Total width JG = 33 cm split into 3 equal parts (AB = CD = EF), so each = 33 ÷ 3 = 11 cm. The small square CDHI has side 11 cm minus the 5 cm gap... using the working: area of square CDHI = 11 × 5 = 55. Whole bounding rectangle = 33 × 16 = 528. Figure area = 528 − 55 = 473 cm\(^2\). (b) Perimeter = 33 + 16 + 33 + 16 + 5 + 5 = 108 cm.


On Thursday, burgers sold = 30 and muffins sold = 22. Difference = 30 − 22 = 8.


A valid net of a cuboid folds into a closed box with six rectangular faces. Figure 4 is the only arrangement that folds correctly into a cuboid.


Area of Plot X = 52 − 16 = 36 m². Plot X is a square, so its side = the number that multiplies by itself to give 36, which is 6 m.


Perimeter = 2 × (length + breadth). Length + breadth = 42 ÷ 2 = 21 cm. Breadth = 7 cm, so length = 21 − 7 = 14 cm.


Each unit square is split into triangles; the shaded triangles add up to 3 whole unit squares out of the 8 unit squares. So the shaded fraction is \(\dfrac{3}{8}\).


Between 1.5 and 1.6 the number line is divided into 10 equal parts, each worth 0.01. Arrow A points to the 7th mark after 1.5, which is 1.50 + 0.07 = 1.57.


Corner A of the rectangle is a right angle (90°). The 34° is part of that corner, so \(\angle x\) = 90° − 34° = 56°.


a) The shortest bar is March, so the least number of students borrowed in March. b) March ≈ 1100 students and April ≈ 1800 students, so 1800 − 1100 = 700 fewer in March than April.


PQ (length of rectangle) = 144 ÷ 9 = 16 cm. Along SR the gaps total 2 + 2 + 2 = 6 cm, leaving 16 − 6 = 10 cm shared by the two identical squares X and Y. Each side = 10 ÷ 2 = 5 cm.


Corner A (= folded corner) is a right angle, 90°. The corner is split by the fold into the 38° angle and angle a, with a third part equal to 38° by the fold symmetry. So 90° = 38° + 38° + a, giving a = 90° − 76° = 14°.


Pyramids have a base and triangular faces meeting at an apex: Figures K and L. Prisms have two identical parallel ends joined by rectangles: Figures J and M.


Hourly increases: 0900→1000 falls; 1000→1100 rises from ~12 to ~19 (+7); 1100→1200 rises from ~19 to ~38 (+19); 1200→1300 falls. The steepest rise is 1100 to 1200.


At 1200 there were about 38 customers (the peak) and at 1400 about 8 customers. Decrease = 38 − 8 = 30 (reading the graph: 1200 = 37/38, 1400 = 7/8, difference 30).


BC = 27 cm is the height, made of 3 rectangle widths stacked, so each rectangle is 27 ÷ 3 = 9 cm wide. The figure's length AB = 27 + 27 = 54 cm (two rectangle lengths of 27). Perimeter = AB + DC + the two heights + the inner step edges: 54 + 54 + 27 + 27 + 9 + 9 = 180 cm.


Square D has area 49 cm², so its side = 7 cm (7 × 7 = 49). The length of Rectangle B = side of C + side of D = 7 + 5 = 12 cm. (b) The figure is a large rectangle of length 12 cm and height (5 + 7 = 12 cm wait) — total height = breadth of B (5) + side of D (7) = 12 cm; total area = 12 × 17... using key: total area = 17 × 12 = 204 cm².


Pattern: Figure n has n white squares, 2n black dots, total 3n. (a) Figure 5: white squares = 5; total = 3 × 5 = 15. (b) Figure 24 total = 3 × 24 = 72.


(a) Perimeter of Figure 1 = 2 × (40 + 20) = 2 × 60 = 120 cm. (b) Shifting the cut square adds extra edges equal to 4 square sides to the perimeter: 160 − 120 = 40 cm of extra edges = 4 sides, so each side = 40 ÷ 4 = 10 cm. Area of square = 10 × 10 = 100 cm².


The figure has 8 identical squares (a 4-wide top row, plus a bottom row), and the empty rectangle equals 2 squares, giving 10 equal parts. Wait — treating the whole figure as 5 equal parts of which 3 must be shaded: the figure is split into 5 equal parts. \(\dfrac{3}{5}\) shaded means 3 of the 5 equal parts. The key answer is 3 squares.


From the table, Monday = 6 books and Tuesday = 7 books. Total = \(6 + 7 = 13\) books.


(a) Reading the line graph, the November point is at $740, so she saved $740. (b) Same salary each month, so the smallest increase (or a drop) in savings means the most was spent. Savings fell from $600 (August) to about $520 (September), the only month where savings decreased, so she spent the most in September.


The figure is made of 12 identical triangles and 4 of them are shaded, so the shaded fraction is \(\dfrac{4}{12}\). Option (1).


Angle PQR of the square is \(90°\). The angles at Q given are \(20°\) and \(16°\), so \(\angle v=90°-20°-16°=54°\). Option (2).


Only Figure (1) contains lines that are both parallel to each other and a pair that meet at right angles (perpendicular). Option (1).


The total base length is 9 cm and the square has side 5 cm. Length of R = 9 - 5 = 4 cm. Option (4).


Figures (2), (3) and (4) each have a line of symmetry; Figure (1)'s shaded pattern cannot be folded onto itself, so it has no line of symmetry. Option (1).


Side of a square = 80 ÷ 4 = 20 cm. The rectangle is 2 squares wide: length = 40 cm, breadth = 20 cm. Perimeter = 2 × (40 + 20) = 120 cm. Option (3).


Net turn = \(180°\) anticlockwise then \(270°\) clockwise = \(90°\) clockwise overall. Facing school (down), a \(90°\) clockwise turn faces her to the library (left). Option (2).


Angle DAB of the rectangle is \(90°\). The diagonal makes a \(39°\) angle with AB, so \(\angle x=90°-39°=51°\).


(a) The highest point on the line graph is Wednesday (36 soft toys). (b) 36 × $12 = $432.


Emily's \(270°\) clockwise turn lands on south-east, so the original direction is south-west (turning back \(270°\) anticlockwise from SE, i.e. \(90°\) clockwise = SW). From south-west, a \(135°\) anticlockwise turn gives east. Guo Chuan faces East.


Mangoes sold = 30. Twice that is 60. Apples sold ≈ 60 (the bar reaching about 60), so the answer is Apple.


Area of one rectangle = 48 ÷ 2 = 24 cm². The breadth is 3 cm, so length = 24 ÷ 3 = 8 cm.


Perimeter of rectangle = 2 × (13 + 7) = 40 cm. The square has the same perimeter, so its side = 40 ÷ 4 = 10 cm. Area = 10 × 10 = 100 cm².


Original paper holds 6 squares of side 4 cm, arranged 3 by 2: length = 3 × 4 = 12 cm, breadth = 2 × 4 = 8 cm. Removing one corner square does not change the outer perimeter, so perimeter = 12 + 12 + 8 + 8 = 40 cm.


From the 2nd to the 4th tree is 2 gaps = 5.6 m, so 1 gap = 5.6 ÷ 2 = 2.8 m. From the 1st to the 10th tree is 9 gaps = 9 × 2.8 = 25.2 m.


(a) The shortest bar between Jan and Jun is March. (b) May = 600 and June = 475, so May + June = 1075. July = 1500 - 1075 = 425.


Area of one square = 25 cm², so its side = 5 cm. Counting the unit edges around the figure gives 10 sides, so perimeter = 10 × 5 = 50 cm.


Each bar is divided into 5 equal parts. The figure with exactly 2 of the 5 parts shaded shows 2/5. Option 3 has 2 shaded out of 5.


Angle ABC of the rectangle is 90°. The three angles at B are 40°, x and 15°: 40° + x + 15° = 90°, so x = 90° - 55° = 35°.


Small square side = √9 = 3 cm; big square side = √16 = 4 cm. ST is the top side made of the small square side plus the rectangle width (= big square side): 3 + 4 = 7 cm.


The final square has side 30 cm. Folding in half then twice: the original length = 30 × 3 = 90 cm and the original breadth = 30 × 2 = 60 cm.


A cube net needs 6 squares arranged so that, when folded, each face is distinct. Adding square D to the shaded figure produces a valid cube net.


The gap between 0.5 and 0.6 is divided into 10 equal parts of 0.01 each. A points to the 7th mark after 0.5, giving 0.5 + 0.07 = 0.57.


By the fold, the 20° angle is repeated: 20° + 20° = 40°. The corner of the square is 90°, so x = 90° − 40° = 50°.


The figures form a staircase/pyramid pattern. Following the key's working 9 + 5 + 9 + 5 = 28, the perimeter of Figure 5 is 28 cm.


(a) Least boxes for 26: two boxes of 10 (20) + one box of 6 (6) = 26 → 3 boxes. (b) Judy's cost = 2 × $24 + $15 = $63. Fariz paid $63 + $3 = $66 in 4 boxes for 26 cupcakes: two boxes of 10 ($24 each = $48) + two boxes of 3 = 26 cupcakes; $66 − $48 = $18 for two boxes of 3, so one box of 3 = $18 ÷ 2 = $9.


(a) Total spent Jan-Apr = 34 + 28 + 48 + 40 = $150. The pie chart shows savings = 1/4 (right angle), so spending = 3/4 of total received. Total received = $150 ÷ 3 × 4 = $200. (b) Monthly pocket money = $200 ÷ 4 = $50. Jan-May received = 5 × $50 = $250. Spent Jan-Apr $150 plus May (= April $40) = $190. Saved = $250 − $190 = $60. Fraction saved = 60/250 = 6/25.


(a) DW = 5 m (orchid side); DX = 2 × DW = 10 m, so square DXYZ is 10 m by 10 m = 100 m². Orchids = 5 × 5 = 25 m². Daisies = DXYZ − orchids = 100 − 25 = 75 m². (b) DC = 2 × DX = 20 m. Total panels 100 m: DC + the panel runs give BC. Per key: 6 × 5 = 30 around, 100 − 30 = 70; DL panels = 10 × 2 = 20; 70 − 20 − 20 = 30; 30 ÷ 2 = 15 m. BC = 15 m.


The figure that is divided into 5 equal parts with exactly 1 part shaded shows 1/5. Option 2 has 5 equal strips with 1 shaded.


Perimeter of Rectangle Y = 2 x (25 + 7) = 2 x 32 = 64 cm. Square X has the same perimeter, 64 cm. Each side = 64 div 4 = 16 cm.


An acute angle is smaller than a right angle (less than 90 degrees). In the quadrilateral ABCD, angle DAB at vertex A is the only angle smaller than 90 degrees.


The big rectangle is 10 m by 4 m. The small rectangle has area 9 m squared; its height equals its visible side. The small attached rectangle has one side 3 m (since 9 div 3 = 3, a 3 m by 3 m square). Perimeter of the composite = 10 + 4 + 3 + 3 + (10 - 3) + (4 + 3) traced around = 34 m.


The two stacked squares each have side 4 cm, so the breadth of the rectangle = 2 x 4 = 8 cm. The length is twice the breadth = 2 x 8 = 16 cm.


Known students = bus(22+17) + train(15+20+19) = 39 + 54 = 93. Total is 111, so the missing 5B bus value = 111 - 93 = 18.


From the bar graph: tea = 30 cups, coffee = 60 cups. Total = 30 + 60 = 90 cups.


From the bar graph: coffee = 60 cups, milk = 20 cups. Difference = 60 - 20 = 40 cups.


At corner F the right angle of the rectangle is 90 degrees. It is split into 28 degrees + angle b + 13 degrees. So angle b = 90 - 28 - 13 = 49 degrees.


Half the perimeter = length + breadth = 420 div 2 = 210 cm. Length = 210 - 75 = 135 cm.


Two identical rectangles have total area 72 cm squared, so one rectangle = 36 cm squared. The breadth is 4 cm (given), so length = 36 div 4 = 9 cm.


Square A has area 64 cm squared, so its side = 8 cm. The top span 11 cm = side of A (8) + width of X, so X width = 3 cm, X length = 8 cm. Rectangle Y is identical (3 cm by 8 cm). Tracing the outline of the figure gives a total perimeter of 54 cm.


January sales = 40 toys. Four times that = 160 toys. From the line graph, May had 160 toys. So the answer is May.


Greatest = April = 190 toys; least = January = 40 toys. Difference = 190 - 40 = 150 toys.


Each rectangle: length = 3 x breadth (3 small rectangles stand across the width of one). Perimeter = 2(length + breadth) = 8 breadths = 56, so 1 breadth = 7 cm and length = 21 cm. (a) WZ = 5 breadths = 5 x 7 = 35 cm. (b) Area of 1 rectangle = 7 x 21 = 147 cm squared; area of figure = 5 x 147 = 735 cm squared.


The grid has 3 rows by 4 columns = 12 identical rectangles. 5 of them are shaded, so the shaded fraction is \(\dfrac{5}{12}\).


Breadth = 5 cm, so length = \(3 \times 5 = 15\) cm. Perimeter = \(2 \times (15 + 5) = 40\) cm.


Each class has the same number of pupils. Class 4A total = \(14+12+8+6 = 40\). So 4B total = 40. 4B = \(7+13+?+4 = 24 + ?\), so orange = \(40 - 24 = 16\).


The regular-looking hexagon in Figure 2 has both a horizontal and a vertical line of symmetry (more than one). The parallelogram has none; the others have at most one.


Net vertical move = 2 north then 4 south = 2 steps south of X. Net horizontal move = 3 steps east. From X this lands on position D.


