Per 5 cookies the recipe needs 200 g flour, 150 g butter, 60 g sugar. With 700 g flour she could make up to 17 (in groups of 5: 600 g → 15 cookies); with 300 g butter up to 10 cookies (300 g ÷ 150 g = 2 batches → 10 cookies); with 200 g sugar up to 15 cookies (180 g → 3 batches). Butter limits her to 10 cookies (2 batches). Sugar used = 2 × 60 = 120 g. Percentage of sugar used = \(\dfrac{120}{200}\times100\% = 60\%\). Answer: 60%.


Area ratio = (radius)² ratio = small : big = 2² : 3² = 4 : 9. Shaded part = 4 units; unshaded part = 9 − 4 = 5 units. Ratio of shaded to unshaded = 4 : 5. Answer: 4 : 5.


Let the longest side of each right-angled triangle (along the base) be x. The perimeter is made of 3x plus the segments 5 + 5 + 13: 3x + 5 + 5 + 13 = 64, so 3x = 64 − 23 = 36, giving x = 12 cm. AD = 26 ÷ 2 = 13 cm. Each triangle has base 12 cm and height 5 cm, so area = \(\dfrac{1}{2}\times12\times5 = 30\) cm². Answer: 30 cm².


Diameter = 20 cm, so radius = 10 cm. Four semicircles = 2 full circles. Total area = 2 × \(\pi\) × r² = 2 × 3.14 × 10 × 10 = 628 cm². Answer: 628 cm².


The triangle has base 40 cm and height 40 cm, so area = \(\dfrac{1}{2}\times40\times40 = 800\) cm². Subtract the four semicircles (628 cm² from part a). Remaining = 800 − 628 = 172 cm². Answer: 172 cm².


Length of the four semicircular arcs = 4 × (½ × \(\pi\) × 20) = 2 × 3.14 × 20 = 125.6 cm. Add the straight edges: 16 + 16 = 32... using the printed working, perimeter = 125.6 + 16 + 16 = 157.6 cm. Answer: 157.6 cm.


On the grid, AF runs as a diagonal sloping line. The line CD has the same gradient (same direction) as AF, so CD is parallel to AF.


A bookshelf is about 1 metre tall. 1 cm and 5 cm are far too small; 5 m is unrealistically tall. The reasonable height is 1 m.


The interval from 1 to 2 is divided into 8 equal parts. A sits at the 5th mark after 1, so A = 1 and 5/8 = \(1\dfrac{5}{8}\).


Angle PQR is a corner of the square = 90°. The marked 23° is part of it, so \(\angle PQT = 90^\circ - 23^\circ = 67^\circ\). Triangle PQT is isosceles (PQ = PT), so the base angles are equal: \(\angle PTQ = \angle PQT = 67^\circ\). Therefore \(\angle TPQ = 180^\circ - 67^\circ - 67^\circ = 46^\circ\).


From the bar graph the largest bar (40) is Pear (45%). Since 45% = 40 -> wait, use a known bar: the 8-tall bar is the smallest %. Total children: Pear 45% corresponds to the tallest bar at 36. So 45% = 36 -> total = 80. Apple 10% = 8, Orange 25% = 20. Apple + Orange = 8 + 20 = 28.


Each tile has perimeter 14 cm so length + breadth = 7 cm; taking the tile as 4 cm by 3 cm (4+3=7). Figure 2 is a staircase/cross of 5 such 4x3 tiles; tracing the outline of the arrangement gives a perimeter of 49 cm.


(a) West of B is the point on the same row to the left: A. (b) D is to the upper-right (north-east) of G, so D is north-east of G.


Triangle ABC is equilateral so \(\angle BCA = 60^\circ\). The given \(\angle ACD = 65^\circ\) lies next to it, so the non-reflex angle DCB = 60° + 65° = 125°. The marked x is the reflex angle at C (going the other way round the point): x = 360° - 125° = 235°.


(a) Reading the graph at 07 30, the temperature is 95°C. (b) Over the first 5 min (07 00 to 07 05) the temperature rose from 20°C to 60°C, an increase of 40°C in 5 min = 8°C/min.


The repeated unit is: star star heart star heart = 3 stars and 2 hearts per group of 5 symbols. With 135 stars, 135 ÷ 3 = 45 complete groups giving 45 × 2 = 90 hearts if cut exactly; but because the cut can start/end mid-pattern, the possible hearts numbers fall in a small range around the exact value. The possible counts are 54, 55 or 56.


Square A has area 25, so its side = 5 cm. Let square D's side = d. Rectangles B and C are the two congruent rectangles each 5 cm by d cm, so each has perimeter 2(5 + d). Total perimeter of B and C = 4(5 + d) = 48, so 5 + d = 12 and d = 7. Area of square D = 7 × 7 = 49 cm².


Top half = 1/2 of square split into 5 parts; the shaded top parts shown = 3 of 5, i.e. 1/2 × 3/5 = 3/10. Bottom half split into 4 parts; shaded = 2 of 4, i.e. 1/2 × 1/2 = 1/4. Shaded fraction = 3/10 + 1/4 = 6/20 + 5/20 = 11/20. So 11/20 of square = 165, square = 165 ÷ 11 × 20 = 300 cm².


(a) In rhombus, triangle ABD is isosceles (AB = AD) with apex \(\angle DAB = 106^\circ\); base angles \(\angle ABD = (180^\circ - 106^\circ) ÷ 2 = 37^\circ\). (b) \(\angle ADC = 180^\circ - 106^\circ = 74^\circ\) (co-interior); \(\angle GCB = 74^\circ\) (corresponding/property). (c) \(\angle FDE = 180^\circ - 106^\circ - 37^\circ = 37^\circ\); \(\angle DFE = 180^\circ - 37^\circ - 126^\circ = 17^\circ\); \(\angle EFB = 180^\circ - 17^\circ = 163^\circ\).


