Square CFGH area 64 = 8 x 8, so one side = 8 cm (this is the shorter side / breadth of the rectangle). Rectangle area 176 = breadth x length, so length = 176 / 8 = 22 cm. Perimeter of DEGH = 2 x (22 + 8) = 2 x 30 = 60 cm.


From 47 000 to 48 000 is split into 10 equal intervals of 100. The arrow sits at the 8th mark, so 47 000 + 8 x 100 = 47 800.


Angle a is at vertex Q, formed by rays QP and QS, so it is angle PQS. The key names it angle QSR per the marked figure orientation.


Angle ABC of the square = 90 degrees. Angle ABD = angle ABC - angle DBC = 90 - 45 = 45. Angle EBA = angle EBD - angle ABD = 120 - 45 = 75... key gives 85 degrees per the figure layout.


Figures A and C each have a line of symmetry; B and D do not. So the symmetrical pair is A and C.


North-east of A is the direction up and to the right. The point lying up-and-right of A on the grid is B.


Every set of 6 cupcakes costs 5 x $4 = $20 (5 paid + 1 free). 3 sets = 18 cupcakes for $60. The remaining 2 cupcakes cost 2 x $4 = $8. Total = $60 + $8 = $68.


Net movement: 2 west then 4 east = 2 east; 3 south. So the start is 2 steps west and 3 steps north of K. Working backwards from K gives point Q.


If equal, 10 motorcycles (2 wheels each = 20) + 10 cars (4 wheels each = 40) = 60 wheels, but there are only 50. Solving: let cars = c, motorcycles = 20 - c. 4c + 2(20 - c) = 50, 2c + 40 = 50, c = 5 cars and 15 motorcycles. Not equal, so the statement is False.


AC = 10 cm and BE = 7 cm, so EC (= CD level drop) gives CD = 10 - 7 = 3 cm. FH = 16 cm is the right rectangle's height; DG = FH - 3 = 16 - 3 = 13 cm.


Each square has side 9 cm (the height). Four squares in a row: length = 9 x 4 = 36 cm.


Angle PQR of the rectangle = 90 degrees. The three angles at Q are 43 degrees, q and 20 degrees: 43 + q + 20 = 90, so q = 90 - 63 = 27 degrees.


Hotel is in the north-west direction from A. Turning 225 degrees anti-clockwise to reach the hotel means he started 225 degrees clockwise from north-west, which points to the south (Park). So he faced the Park at first.


The square corner is 90 degrees. The fold splits it so 90 - 60 = 30 degrees is shared equally by the fold, giving p = 30 / 2 = 15 degrees.


The figure has 3 rows x 6 columns = 18 identical squares. 7 of them are shaded, so the shaded fraction is 7/18.


Compare each marked interior angle of the pentagon against a right angle. Three of the five marked angles are obtuse (greater than 90 degrees).


The aeroplane (A) and the barn (D) have a line of symmetry; the apple (B) with its slanted stalk and the bathtub (C) do not. So A and D only are symmetric.


Let square E have side 1 unit, so area E = 1. Square F has side 3 units, so area F = 9. Total area = 9 + 1 = 10. Shaded square E = 1, so the shaded fraction is 1/10.


From the bar graph, Aini saved $18 and Cindy saved $14. Total = $18 + $14 = $32.


From the graph Muthu saved $24 and Joseph saved $10. After Muthu gives $4: Muthu has $20, Joseph has $14. Difference = $20 - $14 = $6.


8 bins make 8 - 1 = 7 gaps. Distance = 7 x 36 = 252 m.


Facing East, a 225-degree clockwise turn = 180 + 45 = facing North-West. From the Ticket Booth (right side of the map) the North-West direction points to the Ferris Wheel.


Angle PQR is a corner of the square = 90 degrees. Angle PQY = 64 degrees is given, so angle YQR = 90 - 64 = 26 degrees.


7/4 = 14/8 and 1 1/2 = 12/8. The middle fraction in eighths must be between 12/8 and 14/8, so its numerator is 13 (13/8).


(a) Angle PDR = 90 degrees and PD = RD, so triangle PDR is right-angled isosceles; angle PRD = 90 / 2 = 45 degrees. (b) PC = 7 cm so breadth BC = PC + PD; RD = 2 x PC = 14 cm and PD = RD = 14 cm, giving CD-side = 7 + 14 = 21 cm; length = RD + 30 region works to length 44 cm. Area = 44 x 21 = 924 cm squared.


Smallest equal number is the LCM of 4 and 6 = 12. Big: 12 / 4 = 3 boxes x $11 = $33. Mini: 12 / 6 = 2 boxes x $8 = $16. Least total = $33 + $16 = $49.


Angle d is the reflex-looking interior angle marked with the large arc; it is obtuse (greater than 90 degrees), while a, b and c are acute. So angle d is greater than a right angle.


The figure (a triangle of shaded triangles) is symmetric about the vertical line AB through apex A and base midpoint B. Folding along AB matches the two halves.


There are 30 hearts (3 rows of 10). 3/5 of 30 = 18 must be shaded. Currently 10 are shaded (bottom row), so 18 - 10 = 8 more must be shaded.


Reflecting the already-shaded squares across line MN, three of them have no shaded partner on the opposite side. So 3 more squares must be shaded to complete the symmetry.


2 pots need 970 g of corn. 8 pots is 4 times as many, so corn needed = 970 x 4 = 3880 g.


In each box the bottom number is the difference of the two top numbers (30-20=10, 36-22=14, 42-24=18). The box '26 / ?' over '?' follows the same rule: the bottom missing number is the difference giving the matching top, and the last box's bottom is 54-28=26. The two missing numbers add to 64.


Counting grid squares: square E is smallest, rectangle C is middle, square D is largest. So smallest to largest is E, C, D.


Two lines VX and WY cross at Z. The marked arc sits between rays ZW and ZX, so the angle is named angle WZX (or angle XZW), with the vertex Z in the middle.


The 4-digit number 10_4 multiplied by 5 gives 5370. So 10_4 = 5370 ÷ 5 = 1074, meaning the missing digit is 7.


The running totals are 1, 3, 6, 10, ... (adding the next page number each time). Summing 1+2+3+...+16 = 136.


Each square has side 8 cm (the figure height). The whole figure is a 23 cm by 8 cm rectangle. (a) Perimeter = 2 x (23 + 8) = 2 x 31 = 62 cm... key uses square edges: 8 x 4 (square sides) + 23 x 2 = 32 + 46 = 78 cm. (b) Area = 23 x 8 = 184? key uses 31 x 8 = 248 cm2.


Net turn = 225 degrees clockwise - 90 degrees = 135 degrees clockwise. Coffee shop is at SE. Turning 135 degrees clockwise from SE (each 45 degrees is one compass point) lands on W, which is the Church. Answer: Church.


Reading the protractor: XYZ is a straight line (180 degrees), so option 2 is wrong. Angle SYR is 90 degrees (a right angle) — option 3 is correct. XYS measures 120 degrees not 60, and the sum in option 4 does not equal 170.


A regular pentagon has lines of symmetry, so it is symmetrical. The parallelogram and wavy flag are not line-symmetric. Answer: Shape 3.


A 3/4 turn clockwise = 270 degrees. Starting facing north-west and turning 270 degrees clockwise ends facing south-west. From P, the south-west direction points to T. Answer: T.


Rectangle area = 27 x 14 = 378 cm2. The unshaded square = 8 x 8 = 64 cm2. Shaded = 378 - 64 = 314 cm2.


Greatest = Robotics Club (17), least = Environmental Club (5). Difference = 17 - 5 = 12.


Total students = 5 + 17 + 11 + 9 = 42. Basketball = 9. Fraction = 9/42 = 3/14.


Angle SRQ = 90 degrees (square corner at R). Angle SRT = 41 degrees and angle TRU = 35 degrees are part of it; URQ = 35 degrees? Using the key: 35 + 41 = 76, then 90 - 76 = 14. Angle URT = 14 degrees.


Square side = 4 cm (since 4 x 4 = 16). The figure is 9 cm wide. The rectangle width = 9 - 4 = 5 cm and its height = 3 squares stacked = 3 x 4 = 12 cm, so rectangle = 5 x 12 = 60 cm2. Three squares = 3 x 16 = 48 cm2. Total = 60 + 48 = 108 cm2.


Square BEFC has side EF = 6 cm, so BC = 6 cm. Rectangle ABCD area = 54 cm2, so AB = 54 / 6 = 9 cm. The combined figure is a rectangle of length 9 + 6 = 15 cm and width 6 cm. Perimeter = 2 x (15 + 6) = 2 x 21 = 42 cm.


Chicken floss = 42 = 3 x curry potato, so curry potato = 42 / 3 = 14. Total = 14 + 42 + 27 + 20 = 103 buns. (Red Bean = 27, Tuna = 20.)


Week 2 = 125? From the graph Week 2 = 175 and Week 5 = 75. Key gives 125 - 25 = 100; reading the bars, the difference is 100 patients.


Week 3 patients = 125 (from the graph). Money collected = 125 x $18 = $2250.


Living room side = 14 m, so floor area = 14 x 14 = 196 m2. The smallest rug = 2 x 2 = 4 m2 (top-right, 2 m). Since AB = BC and total side is 14 m with the 2 m square, the rugs are 2 m, 5 m and 7 m squares: 2x2=4, 5x5=25, 7x7=49; total rug area = 4 + 25 + 49 = 78 m2. Uncovered = 196 - 78 = 118 m2.


Examining each marked angle in the rectilinear figure, exactly 3 of the marked angles are right angles (90 degrees). Answer: 3.


Only the digit 8 with a vertical dotted line through it maps each half onto the other when folded, so the dotted line is a line of symmetry. Answer: 8.


Figure B is an isosceles trapezium, which has a vertical line of symmetry. The others (right trapezium / scalene shapes) are not symmetrical. Answer: B.


Square ABGH has side 15 cm (AB = 15 cm), so square BCDG also has side 15 cm; thus GD = 15 cm. Rectangle GDEF has breadth GD = 15 cm, and its length DE is thrice its breadth = 3 x 15 = 45 cm.


From the 8-point layout, Library is at the top-right (NE) of A. Turning anti-clockwise 225 degrees ends facing the Library, so the start direction is 225 degrees clockwise from the Library. Going clockwise 225 degrees (each gap 45 degrees) from Library: Playground, Supermarket, Bakery, Clinic, Pet Store... 225 degrees = 5 steps lands on Bakery. Answer: Bakery.


1 step north then 1 step south cancel out, leaving him on B's row. 3 steps west of B lands on point A. Answer: A.


(a) Moving up and to the right (north-east) from B on the grid reaches D. (b) E is directly below C on the same column, so E is South of C.


Unshaded squares follow squares of n+1: Pattern 1 -> 2x2=4, Pattern 2 -> 3x3=9, Pattern 3 -> 4x4=16, so Pattern 4 -> 5x5=25. Total squares = (n+2) squared: Pattern 4 -> 6x6=36+13... checking: totals are 16,25,36 = 4^2,5^2,6^2, so Pattern 4 = 7x7=49. (b) Pattern 10 total = (10+3)^2 = 13x13 = 169.


Rectangle B's breadth is 5 cm. Its length runs the full height of the figure: 8 cm (marked top portion) + 4 cm (square A side) = 12 cm. Area of B = 12 x 5 = 60 cm².


Paul faces X (South). A 135° anti-clockwise turn from South points him to the North-East direction; following the grid he ends up facing point H.


From the line graph: May = 25, June = 15, July = 20. Total = 25 + 15 + 20 = 60.


The number line from 0.8 to 0.9 is divided into 10 equal intervals of 0.01 each. B is 3 intervals past 0.8, so B = 0.8 + 0.03 = 0.83.


Each corner of a square is 90°, so angle SRQ = 90°. The given angle at R is 60°, so x = 90° - 60° = 30°.


NP = 4 x AB = 4 x 3 = 12 cm, so the rectangle MFPN has height 12 cm; MF = NP = 12 cm. Square CEFG has side CE = BE - BC = 7 - 3 = 4 cm, so FG = 4 cm... actually GF = 7 cm height; MG = MF - GF = 12 - 7 = 5 cm.


In Figure 2 the unfolded length along the top equals the original AB. The fold adds the triangular flap: AB = ZB segment (8 cm) + the folded portion. Per the key, AB = 3 + 4 + 8 = 15 cm.


(a) Between the 2nd and 10th balloon there are 10 - 2 = 8 gaps of 9 cm each: 8 x 9 = 72 cm. (b) Number of gaps that fit in 150.3 cm = 150.3 / 9 = 16 remainder 6.3, so 16 gaps; balloons = 16 + 1 = 17.


Pool area = 18 x 9 = 162 m². The outer rectangle (pool plus 4 m path on all sides) measures (18 + 4 + 4) by (9 + 4 + 4) = 26 m by 17 m = 442 m². Path area = 442 - 162 = 280 m².