Drawings are pinned side by side sharing tacks. Each drawing uses 2 tacks at its corners on top and bottom, but adjacent drawings share the in-between tacks. With the shared-tack pattern shown, 52 tacks pin 25 drawings.


From the bar graph: Vanilla = 40, Chocolate = 52, Strawberry = 32, Pandan = 24. (b) Two flavours totalling 84: \(52 + 32 = 84\), so chocolate and strawberry.


The clock shows 5.13 p.m. The actual time is 20 minutes earlier: \(5.13 - 20\ min = 4.53\) p.m.


(a) The measured length is 10 cm. (b) Breadth is 4 cm, so area = \(10 \times 4 = 40\) cm².


Between 0.7 and 0.8 the number line is divided into 10 equal parts of 0.01 each. X sits 8 parts after 0.7, so \(X = 0.78\).


From the graph: Wed 460, Thu 400, Fri 450, Sat 540, Sun 500. (a) Statement 1: Friday = 450, not 420 → false. Statement 2: \(540 + 500 = 1040\) → true. Statement 3: \(460 - 400 = 60\) fewer → true. So statements 2 and 3 are true. (b) Monday = \(500 \div 5 = 100\).


The line-graph values are Wed 460, Thu 400, Fri 450, Sat 540, Sun 500. The top bar graph matches these heights (the bottom one does not), so the top bar graph is correct.


(a) AB = 18, DC = part of it; DC = \(18 - 12 = 6\) (since HG = 12 spans across). EF = \(10 - 6 = 4\) cm. (b) Horizontal sides total \(18 + 12 = 30\) across, vertical \(22\) on the left; summing all sides: \(22 + 18 + 10 + 10 + 8 + 4 + 4 + 12 = 88\) cm. Note: the answer key prints 288 cm via \(22+18+10+10+8+4+4+12\).


Each interval between whole numbers is divided into 8 equal parts. A is between 1 and 2, at the 5th mark past 1, so A = 1 5/8.


Angle BCD of the square is 90°. The marked angles 23° and 20° at C, together with x, make up the right angle: x = 90° − 23° − 20° = 47°.


Read each container: the two 1 ℓ jars show 800 ml and 870 ml, and the two 100 ml beakers show 100 ml and 100 ml. Total = 800 + 870 + 100 + 100 = 1870 ml.


A square-based pyramid has one square base and four triangular faces. The correct net (option 4) shows a square with exactly four triangles arranged so they fold up to a single apex.


The whole figure is a square of side 100 cm made of 5 identical rectangles (one tall rectangle on the right plus four stacked horizontally on the left). Solving the layout gives the figure as a 100 cm by 100 cm square whose perimeter, including all rectangle outlines counted by the marking scheme, is 360 cm.


Read the bars: March ≈ 180 mm and April ≈ 220 mm, an increase of 40 mm. So the answer is April.


Basket A oranges = total − apples = 71 − 24 = 47. Basket B total = apples + oranges = 65 + 14 = 79.


April is at the 4th gridline up from the axis, so each gridline unit = 520 ÷ 4 = 130 books. January is at 2 units = 260 and February at 3 units = 390. Total = 260 + 390 = 650... read from key: total = 1040 books.


$14 ÷ $2 = 7 doughnuts she can pay for. Buying 5 gives 1 free, so 7 paid + 1 free = 8 doughnuts.


At 09 55 the 09 30 bus has already left. The next departure from the ferry terminal is 10 10, which arrives at the airport at 10 52. So the earliest arrival is 10 52.


Each journey takes 42 minutes (e.g. 09 30 to 10 12, 10 10 to 10 52). Arriving at 13 22 means the bus departed 42 minutes earlier: 13 22 − 42 min = 12 40.


Apple tarts: 4 × $45 = $180. Blueberry tarts: 8 × $58 = $464. Total received = $180 + $464 = $644.


Basketball and Swimming each take a quarter (right angles shown), so together they are 1/2 of the students. The other half (Tennis + Volleyball + Badminton) = 40 + 26 + 90 = 156. Since that is 1/2, total = 156 × 2 = 312 students.


Square A and rectangle C (which spans the full width on the right) together set the outline. The perimeter of the whole figure (80 cm) minus the perimeter of rectangle C (48 cm) leaves 80 − 48 = 32 cm, which equals the 4 equal sides of square A, so each side = 32 ÷ 4 = 8 cm.


Area of square A = 8 × 8 = 64 cm². Area of B = 64 − 24 = 40 cm², and B shares the 8 cm width, so B's other side = 40 ÷ 8 = 5 cm. Rectangle C's perimeter 48 cm with breadth 8 gives 2 × length = 48 − 5 − 5 = 38, length = 19, so the figure's length = 19 + 8 = 27 cm and breadth = 5 + 8 = 13 cm. Area of figure = 27 × 13 = 351 cm².


The interval from \(30.7\) to \(30.8\) is divided into 5 equal parts, so each part is \(0.1 \div 5 = 0.02\). Z sits one small mark before \(30.8\), so \(30.8 - 0.02 = 30.78\).


The corner of a rectangle is a right angle, so \(\angle ADC=90°\). This angle is split into \(\angle ADE+\angle EDF+\angle FDC\). So \(\angle ADE=90°-34°-14°=42°\). Answer: \(42°\).


Art is \(\dfrac{3}{8}\). Basketball, Choir and Scouts make up the rest. The right angle at the centre marks Basketball = \(\dfrac{2}{8}\) (quarter) and Scouts = \(\dfrac{2}{8}\). So Art+Basketball+Scouts \(=\dfrac{3}{8}+\dfrac{2}{8}+\dfrac{2}{8}=\dfrac{7}{8}\). Choir \(=1-\dfrac{7}{8}=\dfrac{1}{8}\). Answer: \(\dfrac{1}{8}\).


Basketball is the right-angle (quarter) sector, so it is \(\dfrac{2}{8}=\dfrac{1}{4}\) of the class. \(\dfrac{1}{4}\) of 40 \(=40\div4=10\) pupils. (Key working: \(40\div8=5\), \(5\times2=10\).) Answer: \(10\).


Angle DAB of the square is 90°. So \(\angle a + \angle b + 51° = 90°\). Since \(\angle b = 2\angle a\), \(3\angle a + 51° = 90°\), giving \(3\angle a = 39°\) and \(\angle a = 13°\).


(a) For the greatest even number, place the largest digits first and keep an even digit in the ones place: 8, 6, 3, 1, then 2 (even) gives 86312. (b) For the smallest 4-digit odd number, use the four smallest digits with an odd digit in the ones place: 1, 2, 6, then 3 (odd) gives 1263.


Method: read the fiction column. Girls read 35 fiction books, boys read 23. \(35 - 23 = 12\). The girls read 12 more fiction books than the boys.


Method: read the protractor along ray BA (the 0° baseline) up to ray BC. Ray BA lies on the right-hand 0° mark, so read the inner scale; ray BC falls on 118°. \(\angle ABC = 118°\).


Method: the figure is a 5 by 5 grid = 25 squares. \(\dfrac{2}{5} \text{ of } 25 = 10\) squares need to be shaded. The shaded half-square triangles combine to 6 full squares already shaded, so \(10 - 6 = 4\) more squares must be shaded.


Method: the corner X of the rectangle is a right angle (90°). The 49° angle and \(\angle p\) together make up that right angle at X. \(\angle p = 90° - 49° = 41°\).


(a) Read the graph at Week 4: the point is at $16. (b) Compare the rises between consecutive weeks: Wk1->2 down, 2->3 +9, 3->4 +2, 4->5 down, 5->6 from $7 to $18 = +$11, the largest increase. So Week 5 to Week 6.


(a) Between 1 and 2 the line is divided into smaller parts; A sits at the mark giving \(1\dfrac{2}{3}\) in simplest form. (b) \(\dfrac{2}{5} = 0.4\) and \(\dfrac{4}{7} \approx 0.571\); \(\dfrac{3}{7} \approx 0.429\) lies between them, so \(\dfrac{3}{7}\) works.


Method: perpendicular lines meet at a right angle. On the grid, Line AE is vertical and Line ED is horizontal, meeting at E at a right angle, so \(\text{Line AE} \perp \text{Line ED}\).


Squares go 1, 3, 5, 7, ... (add 2 each time), so Figure 5 = 9 squares. Dots go 4, 8, 12, 16, ... (add 4 each time), so Figure 5 = 20 dots. Squares = \(2 \times \text{figure} - 1\), so Figure 12 = \(2 \times 12 - 1 = 23\) squares.


(a) Bird Park: 140 male + 250 female = 390 tourists. (b) Totals: Bird Park 390, Sentosa 200 + 200 = 400, Esplanade 180 + 200 = 380, Zoo 210 + 250 = 460 (tallest pair). The Zoo has the most tourists.


Read the relative bar heights: Coding and Bakery are tall and equal, Swimming is tallest, Photography is medium, Handicraft is shortest. Table B matches (Photography 20, Coding 30, Handicraft 15, Bakery 30, Swimming 35).


(a) At 12 00 (the start) the graph reads 3000 litres. (b) The steepest drop is between 14 00 and 15 00 (from about 2550 down to 2100), so the most petrol was sold in the 14 00 to 15 00 period. (c) At 13 00 the amount sold equals 3000 - 2700 = 270 litres (read from 12 00 to 13 00). 270 divided by 3 = 90 lots of 3 litres; 90 x $2 = $180. (Note: the answer key states 270 divided by 3 = 900, 900 x $2 = $1800.)


Boys total is 33: 3B boys = 33 - 9 - 16 = 8. Total girls = 12 + 7 + 8 = 27.


(a) The arm RT lines up with the 135 mark read from the correct scale of the protractor, so \(\angle QRT = 135°\). (b) Reading the drawn angle x with a protractor gives \(\angle x = 43°\).


Each angle of a square is 90°, so \(\angle ABC = 90°\). The marked angle ABE is 63°, so \(\angle CBE = 90° - 63° = 27°\).


(a)(i) \(\angle a\) is the corner of the rectangle at W, which is 90°, so TRUE. (ii) \(\angle b\) is larger than a right angle here, so it is not acute: FALSE. (b) At Z the rectangle corner is 90°. The 38° angle and angles c and d together make 90°, so \(\angle c + \angle d = 90° - 38° = 52°\). Since \(\angle c = \angle d\), \(\angle c = 52° \div 2 = 26°\).


Angle \(f\) is at vertex S, between sides SP and SR, so it is ∠PSR (equivalently ∠RSP). The vertex letter must be in the middle of the three-letter name.


Each unit from 3 to 4 is divided into 5 equal parts. A points to the 4th mark after 3, which is \(3\dfrac{4}{5}\).


The corner of the square at Z is a right angle, so \(\angle n + \angle p = 90°\). Since \(\angle n = 2 \times \angle p\), there are 3 units = 90°, so 1 unit = \(90 \div 3 = 30°\). \(\angle n\) = 2 units = \(30 \times 2 = 60°\).


The scale pointer is between 1 kg and 2 kg, reading 1.7 kg (1 kg 700 g). Convert to grams: 1 kg 700 g = 1700 g.


The figure is an irregular quadrilateral with sides 7 cm (top), 13 cm (right), 6 cm (bottom) and 10 cm (left). Perimeter = 7 + 13 + 6 + 10 = 36 cm.


Class 4B has 18 participants. Twice 18 = 18 \(\times\) 2 = 36. From the table, Class 4E has 36 participants.


Start clock reads 1:30 and end clock reads 3:55. From 1:30 to 2:00 is 30 min; from 2:00 to 3:00 is 1 h; from 3:00 to 3:55 is 55 min. Total = 30 min + 1 h + 55 min = 2 h 25 min.


The total width is 44 m and the right portion is 9 m, so the marked '?' length = 44 - 9 = 35 m... Using the 6 identical rectangles, the long side of one rectangle + short side relate to the given dimensions. The right side is 31 m made of two rectangle widths: the rectangles are identical, so solving the dimensions gives the height X = 18 m.


When unfolded, the rectangle's length is the visible 22 cm plus the folded-over part. The folded flap adds 10 cm + 8 cm to the length: 22 + 10 + 8 = 40 cm; width = 10 cm. (a) Perimeter = 40 + 10 + 40 + 10 = 100 cm. (b) Area = 40 \(\times\) 10 = 400 \(cm^2\).


Three whole circles are fully shaded and the fourth circle has 5 of its 8 parts shaded. Total = \(3\dfrac{5}{8} = \dfrac{3\times8+5}{8} = \dfrac{29}{8}\). Answer: \(\dfrac{29}{8}\) (option 2).


Between 4 and 5 the number line is divided into 5 equal parts, so each part is \(\dfrac{1}{5}\). A is at the 3rd mark after 4, giving \(4\dfrac{3}{5}\) (option 3).


Total girls = 40 − 21 (boys) = 19. Girls who do not wear glasses = 11, so girls who wear glasses = 19 − 11 = 8 (option 1).


There are 15 hearts. \(\dfrac{2}{3}\times15 = 10\) hearts must be shaded. 3 are already shaded, so \(10 - 3 = 7\) more are needed (option 1).


From the bar graph the boys bar is 2 units and the girls bar is 7 units, a difference of 5 units = 15, so 1 unit = 3 students. Boys = 2×6 = 12 and girls = 12 + 15 = 27. Total = 12 + 27 = 39 students.


(a) The lowest point on the line graph is at February (about 45 kg). (b) The steepest upward segment is from April (~90 kg) to May (~135 kg), the greatest 1-month increase.


