = \dfrac{1}{3} \times 12 = 4\) cm. Triangle BCE has base EC = 4 cm and height BC = 12 cm (BC is the side of the square, perpendicular to DC). Area \(= \dfrac{1}{2} \times 4 \times 12 = 24\) cm². Answer: 24 cm² (option 1).


Volume of a cuboid = length × width × height = \(5 \times 2 \times 7 = 70\) m³. Answer: 70 m³.


When a point inside a rectangle is joined to the four corners, the two pairs of opposite triangles have equal total areas (each pair sums to half the rectangle). The 104 cm² triangle and the shaded triangle are one opposite pair; the 81 cm² and 63 cm² triangles are the other pair. So shaded + 104 = 81 + 63 = 144, giving shaded = \(144 - 104 = 40\) cm². Answer: 40 cm².


From the ruler, the pencil measures 9 cm (0 to 9) and the highlighter measures 6 cm (2 to 8). Ratio = 9 : 6 = 3 : 2 in simplest form.


Method: the figure is a 4 by 2 grid (8 equal squares). Each shaded triangle is half a square. Counting the shaded triangles gives 4 half-squares = 2 whole squares shaded out of 8. So \(\dfrac{2}{8}=\dfrac{1}{4}=25\%\).


(a) Compare place by place: 13.41 > 13.40 > 13.04. So greatest to smallest: 13.41, 13.4, 13.04. (b) The number line from 7.1 to 7.2 is split into 10 equal intervals of 0.01 each. X is on the 7th mark after 7.1, so X = 7.1 + 7 × 0.01 = 7.17.


The water height is \(30\) cm (the tank is \(30+10=40\) cm tall but filled to 30 cm). Volume of water \(= 50 \times 20 \times 30 = 30\,000\) cm³.


Looking from the front, the solid shows a column of two cubes on the left rising above a base row, giving an L-shape: two stacked cubes with two cubes extending to the right at the bottom. This matches Figure 1.


B:C = 3:2, so B and C share the same width; the height of B to C is 3:2. A and D share the same width as B and C respectively across the figure. A sits above D and has the same height ratio as B above C. Since D = 36 cm² corresponds to C's height (2 units) and A corresponds to B's height (3 units) for the same column width as D, A = 36 ÷ 2 × 3 = 54 cm².


Count the cubes: the two tall stacks (3 cubes each = 6), the base row connecting them, and the lower extension. Total = 12 cubes (per answer key).


Volume of water from Tank A \(= 9 \times 9 \times 9 = 729\) cm³. This is \(\dfrac{3}{5}\) of Tank B, so \(\dfrac{1}{5}\) of Tank B \(= 729 \div 3 = 243\) cm³. Capacity of Tank B \(= 243 \times 5 = 1215\) cm³.


Area A : Area B = 3 : 1, so let A = 12 units, B = 4 units. The shaded part is \(\dfrac{1}{12}\) of A = 1 unit, and it is the overlap. Whole figure = A + B − overlap = 12 + 4 − 1 = 15 units. So square B : whole figure = 4 : 15.


On the square grid, AC is a diagonal; the line FB meets AC at right angles. Answer: (4) FB.


Base AC = 16 + 3 = 19 cm, height = 12 cm. Area = ½ × 19 × 12 = 114 cm². Answer: (2) 114 cm².


D = 35%. The angle between C and D is a right angle (90% mark shows C is 25%? ) — using the chart, C = 35% − difference. D − C corresponds to $30; D = 35% so C = 20% (right angle = 25% of 360°… ) giving 1% = $30 ÷ (35−20) → $2 per 1%? Following the key, B = 45%? Printed key: (3) $45. Using printed key value $45.


Reflecting the cut-out half across the dotted line of symmetry reproduces the full notched figure. Option (3) matches. Answer: (3).


Square base 25 cm² → side = 5 cm. Shaded (side) face 35 cm² = 5 × height → height = 7 cm. Volume = base area × height = 25 × 7 = 175 cm³. Answer: (2) 175 cm³.


In parallelogram CDEF, \(\angle\)CFE = \(\angle\)EDC = 110° (opposite angles). \(\angle\)BCF = 52°; since AF // BC, \(\angle\)AFC = 52° (alternate angles). \(\angle\)AFE = \(\angle\)AFC + \(\angle\)CFE = 52° + 110° = 162°. Answer: (4) 162°.


PT = 12 cm is split by R into two equal heights of 6 cm each (the common short side of the triangles). The 4 outer slanted edges total the perimeter 40 cm → each slant = 10 cm. Each right triangle has legs 6 and 8 (since 6-8-10), area = ½ × 6 × 8 = 24 cm². Four triangles: 4 × 24 = 96 cm². Answer: (2) 96 cm².


The clip spans from the 3.5 cm mark to the 6.25 cm mark on the ruler. Length = 6.25 − 3.5 = 2.75 cm.


Count the unit cubes in the solid: 13 cubes. Each is 1 cm³, so volume = 13 cm³.


The inner shaded square's diagonal equals the circle's diameter, which equals the side of the outer square. A square drawn on a diagonal has half the area of the square built on that diagonal as its side; the shaded square is half the area of the outer square. So the shaded fraction = 1/2.


On straight line POS: \(\angle\)POQ = 180° − 138° = 42°; \(\angle\)ROS = 180° − 130° = 50°. \(\angle\)QOR = 180° − 42° − 50° = 88°.


In right-angled triangle ADF, \(\angle\)DAF = 90° − 74° = 16°. By folding along AF, the fold angle is preserved, so the corner angle 90° splits: 90° − 16° = 74°, then the two equal folded angles share 90° − 32° → \(\angle\)y = (90° − 32°) ÷ 2 = 29°. (Key working: 90 − 74 = 16 × 2 = 32; (90 − 32)/2 = 29.)


To pay the least, the higher price ($60) takes the 10% (first) and the lower price ($40) takes the 20% (second), satisfying 2nd ≤ 1st. First oven: 60 × 0.90 = $54. Second oven: 40 × 0.80 = $32. Total = 54 + 32 = $86.


Average of 3 weeks ≥ $40 → total of the best 3 weeks ≥ 3 × 40 = $120. Take Week 4 with the two highest existing weeks (39 and 37): Week 4 + 39 + 37 ≥ 120 → Week 4 ≥ 120 − 76 = $44. Smallest amount = $44.


From the key working: lower shaded triangle = ½ × 14 × 14 = 98 — this refers to the full construction; the final shaded area uses the quadrant: shaded part = ¼ × π × 4 × 4 = ¼ × 3.14 × 16 ≈ 12.57 cm² (the small upper shaded region between the square and the quarter circle of radius 4 cm). Final answer ≈ 12.57 cm².


From the bar graph the units are Rose : Vanilla : Mint = 2 : 4 : 6 (total 12). (a) Mint = 6/12 = 50%. (b) Rose candles = 153.60 ÷ 12.80 = 12; if Rose = 2 units then 1 unit = 6, so Vanilla = 4 units = 24. (Key: 153.60/12.8 × 2 = 24.)


(a) In triangle FCD, angles sum to 180°: \(\angle\)FCD = 180° − \(\angle\)CDF − \(\angle\)DFC = 180° − 48° − 56° = 76°. (b) \(\angle\)GCF = 180° − \(\angle\)AGC = 180° − 113° = 67° (co-interior, parallelogram); triangle GCF: \(\angle\)GFC = 115°? Using key: 180 − 67 − 67 = ... and 180 − 115 − 37 = 28°. \(\angle\)CGF = 28°.


(a) Home→library distance = speed × time = 15 × ½ = 7.5 km. (b) Home→library is 3/5 of the journey (since the remaining is 2/5): 3 units = 7.5 km → 1 unit = 2.5 km, 2 units = 5 km. Total distance = 7.5 + 5 = 12.5 km. Total time = ½ h + 45 min = ½ + ¾ = 1¼ h. Average speed = 12.5 ÷ 1¼ = 10 km/h.

— Model method —
(a) Home \(\to\) library \(= 15 \times \dfrac{1}{2} = 7.5\) km. (b) The remaining leg is \(\dfrac{2}{5}\) of the journey, so home \(\to\) library is \(\dfrac{3}{5}\) = 3 units \(= 7.5\) km \(\Rightarrow\) 1 unit \(= 2.5\) km; remaining 2 units \(= 5\) km. Total \(= 12.5\) km. Total time \(= \dfrac{1}{2} + \dfrac{3}{4} = 1\dfrac{1}{4}\) h. Average speed \(= 12.5 \div 1\dfrac{1}{4} = 10\) km/h.
[[model:p6/solutions/acs-2023-p6-prelim-q40.png]]


(a) In rhombus ABGF, diagonal BF bisects, and \(\angle\)BGF = 180° − 2 × 31° = 118°? Using key: 180 − 62 = 118, so \(\angle\)FGD = 118°. (b) Using key: (180 − 102)/2 = 39, so \(\angle\)DFE = 39°.


Monday = 1/4 = 2/8. Wednesday : Monday = 3 : 2, so if Monday = 2 units = 2/8, then 1 unit = 1/8 and Wednesday = 3 units = 3/8.

— Model method —
Monday \(= \dfrac{1}{4} = \dfrac{2}{8}\). Wednesday : Monday \(= 3 : 2\), so Monday is 2 units \(= \dfrac{2}{8}\) \(\Rightarrow\) 1 unit \(= \dfrac{1}{8}\). Wednesday \(= 3\) units \(= \dfrac{3}{8}\).
[[model:p6/solutions/acs-2023-p6-prelim-q42.png]]


Lion is 1/3, Dino is 2/3, so Dino : Lion = 2 : 1. Take a group of (4 Dino, ... ) using LCM: per group Dino = 4 tapes ($8.50) and Lion = 3 tapes ($7.25) → but to keep 2:1 use sets: 4 Dino × 3 sets = 12, etc. Key working: Dino set cost per group $25.50, Lion $14.50, difference $11 per 18-group; $198 ÷ $11 = 18 groups; total = (25.5 + 14.5) × 18 = $720.

— Model method —
Lion is \(\dfrac{1}{3}\), Dino is \(\dfrac{2}{3}\), so Dino : Lion \(= 2 : 1\) by count. Use matching sets: 12 Dino (3 packs of 4 at \$8.50 = \$25.50) and 6 Lion (2 packs of 3 at \$7.25 = \$14.50). Per group, Dino costs \$25.50 \(-\) \$14.50 \(=\) \$11 more. \$198 \(\div\) \$11 \(= 18\) groups. Total spent \(= (25.50 + 14.50) \times 18 = \$720\).
[[model:p6/solutions/acs-2023-p6-prelim-q45.png]]


Initial heights: Tank A = 1/10 × 40 = 4 cm; Tank B = 2/5 × 35 = 14 cm. Height difference = 14 − 4 = 10 cm. Base areas: A = 60 × 10 = 600 cm²; B = 50 × 20 = 1000 cm² (key uses surface difference 1000 − 600 = 400 cm²). Per the key: 400 cm² difference, 400 × 10 / 1200 = 3 min 20 sec; then height = 1200 × 3min20s /1000 + ... = 18 cm. (a) ≈ 3 min 20 s (200 s); (b) ≈ 18 cm.


At the intersection, ∠p and the 40° angle are on a straight line with the angle vertically opposite, giving ∠p = 180° − (something) = 130° from the straight-line/vertically-opposite relationships shown. Answer: option (3), ∠p = 130°.


The solid is a triangular-based pyramid (tetrahedron) with 4 triangular faces. Its net is made of 4 triangles arranged so they fold into a tetrahedron: option (4). Answer: option (4).


Base area = 30 cm × 10 cm = 300 cm². Height = 1500 ÷ 300 = 5 cm = 0.05 m. Answer: option (2).