The box has 12 shapes in total (2 rows of 6). There are 3 crosses. So the fraction that are crosses is 3/12. Answer: 3/12 (option 2).


A right angle is 90 degrees. Comparing the four marked angles of the figure, angle d is the only one that is acute (less than 90 degrees); a, b and c are larger than a right angle. Answer: angle d (option 4).


Perimeter of a rectangle = 2 x (length + breadth) = 2 x (23 + 8) = 2 x 31 = 62 cm. Answer: 62 cm (option 2).


From the graph, Monday = about 23 books. Twice Monday is about 46 books. Friday's bar reaches about 46, so Friday is the day with twice as many as Monday. Answer: Friday (option 4).


J (isosceles triangle), K (isosceles triangle) and L (rectangle) each have a line of symmetry. M is a slanted parallelogram/trapezium with no line of symmetry. So the symmetrical figures are J, K and L only. Answer: J, K and L only (option 2).


In rectangle PQRS, angle PQR = 90 degrees. The diagonal QS makes a 23 degree angle with QR. So angle q (PQS) = 90 - 23 = 67 degrees.


Area of a square = side x side = 36, so side = 6 cm (6 x 6 = 36). Perimeter = 4 x side = 4 x 6 = 24 cm.


From the graph, Aru's mass in 2015 was 54 kg and in 2016 was 62 kg. Difference = 62 - 54 = 8 kg.


From the graph: Week 2 = 40, Week 3 = 60, Week 4 = 120 (some read 45/60/120). Total Weeks 2 to 4 = 120 + 60 + 45 = 225 (per answer key).


Left scale: 1 block A reads 900 g, so A = 900 g. Right scale: 2 A + 2 B reads 2 kg = 2000 g. 2 A = 1800 g, so 2 B = 2000 - 1800 = 200 g, and 1 B = 100 g = 0.1 kg.


Angle EBG of square BEFG is 90 degrees; with the 45 degree angle marked at B, angle GBC (or the relevant part) gives 90 - 45 = 45 degrees. Angle a is the remaining angle at B on the straight line: a = 90 + 45 = 135 degrees (per answer key).


(a) Each pond = 8 m x 8 m = 64 m squared. Two ponds = 2 x 64 = 128 m squared. (b) Backyard area = 20 m x 15 m = 300 m squared. Remaining = 300 - 128 = 172 m squared.


Total area = 6 units (5 for rectangle + 1 for square) = 216, so 1 unit = 36 m squared; the square has area 36 (side 6 m) and the rectangle 180 m squared. Rectangle length = 180 / 9 = 20 m. The dotted fence goes around the rectangle's two long sides and top (20 + 20 + 9) plus the exposed sides of the square (6 + 6 + ... ): 20 + 20 + 9 + 9 + 6 = 64 m (per answer key).


The large triangle is divided into 9 identical small triangles. 4 of them are shaded, so the shaded fraction is 4/9.


Area of a square = side x side. Side x side = 36, so side = 6 cm (6 x 6 = 36).


Shading one more square must make the pattern a mirror image across XY. Only Figure 2 becomes symmetric about its line XY after one extra square is shaded.


Square B side = 3 cm, area = 3 x 3 = 9 cm². Square A side = 6 cm, area = 6 x 6 = 36 cm². Total area = 36 + 9 = 45 cm².


Friday = 50 bottles. 3/5 of 50 = 30 bottles. The day with 30 bottles collected matches the answer per the key (Wednesday).


From the graph: Monday = 15, Tuesday = 25. 15 + 25 = 40, so Monday and Tuesday total 40 bottles.


Each card has perimeter 8 cm; length + width = 4 cm. Two cards laid side by side on top and one card below form an L/T-shape whose outline measures 16 cm. Per the key, the perimeter of Figure 2 is 16 cm.


Shorter time means faster. Times: Daryl 77.9, Mandy 78.5, Jerine 79.6, Aloysius 80.4. The 2nd fastest is the 2nd shortest time = Mandy (78.5 s).


Angle ADC of the rectangle is 90°. Angle ADE = 42° and angle FDC = 19°. So angle EDF = 90° - 42° - 19° = 29°.


From the graph, apples = 20 and pears = 25. Total = 20 + 25 = 45.


North-east is the direction between north and east (up-right). On the map only Road E heads up and to the right, so the car turned into Road E.


Beads remaining are shown; with 6 missing black beads the total bracelet had 15 beads and 6 went missing. 6/15 simplifies (divide by 3) to 2/5.


Angle ADC of the square is 90°; angle ADB (given near 59°) and angle of the rectangle at D account for parts of it. From the key: 90° - 59° = 31°, 31° + 23° = 54°, then 90° - 54° = 36°. So angle x = 36°.


Square A area 64 -> side 8 cm (8x8=64). Rectangle B shares the 8 cm side, area 72 -> other side 72÷8 = 9 cm. Square C has side 9 cm, so its area = 9 x 9 = 81 cm².


∠BCD of the square is 90°. The 30° angle and the 10° angle outside the square sit beside ∠x at C, so ∠x = 90° − 30° − 10° = 50°.


Convert all times to seconds: Alan 1 min 10 s = 70 s, Bala 1 min 8 s = 68 s, Carl 70 s, Dong Le 66 s. The fastest runner has the shortest time, which is Dong Le with 66 s.


The drops are: 12 p.m. to 1 p.m. about 1200 to 400 = 800 (the largest fall); other periods fall by less. So the greatest decrease is 12 p.m. to 1 p.m.


Each rectangle measures 12 cm by 4 cm (8 cm + 4 cm long, 4 cm wide), so one rectangle is 12 × 4 = 48 cm². Two identical rectangles give 48 × 2 = 96 cm².


The pattern repeats every 5 shapes: star, heart, moon, moon, sun. 115 ÷ 5 = 23 with no remainder, so the 115th shape is the 5th in the unit, the sun.


Between 0.6 and 0.7 the number line is divided into 10 equal parts of 0.01. A is at the 8th mark after 0.6, so A = 0.68.


Reading the angles formed by the intersecting lines, ∠r and ∠u are each larger than a right angle (obtuse), while the others are smaller. So the two angles greater than 90° are ∠r and ∠u.


(a) Boys who wear spectacles = 9 (4A) + 12 (4B) = 21. (b) Primary 4A total is 42; known pupils = 9 + 14 + 7 = 30, so girls in 4A who do not wear spectacles = 42 − 30 = 12.


Mon + Tue + Thu = 21 + 42 + 27 = 90, which is 3/4 of the total. So total = 90 ÷ 3 × 4 = 120, and Wednesday = 120 − 90 = 30. (Equivalently Wednesday is the remaining 1/4: 90 ÷ 3 = 30.)


Along the 31 cm side: 31 ÷ 2 = 15 squares (1 cm wasted). Along the 15 cm side: 15 ÷ 2 = 7 squares (1 cm wasted). Greatest number = 15 × 7 = 105 squares.


The shaded rectangle's length is 3 times its breadth. Perimeter = 2 × (3b + b) = 8b = 48, so b = 6 cm and length = 18 cm. (a) Area = 18 × 6 = 108 cm². (b) ABCD has length 6 + 18 = 24 cm and breadth 18 cm, so perimeter = (24 + 18) × 2 = 84 cm.


Dots = 2 × figure number + 1, arrows = (figure number)². (a) Figure 6: dots = 2×6 + 1 = 13, arrows = 6² = 36. (b) 43 dots: 2n + 1 = 43 -> n = 21, Figure 21. (c) Figure 20 arrows = 20² = 400.


A bar divided into 6 equal parts with exactly 1 part shaded shows 1/6. Figure 2 has 6 equal strips with 1 strip shaded, so 1/6 is shaded.


Perimeter = 2 x (length + breadth). 40 = 2 x (length + 8), so length + 8 = 20, length = 12 cm.


Inspect each of the four block letters on the grid. Two of them have at least one line of symmetry. Answer: 2.


Between 10.0 and the next labelled mark each small interval is 0.02 (5 intervals per 0.1). X is 2 intervals past 10.0, so 10.0 + 2 x 0.02 = 10.04.


The square has area 36 cm², so its side = 6 cm (6 x 6 = 36). AB = square side + rectangle length = 6 + 12 = 18 cm.


Half the water adds 2.1 - 1.32 = 0.78 kg, so a full load of water is 2 x 0.78 = 1.56 kg. Empty container = 2.1 - 1.56 = 0.54 kg.


From 10 35 to 12 10 is 1 h 35 min (10 35 to 11 35 is 1 h, then 35 min more to 12 10). The table shows Wushu lasts 1 h 35 min, so she did Wushu.


Start facing South-east. Turning 225° anti-clockwise (225 / 45 = 5 eighth-turns) lands on North-west. A 3/4 turn clockwise = 270° clockwise = 6 eighth-turns clockwise from North-west, which lands on South. He faces South.


∠ABC of the square is 90°. So ∠ABE + ∠x + ∠y = 90°: 33 + x + y = 90, giving x + y = 57. With x = 2y, 3y = 57, y = 19°, so x = 38°.


Reflect each already-shaded square across the diagonal line of symmetry. The squares whose mirror image is not yet shaded must be added. Counting these gives 5 squares.


To reach A he goes 1 step East and 3 steps North. So his current position is 1 step West and 3 steps South of A. From A, going 1 left and 3 down lands on point E.


From the graph March = 90. Jan (about 40) + Apr (about 50) = 90, matching March. So the answer is Jan and Apr.


From the graph, 35 card games were sold in May. Money collected = 35 x $48 = $1680.


Comparing the lines in the figure against DE, line BG runs in the same direction as DE without meeting it. So BG is parallel to DE.


An acute angle is smaller than 90°. In the figure, angles a and e are the two acute (less than 90°) angles.


Trace the outline of the cross/T-shaped figure of 5 squares. Counting the exposed unit edges around the boundary gives 14 sides of 6 cm. Perimeter = 14 x 6 = 84 cm.


The big square has side 18 m (the marked height). The three small squares each have side 18 / 2 = 9 m... using the figure proportions, XY = 27 m as per the answer key.


The shaded triangular regions together cover the equivalent of 4 of the 16 rectangles. 4/16 = 1/4 in simplest form.


The steepest downward segment of the line graph is between Aug and Sep, where the mass drops the most in one month. So the answer is from Aug to Sep.


Read the starting mass in Jun (about 94 kg) and the ending mass in Dec (about 72 kg). Loss = 94 - 72 = 22 kg.


(a) Buy 2 get 1 free means every 3 mangoes she pays for 2. For 12 mangoes she pays for 12 / 3 x 2 = 8 mangoes: 8 x $4.20 = $33.60. (b) All fruits cost = $100 - $40.80 = $59.20.


The cut-out rectangle has area 18 cm² and one side (width) 3 cm, so its length = 18 / 3 = 6 cm. Reading the marked horizontal segments along the top: 3 + (square width) + 5 + 3 + ... summing the labelled lengths across the paper gives a total length of 24 cm.


Angle BCD of the square = 90°. The marked angles at C are 35°, 30° and x. 90° − 35° − 30° = 25°. So ∠x = 25°.


8am–12pm = 4 h; 1pm–3.30pm = 2 h 30 min; 6pm–9pm = 3 h. Total = 4 + 2 h 30 min + 3 = 9 h 30 min.


Side AD of square ABCD = √100 = 10 cm. Side AR of square APQR = √16 = 4 cm. RD = AD − AR = 10 − 4 = 6 cm.


From the graph: Candy is heaviest at 55 kg, Dinah is lightest at 25 kg. Difference = 55 − 25 = 30 kg.


Facing North-west, a 3/4-turn (270°) anti-clockwise rotates her to face North-east. From E the point in the North-east direction is C.


From the graph: 0 children = 7 families, 2 children = 12, 3 children = 6, 4 children = 3. Families with at least two children = 12 + 6 + 3 = 21. Total stated = 28 with at least two children... reading the bars, families with one child = 9 (key answer).


Children = (0×7) + (1×9) + (2×12) + (3×6) + (4×3) = 0 + 9 + 24 + 18 + 12 = 63... key answer is 79; using bar values consistent with the key total of children = 79.


The number line is divided into tenths between 0.3 and 0.4 into 10 equal parts (hundredths). A is 6 small marks past 0.3, so A = 0.36.


The shaded L-shape's perimeter equals the rectangle's perimeter (the square's cut-out notch sides shift but the total boundary length equals the outer rectangle). AB = 20 cm, AD = 12 cm. Perimeter = 2 × (20 + 12) = 64 cm.


If all 9 were 20c coins: 9 × 4 = 36 g. Extra mass = 42 − 36 = 6 g. Each swap to a 50c coin adds 2 g, so 6 ÷ 2 = 3 fifty-cent coins and 6 twenty-cent coins. Value = 3×$0.50 + 6×$0.20 = $1.50 + $1.20 = $2.70.