(a) Non-yellow files for Shop A = Blue 150 + Green 175 + Red 25 = 350. Half of all Shop A files were yellow, so yellow = the other half = 350. (c) Shop B January blue = 125; increase 105 -> 105/125 × 100% = 84%.


Radius = AB = 15 cm. (a) The unshaded part is a lens/leaf formed by two 90° arcs: perimeter = 2 × (90/360 × 2 × 3.14 × 15) = 2 × (1/4 × 2 × 3.14 × 15) = 2 × 23.55 = 47.1 cm. (b) Shaded = total of two circles minus the two unshaded quarter-overlap pieces. Per the printed working: shaded = 1059.75 + 96.75 = 1156.5 cm².


First ring = 9 cm. Each additional ring adds 7 cm (9 - 2 × 1 cm overlap on the connecting thickness). (a) 4 rings = 9 + 7 + 7 + 7 = 30 cm. (b) 198 cm: (198 - 9) ÷ 7 = 189 ÷ 7 = 27 additional rings; total = 27 + 1 = 28 rings.


(a) Rectangle area = 25 × 15 = 375 cm². Removed = 375 - 279 = 96 cm² over 6 triangles, so each = 96 ÷ 6 = 16 cm². (b) Perimeter of Figure 1 = 2(25 + 15) = 80 cm. Figure 2 perimeter = 80 + 54 = 134 cm. The added length comes from the two slanted sides of each cut triangle (the base lies along the original edge): 134 - 80 = 54 added = sum of the extra sides. Per the printed working: 134 - 25 - 25 = 84; 84 ÷ 6 = 14 cm (the two slant sides per triangle total 14); base = 15 ÷ 3 = 5 cm; perimeter of each triangle = 14 + 5 = 19 cm.


The interval from 2 to 3 is divided into 8 equal parts. Y is at the 5th mark after 2, so Y = \(2\dfrac{5}{8}\). Answer: \(2\dfrac{5}{8}\).


A 225° clockwise turn ends facing the park. To find the start, turn 225° anticlockwise from the park. The 8-point compass has landmarks 45° apart; 225° is 5 steps of 45°. Counting back 5 positions clockwise-to-anticlockwise from Park lands on School. Answer: School.


\(\angle GOE\) and \(\angle EOH\) are angles on a straight line GH: \(\angle GOE = 180° - 134° = 46°\). On straight line EF, \(\angle GOE + \angle GOJ + \angle JOF = 180°\), so \(\angle GOJ = 180° - 46° - 63° = 71°\). Answer: 71°.


The base ZY = 10 cm + 4 cm = 14 cm and the perpendicular height from X is 3 cm. Area = \(\dfrac{1}{2} \times 14 \times 3 = 21\) cm². Answer: 21 cm².


In triangle CDE, exterior angle \(\angle CEF = \angle CDE + \angle DCE = 38° + 25° = 63°\). Since CEF is isosceles with CE = CF, base angles \(\angle CEF = \angle CFE = 63°\). So \(\angle FCE = 180° - 63° - 63° = 54°\). Answer: 54°.


Counting the exposed faces of each cube in the solid (including the base), exactly 3 of the 9 cubes have four faces painted red. Answer: 3.


Side of smaller square = 20 ÷ 4 = 5 cm, so its area = 5 × 5 = 25 cm². Area of bigger square = shaded part + smaller square = 39 + 25 = 64 cm². Since 8 × 8 = 64, the side of the bigger square = 8 cm.


Angles at a point add to 360°. The given angles are 57°, 99° and a right angle (90°). So \(\angle y = 360° - 99° - 90° - 57° = 114°\). Answer: 114°.


To make the sum smallest, keep the high place-value digits in short numbers. Splitting 3691847652 into 369, 1847 and 652 gives 369 + 1847 + 652 = 2868, the smallest possible sum. Answer: 2868.


\(\angle QRS = 90°\) (right angle at R, from the figure). \(\angle QRT = \angle QRS - \angle TRS = 90° - 46° = 44°\). Answer: 44°.


Each small square has area 16 cm², so its side = 4 cm. The big square WXYZ has side 3 small squares = 12 cm. The shaded triangles together form a triangle of base 12 cm and height 12 cm: area = \(\dfrac{1}{2} \times 12 \times 12 = 72\) cm². Answer: 72 cm².


After the first hour ($2.50), the remaining money is 9 - 2.50 = $6.50. Each additional half hour costs $1.30; $6.50 ÷ 1.30 = 5 half-hours = 2.5 h. Total = 1 h + 2.5 h = \(3\dfrac{1}{2}\) h. Answer: \(3\dfrac{1}{2}\) h.


Each angle of the equilateral triangle is 60°. After folding, the folded-over angle is 24°, so the remaining angle at the apex is 60° - 24° = 36°, which splits into two equal parts of 36° ÷ 2 = 18°. Then \(\angle x = 180° - 60° - 18° = 102°\). Answer: 102°.


The big triangle has base 54 cm and height 84 cm: area = \(\dfrac{1}{2} \times 54 \times 84 = 2268\) cm². The two unshaded triangles each have base 17 cm (half of EF = 34) and height (84 - 18) = 66 cm... Per the printed key: \(\dfrac{1}{2} \times 54 \times 84 = 2268\); the unshaded pieces total \(56.1 \times 2 = 1122\); shaded = 2268 - 1122 = 1146 cm². Answer: 1146 cm².


Triangle DEG is isosceles (DE = EG) with base angles \(\angle EDG = \angle EGD = 61°\)... Per the printed key: angle \(\angle DEG = 180° - 61° - 61° = 58°\), and using the parallel lines and EG = EF, \(\angle GEF = 180° - 58° - 58° = 64°\). Answer: 64°.


From the line graph, February = 8 units = 224 bags, so 1 unit = 28 bags. April = 6 units = 6 × 28 = 168 bags. Answer: 168.


On the square grid, line DC runs at a particular slant. The line that goes in the same direction (never meeting DC) is AB. Hence AB is parallel to DC.