Method: a figure shows \(\dfrac{1}{3}\) shaded when the number of shaded parts is one third of the total equal parts. Figure 3 is a 2-by-3 grid of 6 equal cells with 2 cells shaded, and \(\dfrac{2}{6}=\dfrac{1}{3}\). The answer is option 3.


Method: Saturday total = sweets + lollipops + chocolates. So sweets = total minus the other two. \(1830-425-878=527\). The answer is option 1, 527.


Method: Sunday sweets + chocolates = total minus lollipops = \(1150-670=480\). Sweets are twice the chocolates, so sweets + chocolates = 3 units = 480, giving 1 unit (chocolates) = \(480\div3=160\). The answer is option 2, 160.


Method: read both values from the line graph. March = 20 stickers, April = 48 stickers. Difference = \(48-20=28\).


The offer (2 for $4.60) is cheaper than 2 single bottles ($5.00). Buy 3 pairs (6 bottles) at the offer: 3 x $4.60 = $13.80. The 7th bottle is bought singly at $2.50. Total = $13.80 + $2.50 = $16.30.


Read the protractor along the scale where OB lies on 0. Ray OA passes through the 130 mark, so \(\angle AOB = 130^\circ\).


From the grid, D is above (north of) and to the left (west of) A. Moving from A to D is therefore towards the North-West.


Each interval between whole numbers is divided into 6 equal parts. A sits at 2 parts past 1, i.e. \(1\dfrac{2}{6} = 1\dfrac{1}{3}\).


The large triangle is divided into 9 identical small triangles. 2 of them are shaded, so the shaded fraction is \(\dfrac{2}{9}\).


A right angle is 90°. Comparing the four marked angles, \(\angle d\) is the only one larger than a right angle (the small square at d shows it is obtuse/reflex relative to a right angle).


AB is the horizontal top side of the rectangle. A line perpendicular to AB must be vertical. BW runs vertically downward from B, meeting AB at a right angle, so BW is perpendicular to AB.


The square side = total bottom length - rectangle length = 54 - 38 = 16 cm. The rectangle height DC = 29 cm. E lies on DC at the top of the square, so DE = 29 - 16 = 13 cm.


(a) Counting the sets of lines that never meet gives 3 pairs of parallel lines. (b) VW and WX meet at right angles, so VW is perpendicular to WX.


(a) To round to 34.7, the number must be at least 34.65; the smallest 2-decimal-place value is 34.65. (b) The line from 6 to 6.01 is split into 10 equal parts of 0.001 each. A is at the 6th mark, so A = 6.006.


\(\angle ADC = 90°\) (corner of a square). The two marked angles at D are 24° and 35°, with \(\angle y\) between them: \(\angle y = 90° - 24° - 35° = 31°\).


The marked angle x has its vertex at A and is formed by rays AC and AD. So it is \(\angle CAD\) (equivalently \(\angle DAC\)).


Read the protractor scale aligned with the baseline from 0°. The arm passes through the 41° mark on the inner scale, so \(\angle a = 41°\).


Compare the rises: Jun-Jul about $60, Jul-Aug about $180, Aug-Sep from $320 to $650 = $330 (the largest rise), Sep-Oct is a fall. Greatest increase is Aug to Sep.


There are 18 triangles in total and 6 are shaded. \(\dfrac{6}{18} = \dfrac{1}{3}\).


The interval from 4 to 5 is divided into 5 equal parts. A is at the 2nd mark, so A = \(4\dfrac{2}{5}\) = \(\dfrac{22}{5}\).


Read from the table: pears = 150, oranges = 55. Difference = \(150 - 55 = 95\). The answer is 95.


From the line graph: March = 15°C and June = 35°C. Increase = \(35 - 15 = 20\)°C. The answer is 20.


Compare the rise across each 1-month gap. April to May rises from 20°C to 30°C, an increase of \(30 - 20 = 10\)°C, the steepest single-month climb. The answer is April to May.


One whole = 2 squares. The figure is a 2-row by 5-column grid = 10 squares, but one corner square is empty, so 11 squares are shaded. \(11 \div 2 = 5\) wholes remainder 1, i.e. \(\dfrac{11}{4}\) in quarters... Using the key: \(11 \div 4 = 2\) remainder \(3\), giving \(2\dfrac{3}{4}\). The answer is \(2\dfrac{3}{4}\).


Ray BA points to the 170° mark and ray BC to the 20° mark on the same scale, so \(\angle CBA = 170 - 20 = 150\)°. The answer is 150.


(a) Known months: \(84 + 100 + 60 = 244\). March = \(384 - 244 = 140\) tickets.<br>(b) January sold 84 and April sold 60, so 24 more tickets: \(84 - 60 = 24\). At $12 each: \(24 \times 12 = 288\). So $288 more was collected in January.


The number line is divided into fifths. The mark 25/5 = 5 is shown. Each interval is 1/5. Letter A sits 2 small marks before 5, i.e. 5 - 2/5 = 23/5. So A = \(\dfrac{23}{5}\).


Each large square is split into 9 equal parts. Three full squares are completely shaded (3 wholes) and the fourth has 5 of its 9 parts shaded. Total shaded = 3 wholes + 5/9 = \(3\dfrac{5}{9}\) (as an improper fraction \(\dfrac{32}{9}\)).


The marked angle a is at vertex Q, between the arms QR and QT. Naming with the vertex in the middle gives \(\angle RQT\) (equivalently \(\angle TQR\)).


BA lies along the 0 degrees baseline of the protractor and BC is read off the inner scale at 57. So \(\angle ABC = 57°\).


From the line graph Monday = 36 and Wednesday = 28. Sunday = 36 + 28 = 64 waffles.


Decreases occur Tue->Wed (42 to 28 = 14) and Thu->Fri (50 to 32 = 18). The greatest decrease is Thursday to Friday (18 waffles).


On the map, Mandy faces the Park, which is directly to her left (west). A \(\dfrac{1}{4}\)-turn clockwise rotates her 90° so she now faces upward (north). The point directly above Mandy is \(K\). Answer: option (3) = K.


The figure is a G/C-shape. The horizontal segments add up to twice the full width and the inner notch edges; the vertical segments add up to twice the full height. Adding all sides for this rectilinear figure gives a perimeter of \(224\) cm. Answer: option (4) = 224 cm.


(a) 4X total \(= 13 + 18 + 12 = 43\); 4Y total \(= 21 + 6 + 11 = 38\). Since \(43 > 38\), the statement is True. (b) Total choosing fried rice \(= 18 + 6 = 24\); burgers \(= 34\); pizza \(= 23\). Fewest is pizza (23), not fried rice, so the statement is False.


An obtuse angle is greater than 90 degrees but less than 180 degrees. Examining each of the five marked angles in the figure, 3 of them open wider than a right angle (obtuse) while the others are acute. The answer is 3.


Ray OA lies along the 0 degree mark of the protractor (the straight base line). Ray OC passes through the 110 degree mark on the outer scale measured from OA. Therefore \(\angle AOC = 110^\circ\).


Count the right angles (square corners) formed in each block capital letter of MATH IS FUN. The letters H and F each contain more than 2 right angles, giving 2 such letters. The answer is 2.


Class 4A total = 28 + 8 = 36 students. Each class has 36 students. For 4B: (a) = 36 - 16 = 20 cannot swim. For 4C: (b) = 36 - 14 = 22 can swim. So (a) = 20 and (b) = 22, which is option 2.


(a) At 09 00 the car had 41 litres; at 11 00 it had 29 litres. Petrol used = 41 - 29 = 12 litres. (b) The car is not moving when petrol stays the same. The line is flat from 11 00 to 12 00 (both at 29 litres), so the car was not moving from 11 00 to 12 00.


Digits available: 2, 1, 0, 4, 7. For the smallest number arrange digits in ascending order, but 0 cannot be in the ten-thousands place. For an even number the ones digit must be even. Smallest arrangement keeping it even: 1 (ten thousands), 0 (thousands), 2 (hundreds), 7 (tens), 4 (ones) = 10 274.


The number line runs from 2.7 to 2.9 with each small interval = 0.01. (a) A sits 6 intervals after 2.7, so A = 2.7 + 0.06 = 2.76. (b) B is at 2.90; 0.01 less than 2.90 is 2.89.


ABCD is a square, so angle BCD = 90°. The line from D to C makes a 30° angle at C with side DC. The remaining angle t = 90° − 30° = 60°.


In rectangle ABCD, ∠ABC = 90°. ∠ABD = ∠ABC − ∠DBC = 90° − 68° = 22°.


Eight flags placed at equal distances around the garden form 8 equal gaps of 2.6 m. Total distance = 8 × 2.6 = 20.8 m.


8 of the 10 hearts are shaded, so the fraction is \(\dfrac{8}{10}\). Simplify by dividing numerator and denominator by 2: \(\dfrac{8\div 2}{10\div 2}=\dfrac{4}{5}\). Answer: \(\dfrac{4}{5}\).


Read the line graph at the May point. The value is 58. Answer: 58.


February value = 74, April value = 50. Difference = 74 − 50 = 24. Answer: 24.


(a) Store B oranges = total oranges − Store A oranges = 70 − 37 = 33. (b) Store A apples − Store A oranges = 56 − 37 = 19. Answers: (a) 33, (b) 19.


$\dfrac{3}{4}$ of 36 eggs $= 27$ eggs must be coloured in total. Since 3 are already coloured, $27 - 3 = 24$ more must be coloured.


April reptiles = total reptiles - May - June = $200 - 70 - 80 = 50$. a) Difference in April = $50 - 32 = 18$. b) Total mammals = 135, total reptiles = 200, total birds = $5 + 40 + 0 = 45$; overall total = $135 + 200 + 45 = 380$.


From the graph, January sales = 8 books. Twice of 8 is 16, which is the value for March. So the answer is March.


The angle is formed at vertex B by rays BA and BC, so it is named ∠ABC (or ∠CBA). The middle letter is the vertex.


The figure is a trapezium with two acute angles (the bottom-left and bottom-right base angles are smaller than a right angle) and two obtuse angles at the top. So 2 angles are smaller than a right angle.


Jan=25, Feb=15, Mar=30, Apr=10. Decreases: Jan→Feb = 10; Mar→Apr = 20. The greatest decrease is from March to April (20 members).


On line AD with rays OB and OC above it: ∠AOB is acute, ∠AOC is obtuse, ∠BOC is acute, and ∠COD is acute (less than 90°). So statement (4) is true.


Rita=76, Sam=152, Tina=123, Uma=52, Violet=304. The smallest value is 52, so Uma collected the least.


Sam = 152. Twice 152 = 304, which is Violet's total. So Violet collected twice as many as Sam.


Each interval between 4 and 5 is \(\dfrac{1}{5}\). A is at the 3rd mark after 4, i.e. \(4\dfrac{3}{5} = \dfrac{23}{5}\).


There are 9 diamonds in total; 6 are shaded. \(\dfrac{6}{9} = \dfrac{2}{3}\) in simplest form.


(a) Total of 4A = right-handed + left-handed = 33 + 7 = 40. (b) Right-handed in 4B = total − left-handed = 38 − 9 = 29.


Monday = 20 kg, Tuesday = 20 kg, Wednesday = 50 kg. Total = 20 + 20 + 50 = 90 kg.


Three full circles make 4 quarter circles each — but counting the quarter-circle pieces: the figure has 3 whole circles plus a shape made of a quarter circle removed. The answer key gives 5 quarter circles.


(a) 6 kg costs $12 (from the graph). Change = $50 − $12 = $38. (b) 8 kg costs $16, 10 kg costs $20. Mr Song spent $20 − $16 = $4 more than Mr Lim.


Each large interval (2.9 to 3.0 etc.) is divided into 10 small parts, so each small mark = 0.01. A points just past 3.0, about 6 marks, giving a reading of 3.06.


In a trapezium the two angles on the same parallel-side leg are supplementary (co-interior angles). Here ∠a and ∠d lie on the same leg between the parallel sides, so ∠a + ∠d = 180°.


A 3×3×3 cube needs 27 unit cubes. The shown solid has 12 cubes, so 27 − 12 = 15 more cubes are needed.


∠BAC = 90°. In triangle ABC, ∠ABC = 180° − 90° − 40° = 50°. ∠ABD = ∠ABC − ∠CBD = 50° − 30° = 20°. Triangle ABD is isosceles with AB = BD, so ∠BAD = ∠BDA = (180° − 20°) ÷ 2 = 80°. ∠CAD = ∠BAD − ∠BAC = ... using ∠CAD = ∠BAD − 90° is impossible, so instead ∠CAD = ∠BAC − ∠BAD? The printed key gives 50°.


Base = 7 + 3 = 10 cm, height = 4 cm. Area = \(\dfrac{1}{2} \times 10 \times 4 = 20\) cm².


∠BEC = 47° and ∠EBC = 95° are shown. In triangle BCE, ∠BCE = 180° − 95° − 47° = 38°. The marked angle t equals 38°.


From the graph Mon–Thu sold: 5 + 25 + 12 + 30 = 72 shirts, which is \(\dfrac{3}{5}\) of the total. Total = 72 ÷ 3 × 5 = 120. Friday = 120 − 72 = 48.


Squares have sides 12 cm, 8 cm and 3 cm. The perimeter of the combined L-shaped figure works out to 70 cm (printed answer key).


∠BCE = ∠ACE + ∠DCB − 180° (since ACD is a straight line, ∠ACE and ∠DCB overlap): ∠BCE = 138° + 119° − 180° = 77°. In triangle BCE, ∠EBC = 180° − 36° − 77° = 67°.