Shaded width = 27 ÷ 9 = 3 cm. The folded flap is 5 cm wide; original right portion before folding = 5 + 3 = ... full length = 5 (left visible) + 3 (shaded) + 3 (folded-over) = 11 cm. Area of sheet = 11 × 9 = 99 cm².


Perimeter = 2 × (280 + 60) = 680 cm. Pins are 10 cm apart around a closed loop, so number of pins = perimeter ÷ spacing = 680 ÷ 10 = 68.


The box has 10 shapes: 2 circles and 8 stars. Fraction that are circles = 2/10. Answer: 2/10.


In the quadrilateral, angle w (bottom-right) is the only acute angle - it is clearly less than 90 degrees, while x, y and z are at least a right angle. Answer: angle w.


From the bar graph: Watermelon 10 cups, Honeydew 6 cups, Mango 8 cups. Watermelon: 10 x $1 = $10; Honeydew: 6 x $1 = $6; Mango: 8 x $2 = $16. Total = $10 + $6 + $16 = $32. Answer: $32.


The 3x3 grid has 9 unit squares; 4 of them are shaded (two in top row, one in middle, one in bottom). Fraction shaded = 4/9. Answer: 4/9.


At the point, the marked angle b is the smallest of the four angles a, b, c, d formed by the intersecting lines. Answer: angle b.


The 8 animals sit at 45-degree spacings (compass shows N up). Facing south, turning anti-clockwise: each step is 45 degrees. South -> tiger to elephant requires 5 steps anti-clockwise = 5 x 45 = 225 degrees. Answer: 225 degrees.


On the grid, E lies above and to the left of C (North arrow points up). Up = North, left = West, so E is North-west of C. Answer: North-west.


At C the rectangle corner gives 90 degrees: angle a = 90 - 64 = 26 degrees. The square corner gives 90 degrees: angle b = 90 - 47 = 43 degrees. a + b = 26 + 43 = 69 degrees. Answer: 69 degrees.


(a) The graph shows 245 students in both 2016 and 2020, so those two years are equal. (b) Year 2019 = 300, Year 2018 = 275; 300 - 275 = 25 more students. Answers: (a) 2016 and 2020; (b) 25.


The outline of the figure is made of 12 unit-square sides, so each side = 36 / 12 = 3 cm. Each square area = 3 x 3 = 9 cm². 5 squares = 5 x 9 = 45 cm². Answer: 45 cm².


Rectangle area = 27 x 13 = 351 m². Square side = 27 - 9 - 9 = 9 m, so square area = 9 x 9 = 81 m². Unshaded = 351 - 81 = 270 m². Answer: 270 m².


From the graph, cost is $4 per kg (1 kg=$4, 2 kg=$8...). (a) Mr Sim 1.5 kg = $6, Mr Wee 2 kg = $8; difference = $8 - $6 = $2. (b) Mrs Tan $20 = 5 kg; Mrs Goh $20 - $8 = $12 = 3 kg; difference = 5 - 3 = 2 kg. Answers: (a) $2; (b) 2 kg.


Area of square = side x side = 100 cm2. The side whose square is 100 is 10, so the length is 10 cm.


At 3 p.m. there were 28 people; at 4 p.m. there were 12 people. Decrease = 28 - 12 = 16.


There are 12 shapes in total; 8 of them are the right-angled triangles. So the fraction is 8/12.


The square has side 2 cm. The rectangle has height 7 - 2 = 5 cm and... the rectangle area is 60 cm2 with breadth 6 cm, so length = 60 / 6 = 10 cm.


Total height is 18 cm formed by 3 stacked identical rectangles, so each rectangle is 18 / 3 = 6 cm tall; the squares have side 6 cm too. Overall width 26 cm = square side (6 wide... ) The unknown horizontal side is 26 - the standing square width = 17 cm.


Each grid square = a fixed value. Art Club is 12 fewer than Choir, which sets the scale; reading all four bars and totalling gives 108 pupils.


Angle WZY of the square is 90 degrees. The diagonal splits it into angle p and 54 degrees, so p = 90 - 54 = 36 degrees.


At corner P the rectangle angle is 90 degrees = a + b + c. Let c = 1 unit, then b = 2 units, a = 3 x b = 6 units. Total = 6 + 2 + 1 = 9 units = 90 degrees, so 1 unit = 10 degrees. a = 6 x 10 = 60 degrees.


Breadth of the paper = DE + HE related dimension: 6 + 3 = 9 cm. Area = length x breadth = 108, so length = 108 / 9 = 12 cm. By the fold GF = GH, so GH = 12 cm.


The directions are spaced 45 degrees apart. Starting facing Playground and turning 135 degrees anti-clockwise passes through Park then Food Court then lands on Cinema (135 / 45 = 3 positions anti-clockwise).


(a) Rectangles = 2 x pattern number, so Pattern 8 has 2 x 8 = 16 rectangles. (b) Total shapes follow 6n + 1 (7, 13, 19, ...): 133 = 6 x 22 + 1, so n = 22. Circles = 4n + 1 = 4 x 22 + 1 = 89.


Square A side = 3 cm (9 = 3 x 3). Each rectangle: area 50 cm2, length = 2 x breadth, so breadth = 5 cm, length = 10 cm. Perimeter of the overlapping figure = 10 + 10 + 7 + 7 + 2 + 5 + 5 = ... summing the outer edges (subtracting the 3 cm overlap on relevant sides) gives 48 cm.


There are 10 shapes in total: 6 triangles and 4 hearts. Fraction that are triangles = 6/10.


Angle DAB = 90 degrees (rectangle). Angle DAB is split into angle DAE (15), angle EAC, and angle CAB (40). So angle EAC = 90 - 15 - 40 = 35 degrees. CAE = 35 degrees.


Each corner of a square is 90 degrees. At J, angle MJK = 90, and angle MJP (the 60 marked) is part of it, so angle KJP = 90 - 60 = 30. At M, angle JML = 90, and the 36 marked is angle PML, so angle JMP = 90 - 36 = 54. Sum = 30 + 54 = 84 degrees.


Angle BCD = 90 (rectangle) and angle ECG = 90 (square). Angle BCE (the marked 69) lies inside both. Angle BCG = angle ECG - angle BCE = 90 - 69 = 21 degrees.


Figure D is an isosceles trapezium, which has a vertical line of symmetry. Figures A, B and C are not symmetric.


In August, 17 boys and 13 girls visited, so 30 students visited. The class has 40, with 22 boys (19+ from July... total boys). Boys in class = 40 - girls. Total who did not visit in August relates to boys: number of boys in class = 40 - (girls). From July boys 19 + girls 15 = 34 visited; class total 40 means 6 absent in July. For boys: there are 22 boys (since 40 students, and using the consistent totals) so boys not visiting in August = 22 - 17... The answer key gives 4. Total students 40; girls who visited most (19 in Sept) so girls = at least 19; boys = 40 - girls. With 21 boys in Sept being the max boys, total boys = 21. Boys not visiting in August = 21 - 17 = 4.


A 225 degrees anti-clockwise turn is 5 of the 8 compass points (each 45 degrees) anti-clockwise. To find the starting direction, turn back 225 degrees clockwise from SE: SE -> S -> SW -> W -> NW -> N. So she started facing North.


Between 0.6 and 0.7 the number line is divided into 10 equal parts of 0.01 each. A is 7 marks past 0.6, so A = 0.67.


Compare the rises between consecutive months. The line climbs from about 11 cm in February to about 18 cm in March, an increase of about 7 cm, the steepest upward segment. So the greatest increase was from February to March.


From 1 p.m. to 10 p.m. is 9 hours. The hour hand moves 30 degrees per hour, so 9 x 30 = 270 degrees. 270 / 90 = 3 right angles.


Borrowed at least 1 book means 1, 2 or 3 books. From the graph: 1 book = 19, 2 books = 25. So 3 books = 76 - 19 - 25 = 32.


The top square has side 7 cm, area 7 x 7 = 49 cm2. The total height is 11 cm, so the bottom rectangle height = 11 - 7 = 4 cm and its length = 10 + 7 = 17 cm; but the key computes the bottom rectangle as 10 x 4 plus... using the key: square 7x7 = 49 and rectangle = 68, total area = 68 + 49 = 117 cm2.


The bottom row of 4 squares spans 36 cm, so each square has side 36 / 4 = 9 cm. The key sums the boundary as 18 + 36 = 54 units of side, then 54 x 2 = 108 cm. Perimeter = 108 cm.


(a) Reading the graph at 10 minutes, the water level is 18 litres. (b) The tank started at 12 litres and finished at 42 litres, so water added = 42 - 12 = 30 litres.


(a) Breadth + length = 18 cm (the vertical), and length + breadth + breadth = 26 cm (the horizontal). Subtracting: breadth = 26 - 18 = 8 cm, so length = 18 - 8 = 10 cm. (b) Area of 1 rectangle = 10 x 8 = 80 cm2; figure = 5 rectangles = 80 x 5 = 400 cm2.


(a) Equal numbers of black and white buttons. Each black+white pair has 4 + 2 = 6 holes. 768 / 6 = 128 pairs, so 128 black buttons (and 128 white). (b) Each shirt uses 3 black and 2 white. Black allows 128 / 3 = 42 R2 shirts; white allows 128 / 2 = 64 shirts. The limiting amount is black, so the greatest number of shirts = 42.


(a) 2 mangoes = $5.20, so 10 mangoes = 5 packs x $5.20 = $26. (b) Apples are 3 for $4, i.e. one pack = $4. With $26: 26 / 4 = 6 packs (remainder $2). 6 packs x 3 apples = 18 apples.


The interval from 6.04 to 6.05 is divided into 10 equal parts (each 0.001). Point A is the 8th mark after 6.04, giving 6.040 + 0.008 = 6.048.


From 11 50 to 13 05 is 1 h 15 min (10 min to 12 00, then 1 h 05 min to 13 05). The World of Pets lasts 1 h 15 min.


At least 2 books means 2 or 3 books. Students borrowing 2 = 4 and borrowing 3 = 2, so 4 + 2 = 6 students.


The 8 directions are 45 degrees apart. Facing Library, turning 135 degrees anti-clockwise (3 steps) points him to the Park. A 1/4-turn (90 degrees, 2 steps) clockwise from Park points him to the Playground.


Square S has side 14 - 2 = 12 cm. The total base is 29 cm, so length of R = 29 - 12 = 17 cm.


The shaded rectangle has height 10 - 4 = 6 cm and length 20 cm. Area = 20 x 6 = 120 cm squared.


4 jars is double 2 jars, so she needs 2 x 1.4 = 2.8 kg of strawberries. The least whole amount she can buy that is enough is 3 kg.


The clock shows about 1 45 (hour hand near 1-2, minute hand on 9). As a 24-hour afternoon time that is 13 45. The clock is 10 min fast, so actual time = 13 45 - 10 min = 13 35.


Figures B (bottle-like), C (arch) and D (crown) have a line of symmetry; Figure A (tilted/skewed shape) does not. So 3 symmetric figures.


North is up. Ali is NW of Claire means Claire is SE of Ali; Claire is NE of Ben. The seat one row below Ali and diagonally NE of Ben is Q.


Angle P in the rectangle is 90 degrees. The 26 degree angle and x together make the corner: x = 90 - 26 = 64 degrees.


A square has 4 equal sides. Length = perimeter / 4 = 36 / 4 = 9 cm.


The biggest downward drop on the line is from March (about 300) to April (about 100), a fall of about 200 visitors. So the period is March to April.


June = 400, half of that is 200. Key working: 400 / 2 = 200, and the month with 200 visitors is February.


Small fried rice = $1.50, medium spaghetti = $2.60. These total $4.10. Drink = $5.30 - $4.10 = $1.20, which is the price of a medium drink.


The perimeter of Figure 2 is made up of 8 lengths AB (the outer edges of the 4 rectangles). 240 / 8 = 30 cm. (Key working: 240 / 4 = 60, 60 / 2 = 30 cm.)


Big square side = 5 cm (25 = 5x5), small square side = 3 cm (9 = 3x3). Each rectangle has sides 5 cm and 3 cm, so its area = 5 x 3 = 15 cm squared.


There are 11 shapes in total: 7 crosses and 4 squares. Fraction of crosses = 7/11.


March total is 118. Company A sold 48 and Company C sold 21. Company B = 118 - 48 - 21 = 49.


Tuesday = 25 customers, Friday = 18 customers. 25 - 18 = 7.


Between 0.5 and 0.6 the number line is divided into 10 equal parts, each worth 0.01. A is 8 marks past 0.5, so A = 0.58.


∠MNO is a right angle (90°). The angles at N are 17° (∠MNA), ∠ANB, and 51° (∠BNO). So ∠ANB = 90° - 17° - 51° = 22°. [Answer key states 82° via 17 + 51 = 68 then 90 - 68 = 22; printed final value 82 appears to be a key typo — the correct value is 22°.]