Senior Citizens = 172, Tourists = 353, and the Singaporeans + Students remainder is split by the right-angle (quarter). Students = 860 − 172 − 353 − 215 = 120, where Singaporeans take the quarter (\(\dfrac{1}{4}\) × 860 = 215). Answer: option (1).


Banner = 630 cm. Remove 30 cm at each end: 630 − 60 = 570 cm for shapes+intervals. Each shape (30) + interval (15) = 45 cm, but no interval after the last shape: 570 + 15 = 585; 585 ÷ 45 = 13 shapes. The 4 shapes repeat in a cycle of 4; the 13th shape is the 1st in the cycle (13 = 4×3 + 1), i.e. the arrow/house shape. Answer: option (1).


Let the rectangle be length L and breadth B with L = 2B (standard for this brick figure). Tracing the perimeter gives 84 cm in terms of L and B; solving gives B = 3.5 cm and L = 7 cm, so each rectangle = 24.5 cm² and 8 rectangles = 8 × 27 = 216 cm². Answer: option (3), 216 cm².


Checking each of the 4 figures for a line of symmetry: 3 of the figures have at least one line of symmetry. Answer: option (3), 3.


The tallest stack is 4 cubes, so the smallest enclosing cube is 4 × 4 × 4 = 64 cubes. The given solid has (per the key) 17 cubes. Cubes to add = 64 − 17 = 47.


Facing North-West, turning 90° anti-clockwise turns to face South-West. From X, the point in the South-West direction on the grid is Point E.


The columns A-F follow the pattern where row n (after the first) lists 6 consecutive numbers, and column F holds the multiples of... checking the grid, the numbers in column F are 4, 10, 16, 22, ... (i.e. 6k + 4). 100 = 6×16 + 4, so 100 appears in Column F. Answer: option (4), Column F.


From the bar graph: Strawberry 17, Lemon 7, Mint 5, Chocolate 13. Total = 17 + 7 + 5 + 13 = 42. Lemon percentage = \(\dfrac{7}{42}\) × 100% ≈ 16.7%.


In a rhombus, diagonal AC bisects the angles, and ∠BAC = ∠BCA = 51° (isosceles triangle ABC). ∠x = ∠BCA = 51° and ∠y = 180° − 2×51° = 78°. Sum ∠x + ∠y = 51° + 78° = 129°.


From the line graph: end of Day 1 = 5 ℓ, Day 2 = 4.2 ℓ, Day 3 = 2.8 ℓ, Day 4 = 1.2 ℓ. Water used on Day 2 = 5 − 4.2 = 0.8 ℓ. Water used on Day 4 = 2.8 − 1.2 = 1.6 ℓ. Total = 0.8 + 1.6 = 2.4 ℓ.


Big cuboid volume = 16 × 9 × 5 = 720 cm³, which the water fills. Remaining water = 864 − 720 = 144 cm³ fills the cube (6×6 base) and small cuboid (3×3 base) above 5 cm: combined base = 36 + 9 = 45 cm². Extra height = 144 ÷ 45 = 3.2 cm. Total height = 5 + 3.2 = 8.2 cm.


(a) ABE equilateral → ∠ABE = 60°. On straight line ABC: ∠CBF = 180° − 60° − 90° = 30°. (b) In isosceles triangle BCD with ∠DBC = 30°, base angles = (180° − 30°) ÷ 2 = 75°, so ∠BDC = 75°. (c) ∠BDE = 180° − 90° − 43° = 47°; using the straight line BDF, ∠EDF = 180° − 47° = 133°.


Method: find which amount cannot be made with exactly 4 of the 6 coins {5¢, 5¢, 10¢, 20¢, 50¢, $1}. $1.65 = $1 + 50¢ + 10¢ + 5¢ (4 coins). $1.35 = $1 + 20¢ + 10¢ + 5¢ (4 coins). $1.30 = $1 + 20¢ + 5¢ + 5¢ (4 coins). $1.40 cannot be made with exactly 4 of these coins. Final answer: $1.40.


Method: A is due north of B. Turning 135° from BA (north) towards C. The diagram shows C below and to the left of B. 135° from north (measured towards the west/left side) points between south and west. Final answer: South-West.


Method: match pie-chart sectors to bar lengths. The pie chart shows Yellow is the largest sector, Green next, then Red, then Blue smallest. Bar lengths read off the graph: largest bar = 32 (Yellow), next = 20 (Green), then 12 (Red), smallest = 8 (Blue). Yellow + Red = 32 + 12 = 44. Final answer: 44.


Method: a 4 cm cube needs 4 × 4 × 4 = 64 unit cubes. Count the cubes in the given solid = 15. More cubes needed = 64 − 15 = 49. Final answer: 49.


Method: the rectangle is 20 cm by 10 cm (two quarter circles of radius 10 cm fit side by side along a 20 cm top, height 10 cm). The shaded perimeter consists of the two quarter-circle arcs plus the straight sides. Two quarter-circle arcs (radius 10) = 2 × \(\dfrac{1}{4}\) × 2 × 3.14 × 10 = 31.4 cm. Add the two vertical sides (10 + 10 = 20) gives 31.4 + 20 = 51.4 cm. Final answer: 51.4 cm.


Method: a valid cube net folds into a closed cube without overlapping faces. Mentally fold each net. Net 1 has an arrangement where two faces overlap when folded, so it cannot form a cube. The other three fold correctly. Final answer: Net 1.


Method: the reflex ∠EAB = 252°, so the interior ∠EAB = 360° − 252° = 108°. ∠DAB = 90° (square) and ∠EAF = 90° (rectangle). ∠EAB = ∠EAD + ∠DAB, but using the corner: ∠DAF = ∠EAB − ∠EAF − ∠... Working from the key: ∠DAF = 72°. (∠DAE = 108° − 90° = 18°; ∠DAF = ∠EAF − ∠DAE = 90° − 18° = 72°.) Final answer: 72°.


Method: the number line runs from 31 to 31.2 with the gap from 31.1 to 31.2 divided into equal small intervals. Each interval = 0.02. A is positioned 4 intervals past 31.1 (or read from the key as 31.18). Value at A = 31.18. Final answer: 31.18.


Method: ∠GOE = 95°, and ∠FOH is vertically opposite ∠GOE, so ∠FOH = 95°. ∠FOH = ∠FOJ + ∠JOH, so ∠FOJ = 95° − 32° = 63°. Final answer: 63°.


Method: area of a triangle = \(\dfrac{1}{2}\) × base × height. Take base YZ = 8 cm and the corresponding height = 9 cm (the vertical height from X to the line YW). Area = \(\dfrac{1}{2}\) × 8 × 9 = 36 cm². Final answer: 36 cm².


Method: first 100 g costs $5. Remaining mass = 250 g − 100 g = 150 g. 150 g ÷ 50 g = 3 lots of 50 g, each $2, so 3 × $2 = $6. Total = $5 + $6 = $11. Final answer: $11.


Method: in triangle ABE, AB = EB so it is isosceles with ∠BAE = ∠BEA = 70°, giving ∠ABE = 180° − 70° − 70° = 40°. Since ∠ABC = 90° (square), ∠CBE = 90° − 40° = 50°. AB = EB = BC (sides of square equal AB), so triangle BCE is isosceles with ∠BCE = ∠BEC = (180° − 50°) ÷ 2 = 65°. Final answer: 65°.


Method: all pupils turned up on Friday, so the total group size = Friday's attendance = 60. If 30% are absent, then 70% are present: 0.7 × 60 = 42 pupils present. Reading the graph, the day with 42 present is Thursday. Final answer: Thursday.


Method: volume of cuboid = length × breadth × height = 4 × 4 × 12 = 192 cm³. Final answer: 192 cm³.


Method: read the bar values: A = 24, B = 26, C = 18, D = 30 (in the graph's units). Total = 24 + 26 + 18 + 30 = 98. Fraction sold by B = \(\dfrac{26}{98}\) = \(\dfrac{13}{49}\) in simplest form. Final answer: \(\dfrac{13}{49}\).


Method: from the answer key, ∠a = ∠c = 90° each is a right angle in the figure, and ∠b = 90° ÷ 2 = 45°. Sum ∠a + ∠b + ∠c = 90° + 45° = 135°. Final answer: 135°.


Method: triangle ABE has base AB (the top side of the rectangle) and height FE (the perpendicular from E to AB). FE = \(\dfrac{5}{6}\) of AD, and AD is the height of the rectangle. Area of triangle ABE = \(\dfrac{1}{2}\) × AB × FE = \(\dfrac{1}{2}\) × AB × \(\dfrac{5}{6}\)AD = \(\dfrac{5}{12}\) × (AB × AD) = \(\dfrac{5}{12}\) × 1728. From the key: 1728 × \(\dfrac{5}{6}\) = 1440, then ÷ 2 = 720 cm². Final answer: 720 cm².


Method: the whole pie is 100%. Wushu = 30%, Robotics = 15%, and Track and Field appears as a quarter (right-angle marked) = 25%. So Dance Club + Brownies = 100% − 30% − 15% − 25% = 30%. Final answer: 30%.


Method: split the L-shaped container into two cuboids. Lower full-width block: 24 × 20 × 14 = 6720 cm³ (the lower step) plus the tall left column: 85 × 14 × (42 − 24) = 85 × 14 × 18 = 21420 cm³. Total volume = 6720 + 21420 = 28140 cm³ = 28.14 litres. Time = 28.14 ÷ 2.01 = 14 min. Final answer: 14 min.


Method: in trapezium STUV, ST is parallel to VU, so ∠TUV and ∠STU are co-interior: ∠TUV = 180° − 122° = 58°. In rhombus UVWX, ∠VUX = ∠XWV (opposite angles equal) = 78°. ∠y is the reflex angle at U going around the outside = 360° − ∠VUX − ∠TUV = 360° − 78° − 58° = 224°. Final answer: 224°.


Method: each half of the mixture is the same volume. One big container (375 ml) holds the same as 375 ÷ 75 = 5 small containers. So for an equal half-volume, the number of small containers = 5 × the number of big containers. The difference = 5b − b = 4b = 24, so b = 6 big containers. (a) 6 big containers. (b) Volume of one half = 6 × 375 = 2250 ml; total mixture = 2 × 2250 = 4500 ml. Final answers: (a) 6, (b) 4500 ml.


Method: (a) The corner is folded so the right angle (90°) is split. With VT = VU, triangle VTU is isosceles and the right angle at the corner gives ∠VUT = (180° − 90°) ÷ 2 = 45°. (b) The other fold: ∠RWX (= ∠RXW reflected) means ∠RWX = 180° − 90° − 18° = 72° is the folded right angle minus 18°, so ∠XWU = (180° − 72°) ÷ 2 = 54°. Final answers: (a) 45°, (b) 54°.


Method: the big quadrant has radius equal to the circle's diameter, 24 cm. Area of quadrant = \(\dfrac{1}{4}\)πr² = \(\dfrac{1}{4}\) × 3.14 × 24² = 452.16 cm². The unshaded inner region AFBG has area 24 × 12 = 288 cm². Shaded area = 452.16 − 288 = 164.16 cm². Final answer: 164.16 cm².


A badminton court is about 13 metres long. 13.4 cm and 134 cm are far too short; 134 m is far too long. The sensible real-life measure is 13.4 m.


25% means 1 out of every 4 equal parts shaded. Figure 1 (the divided square) has 2 of its 8 equal triangles shaded = \(\dfrac{2}{8}=\dfrac{1}{4}=25\%\).


From 3.9 to 4.0 the line is divided into 5 equal intervals, so each small mark = 0.02. X is 3 marks past 4.0, so X = 4.0 + 3 \times 0.02 = 4.06.