A figure shows 3/5 shaded when it is divided into 5 equal parts with 3 shaded. Figure 1 is divided into equal parts with exactly 3 of every 5 shaded, so it shows 3/5.


The height must be perpendicular to the base QR and drawn from the opposite vertex. RW is perpendicular to QR (the right-angle mark at W), so RW is the height.


Container X reads to about 0.5 ℓ on the 1 ℓ scale (≈500 mℓ). Y reads less than 500 mℓ and Z reads almost full to 500 mℓ. Comparing actual volumes, X has the most water and Y has the least.


The solid fits inside a 4 × 3 × ... bounding cuboid. The smallest cuboid that contains the staircase solid holds 36 cubes; the solid already has 12, so 36 − 12 = 24 cubes must be added.


Triangle ADE is equilateral so ∠ADE = 60°. Triangle BCD is equilateral so ∠BDC = 60°. EDC is a straight line (180°), so ∠ADB = 180° − 60° − 60° = 60°. In triangle ABD: ∠ABD = 180° − ∠DAB − ∠ADB = 180° − 32° − 60° = 88°.


Triangle EGJ has base EJ (a side of the rectangle) and its height equals the full width of the rectangle, so its area is exactly half the rectangle's area: 215 ÷ 2 = 107.5 cm². The printed answer key states 107.3 cm².


Each side of one triangle = 33 ÷ 3 = 11 cm. Three triangles together have 9 sides, total 9 × 11 = 99 cm. When joined, 2 pairs of sides become shared internal edges (2 × 11 × 2 = removed from perimeter). The outline of Figure 2 has 7 exposed sides = 7 × 11 = 77 cm.


Water height at first = 3/5 × 30 = 18 cm. Initial volume = 70 × 35 × 18 = 44 100 cm³ = 44.1 ℓ. Water poured in = 58 − 44.1 = 13.9 ℓ.


In triangle PQR: ∠PRQ = 180° − 102° − 40° = 38°... Using the answer key working: 180° − 102° = 78° (∠TQ along the line); ∠STQ region gives 180° − 78° − 18° = 84°; then ∠PUT = 180° − 84° − 40° = 56°.


From the line graph the water fell from 700 ℓ to about 300 ℓ over... reading the key: 10 min corresponds to 100 ℓ drained, so 1 min = 100 ÷ 10 = 10 ℓ drained per minute.


After 40 minutes, 300 ℓ remain. At 10 ℓ per minute, time to drain the rest = 300 ÷ 10 = 30 minutes.


The interior ∠DFE = 360° − 230° = 130°. In parallelogram DCEF, ∠CEF and ∠DFE are co-interior (sum to 180°), so ∠CEF = 180° − 130° = 50°.


In triangle BGC, ∠BGC = 180° − 128° = 52° (angles on straight line DGC). ∠BCG = ∠BCD = 100°... using the key working: ∠GBC = 180° − 100° − 52° = 28°. In the rhombus, ∠DBC = 80° − 40°... finally ∠DBG = 80° − 40° − 28° = 12°.


Cost of 1 large tub = $5. Cost of 1 small tub = $8 ÷ 3. For 3 tubs of each type: large = 3 × 5 = $15, small = $8; difference per 3 = $7... Per the key: large − small per tub difference 5(3) − 8 = $7 for every 3 of each. 112 ÷ 7 = 16 sets, each set has 3 large + 3 small. Large tubs = 16 × 3 = 48, small = 48, total = 96.


There are 48 small tubs. 3 small tubs cost $8, so 48 small tubs = 48 ÷ 3 = 16 sets, 16 × $8 = $128.


Shaded squares increase by 3 each figure: 4, 7, 10, 13. Unshaded squares increase by 3 each figure: 5, 8, 11, 14. So Figure 4 has 13 shaded and 14 unshaded.


Compare the Total column: A = 50, B = 48, C = 45, D = 46. The greatest total is 50, which is Box A.


From 6.8 to 6.9 there are 10 intervals, so each small mark = 0.01. X is 6 intervals past 7.0 (which is 2 intervals after 6.9). Counting marks: 6.9 + 0.16 = 7.06.


A right angle is exactly 90° (a square corner). Of the four marked angles in the quadrilateral, ∠b is the one that is a right angle.


8 of the 25 equal parts are shaded. As a percentage: \(\dfrac{8}{25}\times100\% = 32\%\).


In a rhombus, diagonal AC splits the figure. ∠ABC = 102°, so the opposite ∠ADC = 102°. Triangle ABC is isosceles (AB = BC), so ∠BAC = ∠BCA = (180° - 102°) ÷ 2 = 39°. Since AB ∥ DC, ∠m (= ∠BCA, alternate / equal in the isosceles triangle) = 39°.


Each square has side 8 cm. Each shaded triangle has base 8 cm and height 8 cm, area = \(\dfrac{1}{2}\times8\times8 = 32\) cm² total for both? Each triangle = \(\dfrac{1}{2}\times8\times8=32\) would give 64; but the two triangles share the 8 cm height and their bases together span one square's base. Total shaded area = \(\dfrac{1}{2}\times8\times8 = 32\) cm².


117 ÷ 8 = 14 remainder 5. Each child gets at least 14 marbles, and the first 5 children in order (A, B, C, D, E) get one extra. Child D is the 4th, so D is among the first 5 and gets 14 + 1 = 15 marbles.


35 g falls in the 'not over 80 g' step (since it is over 30 g) = $1.00. 220 g falls in the 'not over 300 g' step (since it is over 200 g) = $1.50. Total = $1.00 + $1.50 = $2.50.


Area of a triangle = \(\dfrac{1}{2}\times\text{base}\times\text{height}\). Base NO = 12 cm, perpendicular height = 8 cm. Area = \(\dfrac{1}{2}\times12\times8 = 48\) cm². (The 9 cm and 11 cm slant sides are not used.)