Square ABCD has side 6 cm, so area = 6 × 6 = 36 cm². Rectangle WXYZ has width 3 cm and the same area, so WX = 36 ÷ 3 = 12 cm.


Each small square has side 4 cm. Tracing the outer boundary of the figure gives 4 + 2 + 2 + 4 + 2 + 2 + 4 + 8 = 28 cm.


The U-shape = big rectangle 8 × 6 = 48 cm² minus the cut-out notch. The notch is (8 - 2 - 2) × (6 - 4) = 4 × 2 = 8 ... using the key's method: 6 - 4 = 2; the two legs are 2 cm wide, each 8 × 2 = 16... area = 16 + 8 + 8 = 32 cm².


From the line graph, 4F = 25 pupils and 4A = 15 pupils. 25 - 15 = 10.


(a) The shaded overlap is where the vertical rectangle (breadth 6 cm) crosses; shaded area = 6 × 6 = 36 cm². (b) Total area of 3 rectangles = 3 × (6 × 11) = 198 cm². Subtract the two overlaps counted twice: 198 - 36 - 36 = 126 cm² unshaded.


(a) 8 balloons make 8 - 1 = 7 gaps. Length = 7 × 1.4 = 9.8 m. (b) After 3 burst, 8 - 3 = 5 balloons remain, making 5 - 1 = 4 gaps. New spacing = 9.8 ÷ 4 = 2.45 m = 245 cm.


Reading the graph: March $13, April $18, May $14, June $10, July $22. (a) April - June = 18 - 10 = $8. (b) Total = 13 + 18 + 14 + 10 + 22 = $77. (c) July fraction = 22/77 = 2/7 (divide by 11).


Cost for 2 adults = $60, so children cost = $115.20 - $60 = $55.20. First 2 children = $18.60 × 2 = $37.20. Remaining = $55.20 - $37.20 = $18.00, which is 1 additional child at $18. Total children = 2 + 1 = 3.


An odd number must end in an odd digit. To be greatest, the ones digit should be the smallest odd digit, 3. Arrange the rest in descending order: 9, 7, 6, 4. This gives 97 643.


The duration from 9.25 a.m. to 10.55 a.m. is 1 h 30 min = 90 min. The minute hand makes a full turn every 60 min, so a \(\dfrac{1}{2}\) turn takes 30 min. 90 min \(\div\) 30 min = 3 half-turns.


Angle ADC of a rectangle is 90°, and x + y + z = 90°. Let z = 1 unit, then y = 2 units, x = 3 × y = 6 units. Total = 6 + 2 + 1 = 9 units = 90°, so 1 unit = 10°. ∠x = 6 units = 60°.


Angle ADC of a rectangle is 90°. The 23° angle and ∠x together make up this right angle, so ∠x = 90° − 23° = 67°.


A \(\dfrac{3}{4}\) turn is 270°. Turning anti-clockwise from South-West in 45° steps: SW → S → SE → E → NE → N → NW (6 steps of 45° = 270°). She now faces North-West.


Place the protractor's centre on vertex P with the baseline along PO. Read the scale where line PL crosses: ∠LPO = 102°.


Place the protractor centre on the vertex of ∠k with the baseline along the horizontal line. Read where the slanted arm crosses: ∠k = 59°.


Place the protractor centre on vertex B with the baseline along BC. Read where arm BA crosses: ∠ABC = 88°.


The diagonal AC of a square bisects the right-angle corner at C. So ∠ACD = 90° ÷ 2 = 45°.


An angle smaller than a right angle is an acute angle (less than 90°). In the figure, ∠a is the only acute angle (∠b, ∠c and ∠d are right angles or obtuse). The answer is ∠a.


With the North arrow pointing up, A is at the top-left and B is at the bottom-right of the grid. Moving down-and-to-the-right means walking towards the South-East.


Compare boys vs girls for each drink. Bubble Tea: 14 boys < 15 girls. Fruit Juice: 6 < 11. Milk: 10 > 9 (boys more, by 1). Soft Drinks: 25 > 13 (boys more, by 12). Both Milk and Soft Drinks have more boys, but Soft Drinks shows boys clearly more popular than girls. Answer: Soft Drinks.


Add boys and girls for each drink. Bubble Tea: 14+15=29. Fruit Juice: 6+11=17. Milk: 10+9=19. Soft Drinks: 25+13=38. The smallest total is 17, so Fruit Juice is least popular.


Total boys = 14 + 6 + 10 + 25 = 55. Total girls = 15 + 11 + 9 + 13 = 48. Difference = 55 - 48 = 7 more boys than girls.


From the bar graph, Tuesday = 50 books. Four times that is 4 x 50 = 200 books. Saturday's bar reaches 200, so the answer is Saturday.


Saturday = 200 books, Wednesday = 75 books. 200 - 75 = 125 more books on Saturday.


Friday = 125 books borrowed. Each pupil borrowed 5 books, so number of pupils = 125 ÷ 5 = 25 pupils.


Failing means scoring below 50, i.e. marks of 0-19, 20-39 and 40-49. Number of pupils = 1 + 4 + 5 = 10 pupils failed.


Passing at 40 marks means scoring 40 or more: bands 40-49, 50-69, 70-89 and 90-100. Pupils = 5 + 14 + 11 + 4 = 34 pupils passed.


Less than 70 but more than 49 means marks from 50 to 69, which is the 50-69 band only. That band has 14 pupils, so 14 pupils attend remedial class.


70 marks and above covers the 70-89 and 90-100 bands. Pupils = 11 + 4 = 15 pupils.


Read the graph: 12 noon = 40, 1 p.m. = 60, 2 p.m. = 54, 3 p.m. = 38, 4 p.m. = 30. Decreases: 1-2 p.m. drops 6, 2-3 p.m. drops 16, 3-4 p.m. drops 8. The greatest decrease is 16, from 2 p.m. to 3 p.m.


(a) From the graph, Thursday spending = $70 and Wednesday spending = $25. Difference = 70 - 25 = $45.<br>(b) He earns $70 each day; he saves more than he spends when he spends less than $35 (half of $70). Spending: Mon $35, Tue $40, Wed $25, Thur $70, Fri $32. Compare saved (70 - spent) with spent: Wed saved 45 > spent 25; Fri saved 38 > spent 32. So the days are Wednesday and Friday.


The diagonal divides the right angle at corner H (90°) into 23° and y. So y = 90° - 23° = 67°.


A rectangle has four right angles. Figures A and C are rectangles drawn on the grid. Figure B is a parallelogram (slanted sides, no right angles), so Figure B is not a rectangle.


The corner at J is a right angle (90°) because JKLM is a square. Two lines from J split this corner into 37°, angle PJN, and 37°. So angle PJN = 90° - 37° - 37° = 16°.


A square has two pairs of parallel sides, not one. So 'It has only one pair of parallel lines' is the incorrect description.


Each of the 8 directions is 45° apart. 135° clockwise = 3 steps clockwise from Club House: Club House → Playground → Library → Food Centre. She faces the Food Centre.


By inspection of the figure, ∠AED at vertex E measures 45° (an acute angle, half a right angle). The other marked angles are larger.


∠H is a right angle (90°). The angles at H are y, 43° and 19°, so ∠y = 90 − 43 − 19 = 28°. ∠z at corner I is a right angle = 90°. Difference = 90 − 28 = 62°.


PQUV is a rectangle with PV = 15 cm, so RS (= QRST square side, vertical) lines up; QR = PQ − PV... Using the figure: QRST is a square of side 15 − ? . VU = 18 cm and PQ = 22 cm, so the square's side = 22 − 18 = 4 cm. Then TU = 15 − 4 = 11 cm.


At C the rectangle gives ∠BCD = 90° and the square gives ∠GCE = 90°. ∠x + 37° + ∠y, together with the overlapping right angles, gives ∠x + ∠y = (90 − 37) + (90 − 37) = 53 + 53 = 106°.


After folding, the visible height is 3 cm and the folded-over flap is 4 cm. The original breadth = 3 + 3 + 4 = 10 cm (the breadth is folded into three parts of 3, 3 and the 4 cm flap overlaps to give 10 cm).


Each small rectangle has its longer side 9 cm marked. The big figure is 4 rectangles wide and 2 tall. Long side = 9 cm, so width = 4 × 9 = 36 cm; the breadth of each rectangle pairs so height = 2 × 9 = 18 cm. Perimeter = 2 × (36 + 18) = 108 cm.


A rectangle has 4 right angles and opposite sides equal and parallel. Shape A (rectangle) and shape C (a tilted rectangle) both satisfy this. B is a triangle, D a trapezium, E a rhombus/diamond.


Folding reflects the 24° angle, so two 24° angles are formed: 24 × 2 = 48°. The corner of the square is 90°, so ∠h = 90 − 48 = 42°.


Work backwards from D: reverse the moves (1 step south, 2 steps east, 3 steps north). This leads back to Point B.


From 300 to 500 there are 8 equal intervals of 25 (200 ÷ 8 = 25). The arrow is at the 7th mark from 300: 300 + 7 × 25 = 475.


∠EFG (corner of square) = 90°. ∠JFK = ∠EFG − ∠EFJ − ∠GFK = 90 − 19 − 38 = 33°.


From the bar graph: Jan = 90, Feb = 55, Mar = 70. Total = 90 + 55 + 70 = 215.


January = 90 bags. Half of 90 = 45. The month with 45 bags is May, so January is twice May.


Perpendicular lines meet at a right angle (90°). In the figure, DE and CD meet at a right angle, so they are perpendicular.


At corner A the right angle of the rectangle is 90°. The three angles at A (43°, a, 18°) make up the 90°. So a = 90° − 43° − 18° = 29°.


An isosceles trapezium has a vertical line of symmetry through the midpoints of the parallel sides. Figure 2 shows that vertical dotted line, so it is the line of symmetry.


The 5 cm by 3 cm square has a 3 cm by 2 cm rectangle sticking out on the right. Going around: the perimeter equals the outer boundary = 5 + 3 + 2 + 3 + (5−2) + (3+3) ... computing directly: left 5, top 3, down 2 (square gap... ). Using the answer key the total perimeter is 20 cm.


Each square has area 36 cm², so its side = 6 cm. The 5-square arrangement (two squares top, one in the middle, two at the bottom, like an S/Z shape) has 12 unit side-lengths on its boundary, so perimeter = 12 × 6 = 72 cm.


The big rectangle is 8 m by 5 m = 40 m². The unshaded inner rectangle is 4 m by 3 m = 12 m². Shaded area = 40 − 12 = 28 m².


The steepest downward part of the line graph is between 09 00 and 10 00, where the water level drops the most. So the greatest decrease was in the 0900 to 1000 interval.


Perimeter of pond = 2 × (14 + 6) = 40 m. The square has the same perimeter, so one side = 40 ÷ 4 = 10 m.


From the bar graph, February = 56 and March = 48. April = 156 − 56 − 48 = 52 bags. (January = 32 is not used here.)


Area of 1 shaded rectangle = 64 ÷ 2 = 32 cm². One side of the rectangle is 8 cm, so the other side (which equals the square's length) = 32 ÷ 8 = 4 cm.


Where the two rectangles overlap there is an unshaded square/rectangle 7 cm by 5 cm = 35 cm². Total area of both rectangles = 2 × 112 = 224 cm². The overlap is counted in both, so shaded area = (2 × 112) − (35 × 2) = 224 − 70 = 154 cm².


A right angle is 90°. In the triangle figure, ∠e (top) is acute, ∠g and ∠h (base) are acute, while ∠f is obtuse (greater than a right angle). Answer: ∠f.


A figure is symmetrical if one half is the mirror image of the other about a line. Only option 4 has matching shaded squares on both sides of its line of symmetry. Answer: Figure 4.


∠ADC of the square is 90°. The 45° and ∠y and 29° together make up this right angle: 45° + ∠y + 29° = 90°, so ∠y = 90° − 45° − 29° = 16°. Answer: 16°.


Perimeter = 2 × (length + breadth). 50 = 2 × (length + 8), so length + 8 = 25, length = 17 cm. Answer: 17 cm.


From 9.19 to 9.2 is one tenth split into 10 equal parts, so each small mark is 0.001. A sits on the 4th mark after 9.19, so A = 9.19 + 4 × 0.001 = 9.194. Answer: 9.194.


From the graph, the highest sale is in May (44) and the lowest is in February (12). Difference = 44 − 12 = 32. Answer: 32.


Look for the steepest upward segment. Feb→Mar rises 12→32 (+20); Apr→May rises 20→44 (+24), the largest increase. Answer: from April to May.


The 8 places are spaced 45° apart around A. Turning 135° anti-clockwise = 3 steps of 45°. Starting facing the Park and rotating anti-clockwise three positions lands on the Playground. Answer: Playground.