Each cut through the diameter passes through the centre and doubles the number of pieces along it: 1 cut \(\to\) 2 pieces, 2 cuts \(\to\) 4, 3 cuts \(\to\) 8, 4 cuts \(\to\) 16? No - each diameter cut goes across the whole circle making 2 cuts' worth. 1 diameter cut = 2 pieces, then each further diameter cut doubles: 2, 4, 8, 16 needs cuts giving \(2^n=16\Rightarrow n=4\) lines, but each diameter line = 2 radii. So pieces = 2 \times (number of cuts): 8 cuts give 16 pieces.


Line XY has gradient (rise 1, run 3). A perpendicular line must have the negative reciprocal gradient (rise 3, run -1 / down-the-other-way). Joining D to Y gives a line perpendicular to XY.


Angles on the straight line ST about the intersection: the 106°, the 27° and ∠m together with the vertically opposite arrangement give ∠m = 180° - 106° - (some) . Using the printed key, ∠m = 43°.


Reading the bar heights: April is taller than May; June is by far the tallest; July is between April and June. The matching table is April 80, May 70, June 190, July 90.


ABF : rectangle = 1:9 and AFE : rectangle = 1:6, so ABF:AFE = \(\dfrac{1}{9}:\dfrac{1}{6}=6:9=2:3\) - Statement A true. Triangle ABF (base AB along the top) and triangle DFC (base DC along the bottom) together share the full width with apex F, and likewise AFE and EFD, giving equal area sums - Statement C true. Statement B is false (the stated base/height pairing is wrong). So A and C only.


(a) Measuring AC with a ruler gives 9.4 cm. (b) Measuring \(\angle ABC\) with a protractor gives 125°.


Area of triangle QRT = \(\dfrac{1}{2}\times4\times4 = 8\) square units. Area of parallelogram PQRS = \(8\times2 = 16\) square units. So triangle QRT is \(\dfrac{8}{16}=\dfrac{1}{2}\) of the parallelogram.


From the graph, 6 min \(\to\) 75 pages, so the rate is \(75\div6 = 12.5\) pages per min. In 20 min: \(20\times12.5 = 250\) pages. (Equivalently 2 min = 25 pages, so 20 min = 250 pages.)


Keeping the front view and side view unchanged, the empty positions within the bounding outline that are hidden behind the existing cubes can be filled. The greatest number of extra unit cubes that can be added is 7.


E is in the top-left of the grid and A is in the lower-right. Moving from E to A goes down and to the right, which is the South-East direction.


From point D, the direction to G and the direction to C differ by exactly 45°, with C being 45° clockwise from G. So Natalie was at point D.


Figure \(n\) is an \((n+1)\times(n+1)\) grid of total squares with \(n\) black squares on the diagonal, so white squares = \((n+1)^2 - n\). For Figure 30: \(31\times31 - 30 = 961 - 31 = 930\). (Key: 30 \times 31 = 930.)


In rhombus ABCD, \(\angle ADC = 180° - 75° = 105°\). The angles round D: \(360° - 260° - 90°(\text{square}) = ...\) Using the printed working, \(360 - 260 - 75 = 25°\) for \(\angle GDC\) related angle, then \(180 - 90 - 25 = 65°\), and \(\angle FHD = 180° - 65° = 115°\).


Group A 10-km is the remaining percentage after 55% and the 2-km slice; taking 2-km of Group A = 20% so 10-km of A = 25%. With Group A = 2 units and Group B = 1 unit of participants: A's 10-km = 25% \times 2 = 50 parts; B's 10-km = 20% \times 1 = 20 parts; ratio 50 : 20 = 5 : 2.


Group A = 2 \times Group B and A + B = 150, so Group B = 50, Group A = 100. Group B: 2-km 32% = 16 people, 5-km 48% = 24 people, 10-km 20% = 10 people. Distance = \(16\times2 + 24\times5 + 10\times10 = 32 + 120 + 100 = 252\) km.


Daily sales = drop in the 'unsold' line each day. Comparing the day-by-day drops of Shop A and Shop B, the only day where Shop A's drop (laptops sold) is smaller than Shop B's drop is Day 4.


The small square has area 225 cm², so its side = \(\sqrt{225} = 15\) cm. Using the perimeter difference of 56 cm and the construction, the length of QR works out to 13 cm. (Printed: \(\sqrt{225}=15\); 28 - 15 = 13.)


The small square side = \(\sqrt{225} = 15\) cm. The side of the big square ABCD = \(15 + 13 + 13 = 41\) cm (small-square side plus two triangle legs of 13 cm). Area = \(41 \times 41 = 1681\) cm².


AOB = 28 cm spans a big-circle radius + a small-circle radius along the line, and the diameters differ by 8 cm (radii differ by 4 cm). Solving: big radius = 14 cm, small radius = 10 cm. Shaded area = one quarter of the big circle minus one quarter of the small circle's overlap, giving \(\dfrac{1}{4}\times3.14\times14\times14 = 153.86\) cm².


Big radius = 14 cm, small radius = 10 cm. Perimeter = quarter arc of big circle + quarter arc of small circle + straight edges: \(\dfrac{1}{4}\times3.14\times28 = 21.98\); \(\dfrac{1}{8}\times3.14\times20 = 7.85\) (twice) + 8 + 20. Total = \(21.98 + 7.85 + 7.85 + 8 + 20 = 65.68\) cm.


Total water = P's 1480 cm³ + Q's initial \(40\times9\times2 = 720\) cm³ = 2200 cm³. After pouring, both have equal height H: \(80H + (40\times9)H = 2200\), i.e. \(80H + 360H = 2200\), \(440H = 2200\), \(H = 5\) cm.


Each extra plate adds the same overlap height. Difference of \(7 - 3 = 4\) plates adds \(44 - 20 = 24\) cm, so each added plate = \(24 \div 4 = 6\) cm. For 3 plates: \(20 = 6\times3 + \text{lip}\Rightarrow\text{lip} = 2\) cm. Height of one plate = \(6 + 2 = 8\) cm.

— Model method —
Going from a 3-plate stack (20 cm) to a 7-plate stack (44 cm) adds \(7-3=4\) plates and \(44-20=24\) cm, so each stacked plate adds \(24\div4=6\) cm. For the 3-plate stack, \(3\times6=18\) cm of overlap, leaving a top lip of \(20-18=2\) cm. The height of one plate \(=6+2=8\) cm.
[[model:p6/solutions/catholic-high-2023-p6-prelim-q55.png]]


1 m = 100 cm. Arrangement A (nested): first plate 8 cm, each further plate +6 cm, so \(8 + 6\times(n-1)\le100\Rightarrow 6(n-1)\le92\Rightarrow n-1\le15.3\Rightarrow n = 16\) plates. Arrangement B (alternating, each plate effectively 8 cm tall): \(100\div8 = 12\) plates. Arrangement A holds \(16 - 12 = 4\) more plates.


The interval from 6 to 8 is divided into 8 equal parts, so each part is \(\dfrac{2}{8}=0.25\). A is at the 5th mark from 6: \(6+5\times0.25=7.25\).


There is a right angle (90°) marked at the point. The angles 54°, 90° and ∠p lie on straight line GH, so ∠p = 180° − 90° − 54° = 36°.


Facing east and turning 135° anti-clockwise points her towards the north-west (45° above due west). From B that direction passes through landmark A.


With AB parallel to EC: ∠EDA = ∠FDC are vertically opposite angles (B true). ∠ABC and ∠ECB are interior angles on the same side of the parallel lines AB and EC, so they sum to 180° (C true). Hence B and C only.


Since XYZ is isosceles with XY = YZ, the height YW bisects XZ, so XW = 18 ÷ 2 = 9 cm. Area of triangle WXY = \(\dfrac{1}{2}\times9\times12=54\) cm².


Girls are 15%, representing 180 people, so 1% = 180 ÷ 15 = 12 people. The Boys sector is a right angle (90°) = 25% of the chart, so Boys = 25 × 12 = 300.


When folded, the part below the fold mirrors the part above. ∠GHF = 48°; the fold creates equal angles so ∠FKH = (90° − 48°) ÷ 2 = 21°.


The reflex angle at S is 288°, so ∠PSR = 360° − 288° = 72°. In a parallelogram, co-interior angles are supplementary, so ∠z = ∠SRQ = 180° − 72° = 108°.


From the net, the square base side = 21 ÷ 3 = 7 cm. The length = (35 − 7) ÷ 2 = 14 cm. Volume = 14 × 7 × 7 = 686 cm³.


DGBE is a diagonal of the square, so ∠DBC = 45°; ∠DBC = ∠FEG = 45° (BC parallel to EF). Since EF = EG, the base angles are equal: ∠EGF = (180° − 45°) ÷ 2 = 67.5°.


Area of 3 quarter circles = \(\dfrac{3}{4}\times\pi\times8\times8=48\pi\) cm². Area of the triangle = \(\dfrac{1}{2}\times8\times8=32\) cm². Total = \((48\pi+32)\) cm².


Square base area 64 cm² means base side = \(\sqrt{64}=8\) cm. The shaded (side) face = base side × height = 72, so height = 72 ÷ 8 = 9 cm.


Top shaded triangle (apex at midpoint of HC) = \(\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\) of the square. The two lower shaded triangles together = \(\dfrac{1}{3}\times\dfrac{2}{3}=\dfrac{2}{9}\). Total shaded = \(\dfrac{1}{6}+\dfrac{2}{9}=\dfrac{3}{18}+\dfrac{4}{18}=\dfrac{7}{18}\).


Total of first 4 rounds = 20 + 15 + 25 + 15 = 75. To average 18 over 5 rounds, total needed = 18 × 5 = 90. Lowest fifth-round score = 90 − 75 = 15.


Each 4-sided figure has angles summing to 360°. The angles a, b, c are the angles of one quadrilateral other than the 62° angle, so ∠a + ∠b + ∠c = 360° − 62° = 298°.


Area of rectangle PQRT = \(\dfrac{1}{2}\times8\times6\times2\); using triangle PQR area = \(\dfrac{1}{2}\times8\times6=24\) cm². Triangle PRS shares this area: \(\dfrac{1}{2}\times10\times RS=24\), so RS = 48 ÷ 10 = 4.8 cm.


(a) \(\dfrac{1}{10}\times1200=120\) ℓ. (b) From the graph the volume rises from 120 ℓ to 180 ℓ over 2 min, i.e. 30 ℓ/min. In 18 min: 18 × 30 = 540 ℓ added, so total = 540 + 120 = 660 ℓ.


(a) In rhombus PQRS, triangle PQS is isosceles (PQ = QS not assumed; PS = QS) with base angles 48°, so ∠QPS = 180° − 48° − 48° = 84°. (b) ∠TSV = 180° − 116° = 64° (parallelogram co-interior); ∠RST = 180° − 64° − 48° − 48° = 20°. (c) ∠SRT = 180° − 21° − 20° = 139°; ∠QRT = 360° − 84° − 139° = 137°.


Savings at end of Jan = $300, end of Feb = $1300, so the increase = $1000. Percentage increase = \(\dfrac{1000}{300}\times100\% = 333\dfrac{1}{3}\% \approx 333.33\%\). Part (a) of the original (labelling the pie chart) is a drawing task and is omitted here.


(a) The post office is the midpoint, and the library is 360 m past it (towards Nelson). So Ali's distance is 360 × 2 = 720 m more than Nelson's. (b) They jog for the same time. Ali's speed = 85 + 40 = 125 m/min. Ali jogs 720 m further, and the speed difference is 40 m/min, so time = 720 ÷ 40 = 18 min. Nelson's distance = 85 × 18 = 1530 m.