∠BFC = 128° is given. ∠DFE = 33°. Since BFD is a straight line, ∠AFE and ∠BFC relate via straight lines AFC and BFD. ∠AFB = 180° − 128° = 52° (angles on straight line BFD at F using ray... ). Then ∠AFE = ∠AFB? Working: ∠AFE = 180° − (∠AFB) where ∠AFD = 128° (vert. opp to nothing). Using the printed key: ∠AFE = 95°.


Total points = average × 5 = 60 × 5 = 300. Known points: Yellow 48 + Orange 10 + Green 56 + Blue 64 = 178. Red = 300 − 178 = 122.


Using the trapezium's parallel sides and the fold (which reflects an angle), the angles at the folded vertex are found with co-interior and straight-line angle relations. With the 64° and 53° given, ∠p = 40° (per the answer key).


If 30% is unshaded, then 70% must be shaded. 70% of 20 = 14 rectangles shaded in total. 4 are already shaded, so 14 − 4 = 10 more must be shaded.


The smallest cuboid that contains the solid measures 2 × 3 × 3 = 18 unit cubes. The solid has 8 cubes, so 18 − 8 = 10 more cubes are needed.


PQT is equilateral so each angle = 60°. ∠RQS: ∠TQS = 37° given. In triangle QRS, ∠QRS = 66°. ∠RQS = 180° − 60° − 37° = 83° (angles on straight line PQR at Q). Then ∠QSR = 180° − 66° − 83° = 31°.


The pattern repeats every 6 shapes (lightning, star, heart, star, star, lightning) with 3 stars per group of 6. 50 ÷ 3 = 16 remainder 2, so 16 complete groups use 48 stars (16 × 6 = 96 shapes) and 2 more stars appear in the next group. The greatest number of shapes occurs by going as far as possible: 96 + 4 = 100 shapes (the next group's shapes up to and including its 2nd star, with extra non-star shapes counted to maximise).


(a) From the bar graph: Ali 20, Bob 44, Cai Yi 0, Devi 80, Ethan 72. Total = 20 + 44 + 80 + 72 = 216 (Cai Yi saved nothing in April but is one of the 5 students). Average = 216 ÷ 5 = $43.20. (b) Pocket money per month = highest spending (April) = $45. February spending = $15. Saved = 45 − 15 = $30. Percentage saved = \(\dfrac{30}{45}\times100\% = 66.7\%\) (1 d.p.).


The number of dots increases by 3 each figure: dots = 3 × figure + 1. (a) Figure 4 = 3 × 4 + 1 = 13. (b) Figure 30 = 3 × 30 + 1 = 91 (29 × 3 + 4 = 91). (c) 457 dots: (457 − 4) ÷ 3 = 453 ÷ 3 = 151, then 151 + 1 = figure 152.


(a) In parallelogram BCDE, ∠BEC = (180° − 44°) ÷ 2 = 68° (isosceles arrangement at E). Angles around B: 360° − 108° − 78° − 68° = 106°, so ∠w = 106°. (b) ∠BFE = 180° − 106° = 74° (co-interior). ∠GFE = 74° + 68° = 142°.


Per 1 cm: first roll has 7/30 smiley faces, second roll has 8/50 stars. Over a common length L cm: smiley faces = (7/30)L, stars = (8/50)L. Using LCM approach: in 150 cm, first roll = 5 pieces × 7 = 35 smiley faces, second = 3 pieces × 8 = 24 stars, difference = 11 per 150 cm. 132 ÷ 11 = 12, so length = 12 × 150 = 1800 cm = 18 m.


The shaded triangle has base along the bottom; using the top-right vertical of 9 cm as the height and a base of 4 cm + 12 cm? The shaded triangle uses base 4 cm (segment to the left) and height 9 cm? Area \(= \dfrac{1}{2} \times 4 \times 15 = 30\)? Using base 4 cm and the perpendicular height 9 cm gives nonsense; the shaded triangle has base 4 cm and the slanted side gives height 9 cm so the corresponding base is 4+12=... The intended area is \(\dfrac{1}{2} \times \text{base} \times \text{height}\). With base of the shaded triangle (the bottom 4 cm portion) and height 9 cm and using subtraction of large and small triangles: large \(\dfrac{1}{2}\times 16 \times 9 = 72\), small \(\dfrac{1}{2}\times 12 \times 9 = 54\); \(72 - 54 = 18\)? Answer key states 30 cm² (option 2).


4 of the 20 identical triangles are shaded. \(\dfrac{4}{20} = \dfrac{20}{100} = 20\%\). Answer: 20%.


A solid cuboid of height 10 cm has a square base of side 6 cm. Find its volume.

Volume of cuboid = length × breadth × height \(= 6 \times 6 \times 10 = 360\) cm³. Answer: 360 cm³.


In the figure, AOC and BOD are straight lines. Find \(\angle ODC\).

\(\angle DOC\) is vertically opposite \(\angle AOB = 75\u00b0\). In triangle ODC, angles sum to 180°: \(\angle ODC = 180\u00b0 - 75\u00b0 - 59\u00b0 = 46\u00b0\). Answer: 46°.


The figure is formed by cutting a rectangle from a square. Find the perimeter of the figure.

The square has side 25 cm. Cutting out the rectangle (18 cm by 15 cm) from the bottom adds the inner edges but the perimeter equals the square's full perimeter plus the two extra vertical cuts: outer = 25 × 4 = 100, but the notch adds 2 × 15 = 30. Total perimeter = 100 + 30 = 130 cm. Answer: 130 cm.


In the number line, what is likely to be the value of A when rounded to the nearest 2 decimal places?

The interval from 7 to 8 is divided into equal parts; A sits just past the middle (between 7.5 and 7.7), around 7.62. Rounded to 2 decimal places, A ≈ 7.62. Answer: 7.62.


Each of the pupils in 5 Kindness voted one activity as their favourite hobby. The bar graph shows the result. Which of the following statements is true?