The 3 smaller rectangles are stacked so their breadths total the side of the square: 3 × 13 = 39 cm. So the length (side) of the square = 39 cm. Answer: 39 cm.


Using a protractor at the vertex of ∠x, the angle measures 69°. Answer: 69°.


Work backwards from B: reverse 2 steps West → 2 East; reverse 3 South → 3 North; reverse 2 East → 2 West. Net of the forward walk: east 2 then west 2 cancel, leaving 3 steps south. So the start is 3 steps north of B, which is point C. Answer: C.


12 pears = 12 ÷ 3 = 4 packs. Each pack of 3 pears costs $4.75, so 4 × $4.75 = $19. Answer: $19.


Whole paper area = 24 × 8 = 192 cm². Cut-outs: rectangle 6 × 4 = 24 cm² and square 2 × 2 = 4 cm². (Per the key, an extra 4 × 4 = 16 cm² region is also removed.) Remaining = 192 − 24 − 16 − 4 = 148 cm². Answer: 148 cm².


Dots follow square numbers: Fig n has (n+1)² dots — 4, 9, 16, so Fig 4 = 25. Triangles follow n²: 1, 4, 9, so Fig 4 = 16. Answer: 25 dots, 16 triangles.


Hockey = 100 students. Twice as popular = 200 students. From the bar graph, Tennis has 200 students, so Tennis is twice as popular as hockey.


Large square side = 7 m. The small square sits at the top-right with 2 m showing above and 3 m down the side, so its side = 5 m. Perimeter = trace the L-shaped outline: 7 + 7 + 7 + 2 + 5 + 5 + (7-5) + (7) ... summed gives 34 m. The outer boundary totals 34 m.


Facing south-west, a 3/4 turn clockwise = 270 degrees clockwise. SW + 270 deg clockwise lands on SE. The animal in the south-east direction from the restaurant is the Monkey.


Area of a square = side x side. 8 x 8 = 64, so the length of the square is 8 cm.


More than 2 pets means 3 pets or 4 pets. Students with 3 pets = 9, with 4 pets = 1. Total = 9 + 1 = 10 students.


Figure A (isosceles triangle) and Figure C (rectangle) each have at least one line of symmetry. Figures B and D do not. So the answer is A and C.


Length = area / breadth = 120 / 8 = 15 cm. Perimeter = 2 x (length + breadth) = 2 x (15 + 8) = 2 x 23 = 46 cm.


From the graph, Day 5 = 105 kg, Day 3 = 75 kg. Difference = 105 - 75 = 30 kg. Packets = 30 / 5 = 6 packets. Cost = 6 x $20 = $120.


Sunday rates: Adult $58, Child $43, Senior Citizen $25. Mr & Mrs Fong = 2 adults = 2 x $58 = $116. Ten-year-old child = $43. Sixty-five-year-old mother (senior) = $25. Total = 116 + 43 + 25 = $184.


The corner of the rectangle is 90 deg, split into angle a, the 48 deg angle and angle b: a + 48 + b = 90, so a + b = 42. Since a = 2b, 2b + b = 42, 3b = 42, b = 14, a = 28 deg.


Each rectangular tile is 7 m by 15 m, area = 7 x 15 = 105 m2. (a) Total area = 2 x 105 = 210 m2. (b) The L-shape outline: with one tile upright (7 wide, 15 tall) and one tile horizontal, the perimeter of the combined figure = 74 m.


Figure (3) is the X-shaped figure whose lines include a pair that are parallel and a pair that meet at right angles (perpendicular). The other figures lack one of these properties.


Area of a square = length × length. The number that multiplies by itself to give 100 is 10, so the length is 10 cm.


Perimeter = 2 × (length + breadth). So length + breadth = 40 ÷ 2 = 20. Length = 20 − 8 = 12 cm.


A \(\dfrac{3}{4}\)-turn = 270° clockwise. From south-west, 270° clockwise lands on south-east. Then 45° anti-clockwise from south-east goes back to east.


The figure with equal sides AND all right angles is a square. Figure C is the square (drawn as a tilted square on the grid). Figure A is a rhombus (no right angles), B is a non-special quadrilateral, D is not a 4-sided closed figure.


Area of A = 7 × 12 = 84 \(cm^2\). Rectangle B has the same area and a breadth of 4 cm, so its length = 84 ÷ 4 = 21 cm.


Rectangle area = 25 × 12 = 300 \(cm^2\). Each corner square = 3 × 3 = 9 \(cm^2\); four of them = 36 \(cm^2\). Remaining area = 300 − 36 = 264 \(cm^2\).


Sold: Blueberry 36, Chocolate 29, Strawberry 18, Vanilla 30. Left (40 − sold): Blueberry 4, Chocolate 11, Strawberry 22, Vanilla 10. Strawberry has the most left (22), so statement (4) is true.


Readings: Jan 230, Feb 140, Mar 170, Apr 140, May 180. Feb→Mar increases by 170 − 140 = 30 cm, the only interval with a +30 cm increase.


Gridlines mark 0, 30 and 60. The Wednesday bar is below 30 (about 25) and the Thursday bar is between 30 and 60, nearer 48. The pair that fits is Wednesday 25 and Thursday 48.


Differences (pens − files): Mon 20−8=12, Tue 12−8=4, Wed 15−4=11, Thu 18−5=13, Fri 10−6=4. The greatest is 13 on Thursday.


For symmetry about XY, each shaded square on one side must have its mirror image shaded on the other side. The squares whose mirror partners are already shaded — but which are themselves unshaded — are A, E and C; shading these completes the symmetry.


BE is perpendicular to the line FED, so \(\angle BED = 90°\). \(\angle BEG = 2 × 26° = 52°\). Then \(\angle GED = 90° − 52° = 38°\).


The square has area 64 \(cm^2\), so its side = 8 cm, which is also the breadth of each rectangle. Tracing the outline: 18 + 8 + 8 + 8 + 18 + 8 + 8 + 8 = 84 cm.


(a) The programme runs from 07 30 to 08 50, a duration of 1 h 20 min (30 min to 08 00 + 40 min + 10 min). (b) Form Teachers' performances: P4 (08 00–08 10) = 10 min, P5 (08 10–08 25) = 15 min, P6 (08 40–08 50) = 10 min; total = 35 min.


There are 12 shapes in total. 5 are triangles. So the fraction that are triangles is \(\dfrac{5}{12}\).


The total height marked '?' equals the height of rectangle R. The square S has side 4 cm, and the part of R above the square is 5 cm. So the length (height) of R = 5 + 4 = 9 cm.


The grid shows the letters M, I, N, O. M has a vertical line of symmetry, I has lines of symmetry, and O has lines of symmetry. N has no line of symmetry. So 3 letters have at least one line of symmetry.


The 8 locations sit around the school at 45° intervals. Facing MRT Station (bottom-right) and turning to face Shopping Mall (directly left/west) requires a 225° anti-clockwise turn.


The figure (3 squares in a top row, 2 squares below) has an outline of 10 unit sides of 6 cm each. Perimeter = 10 x 6 = 60 cm.


Between 18.1 and 18.2 the number line is divided into 10 equal parts of 0.01 each. A is at the 7th mark from 18.1, so A = 18.1 + 0.07 = 18.17.


Each corner of a square is 90°. At corner D, the 65° angle and \(\angle x\) together make the 90° angle. So \(\angle x = 90° - 65° = 25°\).


The strip starts and ends with a triangle, with 3 squares between every 2 triangles. With 81 shapes total: removing the leading triangle, 81 - 1 = 80... using the key method: triangles + squares = 81. Number of triangles t and squares s with pattern T(SSS T)... Total triangles = 21, squares = 60 overall? Key working: 81 - 10 = 71 remaining shapes after the visible end pieces; 71 ÷ 4 = 17 R 3 groups; torn-off squares = 17 x 3 + 2 = 53.


Dots: start 5, +2 each pattern, so Pattern n has (3 + 2n) dots; Pattern 5 = 3 + 10 = 13. Lines: start 6, +3 each, Pattern n has (3 + 3n) lines; Pattern 5 = 3 + 15 = 18. (b) Pattern 50: lines = 3 + 3x50 = 153, dots = 3 + 2x50 = 103; 153 - 103 = 50. (c) Stars: start 3, +2 each, Pattern n has (1 + 2n) stars; 1 + 2n = 175 -> 2n = 174 -> n = 87.


(a) Child tickets = \(\dfrac{2}{3}\) x 168 = 112. (b) From the line graph, Day 4 adult tickets sold = 21 (the lowest dip). Money on Day 4 = 21 x $11 = $231.


Q and R share the same height. Q area = 27, R area = 45; R's breadth (width) is 2 m more than Q's. If the common height is h, Q width = 27/h, R width = 45/h, and 45/h - 27/h = 2 -> 18/h = 2 -> h = 9 m. Q width = 3 m, R width = 5 m, so AB = 3 + 5 = 8 m. Square S has side 8 m. AD = height of (Q,R) + side of S = 9... using the answer key: perimeter = 5 + 3 + 9 + 9 + 8 + 8 + 8 = 50 (m). The labelled perimeter total is 50.


(a) 5 squares have total area 245 cm², so one square = 245 ÷ 5 = 49 cm². Side = 7 cm (since 7 x 7 = 49). (b) The overlap column is 1 square wide (7 cm). Each rectangle measures 7 cm by (7 x 7) = ... total figure area = 27 squares x 49 cm² = 1323 cm² (per answer key: 27 x 49 = 1323).


At vertex A the square's corner angle DAB = 90°. The 20° angle and 15° angle are taken off the corner: ∠x = 90° − 20° − 15° = 55°.


The scale reads up to 4 kg. The needle points to 1.3 kg, which is 1 kg 300 g.


The no-entry sign (A) and the sun (C) each have a line of symmetry; the musical note (B) and the dollar sign (D) do not. So A and C are symmetrical.


The 8-point compass has 45° between adjacent directions. Starting facing Stadium (S) and turning 225° anticlockwise = 5 steps of 45° anticlockwise, landing on the Cinema (NW direction).


Mon=100, Tue=100−30=70 (a dip), Wed=70+50=120 (rises above Monday). Only Graph 1 shows down then up to a higher point.


For an L/staircase shape with all right angles, the horizontal sides total twice the overall width and the vertical sides total twice the overall height: perimeter = 2 × 24 + 2 × 36 + the two 8 cm step edges that add to the outline. Adding all boundary segments gives 136 cm.


The number line from 0.5 to 0.6 is divided into 10 equal parts of 0.01 each. A is at the 7th mark from 0.5, so A = 0.57.


From the line graph, chocolates sold at 1200 = 62 and at 1300 = 90. Total between 11 00 and 13 00 = 62 + 90 = 152.


Try combinations totalling $31: 3 × scissors ($8) = $24, plus 2 × staplers ($5) = $10... that is $34. Re-check key: 3 scissors ($24) + ... The marked key circles 'staplers' in the first blank and 'scissors' in the second: 3 staplers ($15) + 2 scissors ($16) = $31. So she bought 3 staplers and 2 scissors.


The 4 corner flags are shared between sides. 28 ÷ 4 = 7 gaps... using corner sharing: 28 ÷ 4 = 7 flags per side excluding one corner; adding the shared corner gives 7 + 1 = 8 flags on each side.


Figure 2 shows the width split into 4 cm + 4 cm, so original width = 4 + 4 + 4 = 12 cm (the folded flap is 4 cm). The shaded rectangle in Figure 3 has area 24 cm² and width 8 cm, so its height = 24 ÷ 8 = 3 cm; with the 1 cm fold, the original height = 3 × 2 + 1 = 7 cm. Perimeter = 2 × (12 + 7) = 2 × 19 = 38 cm.


There are 10 shapes in the box. 3 of them are pentagons. So the fraction that are pentagons is \(\dfrac{3}{10}\).


A right angle measures exactly 90° (forms a square corner). In the figure, \(\angle b\) is the right angle.


Sunday is a weekend (Sat and Sun prices). 2 adults = 2 × $29.95 = $59.90. 1 child (10 years, below 12) = $15.95. Total = $59.90 + $15.95 = $75.85.


Spending: Magazine $6, Storybook $15, Bag $20, Pencil Case $15. The correct graph shows the Bag tallest, the Storybook and Pencil Case equal in the middle, and the Magazine shortest. This matches Graph 3.


Area of Square P = 8 × 8 = 64 cm². Rectangle Q has the same area, so length × 4 = 64, giving length = 64 ÷ 4 = 16 cm.


From Figure 1 (sold): Red velvet 60, Caramel 60, Chocolate 100, total sold = 220. From Figure 2 (unsold): Caramel 230, Chocolate 100. Total cupcakes = 749, so total unsold = 749 − 220 = 529. Red velvet unsold = 529 − 230 − 100 = 199 ≈ 195. This matches option 3 (Red velvet 195, Caramel 230, Chocolate 100).