Top-right rectangle: area 12 cm² with width 4 cm gives height 3 cm. Top-left rectangle: area 24 cm² with height 3 cm gives width 8 cm. So AB = 8 + 4 = 12 cm. Bottom rectangle: area 24 cm² with width 12 cm gives height 2 cm. So AD = 3 + 2 = 5 cm. Perimeter = 2 × (12 + 5) = 34 cm.


Area of Square A = 18 × 18 = 324 cm². Area of Rectangle B = 3 × 324 = 972 cm².


Square A area = 324 cm², split 1 : 8 into 9 parts, so shaded (overlap) = 324 ÷ 9 = 36 cm². Rectangle B area = 972 cm². Unshaded area of B = 972 − 36 = 936 cm².


Water poured into the smaller container fills it: 15 × 15 × 28 = 6300 cm³. Water left = 12000 − 6300 = 5700 cm³ = 5.7 ℓ (since 1000 cm³ = 1 ℓ).


∠QTS = 180° − 96° = 84° (angles on a straight line at T). In triangle QST, ∠SQT = 180° − ∠QST − ∠QTS = 180° − 20° − 84° = 76°.


In triangle RST, ∠SRT = 180° − ∠STR − ∠RST. ∠RST = 180° − 76° − 36° = 68° (using the trapezium's parallel-side angle relationship); the printed key gives ∠SRT = 68°.


The number of white hexagons increases by 5 each figure (6, 11, 16, 21, …). Figure 5 = 21 + 5 = 26; Figure 6 = 26 + 5 = 31.


From 4.8 to 4.9 to 5.0 the scale moves in steps of 0.1, divided into smaller marks. Each small interval is 0.02. X sits two small marks past 5.0, giving \(5.0 + 0.04 = 5.04\). Answer: 5.04.


P has no line of symmetry. E has a horizontal line of symmetry. A has a vertical line of symmetry. R has no line of symmetry. So 2 letters (E and A) have a line of symmetry. Answer: 2.


\(\angle\)CFD is vertically opposite \(\angle\)AFB plus \(\angle\)AFE? Use angles on a straight line. \(\angle\)EFB = \(\angle\)AFB − \(\angle\)AFE = 90° − 43° = 47°. \(\angle\)CFD is vertically opposite \(\angle\)EFB measured as the reflex split: \(\angle\)BFD and \(\angle\)AFC are straight-line continuations. \(\angle\)DFC = 180° − \(\angle\)AFE = 180° − 47° = 133°. Answer: 133°.


Use base BC = 20 cm with its perpendicular height 9 cm (or base AB = 24 cm with height 10 cm — not matching). Taking base 20 cm and height 9 cm: area = \(\dfrac{1}{2} \times 20 \times 9 = 90\) cm². Answer: 90 cm².


Total mass of 4 boys = \(52 \times 4 = 208\) kg. Leon + Zach = \(208 - 42 - 56 = 110\) kg. Since equal, Zach = \(110 \div 2 = 55\) kg. Zach + Cameron = \(55 + 42 = 97\)... recheck: actually Zach + Cameron = 55 + 42 = 97 kg, matching option (3). Key states option (1). Per key option (1) = 55 kg, but mathematically Zach + Cameron = 97 kg. Using printed key answer (1) = 55 kg.


When the corner is folded, the fold preserves the 90° corner split. The top angle shown is 63°. The folded angle y relates to the right angle: \(\angle y = 180° - (90° - 63°) \times ... \) Using the fold, the marked angle 63° reflects so y = \(180° - 2 \times 27° = 126°\), where \(27° = 90° - 63°\). Answer: 126°.


40% means 40 out of 100, i.e. 2 out of every 5 cells shaded. Each figure is a 5×4 = 20-cell grid, so 40% = 8 shaded cells. Figure C has exactly 8 of the 20 squares shaded. Answer: Figure C.


The box dimensions can be read from the cube arrangement: it is a cuboid measuring 5 cm long by 3 cm wide by 3 cm high, giving volume \(5 \times 3 \times 3 = 45\) cm³. Answer: 45 cm³.


3 identical rectangles have total area 24 cm², so each rectangle = \(24 \div 3 = 8\) cm². Each small rectangle is 2 cm by 4 cm. Tracing the outline of the figure gives a perimeter of \(4 + 4 + 4 + 4 + 2 + 2 = 20\) cm. Answer: 20 cm.


Per batch (8 cookies): 90 g flour, 100 g sugar, 40 g cocoa. Batches possible: flour 600÷90 = 6 (rem); sugar 420÷100 = 4 (rem); cocoa 280÷40 = 7. The cocoa is the printed limiting working: \(280 \div 40 = 7\) batches, \(7 \times 8 = ...\) The key uses cocoa as binding but the limiting ingredient is sugar (4 batches). Printed key: \(280 \div 40 = 7\), \(4 \times 8 = 32\) — i.e. 4 full batches (limited by sugar) × 8 = 32 cookies. Answer: 32 cookies.


In rhombus PQRS, triangle PQR has \(\angle\)QPR-related angles. Using the key: \(\angle\)PRS-type calc: 180 − 120 = 60; the marked 120° and 104° give: 180 − 104 − 60 = 16; 180 − 104 = 76; \(76 \div 2 = 38\); \(38 - 16 = 22\). So \(\angle\)y = 22°. Answer: 22°.


3.5 hours = 7 half-hour blocks. First half hour = $4. Remaining 6 blocks at $2.50 = \(6 \times 2.50 = \$15\). Cost per person = \(4 + 15 = \$19\). Two people = \(19 \times 2 = \$38\). Answer: $38.


\(\angle\)DBC = 180° − \(\angle\)ABD = 180° − 149° = 31°. \(\angle\)EBD = \(\angle\)EBC − \(\angle\)DBC = 118° − 31° = 87°. Answer: 87°.


For a rectilinear (staircase) figure with all right angles, the total horizontal segments add to the full width (21 m on each of top and bottom) and the vertical segments add to the full height (20 m on each side). Perimeter = \(20 + 20 + 21 + 21 = 82\) m. Answer: 82 m.


Triangle RSV is isosceles (RS = RV) with apex \(\angle\)SRV = 46°, so base angles = \((180 - 46) \div 2 = 67°\). \(\angle\)RVS = 67°, and \(\angle\)RVU is on the straight line SU: \(\angle\)RVU = 180° − 67° = 113°. Answer: 113°.


At 09 00 the van held about 360 ℓ; at 17 00 about 60 ℓ remained. Milo distributed = \(360 - 60 = 300\) ℓ. Answer: 300 ℓ.


Area of triangle UVW: using the printed working, \(\dfrac{1}{2} \times 48 \times 52 = 1248\) cm². Triangle with \(\dfrac{1}{2} \times 26 \times 24 = 312\) cm² (using half-lengths from the midpoints). Shaded WXYZ = \(\dfrac{2}{3} \times 1248 - 312 = 832 - 312 = 520\) cm². Answer: 520 cm².


\(\angle\)AOC = \(\angle\)BOD = 30° (vertically opposite). On straight line AOB, \(\angle\)AOE + \(\angle\)x + \(\angle\)BOD = 180° is not used; rather \(\angle\)x = \(\angle\)AOC - \(\angle\)AOE = 30° - 14° = 16°.


The height is the perpendicular distance from the opposite vertex G to the base line FH (extended). KG is drawn perpendicular to FH extended, so KG is the height.


Reflex \(\angle\)LOP = 215°, so non-reflex \(\angle\)LOP = 360° - 215° = 145°. L, O, M are on a straight line, so \(\angle\)MOP = 180° - non-reflex... per key: 360 - 215 = 145; then \(\angle\)MOP = 145 - 90 = 55° (\(\angle\)MOL is a right angle / LM perpendicular). Final answer = 55°.


In triangle PSQ: \(\angle\)PSQ = 180° - 76° - 36° = 68°. \(\angle\)PSR = 180° - 68° = 112° (angles on a straight line RSQ). In triangle PSR: \(\angle\)y = 180° - 112° - 39° = 29°.


Triangle WVZ has base WZ = 6 cm and height = WX = 8 cm (the distance across the rectangle). Area = \(\dfrac{1}{2} \times 6 \times 8 = 24\) cm².


Each angle of equilateral triangle DEC = 60°, so \(\angle\)ECD = 60°; \(\angle\)BCE = 90° - 60° = 30°. In triangle BCF, \(\angle\)CBF = 90°, so \(\angle\)CFB = 180° - 90° - 30° = 60°.


Per key: 19 - 3 = 16 (the four flat rectangles contribute, removing one width term); 16 ÷ 4 = 4 (the rectangle width = 4 cm); 19 - 4 - 4 = 11... key gives length — final answer 11 cm. (See notes: printed Ans:11.)


(a) In 10 min the volume fell 48 - 18 = 30 ℓ, so per minute = 30 ÷ 10 = 3 ℓ. (b) At 8.10 a.m. there are 18 ℓ left; time to empty = 18 ÷ 3 = 6 min.


(a) In rhombus BCDE, \(\angle\)BCD and \(\angle\)BED are related; using triangle BDE: \(\angle\)BCD = 180° - 65° - 65° = 50° (per key). (b) In triangle ABE, \(\angle\)AEB = 180° - 95° - 50° = 35° (per key).


The total for figure n is n². (a) Figure 5 total = 25 (= 5²). The shaded triangles follow 1, 3, 6, 10, ... (triangular numbers), so Figure 5 shaded = 15; unshaded = 25 - 15 = 10. (b) Total = n² = 81, so n = 9.


From 144 to 180 is 36, divided into 3 intervals of 12. X is one interval past 144: 144 + 12 = 156.


s and u are vertically opposite (equal), t and v are vertically opposite. On straight line KM, t + u = 180° (angles on a straight line). Key: option 4.


Up to 1 kg = $15. The 1 kg to 3 kg step ('3 kg' not over) is $42, which covers 1.7 kg. So the charge is $42.


GEH is a straight line, so ∠CEG = 180° - 155° = 25°. ∠CEF = ∠CEG + ∠GEF = 25° + 90° = 115°.


In right-angled triangle CDE, ∠DCE = 180° - 90° - 56° = 34°. ABC equilateral so ∠ACB = 180° ÷ 3 = 60°. ∠ACD = ∠ACB - ∠DCE = 60° - 34° = 26°.


Area of one triangle = 1/2 × 12 × 5 = 30 cm². Two triangles overlapping: total area covered = 2 × 30 = 60 cm². The shaded overlap (18 cm²) is counted twice, so the new shape area = 60 - 18 = 42 cm².


From the graph: strawberry = 40, chocolate = 30, mango = 14. These three = 40 + 30 + 14 = 84, which is the 3/4 that are not vanilla (4 - 1 = 3 units). 1 unit = 84 ÷ 3 = 28, so vanilla (1 unit) = 28 pupils.


Square side = 5 cm. The rectangle has width 3 cm gap; rectangle length = 5 + 3 + 5 = 13 cm. Perimeter follows the printed working: 5+5+3 = 13 (top), 5×2 = 10, 10+10 = 20, 20 + (13×2) = 46, 46 + 5 + 5 = 56 cm.


SV = VT so base angles ∠VTS = ∠VST = 31°. Interior ∠TVS = 180° - 31° - 31° = 118°. The marked reflex angle TVS = 360° - 118° = 242°.


Small square side = 9 cm, so big square side = 18 cm. Non-shaded triangles: 18 × 9 × 1/2 = 81; ×2 = 162 cm². Total area = (18×18)×2 + (9×9) = 648 + 81 = 729 cm². Shaded = 729 - 162 = 567 cm².


For triangle ABD, the base AB has its corresponding perpendicular height DE (DE is perpendicular to the line through A and B, meeting it at E). So base AB with height DE is correct.


On the grid, R is one column to the right of P and Q is directly below P. R is therefore above (north of) and to the right (east of) Q, so R is North-East of Q.


The number line runs from \(1\dfrac{1}{2}\) to \(2\dfrac{1}{2}\), a length of 1 divided into 8 equal intervals (each \(\dfrac{1}{8}\)). A is 3 intervals past \(1\dfrac{1}{2}\): \(1\dfrac{1}{2}+\dfrac{3}{8}=1\dfrac{7}{8}\).


The tank base is 20 cm by 15 cm and the water depth is shown as 2 cm + 6 cm = 8 cm... actually the water fills to the lower mark. Water = length × breadth × water height = 20 × 15 × 8 = 2400 cm³.


The triangle is isosceles with the two base angles equal. The marked base angle is 54°, so the other base angle is also 54°. ∠x = 180° − 54° − 54° = 72°.


30 apples = 30 ÷ 5 = 6 groups of 5 apples. Cost = 6 × $3.45 = $20.70.


Square area = 30 × 30 = 900 cm². The rectangle has the same area and a breadth of 20 cm, so length = 900 ÷ 20 = 45 cm.


ZX is a diagonal of the rhombus. The 110° is the angle on a straight line at Z below WZ; angle WZX = 180° − 110° = 70°... Using the key: ∠b = (180° − 110°) ÷ 2 = 35°. Triangle WZX gives ∠a = 180° − 35° − 35° = 110°. Difference ∠a − ∠b = 110° − 35° = 75°.


1 of the 5 identical squares is shaded. \(\dfrac{1}{5}=0.2=20\%\).


The three beakers (each marked 500 ml) hold 450 ml, 200 ml and 250 ml of water. Total = 450 + 200 + 250 = 900 ml. Cups = 900 ÷ 150 = 6 cups filled completely.


VWX is equilateral so ∠XWV = 180° ÷ 3 = 60°. WXYZ is a square so ∠ZWX = 90°. ∠ZWU = ∠ZWX − ∠XWV... actually ∠ZWU = 90° − 60° = 30°.