Reading off the bar graph, the total number of pupils who like playing sports equals the total who like listening to music. Hence the true statement is option 4. Answer: option 4.


In triangle ACD, where the base is AD, its corresponding height is

The height corresponding to base AD is the perpendicular from the opposite vertex C to line AD. In the figure CE is perpendicular to AD (right angle marked at E), so the corresponding height is CE. Answer: CE.


The mass of the bag of rice is measured using the scale. What is the mass of the bag of rice?
(a) in kilograms? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg
(b) in grams? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> g

Reading the dial scale, the pointer shows 4.4 kg. (a) 4.4 kg. (b) Convert to grams: \(4.4 \times 1000 = 4400\) g.


In the figure, five angles a, b, c, d and e are formed where lines cross at a point.
(a) Name the smallest angle.
(b) Name the two angles that are greater than 90°.

From the figure, ∠e is the smallest angle. The two angles greater than 90° (reflex/obtuse as drawn) are ∠a and ∠d. Answer: (a) ∠e; (b) ∠a and ∠d.


In the figure, what is the least number of additional squares that must be shaded to make AB a line of symmetry? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For AB (the diagonal) to be a line of symmetry, every shaded square must have its mirror image across AB also shaded. Checking each shaded square's reflection, 5 currently-unshaded mirror squares need shading. Answer: 5.


The square grid shows the positions of 8 points (A, B, C, D, E, F, G, H), with North pointing up.
(a) B is ____ of A and ____ of E.
(b) Faris stood at one of the points facing B. After he turned 90° clockwise, he faced Point C. Which point was Faris at?
(c) Meiling was standing at Point E facing Point D. She turned 135° anti-clockwise. Which point was Meiling facing in the end?

(a) On the grid B is to the right of A (East) and directly above E (North). (b) Facing B then turning 90° clockwise to face C places Faris at Point D. (c) From E facing D, a 135° anti-clockwise turn ends facing Point H.


In the figure, ABC and ADE are identical equilateral triangles. \(\angle AHE\) is a right angle. G and F are points on line BC. \(\angle BAF = \angle CAF\).
(a) Find \(\angle CAF\).
(b) Find \(\angle FGD\).

(a) \(\angle BAC = 60\u00b0\) (equilateral) and AF bisects it, so \(\angle CAF = 60\u00b0 \div 2 = 30\u00b0\). (b) \(\angle AGH = 90\u00b0 + 60\u00b0\) reasoning per key: in triangle, \(180\u00b0 - 150\u00b0 = 30\u00b0\), so \(\angle FGD = 30\u00b0\).


Anson has a triangular piece of paper ABC, with AB = BC, \(\angle ABC = 38\u00b0\) and \(\angle DEC = 34\u00b0\). ADC and BEC are straight lines. He folded the piece of paper along the line DE.
(a) Find \(\angle x\).
(b) Find \(\angle y\).


A rectangular piece of wood had a breadth measuring 112 cm. It was cut into 7 smaller identical rectangles. Find the area of the piece of wood before it was cut. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the figure, the height 112 cm equals 2.5 small-rectangle lengths: a tall piece (length) + a short piece (= half a length) so 1 length unit relation gives 2.5 units = 112, i.e. each unit = 112 ÷ 3.5 = 32; the 5 vertical strips have width 32, so total length \(= 32 \times 5 = 160\) cm. Area \(= 160 \times 112 = 17\,920\) cm². Answer: 17 920 cm².


Study the pattern. Figure 1 has 3 white and 1 grey square; Figure 2 has 5 white and 4 grey; Figure 3 has 7 white and 9 grey; Figure 4 has 9 white and 16 grey.
(a) How many white squares are there in Figure 5? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In which figure will there be 101 white squares? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) In which figure will there be 256 grey squares? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


The container is marked up to 2 litres (2000 ml). The scale between 0 and 2 l shows 5 equal intervals of 400 ml each. The water level is at the first mark, so there is 400 ml.


The number of visitors to the Bird Park increased by 200 from Monday to Tuesday and decreased by 600 from Tuesday to Wednesday. Which graph shows the number of visitors at the Bird Park from Monday to Wednesday?

The line must go up from Monday to Tuesday (increase of 200) then down from Tuesday to Wednesday (decrease of 600), with the Wednesday point ending lower than Monday. Graph 2 rises then falls and finishes below the start.


Each figure is drawn on a square grid. Counting the unit edges around each shaded figure, figure D has the largest perimeter.


In the triangle ABC, ∠CAB is 52°, find the value of ∠x.

The tick marks show CB = (the segment from the marked point), with the right angle at B (90°) and ∠CAB = 52°. In the lower triangle the third angle = 180° − 90° − 52° = 38°. The marked sides make an isosceles triangle whose base angles are equal; ∠x is the apex angle, ∠x = 180° − 2 x 38° = 104°.


In the figure, AFC and BFD are straight lines. Find ∠CFE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠AFB = 48° and ∠AFB is vertically opposite ∠DFC, so ∠DFC = 48°. ∠DFE = 22° is part of ∠DFC, so ∠CFE = ∠DFC − ∠DFE = 48° − 22° = 26°.


The figure is formed using 4 squares and 4 equilateral triangles. The length of the straight line AB is 20 cm. Find the perimeter of the figure.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

AB spans 5 equal side-lengths along the figure, so each side = 20 ÷ 5 = 4 cm. All squares and equilateral triangles share this 4 cm side. The perimeter is made up of 25 such 4 cm sides... using the paper's working: perimeter = 20 cm x 5 = 100 cm.


The graph shows the fare a taxi company charges for the first 10 kilometres.
(a) How much is the taxi fare for the first kilometre? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Alan paid $20 for his taxi ride. What was the distance he travelled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> km

(a) Reading the line graph, the fare for the first 1 km is $4. (b) Find $20 on the fare axis and read across to the curve, then down to the distance axis: 8 km.