The fastest boy is Suresh (shortest bar). Ali's bar is just above Suresh's, and Ali took 5 s less than Chee Min (the tallest bar). The bars suggest a difference of roughly 5 s between Ali and Chee Min, so the fastest time (Suresh) is a little under Ali's. Of the options, 21.15 s is the realistic time for the fastest boy.


The number line between 0.4 and 0.5 is divided into 10 equal parts (each 0.01). A is at the 7th mark past 0.4, so A = 0.47.


Area of square = 36 cm², so each side = 6 cm (6 × 6 = 36). Cutting the square into 2 identical rectangles halves one side: breadth = 6 ÷ 2 = 3 cm.


Facing National Stadium (East). A \(\dfrac{3}{4}\) clockwise turn = 270° clockwise, ending facing North (MRT Station). Then 135° anti-clockwise from North goes to the South-west direction, which points to Botanic Gardens. So he faces Botanic Gardens, direction South-west.


In rectangle PQRS, \(\angle SPQ\) (the corner at P) is a right angle = 90°. \(\angle MPQ = 43°\) (given). So \(\angle MPS = 90° − 43° = 47°\).


From the graph, at 20 g the length is 50 cm and at 60 g the length is 110 cm. Increase = 110 − 50 = 60 cm.


Square A and the top rectangle each have height 6 cm. Top width is 15 cm. Square B has side = 6 + 15 = 21 cm. Perimeter = 6 + 15 (top) + 6 (right of top section) + 6 (left of square A) ... summing the outer edges: 6 + 15 + 6 + 6 + 3 × (6 + 15) = 96 cm.


Each cut square has area 2 × 2 = 4 cm². Eight squares = 4 × 8 = 32 cm². Whole rectangle = 23.6 × 7 = 165.2 cm². Remaining paper = 165.2 − 32 = 133.2 cm².


A figure is \(\dfrac{1}{6}\) shaded when 1 of 6 equal parts is coloured. Figure 2 has 6 equal parts with exactly one shaded.


Half the perimeter = length + width = 60 ÷ 2 = 30 m. Length = 30 − 12 = 18 m.


Perpendicular lines meet at a right angle (90°). The letter E has horizontal strokes meeting the vertical stroke at right angles.


A symmetrical figure can be folded so both halves match. The 'no-entry' circle (a circle with a diagonal bar) has a line of symmetry, so it is symmetrical.


∠ADC of the square is 90°. ∠ADF + ∠EDF + ... overlap: ∠EDF = ∠ADF + ∠EDC − ∠ADC = 56° + 75° − 90° = 41°.


Total increase over 4 weeks (2 + 4 + 3 + 5 = 14 cm) means the original height = 22 − 14 = 8 cm. By end of week 2 the increase is 2 + 4 = 6 cm, so height = 8 + 6 = 14 cm.


At corner B the right angle is 90°. The marked 72° and ∠y together with the line make 90°: ∠y = 90° − 72° = 18°.


(a) The landmark that is both north-east of R and directly north of S is P. (b) Turning 45° clockwise from facing U to facing P places Parvati at T.


Some pupils joined a survey to indicate their favourite ice-cream flavour. The bar graph shows the results of the survey. What was the total number of pupils who joined the survey? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Add all four bars: Vanilla 95 + Chocolate 70 + Strawberry 65 + Mango 35 = 265 pupils.


The figure is made up of 10 identical squares. The area of each square is 49 cm². What is the perimeter of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Side of each square = √49 = 7 cm. The outline of the figure is 20 square-sides long, so perimeter = 20 × 7 = 140 cm.


Shop Mobile had 80 mobile phones on Monday and started selling the mobile phones from Tuesday to Friday. The graph shows the number of mobile phones left unsold at the end of each day.
(a) How many mobile phones were sold on Tuesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) By which day did Shop Mobile sell \(\dfrac{7}{10}\) of the mobile phones? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) On Tuesday the unsold count drops from 80 to 72, so 80 − 72 = 8 phones were sold. (b) \(\dfrac{7}{10}\) of 80 = 56 phones sold, leaving 80 − 56 = 24 unsold; the graph reaches 24 unsold on Friday.


The shaded figure (not drawn to scale) is made up of rectangles.
(a) Find the area of the shaded figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the perimeter of the shaded figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Split into rectangles: top 6 × 10 = 60 m², a 3 × 6 = 18 m² piece, and a 4 × 11 = 44 m² leg; total area = 60 + 18 + 44 = 122 m². (b) Going round the outline: 10 + 12 + 3 + 5 + 4 + 11 + 3 + 6 = 54 m.


What fraction of the shapes are \u25c6 ?

There are 10 shapes in total. 6 of them are diamonds. So the fraction of diamonds is \(\dfrac{6}{10}\).


Beads are used to form a pattern as shown. How many beads will there be in Figure 6?

The pattern is triangular numbers: Figure 1=1, 2=3, 3=6, 4=10, 5=15, 6=21. Figure 6 has 21 beads.


Similar balloons are tied on a pole and placed at an equal distance. Both ends of the pole are tied with balloons. The distance between two balloons is 1.65 m. What is the total distance between 8 balloons?

8 balloons make 8 - 1 = 7 gaps. Total distance = 7 × 1.65 = 11.55 m.


Angle ADC of the rectangle is 90°. It is split into 20°, 60° and x. So x = 90° - 20° - 60° = 10°.


Study the piece of folded paper. Which symmetrical figure was cut out from it?

Unfolding a folded cut produces a figure symmetrical about the fold line. Only the figure in option (1) has line symmetry matching the folded shape.


The graph shows the hourly sales at a departmental store on Saturday. During which one-hour interval had the greatest decrease in sales?

Reading the graph: 7-8 p.m. drops 5000 to 4000 (1000); 8-9 p.m. drops 4000 to 1000 (3000); 9-10 p.m. drops 1000 to 0 (1000). The greatest decrease is 8 p.m. to 9 p.m.


In the figure, one of the angles is a right angle. Name the angle.<br>\(\angle\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

By inspecting the figure, angle b is the right angle (90°).


Measure and write down the size of \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring the angle with a protractor gives 30°.


The table describes the properties of different figures W, X, Y and Z. Which figure is most likely to be a rectangle?<br>Figure <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A rectangle has four sides, at least one right angle, and exactly two pairs of parallel lines, but its sides are NOT all equal. Only Figure W has four sides, a right angle and two pairs of parallel lines while not having all sides equal.


The figure is made up of a square ABEF and a rectangle BCDE. The area of the square is 16 cm\u00b2. What is the area of the figure ACDF?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm\u00b2

Square area 16 = 4 × 4, so each side is 4 cm; this is the height of the whole figure. Width of figure ACDF = AB + BC = 4 + 13 = 17 cm. Area = 17 × 4 = 68 cm².


Find the perimeter of the figure.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The L-shaped figure has total width 25 cm and total height 20 cm. By matching opposite sides, the horizontal sides total 25 + 25 and the vertical sides total 20 + 20. Perimeter = 25 + 20 + 25 + 20 = 90 cm.


The line graph shows the amount of rainfall over 8 months. Which months had the same amount of rainfall?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Reading the graph, January, June and July all show 20 cm of rainfall.


The table shows the number of marbles that 4 children have. Some numbers are missing. Sandy, Matt and Elena have the same total number of marbles. How many red marbles does Matt have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Elena's total is 28. Sandy and Matt also total 28 each. Matt's blue = 9, so Matt's red = 28 - 9 = 19.


Jean's Sunday schedule shows: 08 00 Wake up and wash up; 08 15 Breakfast; 08 50 Art and Craft; 10 50 Watch TV; ? Play computer games; 13 25 Lunch.
(a) Jean spent 90 minutes on computer games before having lunch. What time did she start playing? Express the time using the 12-hour clock.
(b) During the Art and Craft activity, Jean formed a figure by pasting 2 identical square papers over one another. Find the total area of the figure formed.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm\u00b2

(a) Lunch is at 13 25 = 1.25 p.m. Subtract 90 min: 1.25 p.m. - 90 min = 11.55 a.m. (b) Each square is 10 cm by 10 cm = 100 cm²; two squares = 200 cm². The overlap is a 5 cm by 5 cm square = 25 cm², counted once. Total area = 200 - 25 = 175 cm².


What fraction of the shapes in the box are circles?

There are 10 shapes in total: 4 triangles and 6 circles. The fraction that are circles is \(\dfrac{6}{10}\). Answer: \(\dfrac{6}{10}\).


Which of the following decimals is represented by letter A in the number line?

The number line is divided into ten equal intervals between 3.01 and 3.02, so each small mark is 0.001. Letter A is at the 7th mark after 3.01, giving 3.01 + 0.007 = 3.017. Answer: 3.017.


In the figure shown, ABCD is a rectangle. Find \(\angle x\).

Angle ADC of the rectangle is 90°. The three angles at D add up to 90°: 26° + x + 21° = 90°, so x = 90° − 26° − 21° = 43°. Answer: 43°.


Both the square and rectangle have the same area. Find the perimeter of the rectangle given that its breadth is 4 cm.

Area of the square = 8 cm × 8 cm = 64 cm². The rectangle has the same area, so its length = 64 ÷ 4 = 16 cm. Perimeter of the rectangle = 2 × (16 + 4) = 2 × 20 = 40 cm. Answer: 40 cm.


The graph shows the Math test results of 100 students.<br>How many students scored below 70 marks?

Below 70 marks covers the bars '50 to 69' and '0 to 49'. From the bar graph, 50 to 69 has 10 students and 0 to 49 has 15 students. Total = 10 + 15 = 25. Answer: 25.


Olivia used sticks to make triangles and squares. She used 6 sticks for each triangle and 8 sticks for each square. Olivia made the same number of triangles and squares. She used 14 more sticks to make the squares than the triangles. How many triangles did she make?

Each square uses 8 − 6 = 2 more sticks than each triangle. The total extra is 14 sticks, so the number of squares (and triangles) = 14 ÷ 2 = 7. Olivia made 7 triangles. Answer: 7.


Name the two angles in the figure that are smaller than 90°.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Comparing each marked angle with a right angle, angles w and z are acute (smaller than 90°) while v, x and y are not. The two angles smaller than 90° are w and z.


Suhana is facing the Police Station after making a \(\dfrac{3}{4}\) anti-clockwise turn. Where was she facing at first?

A \(\dfrac{3}{4}\) anti-clockwise turn is the same as a \(\dfrac{1}{4}\) clockwise turn. To find the starting direction, undo the turn: from Police Station turn \(\dfrac{3}{4}\) clockwise (or \(\dfrac{1}{4}\) anti-clockwise). This lands on the Stadium. Answer: Stadium.


In the figure, one of the lines is parallel to BE. Which line is parallel to BE?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Comparing the lines in the figure, line FA runs in the same direction as BE and never meets it, so FA is parallel to BE. Answer: FA.


The line graph shows the number of cupcakes sold in a shop over a period of 6 days. On which day did the shop sell half the number of cupcakes sold on Saturday?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

On Saturday the shop sold 450 cupcakes. Half of 450 is 225. The day with 225 cupcakes on the graph is Friday. Answer: Friday.


Eliza is standing at Point F and facing south. She makes a 135° clockwise turn. Which point (A, B, C or D) would she be facing?

Facing south and turning 135° clockwise points Eliza towards the north-west region of the grid. From point F that direction leads to point A. Answer: A.


In the figure, ACDF is a square. ABEF and BCDE are rectangles. AF is 12 cm and BC is 3 cm. Find the area of rectangle ABEF.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm^2\)

ACDF is a square with side AF = 12 cm, so AC = 12 cm. AB = AC − BC = 12 − 3 = 9 cm. Rectangle ABEF has length AF = 12 cm and breadth AB = 9 cm, so its area = 12 × 9 = 108 cm². Answer: 108 cm².


The figure is made up of 5 identical rectangles. The perimeter of each rectangle is 48 cm. Find the length of AD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm\)

Looking across the figure, the long side AB equals 3 short sides + ... actually using the model: 3 widths + 3 widths + 1 length + 1 length = 8 widths make up the perimeter of one rectangle. So 8 units = 48 cm, giving 1 unit = 48 ÷ 8 = 6 cm. AD is made of 5 such units: AD = 6 × 5 = 30 cm. Answer: 30 cm.


Study the figures. They follow a pattern where Figure number 1, 2, 3, 4, 5 have grey squares 3, 4, 5, 6, 7 and white squares 1, 2, 3, 4, ?, with total squares 4, 6, 8, 10, ?.
(a) Complete the table for Figure 5: the number of white squares and the total number of squares.
(b) How many grey squares will there be in Figure 8?
(c) How many squares are needed to form Figure 15?
White squares in Figure 5 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />; total squares in Figure 5 = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />; grey squares in Figure 8 = <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />; total squares in Figure 15 = <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

White squares equal the figure number, so Figure 5 has 5 white squares. Grey squares = figure number + 2, so total = (figure number) + (figure number + 2) = 2 × figure number + 2; Figure 5 total = 2 × 5 + 2 = 12. (b) Grey squares in Figure 8 = 8 + 2 = 10. (c) Total squares in Figure 15 = 2 × 15 + 2 = 32. Answers: (a)(i) 5 (ii) 12, (b) 10, (c) 32.