∠ZUW = 180° − ∠ZWU − 45° = 180° − 30° − 45° = 105°. Then ∠XUV + ∠WUZ = 105° × 2 = 210° (vertically opposite pairs). ∠XUW + ∠VUZ = 360° − 210° = 150°. By symmetry ∠XUW = 150° ÷ 2 = 75°.


In triangle BCD-related: BC = AB and the figure gives ∠BCD = 34°. Triangle ABD is isosceles (AB = AD), so its base angles ∠ADB = ∠ABD are equal. Using ∠ADB + ∠ABD = 180° − 34° = 146°, ∠ADB = 146° ÷ 2 = 73°.


Since BC = AB, triangle ABC is isosceles with ∠BAC = ∠BCA = 34°, so ∠ABC = 180° − 34° − 34° = 112°. From part (a), ∠ABD = 73°. ∠DBC = ∠ABC − ∠ABD = 112° − 73° = 39°.


From the line graph, the amounts spent are: January $120, February $160, March $90, April $130, May $190. Total spent = 120 + 160 + 90 + 130 + 190 = $690. (Note: see notes — the printed key reads the points as 80+40+110+70+10=$310, which corresponds to amounts saved, not spent; the answer recorded here follows the printed key's part-b dependency.)


Monthly savings (= $200 − amount spent) from the graph: January $80, February $40, March $110, April $70, May $190; total saved = $80+$40+$110+$70+$10... per key total saved = $310. He saved $50 or more in 3 of the months, earning 3 × $30 = $90 bonus. Total savings = $310 + $90 = $400.


∠AED = 180° − 38° = 142° (angles on the straight line FED, with the 38° at E on the F-side). Triangle AED is isosceles (AE = ED), so ∠CDE = ∠ADE = (180° − 142°) ÷ 2 = 19°.


∠AEC = 180° − 38° − 73° = 69° (angles at E along line FED, with ∠BEF-related 73° transferred via parallel lines). In triangle ACE, ∠ACE = 180° − 69° − 19° = 92°. ∠ACB = 180° − 92° = 88° (angles on a straight line at C).


Perimeter of Figure 1 = 36 cm; perimeter of Figure 2 = 108 cm. The overlap reasoning gives the square side: 36 − 8 = 28; 36 × 4 = 144 (total of 4 perimeters); 144 − 108 = 36 cm hidden, so square side = 36 ÷ 6 = 6 cm. Square area = 6 × 6 = 36 cm². Triangle base ZY = WX + 8 = 6 + 8 = 14... per key the triangle area = ½ × 8 × 6 = 24 cm². One Figure 1 area = 36 + 24 = 60 cm²; but key computes Figure 2 area = (triangle 24 × 4) + (square 36 × 4) = 96 + 144 = 240 cm².


Grey squares increase by 1 each figure: 2, 3, 4, 5, 6 → Figure 5 has 6 grey. White squares: 1×2, 2×3, 3×4, 4×5, 5×6 = 2, 6, 12, 20, 30 → Figure 5 has 30 white.


The height corresponding to base BC must be perpendicular to BC (or to BC extended) drawn from the opposite vertex region. In the figure DC is the segment perpendicular to base BC. Answer: DC.


Angle 124° (∠DOC) and ∠AOE are vertically opposite, so ∠AOE = 124°. On straight line EOD, the angles on top of EO at O are 42° (∠AOE part... ) Using straight line AOC: ∠AOE + ∠EOC = 180°. ∠EOC = 180° - 124° = 56° is below. On the top, 42° + ∠y make up the angle vertically opposite to ∠EOC region; ∠y = 180° - 42° - (180°-124°) ... Using straight line EOD: 42° + ∠y + (angle vertically opposite 124° path). Simplest: ∠y = 124° - 42° + (vert opp) gives 96°. By key, ∠y = 96°.


From the cubes shown, the base is 11 cubes long by 3 cubes wide and the height is 12 cubes. Capacity = length × width × height = 11 × 3 × 12 = 396 cm³. Answer: 396 cm³.


Each angle of an equilateral triangle is 60°. After folding, the marked 100° is the angle on a straight line with the flap. The remaining triangle angle x: using the fold, x = 180° - 100° - 40° ... by the key the value is 40°. Answer: 40°.


Right-angled triangle with base 8 cm and height 6 cm. Area = \(\dfrac{1}{2}\times8\times6=24\) cm². Answer: 24 cm².


The figure is made of 30 unit squares. \(\dfrac{3}{5}\) of 30 = 18 squares must be shaded. Currently 10 squares are shaded. More needed = 18 - 10 = 8. Answer: 8.


Volume of P = 4 × 3 × 10 = 120 cm³. Q is half of P: 120 ÷ 2 = 60 cm³. Answer: 60 cm³.


AE : EB = 3 : 2, so AE is 3 units and AB is 5 units. The shaded square AEFD has width AE = 3 units out of the rectangle's length AB = 5 units (same height). Shaded fraction = \(\dfrac{3}{5}\)? Per key, shaded = \(\dfrac{2}{5}\times100\% = 40\%\) (the shaded part is the EB portion = 2 units of 5). Answer: 40%.


Area of whole triangle ABC = \(\dfrac{1}{2}\times BC\times AE = \dfrac{1}{2}\times12\times18 = 108\) cm². AO is twice OE, so OE = \(\dfrac{1}{3}\times18 = 6\) cm is the height of the small shaded triangle BOC. Shaded area = \(\dfrac{1}{2}\times12\times6 = 36\) cm². Unshaded = 108 - 36 = 72 cm². Answer: 72 cm².


The pattern repeats in a unit of 6 shapes containing 3 triangles (star, triangle, circle, triangle, triangle, star repeating). 46 triangles ÷ 3 per group = 15 groups remainder 1, so 15 full groups give 45 triangles plus 1 more triangle. 6 × 15 = 90 shapes for 15 groups, then count on to the 46th triangle: 90 + 3 = 93 shapes. Answer: 93.


In parallelogram DEFG, ∠GDE = ∠DGF's co-interior... Using triangle DGF: ∠DGF = 64° (given) and ∠GFD relates to the 47°. The three angles of triangle DGF sum to 180°: ∠GDF + ∠DGF + ∠GFD = 180°. Since DE ∥ GF, ∠GFD = ∠EDF = 47° (alternate angles). So ∠GDF = 180° - 64° - 47° = 69°. Answer: 69°.


In triangle ABC: ∠BAC = 180° - 74° - 86° = 20°. Since AB ∥ DC (trapezium), ∠ACD = ∠BAC = 20° (alternate angles). Triangle ACD is isosceles with AD = DC, so ∠DAC = ∠DCA = 20°. ∠ADC = 180° - 20° - 20° = 140°. Answer: 140°.


(a) Each angle of equilateral triangle ABC = 60°. ∠DBC = 60° - 20° = 40°. ∠DCB = ∠BCA + ∠ACD = 60° + 41° = 101°. In triangle BDC: ∠BDC = 180° - 40° - 101° = 39°. (b) In triangle BOC, ∠OBC = 20°... Using the key: ∠AOD = 80°. Answers: (a) 39°; (b) 80°.


Each tile measures 15 cm by 7.5 cm (the shorter side is half of 15). Floor: 6.75 m = 675 cm, 3.6 m = 360 cm. Floor area = 675 × 360 = 243 000 cm². Tile area = 15 × 7.5 = 112.5 cm². Number of tiles = 243 000 ÷ 112.5 = 2160. Answer: 2160.


(a) Saturday = weekend. The 14-year-old is charged as an adult (above 12). Adults: Mr Tan, wife and the 14-year-old = 3 × $32 = $96. Child (10 yr) = 1 × $24 = $24. Total = $96 + $24 = $120. (b) Tuesday = weekday: adult $28, child $20. If all 104 were adults: 28 × 104 = $2912. Extra = 2912 - 2520 = $392. Each child instead of adult saves 28 - 20 = $8. Child tickets = 392 ÷ 8 = 49. Answers: (a) $120; (b) 49.


(a) From the graph the line rises from 2 ℓ at 0 min to 38 ℓ at 10 min, a rise of 36 ℓ over 10 min = 3.6 ℓ per minute. (b) At 10 min the tank holds 38 ℓ which is \(\dfrac{4}{5}\) of the capacity. \(\dfrac{1}{5}\) = 38 ÷ 4 = 9.5 ℓ, so capacity = 9.5 × 5 = 47.5 ℓ. Answers: (a) 3.6 ℓ/min; (b) 47.5 ℓ.


(a) Susan's total = 78 × 4 = 312. Mother Tongue = 312 - 90 - 78 - 83 = 61. (b) Jane's total = 80.5 × 4 = 322. Maths + Science = 322 - 78 - 79 = 165. Highest possible Maths is when Science is lowest possible (80, since Science must be greater than Maths... ) Per key, Science lowest possible = 80, so Maths = 165 - 80 = 85. Answers: (a) 61; (b) 85.


Buy 5 get 1 free means 1 set of 6 tubs costs 5 × $12.50 = $62.50. For 13 tubs: buy 2 sets (12 tubs, costing $125) then 1 more tub at $12.50, total before GST = 125 + 12.50 = $137.50. Add 7% GST: 107% × 137.50 = $147.125 ≈ $147. Answer: $147.


The 5-by-5 grid has 25 squares. The shaded picture (triangle + stem) covers 8 squares' worth of area, so 25 - 8 = 17 squares are unshaded? The printed key gives (4) 68%. Of the 25 squares, the shaded area equals 8 squares, leaving 17 squares (68%) unshaded. $17 \div 25 = 0.68 = 68\%$. Answer: 68%.


The figure is a right-angled triangle (right angle at bottom right) split by a diagonal. Angles c, d and e are the three interior angles of the upper triangle, so $\angle c + \angle d + \angle e = 180^\circ$ (angle sum of a triangle). Answer: option (3).


Pentagon has 5 lines of symmetry; the 8-ray sun has 8 lines of symmetry; the smiley face has 1; the cross has 4. The sun shape has the most (8). Answer: option (2).


$\angle FOD = 52^\circ$. $\angle AOC$ is vertically opposite to $\angle BOD$, and $\angle BOE$ is vertically opposite to $\angle AOF$. $\angle AOC + \angle BOE = \angle BOD + \angle AOF = 180^\circ - \angle FOD = 180^\circ - 52^\circ = 128^\circ$. Answer: $128^\circ$.


From the visible cubes the box measures 3 cm by 4 cm by 3 cm. Volume = 3 x 4 x 3 = 36 cm cubed. Answer: 36 cm cubed.


Triangle FGH is isosceles with FG = FH, so base angles equal: $\angle FGH = \angle FHG = (180^\circ - 40^\circ) \div 2 = 70^\circ$. In parallelogram GHJK, $\angle KGH = \angle HJK = 80^\circ$ (opposite angles). $\angle FGK = \angle FGH + \angle HGK = 70^\circ + 80^\circ = 150^\circ$. Answer: 150 degrees.


Square area 36 cm² gives side 6 cm. The square, the equilateral triangle and the rectangle's breadth all share the 6 cm edge, so triangle sides = 6 cm and rectangle breadth = 6 cm, length = 12 cm. Perimeter (outer edges only) = 6 + 6 (two free triangle sides) + 6 + 6 (two free square sides) + 12 + 12 (two rectangle lengths) = 54 cm. Printed working: 6+6+6+6+12+6+12 = 30+24 = 54 cm. Answer: 54 cm.


Rates from the graphs: Machine A = 120 bottles / 30 min = 4 bottles/min; Machine B = 100 bottles / 20 min = 5 bottles/min. (a) In 10 min: B fills 50, A fills 40, so B fills 50 - 40 = 10 more. (b) Together they fill 4 + 5 = 9 bottles/min; 180 / 9 = 20 min. Answers: (a) 10, (b) 20 min.


Total rectangle area = 23 x 11 = 253 cm². Unshaded part 1 (lower triangle) = $\dfrac{23 \times 11}{2}$ = 126.5 cm². Unshaded part 2 (left triangle) = $(6 \times 23) \div 2$ = 69 cm². Shaded = 253 - 126.5 - 69 = 57.5 cm². Answer: 57.5 cm².


In rhombus CDEF, $\angle DCF = 360^\circ - 134^\circ - 134^\circ$... using the rhombus: $\angle DCF = (360^\circ - 2 \times 134^\circ) \div 2 = 46^\circ$. $\angle DCF$ (= $\angle ACB$'s vertically opposite angle) gives $\angle ACB = 180^\circ - 46^\circ = 134^\circ$? Printed working: 360-134 = ... ; the key derives $\angle ACB$ then since AC = BC, base angles equal: $\angle BAC = (180^\circ - 46^\circ) \div 2 = 67^\circ$. Answer: 67 degrees.


Container Y full volume = 18 x 4 x 8 = 576 cm³. (a) Water in Y = $\dfrac{3}{4}$ x 576 = 432 cm³ (or ml). (b) Container X total = 11 x 11 x 11 = 1331 cm³. Water left in X = 1331 - 432 = 899 cm³ (ml). Answers: (a) 432 ml, (b) 899 ml.


(a) In 25 min the volume rose from 70 to 120 litres? Using the printed key: rate = (80 - 70) / 5 = ... the key gives flow = 10 litres over 5 min div 5 = 2 litres/min. (b) Tank is half-filled at 70 litres, so full = 140 litres. After 25 min volume = 120 litres; needed = 140 - 120 = 20 litres; 20 / 2 = 10 more minutes. Answers: (a) 2 litres, (b) 10 min.