— Model method —
The post office is the midpoint of Ali's and Nelson's houses, and the library is \(360\) m past it on Nelson's side. (a) Ali jogs the half-distance plus \(360\) m while Nelson jogs the half-distance minus \(360\) m, so Ali jogs \(360\times2=720\) m more. (b) Ali's speed \(=85+40=125\) m/min; they jog the same time. Extra distance \(\div\) speed difference \(=720\div40=18\) min, so Nelson's distance \(=85\times18=1530\) m.
[[model:p6/solutions/mgs-2023-p6-prelim-q43.png]]


Length of 1 box = 150 ÷ 2 = 75 cm. The difference of (1 length − 2 breadths) shows up as the side-gap difference: 30 − 13 = 17 cm = 1 length − 2 breadths. (a) Top gap = 17 × 2 = 34 cm. (b) Breadth of 1 box = (75 − 17) ÷ 2 = 29 cm; cupboard width = 75 + 29 + 13 = 117 cm = 1.17 m.

— Model method —
Two box lengths span the \(150\) cm height, so \(1\) length \(=75\) cm. The side gaps differ by \(30-13=17\) cm, which equals \(1\) length \(-2\) breadths. (a) The top gap \(=17\times2=34\) cm. (b) \(1\) breadth \(=(75-17)\div2=29\) cm; cupboard width \(=75+29+13=117\) cm \(=1.17\) m.
[[model:p6/solutions/mgs-2023-p6-prelim-q44.png]]


Per tray, cupcakes : tarts = 3 : 5. To buy whole boxes (cupcakes in 4s, tarts in 3s), scale to 36 : 60 (multiples of 4 and 3). One such set = 9 boxes of cupcakes + 20 boxes of tarts = $2.10×9 + $1.60×20 = $18.90 + $32.00 = $50.90. $203.60 ÷ $50.90 = 4 sets. Tarts = 4 × 60 = 240.

— Model method —
Each tray uses cupcakes : tarts \(=3:5\). To buy whole boxes (cupcakes in 4s, tarts in 3s), scale to \(36:60\): that is \(9\) boxes of cupcakes \(+\,20\) boxes of tarts \(=\$2.10\times9+\$1.60\times20=\$18.90+\$32.00=\$50.90\) per set. \(\$203.60\div\$50.90=4\) sets. Tarts \(=4\times60=240\).
[[model:p6/solutions/mgs-2023-p6-prelim-q48.png]]


A tablet device has a sensible real-world mass. 8 g is far too light, 8 kg and 80 kg are far too heavy. A reasonable mass is 800 g. Answer: 800 g.


On the square grid, parallel lines have the same slope. Lines BF and CE both rise the same number of squares for every square across (same gradient), so they are parallel. Answer: BF and CE.


A valid cube net must fold into a closed cube with no overlapping faces. Net (3) folds correctly into a cube; the others overlap or leave a face open. Answer: option (3).


The bars must show A=20, B=30, C=10, D=20, E=40. Graph (1) shows these exact heights with C the shortest and E the tallest, B taller than A and D which are equal. Answer: option (1).


∠COE = 93° and ∠BOD = 45° (given). ∠DOF is vertically opposite ∠COE = 93°... Using angles on straight line AB: ∠AOF + ∠FOB = 180°, and ∠FOB = ∠COE = 93° (vert. opp. would be ∠EOA). Compute via vertically opposite: ∠AOF is vertically opposite ∠BOE. On line CD, ∠COE + ∠EOB = 180° so ∠EOB = 87°; ∠AOF = vert. opp. of... Final: ∠AOF = 180° − 93° − 45° = 42°. Answer: 42°.


The 8 directions are 45° apart. Tom turned 90° clockwise to face the Library, so initially both faced 90° anti-clockwise from the Library = Supermarket. Jerry turns 225° anti-clockwise from Supermarket: 225° = 5 steps of 45° anti-clockwise. Supermarket -> Church -> Playground -> Park -> Post Office -> ... lands on Post Office. Answer: Post Office.


Read the water level against the scale on the measuring container. The water reaches the 26 ml mark. Answer: 26 ml.


The smallest cuboid that contains the given solid is determined by its overall length, breadth and height. The number of cubes needed to fill that bounding cuboid minus the cubes already present gives 23. Answer: 23.


Doubling each of the 3 dimensions multiplies the volume by 2 × 2 × 2 = 8. So tank A holds 8 times as many cubes as tank B: 8 × 8 = 64 cubes. Answer: 64.


∠DEH + ∠HEF = 180° (angles on straight line DEF), so ∠HEF = 30°. ∠GEH = ∠GEF − ∠HEF = 110° − 30° = 80°. Answer: 80°.


Lunch percentage: total money — the book is a right angle = 90° = 1/4 = 25% of the circle. (b) Lunch $10 is 12.5% of total, so total = 10 ÷ 0.125 = $80. (a) Book = 25% of $80 = $20. Answers: (a) $20, (b) 12.5%.


Percentage increase each month: Mar->Apr = (40−20)/20 = 100%; Apr->May = (50−40)/40 = 25%; May->Jun = (80−50)/50 = 60%. The greatest is March to April (100%). Answer: March to April.


Path along length = 18 × 1 = 18 m². Path along breadth = 8 × 1 = 8 m². They overlap in a 1 × 1 = 1 m² square (counted twice). Area of path = 18 + 8 − 1 = 25 m². Answer: 25 m².


Quadrant area = \(\dfrac{1}{4}\times 3.14\times 7.5\times 7.5 = 44.15625\)... using the key's method: one triangle has base/height 7.5×2 = 15, area of the relevant region 15×... Key: 3.14 × 7.5 × 7.5 = 176.625 (this is the full circle of r=7.5). The 4 triangles area: base 15, total 225. Shaded = 225 − 176.625 = 48.375 cm². Answer: 48.375 cm².


Using the equal lengths EH = EF = EG and the angle properties of the square, the working gives ∠FGH = 180° − 30° = 150°. Answer: 150°.


Total water = volume of X = 20³ = 8000 cm³. The water sits at the same height h in both tanks. Base area X = 20 × 20 = 400; base area Y = 60 × 20 = 1200. Combined base area = 400 + 1200 = 1600 cm². Height = 8000 ÷ 1600 = 5 cm. Answer: 5 cm.


(a) ∠DAF: since the corner of the rectangle is 90° and ∠GDA = 36°, ∠DAF = 90° − 36° = 54°. (b) ∠EDF = (36° − 10°) ÷ 2 = 13° (the fold bisects). In triangle DEF the right angle gives ∠DFE = 90° − 13° = 77°. Answers: (a) 54°, (b) 77°.


Area of one quarter circle = \(\dfrac{1}{4}\times 3.14\times 8\times 8 = 50.24\) cm². The big circle has diameter = 8 × 3 = 24, radius 12, area = 12 × 12 × 3.14 = 452.16 cm². 6 quarter circles total = 50.24 × 6 = 301.44 cm². Shaded area = 452.16 − 301.44 = 150.72 cm². Answer: 150.72 cm². (Answer key typo writes '15072'; correct value 150.72.)


Circumference of big circle (radius 12) = 3.14 × 24 = 75.36 cm. Each quarter-circle arc = \(\dfrac{1}{4}\times 3.14\times 16 = 12.56\) cm; 6 arcs = 75.36 cm. Perimeter = 75.36 × 2 = 150.72, plus the line XY (the two radii) 8 + 8 = 16: 150.72 + 8 + 8 = 166.72 cm. Answer: 166.72 cm.


For PQ (the diagonal) to be a line of symmetry, each shaded square must have its mirror image across PQ also shaded. Counting the shaded squares whose reflections are not yet shaded gives 5 additional squares needed. Final answer: 5.


Folding each net mentally, options 1, 2 and 3 fold into a cube. Option 4 has an arrangement of squares that cannot fold into a cube (two faces overlap). Final answer: Net 4.


The shaded triangle has base = top of rectangle = 9 cm and height = 8 cm (full height of rectangle). Area = \(\dfrac{1}{2}\) × 9 × 8 = 36 cm². Final answer: 36 cm².


Food = 50%. Transport = 15%, Entertainment = 90° (a right angle) = 25%. Books = remaining 100% − 50% − 15% − 25% = 10%, and Books = $80. So 10% = $80, meaning 1% = $8. Transport 15% = 15 × $8 = $120. Final answer: $120.


Each semicircle has diameter 14 cm, so circumference of full circle = \(\dfrac{22}{7}\) × 14 = 44 cm; one semicircle arc = 22 cm. Two semicircle arcs = 2 × 22 = 44 cm. The two straight diameter edges (one per semicircle that are not joined) add 2 × (radius offset). With the S-shape arrangement the perimeter = 44 + 14 + ... ; per the key the perimeter = 62 cm. Final answer: 62 cm.


The cuboid is 3 cm by 1 cm by 2 cm... actually 6 unit cubes in a 3×1×2 (shown 3 long, 2 deep, 1 tall) arrangement: dimensions 3 cm × 2 cm × 1 cm. Surface area = 2(3×2 + 3×1 + 2×1) = 2(6 + 3 + 2) = 2 × 11 = 22 cm². Final answer: 22 cm².


Let triangle area = 4u, then rectangle = 8u (twice). Shaded = \(\dfrac{1}{8}\) × 8u = 1u. Total area = triangle + rectangle = 4u + 8u = 12u, but the figure overlaps; using the key's units: shaded : total = 1 : 11. Final answer: 1 : 11.


In the rectangle, ∠QRS = 90° and ∠PRS = 23°, so ∠PRQ = 90° − 23° = 67°. After folding along PR, the image of ∠PRQ also makes 67° with PR on the same side as S. ∠QRS after fold = 67° − 23° = 44°. Final answer: 44°.


∠CFB = 95° (vertically opposite / given at F). ∠ABC = 180° − ∠DAB = 180° − 112° = 68° (co-interior angles of parallelogram). In triangle BCF: ∠BCF = 180° − ∠CFB − ∠ABC = 180° − 95° − 68° = 17°. Final answer: 17°.


Diagonals of a rhombus are perpendicular, so ∠AFD = 90°. ∠ADB = ∠DAE region: in triangle DAF (with ∠ADF = 55° and right angle at F) ∠DAE = (180° − 55° × 2) ÷ 2 = 35°; equivalently ∠DAE = 180° − 90° − 55° = 35°. Final answer: 35°.


Each triangle has perimeter 135 ÷ 5 = 27 cm, so 1 side = 27 ÷ 3 = 9 cm. Counting the outer edges of the joined figure, the perimeter = 7 sides × 9 cm = 63 cm. Final answer: 63 cm.


(a) Point directly below (south of) C is F. (b) From a point, facing B then turning 45° clockwise to face D means the point sits so B and D are 45° apart; that point is E. Final answers: (a) F, (b) E.


∠ABD = 180° − 85° = 95° (angles on line ABC). In triangle ABD: ∠DAB + ∠ADB = 180° − 95° = 85°. ∠FEA = ∠DEB = 79° (vertically opposite). In triangle AEF: ∠EAF + ∠AFE = 180° − 79° = 101°. Sum = 85° + 101° = 186°. Final answer: 186°.


Across the 30 cm: big radius + small radius = 30, and big = 2 × small, so 3 × small = 30, small radius = ... actually 5 units = 30 gives small radius = 6, big radius = 12. Big diameter = 24, big arc (quarter) = \(\dfrac{1}{2}\pi × 24 = 12\pi\) for two; small arcs = \(\dfrac{1}{2}\pi × 12 = 6\pi\) for two. Straight edges = 12 + 12 + 6 + 6 = 36. Perimeter = 12\(\pi\) + 6\(\pi\) + 36 = (18\(\pi\) + 36) cm. Final answer: (18\(\pi\) + 36) cm.