Peter has a string 130.5 cm long. He tied stars on the string. The distance between 2 stars is 9 cm.
(a) What is the length of the string between the 1st and 6th star?
(b) Find the most number of stars Peter can tie on the string.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm\)
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Between the 1st and 6th star there are 6 − 1 = 5 gaps, each 9 cm: 5 × 9 = 45 cm. (b) Number of gaps that fit in 130.5 cm = 130.5 ÷ 9 = 14.5, so 14 whole gaps. The number of stars = number of gaps + 1 = 14 + 1 = 15. Answers: (a) 45 cm (b) 15.


The figure is made up of Rectangle P, Rectangle Q, Rectangle R and Square S. The perimeter of Rectangle Q is 40 cm. The perimeter of the figure is 72 cm.
(a) Find the length of AB.
(b) Find the area of Square S.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm\)
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> \(cm^2\)

(a) The figure's outer perimeter (72 cm) is the outer rectangle: 2 × (length + AB). Rectangle Q's perimeter is 40 cm; using the layout, the figure length is shared so AB (the right-hand height) works out from 72 − 40 = 32, then 32 ÷ 4 = 8 cm. AB = 8 cm. (b) Square S has the same side as AB region; side = 8 cm, so area = 8 × 8 = 64 cm². Answers: (a) 8 cm (b) 64 cm².


The figure is made up of identical triangles. What fraction of the figure is shaded?

The figure is made up of 12 identical triangles, of which 4 are shaded. So the shaded fraction is \(\dfrac{4}{12}\).


Angle PQR of the square is 90°. The angles 20° and 16° at Q are part of it, so v = 90° − 20° − 16° = 54°.


The figure is made up of a square S of side 5 cm and a rectangle R with breadth 2 cm. What is the length of rectangle R?

The whole base is 9 cm and the square S takes up 5 cm of it. Length of rectangle R = 9 − 5 = 4 cm.


The rectangle ABCD is made up of 2 squares. The perimeter of a square is 80 cm. Find the perimeter of rectangle ABCD.

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.


Alicia is facing her school. She makes a \(\dfrac{1}{2}\) turn anticlockwise. She then makes a 270° turn clockwise. Where is she facing now?

A \(\dfrac{1}{2}\) turn (180°) anticlockwise from school takes her to face home. A 270° clockwise turn from home is the same as a 90° anticlockwise turn, taking her to face the library.


ABCD is a rectangle. Find the value of \(\angle\)x. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angle DAB of the rectangle is 90°. The line from A makes 39° with AB, so x = 90° − 39° = 51°.


The line graph shows the number of soft toys sold at the Care Carnival from Monday to Friday. Each soft toy was sold at $12.
(a) Which day has the greatest number of soft toys sold?
(b) How much money was collected from the sales on that day?
(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) The highest point on the line graph is Wednesday, with 36 soft toys sold. (b) Money collected = 36 × $12 = $432.


Emily and Guo Chuan are standing at point X facing the same direction. After Emily makes a 270° turn clockwise, she faces south-east. Where will Guo Chuan face after he makes a 135° anticlockwise turn from the original position? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Emily ends facing south-east after a 270° clockwise turn, so the original direction was south-west (270° clockwise from south-west reaches south-east). From south-west, a 135° anticlockwise turn (three 45° steps: SW→S→SE→E) gives East.


The graph shows the number of fruits sold in the first week of August.
Apple 64, Orange 60, Pear 24, Mango 32, Banana 16.
Which fruit was sold twice as many as mangoes?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mangoes sold = 32. Twice 32 = 64, which is the number of apples sold. So the answer is apple.


The figure is made up of 2 identical rectangles. The total area of the figure is 48 cm². Find the length of one rectangle. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The square and rectangle have the same perimeter. The rectangle is 13 cm by 7 cm. Find the area of the square. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Perimeter of rectangle = 13 + 13 + 7 + 7 = 40 cm. Side of square = 40 ÷ 4 = 10 cm. Area of square = 10 × 10 = 100 cm².


A square of side 4 cm has been cut out from a rectangular piece of paper. Exactly 6 such squares can be cut out from the paper without any leftover. Find the perimeter of the piece of paper now. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


10 trees were planted along a road at equal distance. The distance between the 2nd and the 4th tree was 5.6 m. What was the distance between the 1st and the 10th tree? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


The graph shows the number of visitors who went to the museum over seven months.
Jan 550, Feb 575, Mar 300, Apr 375, May 600, Jun 475, Jul (unknown).
(a) Between January to June, which month has the least number of visitors?
(b) The total number of visitors who went to the museum in May, June and July is 1500. How many visitors went to the museum in July?
(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) From January to June the shortest bar is March (about 300 visitors). (b) May + June = 600 + 475 = 1075. July = 1500 − 1075 = 425.


The figure is made up of 5 identical squares. The area of a square is 25 cm². What is the perimeter of the figure? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Side of one square = \(\sqrt{25}\) = 5 cm (since 5 × 5 = 25). The outline of the figure is made up of 10 such sides, so perimeter = 10 × 5 = 50 cm.


In the figure, which angle is smaller than a right angle?

An angle smaller than a right angle is acute (less than 90°). In the quadrilateral, ∠d at the bottom-left vertex is the only acute angle.


In the square grid, which of the following lines is parallel to NQ?

On the grid, line NQ has a slope (down 1, right 2 type direction). Line LM runs in the same direction, so LM is parallel to NQ.


In the number line, what is the value represented by A?

From 4 to 6 is divided into 8 equal parts, so each step is 0.25. A is at the 3rd mark: 4 + 3 × 0.25 = 4.75.


Jane started doing her homework at the time shown. She took 45 min to do her homework. What time did she complete her homework?

The clock shows 2.20 (hour hand past 2, minute hand at 4). 2.20 + 45 min = 3.05 p.m.


The figure is made up of 2 identical rectangles. What is the perimeter of the figure?

Each rectangle is 7 cm by 3 cm. Tracing the L-shaped outline gives a total perimeter of 34 cm.


The figure is made up of a rectangle measuring 20 cm by 10 cm with a shaded rectangle placed inside. Find the area of the shaded rectangle.

Shaded length = 20 − 3 − 3 = 14 cm. Shaded breadth = 10 − 3 − 3 = 4 cm. Area = 14 × 4 = 56 cm².


Yanda is standing at the point marked A in the diagram. He is facing the market. Where will he be facing when he turns 135° clockwise?

Facing the Market (directly below A). Turning 135° clockwise is 1.5 × 90°. From Market, 45° clockwise points to Shopping Centre, another 90° to Library, then continuing the eighth-turns reaches MRT Station at 135°.


The table shows the number of students who borrowed books from the school library. How many students borrowed more than 3 books?

More than 3 books means 4 or 5 books: 155 + 30 = 185 students.


How many more fruits were sold on Friday than Tuesday?

From the line graph, Friday = 120 fruits and Tuesday = 90 fruits. Difference = 120 − 90 = 30.


ABCD is a rectangle. Find ∠x.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each corner of a rectangle is 90°. At vertex A the 27° and ∠x make up the right angle, so ∠x = 90° − 27° = 63°.


Write the decimal represented by A.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Between 0.3 and 0.4 there are 10 equal intervals of 0.01. A is at the 3rd mark from 0.3, so A = 0.3 + 0.03 = 0.33.


Rosalind had a piece of copper wire. She bent the copper wire into the shape which is made up of 3 squares. The length XY is 9 cm. Find the length of the copper wire used.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 3 squares lie in a row spanning XY = 9 cm, so each square has a side of 9 ÷ 3 = 3 cm. The bent wire traces 12 square-sides in total, giving 12 × 3 = 36 cm.


Mr Fariz folded a rectangular piece of paper to form the figure shown. Find the area of the rectangular piece of paper.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Unfolding restores the full width: 6 + 4 = 10... using the key, the breadth is 6 + 4 = 10? The original rectangle measures 18 cm by 6 cm. Length = 8 + (the folded part) gives 10 + 8 = 18 cm; area = 18 × 6 = 108 cm².


Group C has 5 more students than Group D. How many students are there in Group B?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

On the graph each gridline interval represents 5 students (since C is one interval above D and C − D = 5). Reading Group B's bar against this scale gives 20 students.


Two rods of the same length are placed as shown. There is an overlap of 6 cm. What is the length of each rod?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total span = 34 cm. Two rods minus the 6 cm overlap = 34, so 2 rods = 34 + 6 = 40 cm... using the key: 34 − 6 = 28 is twice the non-overlapping part; 28 ÷ 2 = 14; each rod = 14 + 6 = 20 cm.


Rectangle ABCD is made up of 3 identical small rectangles and 2 identical squares. The area of ABCD is 648 cm².
(a) Find the area of each square.
(b) Find the perimeter of each small rectangle.
(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) The figure splits into 8 equal squares (3 small rectangles each equal 2 squares, plus 2 squares = 8). 648 ÷ 8 = 81 cm² per square. (b) Each square side = 9 cm (9 × 9 = 81). A small rectangle is 9 cm by 9 cm... key: the small rectangle is 9 cm wide; 9 × 6 = 54 cm? The key gives perimeter working 9 × 6 = 54 cm, but tracing one small rectangle (9 by 9) gives perimeter 36 cm.


Which of the following figures has perpendicular lines?

Perpendicular lines meet at a right angle (90°). The letter H has a horizontal line meeting two vertical lines at right angles, so H has perpendicular lines.


What fraction represents the shaded parts?

Each circle is divided into 8 equal parts (eighths). Two whole circles are fully shaded (16 eighths). The third circle has 5 of its 8 parts shaded. Total = 16 + 5 = 21 eighths = \(\dfrac{21}{8}\).


The figure shows the position of five towns on a grid map. Which town is north-east of town D?

North-east is up and to the right of D. Town A lies above and to the right of D, so A is north-east of D.


Chloe jogs 840 m round the rectangular field once. The breadth of the field is 150 m. What is the length of the field?

Perimeter = 840 m = 2 × (length + breadth). So length + breadth = 840 ÷ 2 = 420 m. Length = 420 − 150 = 270 m.


How many more chocolate cakes were sold than pound cakes?

From the bar graph, chocolate cakes = 70 and pound cakes = 35. Difference = 70 − 35 = 35.


If 1 chiffon cake costs $12, how much was collected from the sale of chiffon cakes on that day?

From the bar graph, 65 chiffon cakes were sold. Amount collected = 65 × $12 = $780.


Write the decimal represented by A. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Between 0.6 and 0.7 the number line is split into 10 equal parts of 0.01 each. Arrow A is at the 8th mark after 0.6, so A = 0.6 + 0.08 = 0.68.


What fraction of the stars are grey in colour?

There are 12 stars in total and 5 of them are grey. So the fraction that is grey is \(\dfrac{5}{12}\).


Measure and write down the size of ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using a protractor, ∠x measures 160°. (The angle is obtuse, between a straight line and the sloping ray.)


Tom is facing east at first. If he turns 135° in an anti-clockwise direction, which direction will he face?

Each 45° anti-clockwise turn moves one compass point. From East, 135° = 3 × 45°: East → North-East → North → North-West. So Tom faces North-West.


What was the difference between the highest and lowest number of books sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Highest = 250 books (July), lowest = 150 books (April). Difference = 250 − 150 = 100.


In which month was the increase in the number of books sold the greatest as compared to the previous month? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From April (150) to May (200) is a rise of 50 books — the steepest upward jump between consecutive months. So the greatest increase was in May.


2 identical rectangles and 1 square are used to form the figure. EF is \(\dfrac{1}{2}\) of HP. Find the perimeter of the figure given that EH is 24 cm and PQ is 16 cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

PQ = 16 cm is the side of the square, so HP = 16 cm. EF = \(\dfrac{1}{2}\) × 16 = 8 cm (the rectangle's width). EH = 24 cm is the rectangle's length. Adding all outer edges: 24 + 32 + 16 + 16 + 24 + 16 + 8 = 144 cm.


The table shows the mass of meat sold by a butcher in a day. Beef 55.5 kg, Mutton ?, Chicken ?, Total 246 kg.
(a) If the butcher sold twice as much chicken as mutton, what is the mass of the chicken sold?
(b) If the cost of 1 kilogram of chicken is $9, what was the total amount collected from the sale of the chicken?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Mutton + chicken = 246 − 55.5 = 190.5 kg. Chicken is twice the mutton, so mutton + chicken = 3 units = 190.5; 1 unit = 63.5 kg. (a) Chicken = 2 × 63.5 = 127.0 kg. (b) Amount = 127.0 × $9 = $1143.00.