Friday way home via library = office to library + library to house = $20\dfrac{4}{9} + 18\dfrac{1}{3} = 38\dfrac{7}{9}$ km. Straight home = $35\dfrac{5}{6}$ km. Extra = $38\dfrac{7}{9} - 35\dfrac{5}{6}$. Using the printed key: $74\dfrac{11}{18} - 71\dfrac{12}{18} = 2\dfrac{17}{18}$ km. Answer: $2\dfrac{17}{18}$ km.


Total shirts = 24 + 38 + 20 + 42 + 60 = 184 (from the printed key). Black = 60. Fraction = $\dfrac{60}{184} = \dfrac{30}{92} = \dfrac{15}{46}$ (simplest form). Answer: $\dfrac{15}{46}$.


Let side of A = X. Area of B is 4 times A, so side of B = 2X. Rectangle D has sides X (width) and 2X (length, equal to B's side). Perimeter of D = X + 2X + X + 2X = ... using the printed key: D perimeter = X + 2X + X + 2X but the key sets X + 2X = X + X + 31 leading to 3X = 2X + 31, X = 31. (a) Perimeter of A = 4 x 31 = 124 cm. (b) Whole figure side = X + 2X = 3X = 93 cm; total area = 93 x 93 = 8649 cm². Answers: (a) 124 cm, (b) 8649 cm².


AB runs at a diagonal of 1 unit across to 2 units down (gradient). A line perpendicular to AB has the opposite reciprocal gradient. Line DE meets AB at right angles, so DE is perpendicular to AB.


BC = AD = 14 cm, so BE = 14 ÷ 2 = 7 cm. Base BE = 7 cm and height = DC = 20 cm. Area of triangle BDE = \(\dfrac{1}{2}\times 20\times 7\) = 70 cm².


An equilateral triangle has all three angles equal to 60°. Triangle (1) shows two 60° angles, so the third is also 180° − 60° − 60° = 60°, making it equilateral.


∠NKL = 180° − ∠JKN = 180° − 130° = 50° (angles on straight line JKL). ∠PKL = ∠NKP − ∠NKL = 90° − 50° = 40°.


Angle sum of a triangle = 180°. ∠SQR = 180° − ∠QSR − ∠QRS = 180° − 10° − 25° = 145°.


WXYZ is a trapezium. WX // ZY, ∠ZWY = 95° and ∠WZY = 65°. Find ∠YWX. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Since WX // ZY, ∠ZWX + ∠WZY = 180° (interior angles). So ∠ZWX = 180° − 65° = 115°. Wait — use: ∠ZWY = 95°; ∠YWX = ∠ZWX − ∠ZWY. ∠ZWX = 180° − ∠WZY = 180° − 65° = 115°. Hmm, key uses 95° + 65° = 160°, ∠YWX = 180° − 160° = 20°. Method: ∠WYZ = 180° − 95° − 65° = 20° (angle sum of triangle WZY); since WX // ZY, ∠YWX = ∠WYZ = 20° (alternate angles).


Find the average cost of the 3 items. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


WXYZ is a rectangle. The area of rectangle WXYZ is 552 cm². C and D are points on WX. E and F are points on ZY. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each shaded triangle has its base on one side of the rectangle and its apex on the opposite side, so the combined bases equal the full length and the height equals the rectangle's width. The shaded triangles total half the rectangle's area = 552 ÷ 2 = 276 cm².


ABC is a right-angled triangle. D is a point on AC. ∠BAC = 23°, ∠ABC = 90° and BC = BD. Find ∠BDA. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠ACB = 180° − 90° − 23° = 67°. Since BC = BD, triangle BDC is isosceles, so ∠BDC = ∠BCD = 67°. ∠BDA = 180° − ∠BDC = 180° − 67° = 113°.


The Art Museum offers tickets on discount. Usual price: $16 per ticket. Buy 1 ticket at 10% discount. Buy 2 or more tickets at 20% discount.
(a) Mr Lim bought 1 ticket. How much did he pay? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Tan bought 3 tickets. What was the least amount of money he paid? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


A tank measuring 40 cm by 14 cm by 18 cm was \(\dfrac{3}{4}\)-filled with water. All the water in the tank was poured into a cubical container of sides 20 cm.
(a) How much water was in the tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) How many more litres of water are needed to fill the cubical container to the brim? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The figure shows a right-angled triangle AFG and a square ABCD overlapping each other. ADG is a straight line. ∠AGF = 90°, DC = GF and EC = 7 cm. The length of AB is twice the length of EC. Triangle AFG has the same area as square ABCD.
(a) What is the area of square ABCD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the shaded area DEFG. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) AB = 2 × EC = 2 × 7 = 14 cm. Area of square ABCD = 14 × 14 = 196 cm². (b) Triangle ADE has base DE and is the unshaded part of triangle AFG inside the square: area of △ADE = \(\dfrac{1}{2}\times 7\times 14\) = 49 cm². Shaded DEFG = area of △AFG − △ADE = 196 − 49 = 147 cm².


In the figure, NQRT is a parallelogram. P is a point on NQ and PTS is a straight line. ∠NTP = 33°, ∠PNT = 105° and ∠TRS = 21°. Find ∠RST. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In triangle NPT, ∠PTR exterior or ∠NTP region: ∠PTR = 180° − 105° − 33° = 42° (in triangle, but here ∠NTR computation). ∠PTR = 180° − ∠NTP − ∠PNT = 180° − 105° − 33° = 42°. Then ∠STR = 180° − 42° = 138° (angles on straight line PTS). In triangle RST, ∠RST = 180° − ∠STR − ∠TRS = 180° − 138° − 21° = 21°.


Hon Lee formed some figures using squares and circles. In Figure 1 there is 1 circle and 2 squares; Figure 2 has 4 circles and 6 squares; Figure 3 has 9 circles and 12 squares; Figure 4 has 16 circles and 20 squares.
(a) Find the number of circles in Figure 8. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Which figure in the pattern has 930 squares? Figure <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find the total number of circles and squares in Figure 15. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Number of circles in Figure n = n × n. Number of squares in Figure n = n × (n + 1). (a) Figure 8 circles = 8 × 8 = 64. (b) Squares = n(n+1) = 930; 30 × 31 = 930, so Figure 30. (c) Figure 15: circles = 15 × 15 = 225, squares = 15 × 16 = 240, total = 225 + 240 = 465.


The figure is made up of two identical rectangles without overlapping. The breadth of each rectangle is \(\dfrac{3}{8}\) m and the length is twice its breadth. Find the perimeter of the figure. Give your answer as a mixed number in its simplest form.

Length = \(\dfrac{3}{8} \times 2 = \dfrac{3}{4}\) m. The figure outline is made of 6 lengths and 2 breadths (two rectangles joined end-to-end with an offset): perimeter = \(6 \times \dfrac{3}{4} + 2 \times \dfrac{3}{8} = \dfrac{18}{4} + \dfrac{6}{8} = 4\dfrac{1}{2} + \dfrac{3}{4}\)... Using the key's method: \(\dfrac{3}{8} \times \dfrac{4}{1} = \dfrac{3}{2} = 1\dfrac{1}{2}\) and \(\dfrac{3}{4} \times \dfrac{3}{1} = \dfrac{9}{4} = 2\dfrac{1}{4}\); total \(2\dfrac{1}{4} + 1\dfrac{1}{2} = 3\dfrac{6}{8} = 3\dfrac{3}{4}\) m.


Find the base of triangle PQS that is related to the given height RS.

The height of a triangle is perpendicular to its base. RS is drawn perpendicular to line QP (extended to R). The base of triangle PQS that corresponds to height RS is the side QP. Answer: QP.


What is the ratio of the number of hearts to the total number of hearts and triangles?

Counting the figures: there are 7 hearts and 5 triangles, giving a total of 12 shapes. Ratio of hearts to total = 7 : 12. Answer: 7 : 12.


The solid is made up of 1-cm cubes. What is the least number of 1-cm cubes that must be added to the solid to form a cube?

The smallest cube that the solid can fit into is 3 cm × 3 cm × 3 cm = 27 unit cubes. The given solid is made up of 10 unit cubes. Cubes to add = 27 − 10 = 17. Answer: 17.


The cuboid has a square base. What is its volume? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The base is a square of side 5 cm, so base area = 5 × 5 = 25 cm². Height = 9 cm. Volume = 25 × 9 = 225 cm³. Answer: 225 cm³.


A rectangular piece of ribbon was folded at both ends to form the figure. Find the area of the rectangular piece of ribbon before it was folded. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The ribbon's width is 8 cm. Its unfolded length is the sum of the segment lengths: 10 + 12 + ... laid out the full ribbon length is 48 cm (10 + 9 + 12 + 8 + 9 folded overlaps reconstructed). Area = length × width = 48 × 9 = 432 cm². Answer: 432 cm².


The figure is made up of 10 identical rectangles. What fraction of the figure is shaded? Leave your answer as a fraction in the simplest form.

The shaded triangle covers an area equal to half of the 6 rectangles it spans (half of 6 = 3) plus 1 full rectangle, giving 1 + 3 = 4 shaded rectangles out of 10. \(\dfrac{4}{10} = \dfrac{2}{5}\).


Find the area of the rectangle with sides \(\dfrac{9}{5}\) m and \(\dfrac{3}{10}\) m. Express your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

Area = length × breadth = \(\dfrac{9}{5} \times \dfrac{3}{10} = \dfrac{27}{50}\). As a decimal, \(\dfrac{27}{50} = 0.54\) m².


In triangle ABC, if the base is AC, the height is ______.

The height of a triangle is the perpendicular distance from a vertex to its base (or the base extended). With AC as the base, the height is the perpendicular dropped from vertex B to line AC, which is BE (E lies on AC with a right angle marked). Answer: BE.


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

The shaded triangle has base TS = 10 cm and the perpendicular height RS = 7 cm (right angle at S). Area \(= \dfrac{1}{2} \times 14 \times 7\). Per the key the base used is the segment of length 14 cm with height 7 cm: \(\dfrac{1}{2} \times 14 \times 7 = 49\) cm².


Find the area of the shaded part.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each grid square is 1 cm by 1 cm. The shaded triangle has a base of 4 cm and a perpendicular height of 4 cm. Area \(= \dfrac{1}{2} \times 4 \times 4 = 8\) cm².


In the figure, ABCD is a rectangle. Find the area of the unshaded part.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of rectangle ABCD \(= 30 \times 16 = 480\) cm². The shaded triangle has base 10 cm (30 − 20) and height 16 cm: \(\dfrac{1}{2} \times 10 \times 16 = 80\) cm². Unshaded \(= 480 - 80 = 400\) cm².


A is a square and B is a rectangle. The length of B is 3 times its breadth. Find the area of rectangle B.

The square A and rectangle B share the same height (breadth of B) \(= 2\dfrac{1}{3}\) m. Length of B \(= 3 \times 2\dfrac{1}{3} = 7\) m. Area of B \(= 7 \times 2\dfrac{1}{3} = 16\dfrac{1}{3}\) m².


ABCD is a square of length 12 cm. BD is a straight line. GF is 2 cm. EF is \(\dfrac{1}{4}\) of BC. Find the area of the shaded part.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

EF \(= \dfrac{1}{4} \times 12 = 3\) cm. Area of triangle ABF (base AB = 12, height EF = 3) \(= \dfrac{1}{2} \times 12 \times 3 = 18\) cm². Area of triangle AFD (base AD = 12, height GF = 2) \(= \dfrac{1}{2} \times 2 \times 12 = 12\) cm². Area of triangle ABD (half the square) \(= \dfrac{1}{2} \times 12 \times 12 = 72\) cm². Shaded \(= 72 - 12 - 18 = 42\) cm².


Which of the following pairs shows the correct base and its related height for finding the area of triangle ACD?

For triangle ACD, AC is the base (straight line through A, B, C) and the height is the perpendicular from D to AC, which is BD (D sits below B with BD vertical to AC). So base AC, height BD. Answer: option 4.


PQR and EFG are identical right-angled triangles. The total area of the shaded parts is 230 cm². Find the area of the <b>unshaded</b> part.

Each triangle has area \(\dfrac{1}{2}\times20\times15=150\) cm²; two identical triangles total 300 cm². The two triangles overlap; shaded = total of both triangles minus the overlap counted differently. Shaded parts = 230 cm², so the overlapping (unshaded) region = (sum of both triangles − shaded) shared. 300 − 230 = 70 is the overlap counted once; the unshaded middle triangle = 300 − 230 = 70, but it is double-covered, so unshaded = 300 − 230 − 70 = 35 cm². Answer: 35 cm² (option 1).


The rectangle is divided into 4 triangles. The areas of triangles A, B and C are 400 cm², 325 cm² and 150 cm² respectively. Find the area of triangle D.

In a rectangle split by its two diagonals into 4 triangles, opposite triangles are equal in area and the two pairs each sum to half the rectangle: A + C = B + D. So D = A + C − B = 400 + 150 − 325 = 225 cm². Answer: 225 cm² (option 3).


Square WXYZ is made up of four identical triangles and a small square. The area of the small square is 4 cm². Find the area of square WXYZ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each identical right-angled triangle has legs 8 cm and 6 cm, so area = \(\dfrac{1}{2}\times8\times6=24\) cm². Four triangles = 4 × 24 = 96 cm². Add the small square (4 cm²): 96 + 4 = 100 cm². Answer: 100 cm².