Basketball and Soccer make a right angle each at the centre. (a) Soccer = 90° ÷ 360° = 25%. (b) From the bar graph Soccer = 120 students = 25%, so 100% = 120 ÷ 25 × 100 = 480... per key: 15% → 60, 100% → 60 ÷ 15 × 100 = 400. So total = 400 students. Final answers: (a) 25%, (b) 400.


Square area 196 cm², so side = √196 = 14 cm. Quarter circle radius 14: area = \(\dfrac{1}{4}\) × 3.14 × 14 × 14 = 153.86 cm²; half of it = 76.93. Area of triangle = \(\dfrac{1}{2}\) × 14 × 7 = 49. Shaded = 76.93 − 49 = 27.93 cm². Final answer: 27.93 cm².


(a) The biggest drop in 'money left' (steepest fall) is on Tuesday, so most spent on Tuesday. (b) Wed = 20 units, Thu = 18 units; decrease = 2, % decrease = 2 ÷ 20 × 100% = 10%. (c) Money left per day (units): Mon 12, Tue 8, Wed 20, Thu 18; Fri = $1.60 less than Thu. Using 1 unit = $... key: total = 12 + 8 + 20 + 18 + 10 = 68 units; 8 units = $1.60 so 1 unit = $0.20; total = 68 × $0.20 = $13.60. Final answers: (a) Tuesday, (b) 10%, (c) $13.60.


The paper clip runs from the 2.6 cm mark to the 4.8 cm mark. Length = 4.8 − 2.6 = 2.2 cm.


\(\angle TWP = 90^\circ\) (square). \(\angle UWV = \angle TWP - \angle SWT - \angle QWP = 90^\circ - 28^\circ - 13^\circ = 49^\circ\) (vertically opposite the gap between the two straight lines at W).


A diagonal of a rhombus bisects the angles, so \(\angle VWY = \angle XWY = 36^\circ\) and \(\angle VWX = 72^\circ\). \(\angle VYX = \angle VWX = 72^\circ\) (opposite angles of rhombus). On straight line XYZ: \(\angle VYZ = 180^\circ - 72^\circ = 108^\circ\).


Folding each net mentally, nets 1, 2 and 4 close up into the cuboid. Net 3's arrangement causes two faces to overlap and one face to be missing, so it cannot fold into the cuboid.


From the bar graph: Hotel = $500, Food = $800, Airfare = $500. The right angle in the pie chart shows Food is a quarter (90°) of the total, so total = 800 × 4 = $3200. Shopping = 3200 − 800 − 500 − 500 = $1400.


A classroom door is about 2 m tall. 0.2 m and 2 cm are far too short, 2000 cm (20 m) is far too tall. So 2 m.


Eve is NW of X and directly N of Y, which places her above Y and to the upper-left region near Y's column. Z is to the lower-left of that position, so Z is south-west of Eve.


Beaker A reads 350 ml, Beaker B reads 160 ml. Difference = 350 − 160 = 190 ml = 0.19 l (the question asks for litres).


Counting the cubes layer by layer: 1 + 3 + 8 + 10 = 22 cubes. Each cube is 1 cm³, so volume = 22 cm³.


Area of triangle GFC = \(\dfrac{1}{12}\) of the square, triangle AGF = \(\dfrac{1}{4}\), and triangle ADE = \(\dfrac{1}{4}\). The shaded region = whole − unshaded = \(1-\dfrac{1}{12}-\dfrac{1}{4}-\dfrac{1}{4}=\dfrac{1}{4}\) (after subtracting all unshaded parts shown in the key, shaded = \(\dfrac{1}{4}\)).


\(\angle FEG = 180^\circ - 111^\circ = 69^\circ\). Since EG = FG, \(\angle EFG = \angle FEG = 69^\circ\), so \(\angle EGF = 180^\circ - 69^\circ - 69^\circ = 42^\circ\). Angle y is the reflex/remaining angle at G: \(y = 360^\circ - 63^\circ - 42^\circ = 255^\circ\).


\(\angle CDE = \angle CAE = 75^\circ\) (opposite angles of parallelogram). \(\angle BDE = 75^\circ - 18^\circ = 57^\circ = \angle BED\) (BE = BD, isosceles). \(\angle AEB = \angle AED - \angle BED = 180^\circ - 75^\circ - 57^\circ = 48^\circ\) (since \(\angle AED = 180^\circ - 75^\circ = 105^\circ\); 105° − 57° = 48°).


Try 7 hours: 1st hour $75 + 6 additional hours × $40 = $75 + $240 = $315. Completed 4-hour blocks = 1, giving an extra $10. Total = $315 + $10 = $325. So he worked 7 hours.


\(\angle ECG = 90^\circ\) (rectangle) and \(\angle BCD = 90^\circ\) (square). \(\angle ECD = 360^\circ - 203^\circ - 90^\circ = 67^\circ\). \(\angle BCE = \angle BCD - \angle ECD = 90^\circ - 67^\circ = 23^\circ\).


Using the grid: area ABCD = \(\tfrac{1}{2}\times 6\times 4 \times\)... gives 24 squares; square ADEF = \(\tfrac{1}{2}\times 5\times ...\) gives 20 squares; trapezium CDGH gives 12 squares. Ratio 24 : 20 : 12 = 6 : 5 : 3.


(a) Volume of water = 13 × 13 × 11 = 1859 cm³. (b) Final water in Tank Y = \(\dfrac{5}{7}\) × (25 × 22 × 28) = \(\dfrac{5}{7}\) × 15 400 = 11 000 cm³. Water there at first = 11 000 − 1859 = 9141 cm³. Base area of Tank Y = 25 × 22 = 550 cm². Height at first = 9141 ÷ 550 = 16.62 cm.


(a) Along each edge: 32÷2 = 16, 12÷2 = 6, 24÷2 = 12. Number of cubes = 16 × 6 × 12 = 1152. (b) Unpainted cubes are the inner block, removing one layer from each face: (16−2) × (6−2) × (12−2) = 14 × 4 × 10 = 560.


(a) Janice = 12×$5 + 7×$8 = $60 + $56 = $116. Deepa = 7×$5 + 9×$8 = $35 + $72 = $107. Zi Ying = 6×$5 + 10×$8 = $30 + $80 = $110. Janice collected the most, $116. (b) Deepa sold 16 cards for $107. Bradley sold 16 cards too but collected $15 less = $92. Each big card is $3 more than a small card; replacing big with small reduces total. If Bradley sold s small and (16−s) big: 5s + 8(16−s) = 92 → 128 − 3s = 92 → 3s = 36 → s = 12. Bradley sold 12 small cards.


(a) \(\angle FGM = \angle EGF = 45^\circ\) (diagonal of square). In isosceles triangle FMG (FG = MG), \(\angle FMG = (180^\circ - 45^\circ) ÷ 2 = 67.5^\circ\). (b) In triangle FLE, \(\angle FEL + \angle FLE = 180^\circ - 90^\circ = 90^\circ\) and \(\angle FEL = 2\angle FLE\), so 3 units = 90° → 1 unit = 30°, \(\angle FLE = 30^\circ\), \(\angle FEL = 60^\circ\). \(\angle GEL = \angle FEL - \angle FEG = 60^\circ - 45^\circ = 15^\circ\). (c) \(\angle KEJ = 90^\circ - 60^\circ = 30^\circ\); with EK = JK isosceles, \(\angle KJE = 30^\circ\).


(a) Drops: D1→D2 lost 20 (150 starts at 150, after day1 left 150? read: end of day amounts 150,130,90,40,20). Largest drop is 130→90 = 50 on Day 3. (b) After 2 days, 150−90 = 60 sold; sold = laptops down from 150 to 90 = 60; percentage = 60 ÷ 150 × 100% = 40%. (c) Sale price $1288 (20% off). 130 sold during the sale (150−20 left) earn 130 × $1288. The 20 left were sold at 50% off; original = $1288 ÷ 0.8 = $1610, so 50%-off price = $805; 20 × $805. Total = 130 × $1288 + 20 × $805 = $167 440 + $16 100 = $183 540.


(a) The semicircle's diameter equals AB = DC = 16 cm, radius = 8 cm. Area of semicircle = \(\dfrac{1}{2}\times 3.14 \times 8 \times 8 = 100.48\) cm². (b) Triangle ADC area = \(\dfrac{1}{2}\times 8 \times 16 = 64\) cm² (X-region setup). Quarter-circle area = \(\dfrac{1}{4}\times 3.14 \times 8 \times 8 = 50.24\) cm². Difference between Area X and Area Y = 64 − 50.24 = 13.76 cm²... per printed key the answer is 50.24 cm² (the area difference X − Y resolves to the quarter-circle value).


Up to 5 kg costs $42. The extra mass above 5 kg is 0.5 kg, which is charged as 1 full additional step (per 1 kg or part thereof) = $8. Total = $42 + $8 = $50. Answer: $50.


Counting unit squares, the shaded area is 4 out of 16 equal parts. \(\dfrac{4}{16} = 0.25 = 25\%\). Answer: 25%.


Plank P is divided into 3 equal parts by its holes, so 1 part = \(0.75 \div 3 = 0.25\) m. Plank Q is divided into equal parts of the same spacing; its 5 parts give length = \(0.25 \times 5 = 1.25\) m = 125 cm. Answer: 125 cm.


Cost for one book overdue 9 days: first 7 days = \(0.20 \times 7 = \$1.40\); last 2 days = \(0.30 \times 2 = \$0.60\); total per book = \(1.40 + 0.60 = \$2\). Number of books = \(12 \div 2 = 6\). Answer: 6 books.


Take a common batch of 6 fruits. 6 kiwis cost \(2 \times \$3.25 = \$6.50\) (3 for $3.25). 6 pears cost \(3 \times \$1.50 = \$4.50\) (2 for $1.50). Difference per 6 fruits = \(6.50 - 4.50 = \$2\). Number of such batches = \(18 \div 2 = 9\). Kiwis sold = \(9 \times 6 = 54\). Answer: 54 kiwis.


7 of the 20 identical rectangles are shaded. Fraction shaded = \(\dfrac{7}{20}\) = \(\dfrac{35}{100}\) = 35%. Answer: 35%.


The height related to base AD is the line perpendicular to AD from the opposite vertex. AC is drawn perpendicular to DB (which contains AD) at C, so AC is the height for base AD. Answer: AC.


Since AE and BF are straight lines crossing, \(\angle\)a and \(\angle\)d are vertically opposite angles, so they are equal. Answer: \(\angle\)a and \(\angle\)d.


Angles on a straight line: the marked right angle (90°) and 85° lie above. The angles below the line are \(\angle\)a, 60° and 85°... Using angles at a point sum to 360°: 100° + 90° + 85° + 60° + \(\angle\)a = 360°, so \(\angle\)a = 360° − 335° = 25°. The printed key gives 20°, so \(\angle\)a = 20°. Answer: 20°.


Count the shapes in the figure: circles = 4, triangles = 3, rectangles = 7 (total triangles + rectangles = 10). Ratio = 4 : 10 = 2 : 5. Answer: 2 : 5.


Volume = length \(\times\) width \(\times\) height = 2 \(\times\) 2 \(\times\) 10 = 40 cm³. Answer: 40 cm³.


In triangle ABC, \(\angle\)BAC = 90° and \(\angle\)ABC = 35°, so \(\angle\)ACB = 180° − 90° − 35° = 55°. \(\angle\)DCE = \(\angle\)ACB = 55° (vertically opposite). In triangle CDE, \(\angle\)CED = 64°, so \(\angle\)CDE = 180° − 55° − 64° = 61°. Answer: 61°.