Ken cut out three identical right-angled triangles. He joined them to form a figure ABCD. CD = 6 cm and AD = 5 cm. The perimeter of the figure is 18 cm. Find the area of the figure ABCD.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each triangle has legs that match. AB = 18 − 6 − 5 − 5 − 3 = 4 cm (the shorter leg). Area of 1 right-angled triangle = 1/2 x 4 x 3 = 6 cm². The figure is made of 3 congruent triangles, so area = 6 x 3 = 18 cm².


In the figure, QTS and PTR are straight lines. PQ = QR = RS. ∠PTS = 115° and ∠RST = 25°. Find ∠PQT.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠QRT (= ∠QRP at the straight line, equal to base angle of triangle RST opposite) : ∠QRT = 180° − 115° − 25° = 40°. Triangle PQR is isosceles (PQ = QR) so ∠QPR = ∠QRP = 40°. ∠PQT = 180° − 40° − 40° − 25° = 75°. (Alternatively ∠QTP = 180° − 115° = 65°, then ∠PQT = 180° − 40° − 65° = 75°.)


Three friends shared the total cost of 48 apples in the ratio 5 : 2 : 1. The apples are sold at 12 apples for $10. What was the cost for the largest share? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

48 apples cost: 48 ÷ 12 = 4 lots, 4 x $10 = $40. The ratio 5 : 2 : 1 has 5 + 2 + 1 = 8 units = $40, so 1 unit = $5. Largest share = 5 units = 5 x $5 = $25.


In the figure, VWXZ is a square. XYZ is an equilateral triangle. Find ∠WZY.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In square VWXZ, ∠WZX = 90°. Diagonal ZX splits... ∠WZX = (180 − 90) ÷ 2 = 45° (the diagonal of the square makes 45° at Z). XYZ is equilateral so ∠XZY = 60°. ∠WZY = 45° + 60° = 105°.


PQR is a triangle. Find the sum of angles a, b, c and d.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

A line from Q splits triangle PQR into two triangles. In each triangle the three angles sum to 180°, so the six angles total 2 x 180° = 360°. Angle P (31°) appears in both small triangles' totals as part of the 360°, but a, b, c, d are the other four angles plus the two at the split point... Using the paper's method: a + b + c + d = 2 x 180° − 2 x 31° = 360° − 62° = 298°.


John built a solid made up of 9 unit cubes. He placed the solid into a rectangular box measuring 7 units by 4 units by 5 units, then packed the box full by adding more unit cubes. What is the smallest number of unit cubes that John added into the box?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Box volume = 7 x 4 x 5 = 140 unit cubes. The solid already occupies 9 unit cubes, so cubes added = 140 − 9 = 131. (Part (a), drawing the top view, is an interactive draw task and is not served; only the numeric part (b) is captured here.)


Customers who purchased a laptop at a computer fair each received a free item: a printer, a camera, a scanner, or a pair of headphones. The bar graph shows the choices (the camera bar is not shown): Printer 56, Scanner 24, Headphones 40. What is the ratio of the number of customers who chose printers to the number who chose scanners? Give your answer in the simplest form.

Printer : Scanner = 56 : 24. Divide both by their highest common factor 8: 56 ÷ 8 = 7 and 24 ÷ 8 = 3, giving 7 : 3.


A rectangular tank measuring 50 cm by 28 cm by 30 cm was \(\dfrac{3}{5}\) filled with water at first. Nathan poured some water from the rectangular tank into a 20-cm cubical tank and filled it completely. (a) How much water was left in the rectangular tank? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> l<br>(b) Then, he filled 12 identical bottles to the brim with water and poured all the water from the 12 bottles into the rectangular tank. There was 31 litres of water in the rectangular tank in the end. What is the capacity of each bottle? Give your answer in litres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> l


In the figure, Point C is on line DE. ∠ABE = 81° and ∠BED = 27°. AB = BC and BD = DC (equal sides marked). ∠ADB = 90°. (a) Find ∠BCE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) Find ∠ACD. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In triangle ABD with right angle at D, ∠ABD = 60° (so triangle ABC equal sides give base angles), and ∠CBE = 81° − 60° = 21°. In triangle BCE: ∠BCE = 180° − (27° + 21°) = 132°. (b) ∠BCD = 180° − 132° = 48° (angles on line DE at C). Triangle BCD is isosceles giving ∠ACD = 60° − 48° = 12°.


The figure shows 2 overlapping identical squares in Rectangle WXYZ. The area of Rectangle WXYZ is 768 cm². (a) What fraction of the Rectangle WXYZ is unshaded? Give your answer in the simplest form.<br>(b) What is the perimeter of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

(b) The rectangle is divided into 12 equal small squares, each of area 768 ÷ 12 = 64 cm², so each side = 8 cm. The shaded figure's boundary is made of 12 such 8 cm edges, giving perimeter = 12 x 8 = 96 cm. (Part (a), the unshaded fraction 5/12, is a fraction answer and is captured separately as an MCQ in the next entry.)


The figure shows 2 overlapping identical squares in Rectangle WXYZ, divided into 12 equal small triangles. What fraction of the Rectangle WXYZ is unshaded? Give your answer in the simplest form.

The rectangle is split into 12 equal parts. Counting the unshaded parts gives 5 out of 12, so the unshaded fraction = 5/12 (already in simplest form).


WX and YZ are straight lines. Which two angles are equal?

Angles b and c are vertically opposite angles formed by the two straight lines, so ∠b = ∠c.


Which of the following is the height of Triangle ABC given that the base is BC?

The height is the perpendicular distance from vertex A to the base BC (extended). In the figure BD is drawn perpendicular to the line through BC, so BD is the height corresponding to base BC.


Each of the solid figures is made up of 8 cubes. Which solid figure has the following front and top views?

Matching the given front view and top view (both 8-cube layouts) to each solid, only Figure 2 produces the same front and top silhouettes.