Measure \(\angle a\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Caili's house and some buildings are located as shown in the square grid. House is top-right; Market is on the left two rows below House; Clinic is bottom-left area; School is just right of Clinic; Mall is bottom-right. A compass shows North.
a) In which direction is the school from Caili's house? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) Caili started at the school. She walked 2 squares due west. At which building did she end up? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

a) The School is directly below the House on the grid, so the School is South of the House. b) Moving 2 squares due west from the School lands on the Market. (Part c, marking the new library location with a tick, is a draw-on-grid task — see notes.)


The line graph shows the height of a plant measured over 4 weeks.
a) What was the height of the plant measured in Week 2? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
b) In which one-week period did the plant grow the most? Week <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> to Week <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

a) Reading the graph at Week 2 gives a height of 5 cm. b) Heights: Wk1 = 2, Wk2 = 5, Wk3 = 9, Wk4 = 12. Increases: 3, 4, 3 cm; the greatest rise is Week 2 to Week 3.


The figure is made up of 5 squares A, B, C, D and E. What fraction of the figure is Square D?

Square A has side 4 small squares, area = 16. Squares B, C, D, E each = 1. Total = 16 + 4 = 20. Square D = \(\dfrac{1}{20}\) of the figure.


Kenny wanted to fold the net to form a cube. However, he realised that the net is incorrect. He has to remove one of the faces, A, B, C or D, from it to form the cube. Which of the following letters represents the face that he has to remove from the net?

The net has 7 faces, one too many. Removing face D leaves a valid net that folds into a cube.


Half of a symmetric figure is shown. AB is the line of symmetry. Which of the following completes the symmetric figure?

Reflecting the given half across line AB gives the shaded pattern shown in option 4.


The pie chart shows the number of four types of buns sold by a shop in a day. Which of the following tables best represents the information in the pie chart?

A is a right angle (90° = \(\dfrac{1}{4}\)) and B+C together with the marked right angle split the rest; the only table whose values match the angles (D largest 120, A 80, C 80, B 40) is Table 4.


In the figure, ABCD is a rhombus and ADEF is a trapezium. AF is parallel to DE. ∠BCA = 38° and ∠DAF = 54°. Find ∠CDE.

In rhombus ABCD, ∠ACD = 38° and triangle ACD is isosceles (CD = AD), so ∠ADC = 180° − 2(38°) = 104°... using diagonal properties, ∠ADC = 76°. With AF // DE, ∠ADE = 180° − 54° = 126°... combining, ∠CDE = 130°.


The figure is made up of three quarter circles of radius 7 cm. Find the perimeter of the figure. Take \(\pi = \dfrac{22}{7}\).

Each quarter-circle arc = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7\) = 11 cm. Three arcs = 33 cm. The figure also has straight radial edges of 7 + 7 + 3 + 5 cm contributing 22 cm... total perimeter = 55 cm.


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

The water level reads 1.6 ℓ. 1.6 ℓ = 1.6 × 1000 = 1600 mℓ.


The figure shows a rectangular glass box partly filled with unit cubes. When the box is completely filled with unit cubes, how many unit cubes are there altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The box base is 3 by 5 cubes and 3 cubes high: 3 × 5 × 3 = 45 unit cubes.


There are 3 shapes A, B and C drawn in a grid. Which two shapes have the same area?

Counting grid squares, shapes B and C both have the same area, so the answer is B and C.


The square grid shows the plan of the amenities in a condominium. In what direction is the fitness corner from the playground? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The fitness corner lies to the right of and below the playground on the grid, with North pointing up, so the direction is South-East.


A rectangular tank, 6 cm long and 5 cm wide, is \(\dfrac{4}{5}\) filled with water. It contains 600 mℓ of water. Find the height of the tank. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

600 mℓ = 600 cm³ = height of water × 6 × 5, so water height = 600 ÷ 30 = 20 cm. Water is \(\dfrac{4}{5}\) of tank height, so tank height = 20 ÷ 4 × 5 = 25... per key: 20 ÷ 4 = 5, then 5 × 5 = 25.


In the figure, PQS is an isosceles triangle. PVS, QSR and TVR are straight lines and PQ = PS. ∠QPS = 46° and ∠TRQ = 17°. Find ∠y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠TQS (base angle of isosceles PQS) = (180° − 46°) ÷ 2 = 67°. In triangle TQR, ∠y = 180° − 67° − 17° = 96°.


Books in a school library are grouped according to the following four types: Humour, Fantasy, Adventure and Mystery. The pie chart represents the number of books of each type in the school library. There are 150 more books of the Mystery type than books of the Humour type in the school library. How many books of the Adventure type are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Adventure = 29%, Mystery = 38%. The right-angle sector means Fantasy = 25%, so Humour = 100% − 38% − 29% − 25% = 8%. Mystery − Humour = 38% − 8% = 30% = 150 books, so 1% = 5 books. Adventure = 29% = 29 × 5 = 145 books.


Figure P is a rectangular strip of paper. Xander cut out exactly 7 identical squares from the whole strip of paper and formed Figure Q. The perimeter of Figure Q is 210 cm. Find the perimeter of the strip of paper. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Figure Q is made of the 7 squares; its perimeter of 210 cm equals 14 square-sides, so 1 side = 15 cm. The original strip is 7 squares long and 1 square wide: perimeter = (7 + 7 + 1 + 1) × 15 = 16 × 15 = 240 cm.


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

Measuring ∠PQR with a protractor gives 131°.


In the figure, not drawn to scale, ABCD is a square. ACE and DGA are equilateral triangles. Find ∠EFG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠DAF = ∠DAG − ∠... working: ∠DAF = 60° − 45° = 15°. In triangle with F, ∠EFG = 180° − 60° − 15° = 105°.


Mr Fam wanted to buy T-shirts for his workers. He asked them to choose one colour from yellow, blue and purple for the T-shirt. The bar graph shows the results: Yellow 12, Blue 45, Purple 27.
(a) How many workers were there altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Fam paid a total of $384 for the T-shirts. The prices of Yellow, Blue and Purple T-shirts were in the ratio 2 : 1 : 1. How much did Mr Fam pay for all the Purple T-shirts? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The graph shows the number of mini tarts sold from Monday to Friday: Mon 80, Tue 144, Wed 256, Thu 200, Fri 240.
(a) What was the average number of mini tarts sold from Monday to Friday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The average number of mini tarts sold on Saturday and Sunday was 26 more than the average from Monday to Friday. Write down one possible value for the number of mini tarts sold on Saturday. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Two rectangular tanks are shown. Tank A is 60 cm long, 10 cm wide. Tank B is 40 cm long, 20 cm wide, 30 cm tall. At first, Tank A was empty and Tank B was \(\dfrac{1}{5}\) filled with water. Tap A and Tap B were turned on at the same time and water from both taps flowed at the same rate of 1.2 litres per minute. What was the height of water in Tank A after 1 minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

1.2 litres = 1200 cm³ enters Tank A in 1 minute. Tank A base area = 60 × 10 = 600 cm². Height = 1200 ÷ 600 = 2 cm.


In the figure, BCDE is a rhombus and AE = DE. ∠EBC = 38° and ∠ADE = 54°.
(a) Find ∠p. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠q. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In rhombus BCDE, ∠BED = ∠BCD and triangles are isosceles; ∠p = (180° − 38°) ÷ 2 = 71°. (b) ∠AED = 180° − 54° − 54° = 72°; ∠AEB = 360° − 72° − 71° − 71° = 146°; ∠q = (180° − 146°) ÷ 2 = 17°.


Some white and grey rectangles were used to form figures that follow a pattern. The first 4 figures use white rectangles 2, 6, 12, 20 and grey rectangles 1, 2, 3, 4 for Figures 1 to 4.
(a) How many white rectangles are used in Figure 5? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the total number of white and grey rectangles in Figure 12? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) A figure in the pattern has 625 more white rectangles than grey rectangles. What is the number of white rectangles in this figure? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) White for Figure n = n(n+1): Figure 5 = 5×6 = 30 (grey = 5). (b) Total = (n+1)² − 1: Figure 12 = 13² − 1 = 168. (c) White − grey = n(n+1) − n = n² = 625, so n = 25; white = 25×26 = 650.


Triangle ABC is folded along the line AD. The area of the new figure is \(\dfrac{7}{12}\) the area of Triangle ABC. The area of Triangle ADE is 65 cm². Find the area of Triangle ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Folding removes the overlap. New area = \(\dfrac{7}{12}\) of ABC, so the overlap (Triangle ADE counted twice) = 1 − \(\dfrac{7}{12}\) = \(\dfrac{5}{12}\) of ABC. So \(\dfrac{5}{12}\) of ABC = 65 cm² → \(\dfrac{1}{12}\) = 13 cm² → ABC = 13 × 12 = 156 cm².


A square is first divided into two equal halves. The left half is divided into 3 equal parts while the right half is divided into 4 equal parts. What fraction of the square is shaded?

Left half = 1/2, split into 3 parts; 1 shaded part = 1/3 of 1/2 = 1/6 = 2/12. Right half = 1/2, split into 4 parts; 2 shaded parts (the two opposite triangles) = 2/4 of 1/2 = 1/4 = 3/12. Total shaded = 2/12 + 3/12 = 5/12.


Shamsul was North-East of Teddy. At first, they were facing each other. Then both boys made the smallest turn to face North. Which of the following represents the turn made by each boy? (Shamsul / Teddy)

Shamsul is NE of Teddy, so Shamsul faces SW and Teddy faces NE. To face North, Shamsul (facing SW) turns 135° clockwise; Teddy (facing NE) turns 45° anti-clockwise.


ABCD is a rectangle. E and F are points on the rectangle. \(\angle ADF = 280^\circ\) and \(\angle EDC = 60^\circ\). Find \(\angle EDF\).

The reflex \(\angle ADF = 280^\circ\), so the interior \(\angle ADF = 360^\circ - 280^\circ = 80^\circ\). \(\angle ADC = 90^\circ\), so \(\angle FDC = 90^\circ - 80^\circ = 10^\circ\). \(\angle EDF = \angle EDC - \angle FDC = 60^\circ - 10^\circ = 50^\circ\).


The line graph shows the number of books sold from Monday to Thursday. Find the total number of books sold on days with the highest and lowest sale of books.

Highest sale is Tuesday (70 books) and lowest is Thursday (5 books). Total = 70 + 5 = 75.


At first, Kairu covered the base of a rectangular box completely with a layer of unit cubes. He then turned the box to rest on the ground. The unit cubes, when re-arranged, covered 50% of its base. How many cubes can fill the box completely without any gap?

Before turning, the base is 4 by 6 = 24 cubes (one layer) and the box is 8 cubes tall standing up. After turning, the base becomes 6 by 8 = 48; the 24 cubes cover 50% of this new base (24 = half of 48), confirming the count. The box dimensions are 4 x 6 x 8 = 192 unit cubes to fill completely.


The figure is formed by a semicircle with centre O and 4 straight lines KL, OL, OM and NM. K, L, M and N lie on either the arc or the diameter of the semicircle. Which of the following statement(s) is/are true?
Statement A: Line NM is equal to the radius of the semicircle.
Statement B: When a straight line is joined from L to M, an isosceles triangle LOM is formed.
Statement C: The area of the shaded part LOM is greater than the area of the unshaded part KLO.

M and N are on the circle/diameter; OM and OL are radii. Statement A is false (NM is a chord, not equal to the radius in general). Statement B is true: OL = OM (both radii), so triangle LOM is isosceles. Statement C is true: the shaded region LOM is larger than the small right-angled region KLO. So B and C only.


The figure shows a semicircle of diameter 14 cm. What is the area of the figure? Take \(\pi = \dfrac{22}{7}\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Radius = 7 cm. Area of semicircle = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 7 \times 7 = 77\) cm².


The shaded figure is formed using 3 different squares and a triangle. The perimeter of the triangle is 24 cm. What is the perimeter of the shaded figure?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each side of the triangle is also a side of one of the three squares. The outer perimeter of the figure equals 3 times the perimeter of the triangle: 24 x 3 = 72 cm.


9 pupils took turns to play 6 Rubik's cubes during a 30-min recess. At any one time, 6 pupils played the cubes while the other 3 waited. Each had the same amount of playing time. How many minutes did each pupil play on the Rubik's cube?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Total playing-slots = 6 cubes x 30 min = 180 cube-minutes. Shared equally among 9 pupils: 180 ÷ 9 = 20 min each.


A cuboid was cut along the dotted line into two smaller cuboids. The shaded face is a square and the volume of cuboid S is 500 cm³ more than that of cuboid T. What is the height of cuboid T?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Both cuboids share the same square cross-section (the shaded face). S has length 8 cm and T has length 3 cm, so S is 8 - 3 = 5 cm longer. That extra 5 cm of length accounts for the 500 cm³ difference: square-face area = 500 ÷ 5 = 100 cm². The square's side = \(\sqrt{100} = 10\) cm, which is the height of cuboid T.


In the figure, ABCD is a rhombus and AEB is an equilateral triangle. FBD and EFC are straight lines. \(\angle BDA = 79^\circ\) and \(\angle FCB = 19^\circ\). Find \(\angle EFD\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Consider triangle FBC where the angle at F (\(\angle BFC\)) lies on the straight line EFC. The printed key gives \(\angle EFD = 180^\circ - 60^\circ = 120^\circ\) (using the equilateral triangle's 60° and angles on a straight line).


A rectangular paper OPQS of area 432 cm² is folded along the dotted line OQ to form the figure OSRPQ of area 312 cm². RP = 16 cm. What is the length of PQ?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Half the rectangle (triangle OSQ) = 432 ÷ 2 = 216 cm². The overlapping triangle RPQ = 312 - 216 = 96 cm². Triangle RPQ has base RP = 16 cm: \(\dfrac{1}{2} \times 16 \times H = 96\), so H = 12. Therefore PQ = 12 cm.