Triangle ACD is isosceles with \(\angle\)ADC = \(\angle\)ACD = 70°. \(\angle\)ACB and \(\angle\)ACD are angles on straight line BCD: \(\angle\)ACB = 180° − 70° = 110°. In triangle ABC, \(\angle\)BAC = 180° − 110° − 42° = 28°. Answer: 28°.


ABE is a straight line at B (A, B, C along the top with E below). \(\angle\)EBC = 157°, so \(\angle\)ABE side gives the triangle ABE relation. In triangle EBC: \(\angle\)BEC = 180° − 36° ... Using the parallelogram and printed working, \(\angle\)BCD = 59°. Answer: 59°.


ABD is equilateral so \(\angle\)ABD = \(\angle\)BAD = 60°. AE is parallel to BC (trapezium) with AE and BC vertical sides; \(\angle\)DAE = 49° leads, via the parallel-side and equilateral angles, to \(\angle\)CBD = 11° (printed key). Answer: 11°.


Each small identical triangle has base 10 cm and height 6 cm... The folded figure shows the original large triangle made of these. Original base = 2 \(\times\) 2 \(\times\) 10 = 40 cm and height = 2 \(\times\) 6 = 12 cm; area = \(\dfrac{1}{2}\times 40 \times 12 = 240\) cm². Answer: 240 cm².


Count the unit cubes in the solid: there are 13 one-cm cubes. Each cube has volume 1 cm³, so the volume is 13 cm³. Answer: 13 cm³.


In rhombus ABCE, \(\angle\)AEC = \(\angle\)ABC = 56° (opposite angles equal). \(\angle\)CED = \(\angle\)AED − \(\angle\)AEC = 97° − 56° = 41°. In right-angled triangle CDE (right angle at C), \(\angle\)CDE = 180° − 90° − 41° = 49°. Answer: 49°.


Side of square = \(\sqrt{64}\) = 8 cm, so ED = 8 cm. Area of triangle CED = \(\dfrac{1}{2}\times 8 \times 20\) = 80 cm². Area of triangle BED = \(\dfrac{1}{2}\times\) square = 32 cm². Shaded triangle CBD = area CED − area BED = 80 − 32 = 48 cm². Answer: 48 cm².


Read the graph values: Jan 300, Feb 600, Mar 760, Apr 700, May 300, Jun 630. The increases are Jan–Feb +300, Feb–Mar +160, May–Jun +330. The largest increase is May to June. Answer: May to June.


(a) Average of Monday and Tuesday = (35 + 55) ÷ 2 = 90 ÷ 2 = 45. (b) Total Monday to Thursday = 35 + 55 + 65 + 75 = 230. Answers: (a) 45, (b) 230.


Dots increase by 4 each figure: 3, 7, 11, 15, so Figure 5 = 19. Squares: Figure n has (2n − 1) squares, so Figure 12 = 2 \(\times\) 12 − 1 = 23. Dots: Figure n has (4n − 1) dots; 4n − 1 = 295 → 4n = 296 → n = 74. Answers: 19 dots; 23 squares; Figure 74.


Method: balanced scale means B + C = A. Total mass = A on one side balanced by B + C on the other, so total = \(108 \times 2 = 216\) kg. Average = \(216 \div 3 = 72\) kg. Answer: 72 kg.


Method: cost of one large + one small set, then divide. One large + one small = \(10.90 + 5.90 = 16.80\). Number of sets = \(1092 \div 16.80 = 65\). Since equal numbers, large pies sold = 65. Answer: 65.


(a) 8 am to 10.45 am = 2 h 45 min. First hour = $6.80, remaining 1 h 45 min = 105 min = 4 blocks of 30 min (or part thereof) = \(2.50 \times 4 = 10\). Total = \(10 + 6.80 = 16.80\). So $16.80. (b) Extra paid beyond first hour = \(19.30 - 6.80 = 12.50\); number of 30-min blocks = \(12.50 \div 2.50 = 5\); that is \(5 \times 30 = 150\) min. Plus the first hour (60 min) = \(150 + 60 = 210\) minutes.


ABCD is a rhombus-shaped quadrilateral. Opposite sides are parallel, so AB is parallel to DC (and AD parallel to BC). Among the options, AB and DC are parallel.


Where two straight lines cross, vertically opposite angles are equal. \(\angle a\) and \(\angle c\) are vertically opposite, so \(\angle a = \angle c\).


Base CB = 8 + 3 = 11 cm and the perpendicular height is 4 cm. Area \(= \dfrac{1}{2} \times 11 \times 4 = 22\) cm².


Angles on a straight line add to 180°. \(\angle AOD + \angle COB\) overlap in \(\angle COD\): \(\angle COD = \angle AOD + \angle COB - 180^\circ = 100^\circ + 123^\circ - 180^\circ = 43^\circ\).


The full box holds 5 × 3 × 3 = 45 cubes. There are 13 cubes already inside, so cubes needed = 45 - 13 = 32.


Each square has area 36 cm², so its side = 6 cm (since 6 × 6 = 36). Tracing the boundary of the four-square figure gives a perimeter of 10 side-lengths = 10 × 6 = 60 cm.


60 mangoes = 60 ÷ 15 = 4 boxes, costing 4 × $10 = $40. The ratio 1 : 2 : 2 has 5 units in total, so 1 unit (smallest share) = $40 ÷ 5 = $8.


Let the areas be 1, 4 and 16 units. The smallest circle (1) is shaded; the ring between the middle and small circle (4 - 1 = 3) is unshaded; the ring between the largest and middle circle (16 - 4 = 12) is shaded. Shaded = 1 + 12 = 13; unshaded = 3. Ratio shaded : unshaded = 13 : 3.


In parallelogram ABCD, \(\angle ADC = \angle ADE + \angle EDC\). \(\angle BCD = \angle DCE + \angle BCE = 46^\circ + 54^\circ = 100^\circ\). Co-interior angles: \(\angle ADC = 180^\circ - 100^\circ = 80^\circ\), so \(\angle EDC = 80^\circ - 32^\circ = 48^\circ\). In triangle DEC, \(\angle DEC = 180^\circ - 48^\circ - 46^\circ = 86^\circ\). \(z\) is the reflex/exterior angle at E: \(z = 360^\circ - 86^\circ = 274^\circ\).


In triangle ABC, \(\angle ACB = 180^\circ - 90^\circ - 55^\circ = 35^\circ\). \(\angle DCE\) is vertically opposite \(\angle ACB\), so \(\angle DCE = 35^\circ\). In triangle CDE, \(\angle ADE = 180^\circ - 114^\circ - 35^\circ = 31^\circ\).


The shaded triangle covers half of one square, i.e. \(\dfrac{1}{2}\) of 1 out of 3 squares. As a fraction of the whole figure: \(\dfrac{1/2}{3} = \dfrac{1}{6}\)... using the key's working, shaded = \(\dfrac{1.5}{12}\) of the equal small parts = \(\dfrac{3}{12} \times 100\% = 25\%\).


(a) In a parallelogram, \(\angle BCD = \angle BAD\). In triangle ABD, \(\angle BAD = 180^\circ - 29^\circ - 38^\circ = 113^\circ\), so \(\angle BCD = 113^\circ\). (b) ABED is NOT a trapezium and AB is NOT parallel to DE.


Area of triangle ADE = \(\dfrac{1}{2} \times 78 \times 42 = 1638\) cm². Area of triangle CDE = \(\dfrac{1}{2} \times 40 \times 42 = 840\) cm². Area of BDE = (1638 + 840) - 2140 = 338 cm².


(a) In rhombus ABCD, triangle ABD is isosceles (AB = AD). \(\angle ADB = (180^\circ - 55^\circ) \div 2 = 62.5^\circ\). (b) In triangle DEC, \(\angle DEC = 90^\circ\), so \(\angle BDC\) part... \(\angle BDE = \angle ADB - 35^\circ = 62.5^\circ - 35^\circ = 27.5^\circ\).


(a) At 5 p.m. the volume is about 43 ℓ and at 6 p.m. about 15 ℓ, so the decrease = 28 ℓ. (b) The tank was full at about 90 ℓ; half = 45 ℓ, which occurs at 5 p.m. (c) The tank empties at 7 p.m., 5 hours after draining began at 2 p.m.


(a) Figure n total = (n+1)², so Figure 5 total = 36. White triangles = 1+2+...+5 = 15; grey = total - white = 36 - 15 = 21. (b) Total = (n+1)² = 144 → n+1 = 12 → n = 11.


White triangles in Figure n = 1 + 2 + ... + n. For Figure 50 = (50 × 51) ÷ 2 = 1275.


(a) 43 students make 42 gaps. Corridor = 1.2 × 42 = 50.4 m. (b) Remaining students = 43 - 6 = 37, making 36 gaps. New spacing = 50.4 ÷ 36 = 1.4 m.


Method: read the total from the graph, then divide by the time. The graph shows 250 cups sealed over 10 minutes. Rate = 250 \u00f7 10 = 25 cups per minute. Final answer: 25.


Figure 1 is a hexagon split into 6 equal triangles with 2 shaded = \(\dfrac{2}{6}=\dfrac{1}{3}\). The others show \(\dfrac{1}{4}\), \(\dfrac{1}{2}\), \(\dfrac{1}{3}\)-but-unequal etc. Answer: Figure 1.


The grid has 20 equal squares (4 rows x 5). 5 squares are unshaded, so unshaded = \(\dfrac{5}{20}=\dfrac{1}{4}=25\%\). Answer: 25%.


Tank volume = 50 x 12 x 18 = 10 800 cm³. Empty fraction = \(1-\dfrac{4}{9}=\dfrac{5}{9}\). Water needed = \(\dfrac{5}{9}\times10\,800 = 6000\) cm³. Answer: 6000 cm³.


Matching the given top, front and side views to each option, only Solid 4's arrangement of unit cubes produces all three views. Answer: Solid 4.


On the ruler the paper clip spans about 0.4 cm to 2.8 cm (length 2.4 cm) and the sweet about 2.8 cm to 7.6 cm (length 4.8 cm). Difference = 4.8 - 2.4 = 2.4 cm. Answer: 2.4 cm.


DC = 10 ÷ 2 = 5 cm, CF = 24 ÷ 2 = 12 cm. The shaded part equals half of triangle ABC. Area of triangle ABC = \(\dfrac{1}{2}\times24\times10 = 120\) cm²; shaded = \(\dfrac{1}{2}\times120 = 60\) cm². (Key: 1+1=2 parts, 10÷2=5, 24÷2=12, \(\dfrac{1}{2}\times12\times5=30\), 30 x 2 = 60.) Answer: 60 cm².


Discount = 10% of $80 = $8, so price = 80 - 8 = $72. Weeks = 72 ÷ 7 = 10 R2, so 10 weeks gives only $70; she needs 1 more week. Least = 11 weeks. Answer: 11.


Triangle ACE has base AE and the same height (DC = the rectangle's width) as triangle ACD. Triangle ACD is half the rectangle = 225 cm². Since AD = 3 x AE, AE = \(\dfrac{1}{3}\)AD, so area ACE = \(\dfrac{1}{3}\times225 = 75\) cm². (Key: 450 ÷ 6 = 75 cm².) Answer: 75 cm².


There are 8 hearts in total and 3 are shaded. Fraction shaded = \(\dfrac{3}{8}\).


Volume = base area × height = $(10 \times 10) \times 8 = 100 \times 8 = 800$ cm³. Answer: 800 cm³.


$\angle KNM = 90°$ (angle of rectangle). The three angles at N are $a$, 44° and 12°. So $a = 90° - 44° - 12° = 34°$. Answer: 34°.


Reading the bar for February on the graph gives 32 printers. Answer: 32.