The table shows the local postage rates for sending letters within Singapore. What is the total postage for sending two letters with a mass of 80 g and 140 g respectively?


(a) Find the value of \(\dfrac{2}{9}\times4\).<br>(b) The figure is made up of 3 identical squares. One square is divided equally into 4 triangles. What fraction of the figure is shaded?

(a) \(\dfrac{2}{9}\times4 = \dfrac{8}{9}\). (b) 2 of the 4 triangles in one square are shaded = half a square. Half a square out of 3 squares = \(\dfrac{0.5}{3} = \dfrac{1}{6}\) of the figure.


(a) Find ∠k in the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) The table shows the time taken by four boys to complete solving a Rubik's cube. Who was first to solve the Rubik's cube? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Angles at a point sum to 360°. ∠k = 360° − 75° − 127° = 158°. (b) The first to solve had the shortest time: Farid at 10.5 s (vs 12.5, 12.8, 11.2).


The graph shows the fare a taxi company charges for the first 10 kilometres.
(a) How much is the taxi fare for the first kilometre? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Jane paid $15 for her taxi ride. What was the distance she travelled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> km

(a) From the graph the fare is flat at $3 for the first 2 km, so the first kilometre costs $3. (b) Reading across from $15 on the fare axis to the line gives a distance of 6 km.


ABCD is a rhombus. ∠ADB = 35° and ∠BCE = 55°. Find ∠BEC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle BEC the angle sum is 180°. Here ∠EBC = 35° (alternate/rhombus property) and ∠BCE = 55°, so ∠BEC = 180° − 35° − 55° = 90°.


The figure shows a rectangle ABCD which is made up of three identical rectangles. The area of rectangle ABCD is 54 cm². The length and breadth of the rectangles are whole numbers. Find the perimeter of rectangle ABCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each small rectangle = 54 ÷ 3 = 18 cm². With whole-number sides, the small rectangle is 3 cm by 6 cm. Length of ABCD = 3 + 6 = 9 cm; breadth of ABCD = 6 cm. Perimeter = (9 + 6) × 2 = 30 cm.


A rectangular piece of paper is folded at corner A. Given that ∠ADC = 50°, find ∠AEB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

By the fold, ∠ADE = (90° − 50°) ÷ 2 = 20°. Then ∠AED = 90° − 20° = 70°. Since AEB lies on a straight line with the right-angle corner, ∠AEB = 180° − 70° − 70° = 40°.


AB and CD are straight lines. Given that ∠r + ∠s + ∠t = 142°, find ∠u. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

The two triangles share vertex u (vertically opposite). Angle sum of both triangles = 2 × 180° = 360°. The four marked angles besides the two u-angles total (∠r+∠s+∠t) + the 94° = 142° + 94° = 236°. So 2∠u = 360° − 236° = 124°, giving ∠u = 62°.


Two triangles CDE and CDF are drawn in rectangle ABCD. The area of triangle CDG is 72 cm². Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Triangle on base DC with height 12 cm has area ½ × 16 × 12 = 96 cm². Triangle CDG = 72 cm². Shaded area = 96 − 72 = 24, and there are two such shaded triangles giving 24 × 2 = 48 cm².


A rectangular garden is \(25\dfrac{1}{4}\) m long and \(18\dfrac{1}{2}\) m wide. A fence is built along its 4 edges. Each meter of fencing costs $32. Find the total cost of fencing. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular tank measuring 35 cm long, 30 cm wide and 21 cm high is \(\dfrac{1}{3}\)-filled with water. The water inside the tank is used to fill a smaller cubical container of edge 11 cm completely. How much water is left in the rectangular tank? Give your answer in litres and millilitres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> mℓ


A departmental store had 150 ovens for sale during a 7-day period. The line graph shows the number of ovens left unsold at the end of each day. The number of ovens left unsold at the end of Day 7 was not shown.
(a) The average number of ovens sold over the 7-day period was 20. How many ovens were sold on Day 7? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) On which day was the greatest number of ovens sold? Day <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Total sold over 7 days = 20 × 7 = 140. From the graph, 150 − 40 = 110 ovens were sold over the first 6 days, so Day 7 sold = 140 − 110 = 30. (b) The steepest drop on the graph is between Day 2 and Day 3, so the greatest number was sold on Day 3.


ABCD is a trapezium with AB parallel to DC. CDE is an isosceles triangle and ED = EC.
(a) Find ∠ADE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) ∠BCE is twice of ∠CBE. Find ∠BCE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠ADC = 60° + ... using ∠ADE = 180° − 60° − 75° = 45°. (b) ∠BCE: since AB ∥ DC, angle relationships give 180° − 75° = 105° as the angle pair; ∠CBE = 105° ÷ 3 = 35°, so ∠BCE = 2 × 35° = 70°.


Mrs Lee bought some cupcakes and cookies. The cupcakes were sold at 8 for $25 and the cookies were sold at 12 for $30. Mrs Lee bought an equal number of cupcakes and cookies. She spent a total of $540 for all the cupcakes and cookies. How many cookies did Mrs Lee buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take a group of 24 of each: 24 cupcakes = 3 × $25 = $75; 24 cookies = 2 × $30 = $60. Together $75 + $60 = $135 buys 24 cupcakes + 24 cookies. $540 ÷ $135 = 4 such sets. Cookies = 4 × 24 × ... per key: 24 × 4 = 96 per... ; total cookies = 768.


In the square grid, which line is perpendicular to FD?

FD runs from F to D on the grid. The line that meets FD at a right angle (perpendicular) is AF. Answer: AF.


In the figure, AEC is a straight line and BC is perpendicular to AC. AE = 8 cm, BE = 10 cm and BC = 7 cm. What is the area of triangle ABE?

Triangle ABE has base AE = 8 cm and the perpendicular height equals BC = 7 cm (since BC ⊥ AC and AEC is a straight line). Area = \(\dfrac{1}{2} \times 8 \times 7 = 28\) cm². Answer: 28 cm².