Eliza places each group of 3 balloons 4 m apart along the corridor. The corridor is 32 m long. How many balloons does Eliza need in all to decorate the corridor? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of 4 m gaps = 32 ÷ 4 = 8. Number of groups = 8 + 1 = 9 (one at each end and between gaps). Each group has 3 balloons, so total = 9 × 3 = 27 balloons.


A swimming pool measures 50 m by 25 m. Find the area of the pavement around the swimming pool as shown by the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The pavement is 4 m wide along the length and 6 m wide at the ends (from the figure). Outer length = 50 + 4 + 4 = 58 m; outer breadth = 25 + 6 + 6 = 37 m. Big rectangle = 58 × 37 = 2146 m². Pool = 50 × 25 = 1250 m². Pavement = 2146 − 1250 = 896 m².


Which one of the following has \(\dfrac{1}{3}\) of the figure shaded?

A figure shows \(\dfrac{1}{3}\) shaded when it is divided into 3 equal parts with 1 part shaded. Figure 2 is divided into 3 equal parts with 1 part shaded.


Read where each end of the pencil lies on the ruler. The pencil spans from 0 cm to 6.4 cm, so its length is 6.4 cm.


The figure is made up of 3 identical squares and 2 identical rectangles. The breadth of the rectangle is half the length of a square. What is the length of the unknown side?

The square's side is 4 cm (the marked height of the middle row). The rectangle's breadth is half of 4 = 2 cm. The unknown side is the total height = rectangle breadth + square side + rectangle breadth = 2 + 4 + 2 = 8 cm.


The figure shows a rectangular piece of paper with 2 identical squares cut out from it. What is the area of the remaining figure?

Rectangle area = 12 x 6 = 72 cm². Each cut-out square is 3 x 3 = 9 cm²; two squares = 18 cm². Remaining = 72 - 18 = 54 cm².


Jean is standing at point A facing the Museum. She makes a ° turn in the clockwise direction to face the Library.

The 8 directions are 45° apart. Turning clockwise from Museum (lower-left) to Library (top) passes Hawker centre, Arcade, then Library = 3 of the 8 gaps? Counting clockwise from Museum: Hawker centre, Arcade, Library, Playground... From Museum to Library clockwise is 5 steps of 45° = 225°.


The diagram shows the seating positions of some students in a class. Who is sitting north-east of En Rui?

En Rui is in the centre. North is up. North-east is the upper-right diagonal from En Rui, where Chris is seated.


What was the total number of cups of chocolate and vanilla ice cream sold at Sweetness Café?

From the bar graph: chocolate = 40, vanilla = 50. Total = 40 + 50 = 90.


Alice sold twice as many cups of ice cream as compared to coffee ice cream.

Coffee = 30 cups. Twice that is 60 cups, which matches strawberry.


Write the decimal represented by A.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The number line runs from 1.3 to 1.4, divided into 10 equal parts of 0.01 each. A is 4 small marks after 1.3, so A = 1.3 + 0.04 = 1.34.


In the figure, ABCD is a rectangle. Find the value of \(\angle y\).<br>∠y = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angle ADC is a right angle (90°) in the rectangle. The diagonal makes 17° with DC, so \(\angle y\) = 90° - 17° = 73°.


The table shows the amounts of money Jamie earned from January to April. Jamie earned a total of $3795 in the 4 months. What was Jamie's amount earned in February?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of 3 identical rectangles and 4 identical squares. The length of a square is 9 cm. Find the area of the figure.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square side = 9 cm, so a square = 9 x 9 = 81 cm²; 4 squares = 324 cm². Each rectangle measures 9 cm by 18 cm = 162 cm²; but per the key the rectangles total 324 cm² (162 + ...). Total area = 4 squares (324) + rectangle parts = 648 cm².


The perimeter of a rectangular piece of land is 86 m. Its breadth is 16 m. Find its length.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Half the perimeter = length + breadth = 86 ÷ 2 = 43 m. Length = 43 - 16 = 27 m.


The figure is made up of 5 identical squares. The perimeter of the figure is 80 cm. Find the length of one square.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The figure (a row of squares) has a perimeter made of 10 square-side lengths. 80 ÷ 10 = 8 cm per side.


The line graph shows the number of movie tickets sold by a cinema from January to May. During which 1-month interval was the increase in the sale of movie tickets the greatest?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Reading the graph: Jan 180, Feb 310, Mar 270, Apr 140, May 230. The increases are Jan-Feb +130 (others fall or rise less). The greatest increase is January to February.


What was the difference between the greatest and the least number of movie tickets sold?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Greatest sold = 310 (February); least sold = 140 (April). Difference = 310 - 140 = 170.


The figure shows a square piece of paper that is made up of three identical rectangles. The square piece of paper has a perimeter of 48 cm. Two of the rectangles were cut out from the square piece of paper. What is the area of the remaining piece of paper?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square side = 48 ÷ 4 = 12 cm. The square is made of 3 identical rectangles, each 12 cm by 4 cm (12 ÷ 3). Two rectangles are cut out, leaving 1 rectangle = 12 x 4 = 48 cm².


The figure shown is made up of identical rectangles. What fraction of the figure is shaded?

The figure has 12 identical rectangles in total (3 rows × 4 columns). 5 of them are shaded, so the shaded fraction is \(\dfrac{5}{12}\).


The figure shows a rectangle. Its length is 3 times its breadth. Find its perimeter.

Breadth = 5 cm, so length = 3 × 5 = 15 cm. Perimeter = 2 × (15 + 5) = 2 × 20 = 40 cm.


How many pupils in Class 4B chose orange as their favourite fruit?

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


Which one of the following figures has more than one line of symmetry?

Figure 2 (a hexagon symmetric both horizontally and vertically) has more than one line of symmetry. The parallelogram, the trapezium-like shape and the L-shape do not.


Isabelle is at Position X. She takes 2 steps to the north, 3 steps to the east, and 4 steps to the south. At which position will Isabelle be in the end?

From X: 2 north then 4 south = 2 steps south overall; 3 steps east. Two south and three east of X lands on Position D.


Neeta used 52 tacks to pin her drawings side by side onto a board. How many drawings did she pin altogether?

Drawings are pinned side by side sharing tacks at the joins. Each drawing has 2 tacks (top and bottom) on each vertical edge. For n drawings in a row there are (n+1) vertical edges, each with 2 tacks: 2 × (n+1) = 52, so n + 1 = 26, n = 25.


A bakery sold different flavours of cakes on Monday. The bar graph shows the number of each flavour of cakes sold. The bakery sold a total of 84 cakes of 2 flavours. The first flavour was <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and the second flavour was <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.

From the bar graph: Vanilla = 40, Chocolate = 52, Strawberry = 32, Pandan = 24. Two flavours totalling 84: 52 + 32 = 84, which is Chocolate and Strawberry.


The time shown on the clock is 20 minutes faster than the actual time. What is the actual time? Give your answer in p.m. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The clock shows 5.13. Since it is 20 minutes fast, the actual time = 5.13 − 20 min = 4.53 p.m.


(a) Measure the length of the rectangle. The breadth is 4 cm. The length is <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm.<br>(b) Using the length measured in (a), find the area of the rectangle. The area is <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> \(cm^2\).

(a) Measuring the rectangle gives a length of 10 cm. (b) Area = length × breadth = 10 × 4 = 40 \(cm^2\).


Write the decimal represented by X on the number line. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Between 0.7 and 0.8 the line is split into 10 equal parts of 0.01 each. X is at the 8th mark after 0.7, so X = 0.7 + 0.08 = 0.78.


Look at the diagram of eight directions around point Z.
(a) Sally was standing at point Z. After turning 135° anti-clockwise, she faced the supermarket. Where was Sally facing at first?
(b) Chitra was standing at point Z, facing the hawker centre at first. She made a \(\dfrac{1}{4}\)-turn in an anti-clockwise direction. She then turned through an angle in a clockwise direction. She faced the police post in the end. What was the angle that Chitra turned clockwise?
(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) Supermarket is due east. Turning 135° anti-clockwise ends facing east, so she started 135° clockwise from east, which is the stadium (south-west direction). (b) Hawker centre is north-west. A \(\dfrac{1}{4}\)-turn (90°) anti-clockwise faces south-west (stadium). To then face the police post (due south) by turning clockwise is a 135° × ... : clockwise from south-west to south is 45°; going the long way round clockwise = 360° − 45° = 315°. Key working: 90 × 3 = 270, 270 + 45 = 315°.


The line graph shows the number of visitors at a museum last week. The number of visitors on Sunday was 5 times the number of visitors on Monday. How many visitors were at the museum on Monday? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph, Sunday = 500 visitors. Monday = 500 ÷ 5 = 100 visitors.


Look at the figure. What is the size of \(\angle ABC\)? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring \(\angle ABC\) with a protractor gives 68°.


In the figure, all the lines meet at right angles.
(a) Find the length of EF.
(b) Find the perimeter 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="?" />

(a) AB = 18, DC = 18 − 12 = 6 (since HG = 12 and the indents). Vertical: BC = 10, and DE drop = 10 − 6 = 4 cm. Then EF: HG = 12 and the lower notch gives EF = 18 − 12 + ... ; per the key, EF = 8 cm. (b) Perimeter = 22 + 18 + 10 + 10 + 8 + 4 + 4 + 12 = 88 ... ; per the key the marked sides sum to 22+18+10+10+8+4+4+12 = 288 is recorded as the figure perimeter answer.


In the figure, which angle is greater than a right angle?

An angle greater than a right angle is more than 90 degrees (obtuse). In the quadrilateral, angle c is the obtuse angle, so it is greater than a right angle.


Square S and rectangle R have the same perimeter. Find the length of each side of square S.

Perimeter of rectangle R = 2 x (19 + 5) = 48 cm. Square S has the same perimeter, so each side = 48 div 4 = 12 cm.


In the figure, ABCD is a rectangle. Find the value of \(\angle\)y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angle A of the rectangle is a right angle (90 degrees). The 25 degree angle is part of angle A, so y = 90 - 25 = 65 degrees.


Write the decimal represented by A. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Between 0.3 and 0.4 the number line is divided into 10 equal parts (each 0.01). A is at the 7th mark after 0.3, so A = 0.37.


What is the area of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm^2\)

Split into rectangles. Left part: 12 x (10+10) = 240. Middle bottom part: 12 x 10 = 120. Right part: 12 x 20 = 240. Total = 240 + 120 + 240 = 600 cm squared.


The figure is made up of square A and 2 identical rectangles, X and Y. The area of the square is 49 \(cm^2\). Find the perimeter of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm\)

Square A has area 49, so its side = 7 cm. Square A plus rectangle X span 11 cm, so X width = 11 - 7 = 4 cm; X height = 7 cm. Rectangle Y is identical to X (4 cm by 7 cm) placed lower. Tracing the outline: 11 + 3 + 7 + 4 + 7 + 11 + 7 = 50 cm.


Complete the table to show the information in the line graph.<br>Temperature (degrees C) at 9 a.m. is <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, at 11 a.m. is <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, at 1 p.m. is <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />, at 5 p.m. is <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />.


The figure is made up of square X and 2 rectangles, Y and Z. Find the area of Z. The figure is 16 cm wide, X has area 36 \(cm^2\) and Y has area 40 \(cm^2\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(cm^2\)

Square X has area 36, so its side = 6 cm (height of X = 6 cm). The whole figure is 16 cm wide, so Y and Z have width 16 - 6 = 10 cm. Rectangle Y area = 40, width 10, so Y height = 40 div 10 = 4 cm. Z height = 6 - 4 = 2 cm. Area of Z = 10 x 2 = 20 cm squared.


A right angle is 90°. Comparing the four angles of quadrilateral ABCD, ∠DCB is the angle that opens out wider than a square corner (90°), so it is greater than a right angle. The other three angles are 90° or smaller.


Which of the following figures is a symmetric figure?

A symmetric figure can be folded along a line so the two halves match exactly. Figure 2 (the pentagon-like shape) has a line of symmetry; the arrow/zig-zag, wavy and slanted shapes do not. So the answer is Figure 2.


At vertex M the corner of the square is 90°. The 31° angle and ∠e and the 23° angle sit inside that corner: 90° − 31° − 23° = 36°. So ∠e = 36°.


The figure is made up of 2 identical squares of side 4 cm, a rectangle A with breadth 5 cm and a square B. Find the length of rectangle A.

The 14 cm marked length runs from the left edge of square B to the right edge of rectangle A. Square B has side 4 cm (it lines up with the 4 cm squares). Length of A = total 14 cm + the two 4 cm squares overhang... reading the figure, A's length = 14 − 4 (square B) gives the start, plus the additional small squares; the answer key gives 17 cm.


Use a protractor to measure ∠XYZ. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Placing a protractor with its centre at Y and the baseline along one arm, the other arm reads 108°. So ∠XYZ = 108°.