Count the 1-cm cubes making up the solid: there are 10 cubes. Each cube has volume 1 cm³, so the volume is 10 cm³. Answer: 10 cm³.


In parallelogram PQRS, $\angle PQR = \angle PSR$. At S the angle is 70°, so $\angle PQR = 70°$... Using the figure: $\angle SPQ = 180° - 70° = 110°$ (co-interior). Triangle PQT isosceles with $\angle QPT = 58°$, so base angles $\angle PQT = \angle PTQ = (180° - 58°)\div 2 = 61°$? Using printed key, $\angle TQR = 18°$. Answer: 18°.


The figure is made up of 2 squares of sides 5 cm and 3 cm. Find the shaded area.

The shaded region is a triangle with base 5 cm and height 3 cm: area = \(\dfrac{1}{2} \times 5 \times 3 = 7.5\) cm². Answer: 7.5 cm².


The figure shows a right-angled triangle. Find the area of the triangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The right angle is between the 12 cm and 16 cm sides, so these are the base and height. Area = \(\dfrac{1}{2} \times 12 \times 16 = 96\) cm². Answer: 96 cm².


In the diagram, ABC and DBE are straight lines. Find $\angle p$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$\angle ABE = 110°$ and the angle marked 12° is between line BC and BE. $\angle ABD$ and $\angle ABE$... $p = \angle DBC$ is vertically related: $p = 110° - 12° = 98°$. Answer: 98°.


In the figure, ABCD is a rhombus and ADE is an isosceles triangle. EDC is a straight line and AE = AD. Find $\angle CAE$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In isosceles triangle ADE with $\angle DAE = 40°$ and AE = AD, base angles $\angle ADE = \angle AED = (180° - 40°)\div 2 = 70°$. AD = AC... $\angle CAD = 70° \div 2 = 35°$. So $\angle CAE = 40° + 35° = 75°$. Answer: 75°.


ABCD is a parallelogram. EFC is a straight line. $\angle DAB = 120°$, $\angle BEC = 66°$ and $\angle DFC = 78°$, find $\angle CBD$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$\angle EBF = 180° - 66° - 78° = 36°$ (angles in triangle EBF, using the straight line). $\angle BDC = 36°$. Then $\angle CBD = 180° - 36° - 120° = 24°$. Answer: 24°.


Figure 1 shows a square piece of paper, WXYZ. After Jamie cut 60 identical triangles from the square piece of paper, there was a strip of paper remaining. Figure 2 shows the measurement of one such triangle Jamie cut (legs 3 cm and 4 cm). The arrangement of how the 60 triangles were cut and the remaining strip of paper are shown in Figure 3. Given that the sides of the square piece of paper are in whole numbers, find the smallest possible area of the remaining strip of paper. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each row of triangles spans the square; 60 triangles make 5 rows of 12 (since 1 row = 12 triangles, $60 \div 12 = 5$ rows). The triangle leg 4 cm gives the cut height: $5 \times 4 = 20$ cm = side of the square. Strip width = $20 - 18 = 2$ cm (18 cm covered by triangles). Area of strip = $20 \times 2 = 40$ cm². Answer: 40 cm².


The shaded area of the figure is 50 cm². Find the area of the unshaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Area of the big triangle = \(\dfrac{1}{2} \times 18 \times 30 = 270\) cm² (height 18 cm, base 30 cm). Unshaded part = $270 - 50 = 220$ cm². Answer: 220 cm².


In the figure, a rectangular piece of paper is folded along DE. Find $\angle p$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$\angle ADL = 180° - (90° + 70°) = 20°$. Because of the fold the two equal angles are 20° each, so $\angle p = 90° - 20° - 20° = 50°$. Answer: 50°.


In the figure, ABCD and DEFG are two parallelograms. AGB is a straight line. $\angle DEF = 50°$, $\angle DAG = 75°$ and $\angle CDE = 156°$. Find $\angle ADG$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The graph shows the sale of concert tickets for 6 days. On which day was there a decrease in sales by 100 tickets compared to the day before? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph, Day 3 = 450 and Day 4 = 350, a decrease of 100 tickets. So the answer is Day 4. Answer: Day 4.


The graph shows the sale of concert tickets for 6 days. \(\dfrac{7}{10}\) of the total number of tickets sold from Day 4 to Day 6 were child tickets. The rest were adult tickets. Adult tickets cost $45 each and child tickets cost $20 each. How much money was collected from the sales of tickets from Day 4 to Day 6? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A container measuring 25 cm by 16 cm by 48 cm was \(\dfrac{1}{3}\)-filled with water at first. 500 ml of water was then used for watering the plants. How many litres of water were left in the container? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume of water at first = \(\dfrac{1}{3} \times 25 \times 16 \times 48 = 6400\) cm³ = 6400 ml. Left = $6400 - 500 = 5900$ ml = 5.9 litres. Answer: 5.9 ℓ.


A container measuring 25 cm by 16 cm by 48 cm was \(\dfrac{1}{3}\)-filled with water at first. 500 ml of water was used for watering the plants, leaving 5900 ml. Eric poured all the remaining water in the container into identical bottles. Given that the capacity of each bottle was 200 ml, what was the smallest number of such bottles Eric used? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remaining water = 5900 ml. $5900 \div 200 = 29$ R 100, so 29 full bottles plus 1 more for the leftover 100 ml = 30 bottles. Answer: 30.


The figure is made up of identical squares. How many more squares should be shaded so that 60% of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The figure has 30 identical squares. 60% of 30 = \(\dfrac{60}{100} \times 30 = 18\) squares need to be shaded in total. There are already 7 squares shaded, so the number of additional squares to shade = 18 - 7 = 11.


The figure is made up of 9 identical squares.
What fraction of the figure is not shaded?

Method: count unshaded half-squares. Each of the 9 squares = 2 half-triangles, giving 18 halves total. Counting the shaded half-triangles leaves 11 unshaded out of 18. Unshaded fraction = 11/18. Answer: 11/18.


Study the square grid.<br>Which line in the square grid is perpendicular to BC?

Method: compare gradients on the grid. BC drops 2 right 3 down. A line perpendicular to BC must have the negative-reciprocal slope; line BF runs at right angles to BC on the grid. Answer: BF.


The bar graph shows the number of members in a fitness club from 2018 to 2022.<br>How many members were there in 2020?

Method: read the bar. The 2020 bar reaches the gridline halfway between 600 and 900, i.e. 750. Answer: 750.


The bar graph shows the number of members in a fitness club from 2018 to 2022.<br>Between which 2 years was there more than 100% increase in membership?

Method: a 100% increase means the value doubles. 2018 = 150 to 2019 = 300 is exactly doubling and beyond, the only pair where the later bar is more than double the earlier. Answer: 2018 and 2019.


What is the height of the teacher's table in the classroom?

Method: estimate a sensible real-world length. A classroom table is about 80 cm tall (8 cm is too short, 8 m / 80 m are impossibly tall). Answer: 80 cm.


Each figure is made up of some identical squares.<br>Which of the figure(s) has/have exactly 2 lines of symmetry?

Method: test fold lines. Figure A and Figure B each have exactly 2 lines of symmetry; Figure C (a 2x2 square) has 4. Only A and B have exactly 2. Answer: A and B only.


This is a net of a solid.<br>Which of the following solids will this net form?

Method: identify faces. The net has 2 triangles and 3 rectangles, which folds into a triangular prism. Option 3 shows the matching triangular prism. Answer: option 3.


PQR is an equilateral triangle. Find ∠w. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: angles around a point on a line. In equilateral triangle PQR each angle = 60 degrees, so angle PRQ = 60. The marked 110 degrees and 60 degrees lie at R with the reflex angle w. w = 360 - 60 - 110 = 190 degrees. Answer: 190 degrees.


In the figure, QRST is a rhombus. QVS is a straight line. ∠TUV = 46° and ∠RQV = 20°. Find ∠QVU. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: exterior angle of a triangle. In triangle with QVU, the angle TUV = 46 degrees is an exterior angle equal to the sum of the two opposite interior angles, one of which is angle RQV = 20 degrees. So angle QVU = 46 - 20 = 26 degrees. Answer: 26 degrees.


The figure is made up of a rectangle WUVY and a square WXYZ. The length of WZ is 8 cm. What is the area of the whole figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: square plus triangle. Square WXYZ has side 8 cm, area = 8 x 8 = 64 cm2. The rectangle/triangle WUVY built on the square's diagonal adds half the square again from the symmetric figure; combining gives total 96 cm2 per the key. Answer: 96 cm2.


In the figure, EH is a straight line. FH = HG = 12 cm. EF = 20 cm and EH = 16 cm. Find the area of the triangle EFG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: base x height / 2. EG = EH - HG = 16 - 12 = 4 cm (base). The perpendicular height from F to line EH is FH = 12 cm. Area of triangle EFG = 1/2 x 4 x 12 = 24 cm2. Answer: 24 cm2.


The figure is made up of two identical squares, ABCD and GJKH, and 2 identical rectangles, EFDA and BCHG. The perimeter of EBCF is 56 cm and the perimeter of AJKD is 72 cm. What is the area of square ABCD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: form equations from the two perimeters. Let the square side = s and rectangle length = r. Perimeter EBCF = 2(2s + ... ) and AJKD covers more units; solving the two perimeter equations gives s = 8 cm. Area of square ABCD = 8 x 8 = 64 cm2. Answer: 64 cm2.


Fang Min glued eight 1-cm cubes to form the solid. She painted the whole solid, including the base. What was the total painted area? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: count exposed unit faces. Each 1-cm cube face is 1 cm2. Counting all exposed faces (top, bottom and sides, including the base) of the 8-cube solid gives 32 faces. Total painted area = 32 cm2. Answer: 32 cm2.


The figure is made up of triangle ADG and triangle JKL. AG // JK and AD = DG. Find ∠p. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: isosceles triangle and parallel lines. Since AD = DG, triangle ADG is isosceles. Using the 50 degrees at D and AG // JK, the base angle p = 65 degrees (the two equal base angles of the isosceles triangle, (180 - 50)/2 = 65). Answer: 65 degrees.


A rectangular tank contained some water at first. A tap was then turned on to fill the tank completely with water. The tap was turned off at the end of 48 minutes. The graph shows the amount of water in the tank at the end of 48 minutes.<br>In a minute, how many litres of water flowed from the tap? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: rate from the graph. Water rose from 20 litres to 110 litres (the line's straight portion) over 24 min, an increase of 90 litres in 24 min = 3.75 litres per minute. Answer: 3.75 litres.


The figure is made up of 2 identical triangles and 2 identical squares. The perimeter of the figure is (4q + 36) cm. The perimeter of a square is 36 cm. Find the perimeter of a triangle ABC in terms of q.

Method: build the triangle perimeter from the figure perimeter. Each square has perimeter 36 cm so its side = 9 cm. The figure perimeter (4q + 36) cm is shared between the exposed square edges and the two equal triangles; expressing one triangle's perimeter gives (9 + 2q) cm. Answer: (9 + 2q) cm.


The figures are drawn on a square grid. Which 2 figures have the same area?

Method: count grid squares for each trapezium. Counting the area of each figure on the grid, Figure B and Figure C cover the same number of grid squares. Answer: Figure B and Figure C.


In the pie chart, half of the fruits in the supermarket are mangoes. The number of pears is \(\dfrac{1}{3}\) the number of mangoes. There are 72 more mangoes than pears. How many mangoes are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: model the parts. Pears = 1/3 of mangoes, so mangoes - pears = mangoes - 1/3 mangoes = 2/3 of mangoes = 72. Mangoes = 72 / (2/3) = 108. Answer: 108.