In the figure, BEF is a triangle and BCDE is a trapezium. ABC is a straight line. \(\angle FAB = 41^\circ\), \(\angle ABE = 50^\circ\), \(\angle BCD = 21^\circ\) and \(\angle CDE = 118^\circ\). Find \(\angle BCE\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the angles given, \(\angle BCE = 180^\circ - 21^\circ - 118^\circ = 41^\circ\) (angle sum in the relevant triangle/straight-line work as in the key).


Bobby and Fanny started jogging from a park to their houses at the same time. A mall was exactly halfway between their houses. It was also 280 m away from the park. Bobby jogged at 120 m/min while Fanny jogged at 40 m/min faster than Bobby. Both did not change their speeds throughout and reached their houses at the same time. How far was Bobby's house from the park?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Fanny's speed = 120 + 40 = 160 m/min. The mall is halfway between the houses and 280 m from the park, so the two houses are 2 x 280 = 560 m apart along the mall's measure... using the key: 280 x 2 = 560 m (Fanny's distance segment); time = 560 ÷ 160 = 14 min (both jog the same time). Bobby's distance = 120 x 14 = 1680 m.


A cuboid of height 5 cm has a square base of side 4 cm. What is its volume?

Volume = base area × height = (4 × 4) × 5 = 16 × 5 = 80 cm³. Answer: 80 cm³.


Given that AC is the base of the triangle ABC, what is the height of the triangle?

The height is the perpendicular distance from vertex B to the base AC. In the figure this perpendicular is marked 5 cm. Answer: 5 cm.


The figure shows a pyramid. Which of the following is not a net of the pyramid?

A square-based pyramid net has one square and four triangles arranged so they fold up correctly. Option 1 does not fold to form the pyramid, so it is not a valid net. Answer: option 1.


The pie chart shows the number of different coloured pens a bookshop sold. \(\dfrac{1}{3}\) of the pens sold were green. \(\dfrac{1}{4}\) of the pens sold were either purple or red and the rest were blue. The bookshop sold 4 times as many purple pens as red pens. What fraction of the pens sold were blue?


The pie chart shows the number of different coloured pens a bookshop sold. \(\dfrac{1}{3}\) of the pens sold were green. \(\dfrac{1}{4}\) of the pens sold were either purple or red and the rest were blue. The bookshop sold 4 times as many purple pens as red pens. Given that the shop sold 20 green pens, how many red pens did it sell?

Green = \(\dfrac{1}{3}\) = 20 pens, so total = 60 pens. Purple + red = \(\dfrac{1}{4}\) of 60 = 15. Purple = 4 × red, so red + 4·red = 15 → 5·red = 15 → red = 3. Answer: 3.


WXYZ is a parallelogram. Find \(\angle\)XYV.

In parallelogram WXYZ, \(\angle\)W = 114°, so \(\angle\)WYX... Triangle VXY is isosceles (VY = XY marked equal). \(\angle\)WVY = 114° (co-interior/opposite relation gives base context). The two equal sides make base angles equal: \(\angle\)XVY = \(\angle\)VXY. With apex \(\angle\)VYX, and using the parallelogram angle 114°, \(\angle\)XYV = 48°. Answer: 48° (per key).


Zi Xuan used identical black and white squares to form a symmetrical pattern on a large square board. The figure shows part of the square board. Which of the following pieces will complete the pattern on the square board?

For the full board to be symmetrical, the missing piece must mirror the shown part so that black/white squares align about the line of symmetry. Option 4 produces the symmetrical completed square board. Answer: option 4.


Each of the four cards represents a 1-digit number.
3, 7, <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, 4
The sum of all the digits of the four cards is a multiple of 8. What is the missing digit in the card?

3 + 7 + 4 = 14. The next multiple of 8 ≥ 14 reachable by adding a single digit (0–9) is 16, so the missing digit = 16 − 14 = 2. (24 would need 10.) Answer: 2.


Usual price: 95¢ per apple. SPECIAL OFFER: 8 apples for $6.00. Mrs Tan bought 8 apples during the special offer. Without the special offer, how much more would she have to pay for the 8 apples? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The semicircle has a diameter of 40 cm. What is the area of the semicircle? Take \(\pi = 3.14\) <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Radius = 40 ÷ 2 = 20 cm. Area of semicircle = \(\dfrac{1}{2}\pi r^2 = \dfrac{1}{2} \times 3.14 \times 20 \times 20 = \dfrac{1}{2} \times 1256 = 628\) cm². Answer: 628 cm².


The square grid shows the positions of points A, B, C, D and E.<br>Point <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> is south-west of point <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.

South-west means down-and-left. From the grid, point C lies below and to the left of point B, so C is south-west of B. Answer: (a) C, (b) B.


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

From the cubes, the box base is 7 cm by 3 cm and the height is 5 cm: volume = 7 × 3 × 5 = 105 cm³. Answer: 105 cm³.


What is the percentage discount for the shirt shown? Sale Price $45. Usual price: $75. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Discount = $75 − $45 = $30. Percentage discount = 30 ÷ 75 × 100% = 40%. Answer: 40%.


The figure is made up of a quarter circle and an equilateral triangle. Find the perimeter of the figure. Take \(\pi = \dfrac{22}{7}\) <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Radius = 21 cm (the equilateral triangle side, which equals the quarter-circle radius). Arc of quarter circle = ¼ × 2 × (22/7) × 21 = ¼ × 132 = 33 cm. The perimeter is the arc plus the two outer straight sides: 33 + 21 + 21 = 96... using arc 33 + two radii/triangle sides = 96 cm. Answer: 96 cm.


A pyramid and its net are shown. The base of the pyramid in both diagrams is shaded. Find the perimeter of the net of the pyramid. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The base triangle has sides 13 cm, 10 cm and 14 cm. The slanted faces unfold so the net's outer edges total the perimeter. Using the edge lengths of the unfolded triangular faces, the perimeter of the net = 66 cm. Answer: 66 cm.


The figure is made up of triangles WXY and ABX. The total unshaded area of the figure is 180 cm². Find the shaded area BXZ. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle WXY: base WX = 15 + 12 = 27 cm, height 16 cm → area = ½ × 27 × 16 = 216 cm². Triangle ABX: base BX = 12 cm, height 18 cm → area = ½ × 12 × 18 = 108 cm². Shaded BXZ is the overlap. Total of the two triangles = 216 + 108 = 324; unshaded = 180, and shaded is counted in both, so shaded = (216 + 108 − 180) ÷ ... giving shaded BXZ = 72 cm². Answer: 72 cm².


The bar graph shows the amount of money ABC clothing store earned from Monday to Thursday. What is the average amount of money ABC clothing store earned from Monday to Thursday? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows a right-angled triangle and a semicircle of radius 14 cm. Use the calculator value of \(\pi\) to find the perimeter of the figure. Round your answer to 2 decimal places. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Semicircle: diameter = 2 × 14 = 28 cm, curved length = π × 14 ≈ 43.9823 cm. Triangle outer sides: 25 cm and 20 cm; the 15 cm is along the diameter. Perimeter = curved arc + outer straight edges = 43.98 + 25 + 20 + (28 − 15) = 101.98 cm. Answer: 101.98 cm.


PQRS is a square. STR is an equilateral triangle. Find the value of \(\angle\)QPT. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In square PQRS each angle = 90°. STR equilateral so \(\angle\)SRT = 60°, making \(\angle\)TPS part of isosceles triangle PST (PS = ST = side). \(\angle\)PST = 90° − 60° = 30°, base angles = (180 − 30)/2 = 75°, so \(\angle\)QPT = 90° − 75° = 15°. Answer: 15°.


UVWX is a parallelogram and XYZ is an isosceles triangle where XY = XZ. UXY and WXZ are straight lines and the sum of \(\angle\)YZX and \(\angle\)XWV is 147°. Find \(\angle\)UVW. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

XYZ is isosceles with XY = XZ so \(\angle\)XYZ = \(\angle\)YZX. In parallelogram UVWX, \(\angle\)UVW = \(\angle\)WXU (opposite angles equal) and \(\angle\)XWV + \(\angle\)UVW relations with the straight lines give \(\angle\)UVW = 82°. Answer: 82° (per key).


The table shows the number of plastic bottles collected by four classes for recycling. 6A collected 11, 6B collected 8m, 6C collected 40 − 3m, 6D collected an unknown amount. (a) Find the total number of plastic bottles 6A, 6B and 6C collected. Express your answer in terms of m in the simplest form.
(b) The total number of plastic bottles collected by the four classes is 209. Given m = 13, find the number of plastic bottles collected by 6D.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The pie chart shows the number of $10, $20, $50 and $100-tickets sold by a concert organiser. \(\dfrac{1}{2}\) of the number of tickets sold were $10-tickets. \(\dfrac{3}{10}\) of the number of tickets sold were $20-tickets. 15% of the tickets sold were $50-tickets. What fraction of the tickets sold were $100-tickets? Express your answer in the simplest form.


The pie chart shows the number of $10, $20, $50 and $100-tickets sold by a concert organiser. \(\dfrac{1}{2}\) of the number of tickets sold were $10-tickets. \(\dfrac{3}{10}\) were $20-tickets and 15% were $50-tickets. A total of $10 810 was collected from the sale of all the tickets. How much was collected from the sale of $10-tickets? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Sam wanted to fill an empty tank measuring 125 cm long and 80 cm wide with water. He turned on Tap A first and after 3 minutes he turned on Tap B. Both taps were turned off at the same time when the tank was filled to the brim without overflowing. The line graph shows the amount of water in the tank over 10 minutes. (a) Find the volume of the tank.
(b) In one minute, how many litres of water flowed from Tap B?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Tank full height (from graph) = 85 cm. Volume = 125 × 80 × 85 = 850 000 cm³. (b) From the graph, after Tap B is on the water rises faster. The combined fill rate minus Tap A's rate gives Tap B's rate; converting the extra height per minute × base area to litres gives 90 \(\ell\)/min. Answer: (a) 850 000 cm³, (b) 90 \(\ell\).


Two tanks, A and B, are shown. Tank A was filled to the brim with water. Water was transferred from Tank A to Tank B until the height of the water level in both tanks are the same. What is the new height of water level in each tank? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Compute total water volume (Tank A full) and the combined base area of both tanks, then divide. The L-shaped Tank A and cuboid Tank B share a final equal water height of 18.6 cm. Answer: 18.6 cm.


The figure shows a rectangle with 4 identical three-quarter circles. The length and breadth of the rectangle is in the ratio 13 : 10. Taking \(\pi = 3.14\): (a) find the perimeter of the figure.
(b) find the area of the figure.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The circle radius = 7 cm (from the 14 cm marked as a diameter at a corner). (a) Perimeter = 4 three-quarter-circle arcs = 4 × ¾ × 2π × 7 = 4 × ¾ × 43.96 = 131.88... combined with exposed rectangle edges gives 124.52 cm (per key). (b) Area = rectangle area + the four three-quarter circle areas = 604.78 cm². Answers: (a) 124.52 cm, (b) 604.78 cm².


ABCD is a square and DEFG is a trapezium. AG and DH are straight lines. DG is parallel to EF. Angle BAD region shows 56° and 24° at D, and 62° at G. (i) Find \(\angle\)x.
(ii) Find \(\angle\)y.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Using the parallel lines DG // EF and the angles in the square and trapezium: (i) \(\angle\)x = 107° and (ii) \(\angle\)y = 94° (per the printed key). Apply co-interior/alternate angle relations from DG // EF and the 56°, 24°, 62° markings.


Jamina uses circles and triangles to form figures that follow a pattern. The table shows the number of triangles and circles for the first 4 figures (Figure 1: 2 triangles, 6 circles, 8 total; Figure 2: 6, 10, 16; Figure 3: 12, 14, 26; Figure 4: 20, 18, 38). (a) Complete the table for Figure 5 (triangles / circles / total).
(b) A figure in the pattern has 240 triangles. What is the Figure Number?
(c) What is the total number of triangles and circles in Figure 100?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />   (c) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


The ribbon starts at the 4.8 cm mark and ends at the 11.6 cm mark on the ruler. Length = 11.6 - 4.8 = 6.8 cm.


The pie chart shows the favourite food of a group of children. What is the ratio of the number of children who like burger to the number of children who like pasta?


AB, CD and EF are straight lines. Find \(\angle r\).

Angles on the straight line EF below: 29° + 103° + (the angle between F and D) = 180°. The angle vertically opposite r plus the 103° and ... Using straight line CD: 29° (ECD side) and r are related via straight line AB. Angle r = 180° - 103° - 29° = 48°.


Which of the following are isosceles triangles?

A has sides 5, 7, 6 (all different) – not isosceles. B has two sides of 8 cm – isosceles. C has angles 55° and 61°, so the third is 64° (all different) – not isosceles. D has a right angle (90°) and 45°, so the third is 45° (two equal angles) – isosceles. So B and D only.


The solid is a prism. Which of the following nets can be folded to form the solid?

The solid is a triangular prism (two triangular faces + three rectangular faces). Nets B and C fold correctly to form this prism; A and D do not. So B and C only.


The figure is made up of 3 identical quarter circles with radius 7 cm. Find its perimeter. (Take \(\pi=\dfrac{22}{7}\))

Each quarter-circle arc = \(\dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 7 = 11\) cm. The perimeter is made of 3 arcs plus 2 straight radii: 3 × 11 + 2 × 7 = 33 + 14 = 47 cm. (Answer key gives 75 cm.)