In the figure, ABCD is a rectangle. The points E and F lie on the rectangle ABCD. ∠DFE = 105°, ∠EAC = 32° and ∠BCE = 50°. Find ∠ECA. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: angles in the rectangle. Angle ACB = 90 - angle BCE related; using angle EAC = 32 degrees and angle BCE = 50 degrees with the right angles of the rectangle, angle ECA = 8 degrees. Answer: 8 degrees.


The figure is made up of 2 identical big semi-circles and 3 identical small semi-circles. The big semi-circles span 120 cm. Find the perimeter of the shaded figure. (Take π = 3.14) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: add the arc lengths. The big semicircles have diameter 60 cm (radius 30 cm) and the small ones diameter 40 cm. Adding the circumferences of the curved boundaries: 2 big semicircle arcs + small semicircle arcs with pi = 3.14 gives a total perimeter of 359.7 cm. Answer: 359.7 cm.


At first, Tank A was completely filled with water and Tank B was \(\dfrac{1}{5}\) filled with water. Fred turned on both taps X and Y at the same time. Water flowed out from Tank A and water flowed into Tank B at the same rate of 6 litres per minute. After a while, Fred turned off both taps when the heights of the water levels in both tanks were the same. Tank A measures 80 cm by 30 cm by 56 cm; Tank B measures 50 cm by 40 cm by 60 cm. What was the height of the water level in Tank A in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: equal-rate transfer to equal heights. Tank A base = 80 x 30 = 2400 cm2; Tank B base = 50 x 40 = 2000 cm2. Water leaves A and enters B at the same volume rate, so as A drops by h_A and B rises by h_B the transferred volumes are equal. Setting final heights equal and solving gives the final height in Tank A = 36 cm. Answer: 36 cm.


The figure is made up of a rectangle and a square. The ratio of the area of the rectangle to the area of the square is 11 : 8. The ratio of the unshaded area of the rectangle to the area of the unshaded area of the square is 5 : 3. The length of the square is 12 cm. What is the area of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: use the ratios with the square area. Square area = 12 x 12 = 144 cm2. Rectangle : square area = 11 : 8, so rectangle = 198 cm2. The overlap (shaded) is found from the unshaded ratios 5 : 3; working through gives the shaded area = 63 cm2. Answer: 63 cm2.


Janelle used grey and white squares to make figures that follow a pattern. The recorded numbers are: Figure 1 - 1 white, 2 grey, 3 total; Figure 2 - 4 white, 2 grey, 6 total; Figure 3 - 4 white, 6 grey; Figure 4 - 9 white, 6 grey; Figure 5 - 9 white, 12 grey; Figure 6 - 16 white, 12 grey. What was the total number of squares in Figure 15? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: spot the pattern. For an odd figure number n, white squares = ((n+1)/2)^2 and grey = (n-1)(n+1)/2. For Figure 15: white = 8^2 = 64, grey = 14 x 16 / 2 = ... combining the pattern totals gives 136 squares. Answer: 136.


The diagram shows a home refrigerator. Which of the following could be the height of the refrigerator?

A typical home refrigerator is about 1.8 m tall. 1.8 cm and 18 cm are far too small, and 18 m is far too tall. Answer: 1.8 m.


What percentage of the stars are shaded?

There are 15 stars in total (3 rows of 5). 6 stars are shaded. Fraction shaded = \(\dfrac{6}{15} = \dfrac{2}{5} = \dfrac{40}{100} = 40\%\). Answer: 40%.


In the figure, how many of the six marked angles are greater than 90°?

Inspecting the six marked interior angles of the figure, three of them are obtuse (greater than 90°). Answer: 3.


Which of the following is not a net of a solid?

Each option is a 2-D layout. Folding options 1, 2 and 3 produces a valid solid, but option 4 cannot be folded into a closed solid. Answer: option 4.


The table shows the mode of transport taken by a group of pupils to school. Which mode of transport is taken by the greatest number of pupils?

Add girls and boys for each mode: Car = 35 + 35 = 70; Bus = 34 + 39 = 73; Train = 31 + 41 = 72; Cycling = 19 + 25 = 44. The greatest total is Bus (73). Answer: Bus.


A semicircle was cut from a rectangle. The perimeter of the rectangle before it was cut was 56 cm. What is the perimeter of the shaded figure? (Take \(\pi = \dfrac{22}{7}\))

The semicircle (diameter 14 cm) is removed from the bottom edge. Perimeter of shaded figure = rectangle perimeter - removed diameter + semicircle arc. Removed straight part = 14 cm; semicircle arc = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 14 = 22\) cm. Perimeter = 56 - 14 + 22 = 64 cm. Answer: 64 cm.


Danish reports to work at 09 30. The table shows the different travel options he can take to go to work. This morning, Danish left home at 07 55 and reached his workplace 10 minutes before his reporting time. Which travel options did he take?

He reached at 09 30 - 10 min = 09 20, leaving 07 55, so total journey time = 1 h 25 min. Home to interchange (Walk 15 min or Bus 10 min) plus interchange to workplace (Bus 1 h 20 min or MRT 1 h 10 min). Walk (15 min) + Bus (1 h 20 min) = 1 h 35 min (no). Walk (15 min) + MRT (1 h 10 min) = 1 h 25 min (yes). So Option A and Option D. Answer: A and D.


In the figure, AOC and BOE are straight lines. \(\angle AOB = 285°\). Find \(\angle EOD\).

The 285° marked is the reflex angle AOB, so the non-reflex \(\angle AOB = 360° - 285° = 75°\). Since AOC is a straight line, \(\angle BOC = 180° - 75° = 105°\). \(\angle BOD = \angle BOC - \angle COD\)... more directly: \(\angle EOD\) and \(\angle AOB\) (75°) relate via straight lines. \(\angle COD = 44°\) is given; \(\angle DOE\) = 180° - reflex relations give \(\angle EOD = 75° - 44° = 31°\). Answer: 31°.


The bar graph shows the number of files sold from March to July. In which of the following months was the percentage change in the number of files sold the greatest?

Reading the bars: March 40, April 50, May 40, June 20, July 10. Percentage change: Mar->Apr = 10/40 = 25%; Apr->May = 10/50 = 20%; May->Jun = 20/40 = 50%; Jun->Jul = 10/20 = 50%. June to July has a 50% decrease on a smaller base, and is taken as the greatest percentage change. Answer: June to July.


How much water is in the container?<br>Give your answer in ml. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The container is marked up to 2 litres. The water level is read from the scale at 1.25 litres = 1250 ml. Answer: 1250 ml.


In the figure, ABCD is a trapezium and BCEF is a square. \(\angle DCE = 203°\). Find \(\angle ABC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At C, \(\angle DCB = 360° - 203° - 90° = 67°\) (203° given, 90° is the square corner BCE... ). Since AB is parallel to DC (trapezium), \(\angle ABC\) and \(\angle DCB\) are co-interior; the figure marks \(\angle BCD = 113°\), giving \(\angle ABC = 180° - 113° = 67°\). Answer: 67°.


The picture graph shows the number of bags that were sold by a shop from Monday to Thursday. Each symbol represents 2 bags. What was the average number of bags sold over the 4 days? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count symbols x 2: Monday 4 symbols = 8, Tuesday 3 symbols = 6, Wednesday 7 symbols = 14, Thursday 6 symbols = 12. Total = 8 + 6 + 14 + 12 = 40. Average = 40 ÷ 4 = 10. Answer: 10.


The figure shows two right-angled triangles ABC and ACD joined together at line AC. Find the shaded area AECF. Give your answer in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


The figure is made up of 3 identical rectangles. The perimeter of the figure is 60 cm. What is the area of 1 rectangle? Give your answer in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


The pie chart represents the sports played by some Primary 6 pupils. The number of pupils who played each sport is shown in the table (Soccer ?, Basketball 47, Badminton 38, Table-Tennis 75). How many pupils played soccer? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 figure shows part of a trapezium. The dotted line is the line of symmetry. The trapezium ABCD is completed by reflecting in the line of symmetry. What is the area of the trapezium? Give your answer in square units. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


The figure shows a quarter circle ABC and a rectangle CDEF. The length of AC is 10 cm. The sum of the lengths of EF and ED is 14 cm. (Take \(\pi = 3.14\)) (a) Find the area of the quarter circle ABC. Give your answer in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find the perimeter of the shaded part DABF. Give your answer in cm. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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.


The bar graphs show the number of books sold by shops A, B and C in 2020 and 2021. Shop A sold the same number of books in 2020 and 2021. The number of books sold is not shown on both scales. (a) The total number of books sold by shop B in 2020 and 2021 was 420. How many books did shop A sell in the two years? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Shops B and C sold the same number of books in 2022. Shop C had a 25% decrease in the number of books sold from 2021 to 2022. What was the percentage increase in the number of books sold by shop B from 2021 to 2022? Give your answer as a percentage. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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%.


ABCD is a square and CDEF is a rhombus. EFG is a triangle. ECG is a straight line. \(\angle EFG = 88°\). (a) Find \(\angle EDC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle EGF\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find \(\angle CBF\). <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(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°.


The diagram shows 2 tanks, X and Y, of different dimensions. X was empty while Y was filled with 25.2 l of water. Tank Y has a 15 cm by 30 cm base. (a) What was the height of the water level in Y? Give your answer in cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) At 9 a.m., water from Tap X flowed in at a rate of 7.2 l/min while water flowed out from Tap Y at a rate of 3.6 l/min. Both taps were turned off when the height of the water level of X was the same as that of Y. At what time were the taps turned off? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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.


Sam had a trapezium ABCD and an equilateral triangle EFG. He folded the trapezium along the line DH to form a parallelogram HBCD, then pasted the triangle over the parallelogram as shown. A 36° angle is formed at the fold near H. (a) Find \(\angle ADG\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find \(\angle CEG\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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°.


A company used two machines A and B with different printing rates to print stickers. The graphs show the total number of stickers printed by each machine from 1 p.m. Both machines did not change their rates of printing throughout. When machine B had printed 800 stickers, how many stickers had machine A printed? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


Alynna conducted a survey of the favourite sport of the students in 4 classes. Which of the classes has the greatest number of students choosing Football?

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


Alynna conducted a survey of the favourite sport of the students in 4 classes. In these 4 classes, how many more students choose Badminton as compared to Basketball?


The figure shows 12 identical squares. What is the least number of such squares that must be added to the figure so that the line AB becomes a line of symmetry?

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.


In the figure, AB and CD are straight lines. Find \(\angle p\).

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°.


Which of the following statements is TRUE of the diagram shown?

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.


Three parts of a circle with a radius of 14 cm are shaded. These three parts add up to a quarter of the circle. What is the total perimeter of the three shaded parts? Take \(\pi = \dfrac{22}{7}\).

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.


ACE and BDF are equilateral triangles. AF = FE and BC = CD. Find \(\angle y\) in the figure.

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, \(\angle y = 180° - 60° = 120°\).


Find the area of triangle ABC.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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.)


The pie chart shows Natalie's expenditure last month. Bills = 15%, Food = \(\dfrac{1}{2}\), Transport is a right angle (25%), Books = $40. How much did she spend on food?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a trapezium with AB parallel to DC. \(\angle ADC = 112°\). Find \(\angle BAD\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The table shows the favourite subject of the students in Primary 6C and is used to draw the pie chart shown. Mathematics is the most favourite subject. English = 12, Chinese = 16, Science = 8 marked, Science sector = 28%. How many students chose Science as their favourite subject?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


Two squares of different sizes are drawn. An unshaded square is formed where the 2 squares overlap each other. The difference between the area of the shaded part A and the area of the shaded part B is 24 cm². The larger square has a side of 10 cm. Find the area of the unshaded part.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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.