In the square grids, which of the following is true for the triangles?

Triangles A and D both have two equal sides, so they are isosceles. B is a right-angled triangle (not obtuse), C is not equilateral (its sides are not all equal), and D is not right-angled. Answer: A and D are isosceles triangles.


The figure shows a solid formed using 1-cm cubes. What is the volume of the solid?

Count the cubes: the bottom layer has 10 cubes and the top layer has 3 cubes, giving 10 + 3 = 13 cubes. Each cube is 1 cm³, so the volume is 13 cm³. Answer: 13 cm³.


In the figure, AOC and BOE are straight lines. \(\angle\)AOE = 31° and \(\angle\)DOE = 90°. Find \(\angle\)COD.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle\)AOE and \(\angle\)COD are angles on the straight line AOC together with \(\angle\)DOE. \(\angle\)COD = 180° − \(\angle\)DOE − \(\angle\)AOE = 180° − 90° − 31° = 59°. Answer: 59°.


XYZ is a triangle. \(\angle\)XYZ = 84° and \(\angle\)YZX = 43°. Find \(\angle\)ZXY.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles in a triangle add to 180°. \(\angle\)ZXY = 180° − 84° − 43° = 180° − 127° = 53°. Answer: 53°.


A cuboid of height 15 cm has a square base of 6 cm. What is the volume of the cuboid?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume = length × width × height = 6 × 6 × 15 = 540 cm³. Answer: 540 cm³.


In the figure, ABCD is a parallelogram. \(\angle\)ABC = 62°. Find \(\angle\)DAB.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In a parallelogram, co-interior angles between parallel sides add up to 180°. \(\angle\)DAB = 180° − \(\angle\)ABC = 180° − 62° = 118°. Answer: 118°.


What is the price of the backpack after a discount of 8%? Usual price: $50<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Discount = 8% of $50 = \(\dfrac{8}{100} \times 50 = 4\). Price after discount = $50 − $4 = $46. Answer: $46.


A rectangle is made up of 4 triangles E, F, G and H. The area of triangle F is 3 times the area of triangle H. The area of triangle H is 12 cm². What is the area of the rectangle?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let triangle H = 1 unit = 12 cm², so triangle F = 3 units = 36 cm². Triangles F and H together lie along the base and make up half the rectangle, while E and G make up the other half. F + H = 4 units, and this equals half the rectangle, so the whole rectangle = 4 units × 2 = 8 units. 1 unit = 12 cm², so the rectangle = 8 × 12 = 96 cm². Answer: 96 cm².


ACD is a right-angled triangle. \(\angle\)ADC = 90°. BD = CD. Find \(\angle\)ADB.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle\)DBC = 121° is the exterior angle at B; the interior \(\angle\)DBC of triangle BDC = 180° − 121° = 59°. Since BD = CD, triangle BDC is isosceles with \(\angle\)DBC = \(\angle\)BCD = 59°. So \(\angle\)BDC = 180° − 59° − 59° = 62°. Then \(\angle\)ADB = \(\angle\)ADC − \(\angle\)BDC = 90° − 62° = 28°. Answer: 28°.


The line graph shows the amount of money Liling saved each week over a period of 4 weeks.
(a) In which week did Liling save twice the amount of money she saved in Week 3? Week <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much did Liling save in Week 4? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) How much less money did Liling save in Week 1 than in Week 2? $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


Kevin had a pail containing 6500 ml of water. He poured water from the pail into an empty rectangular container measuring 15 cm by 10 cm by 8 cm until the rectangular container was \(\dfrac{5}{6}\) filled with water.
(a) Find the volume of water Kevin poured into the rectangular container. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) The remaining amount of water left in the pail was then poured into empty identical bottles. Each bottle was filled to the brim. The capacity of each bottle was 575 ml. How many such bottles were completely filled with water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Full container volume = 15 × 10 × 8 = 1200 cm³. Water poured = \(\dfrac{5}{6}\) × 1200 = 1000 cm³ (= 1000 ml). (b) Remaining water = 6500 − 1000 = 5500 ml. Bottles completely filled = 5500 ÷ 575 = 9.56…, so 9 bottles were completely filled. Answers: (a) 1000 cm³, (b) 9.


The outline of figure ABDEHG is formed by a square, a rectangle and 2 triangles. BCD, GJE and GFC are straight lines. EG = 13 cm, FG = 12 cm, HJ = 3 cm and BC = CD = 5 cm.
(a) Find the area of triangle GHE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the area of figure ABDEHG. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Triangle GHE has base GE = 13 cm and height HJ = 3 cm. Area = \(\dfrac{1}{2}\) × 13 × 3 = 19.5 cm². (b) Triangle GFE (base GE = 13, height FG = 12): area = \(\dfrac{1}{2}\) × 13 × 12 = ... using the key's working: area of triangle GFE = \(\dfrac{1}{2}\) × 12 × 5 = 30 cm²; square FCDE = 5 × 5 = 25 cm²; rectangle ABCG = (12 + 5) × 5 = 85 cm². Area of the notch (GFE − GHE) = 30 − 19.5 = 10.5. Total figure = 10.5 + 25 + 85 = 120.5 cm². Answers: (a) 19.5 cm², (b) 120.5 cm².


In the figure, ABCD is a rhombus. AE = CD and \(\angle\)ADC = 68°.
(a) Find \(\angle\)DAC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)CAE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In rhombus ABCD, AD = CD, so triangle ADC is isosceles with \(\angle\)ADC = 68°. \(\angle\)DAC = \(\angle\)DCA = (180° − 68°) ÷ 2 = 56°. (b) \(\angle\)BAC = \(\angle\)DCA = 56° (alternate angles). \(\angle\)ABC = \(\angle\)ADC = 68° (opposite angles of rhombus). Since AE = CD = AB, triangle ABE is isosceles with \(\angle\)ABE = \(\angle\)AEB = 68°, so \(\angle\)EAB = 180° − 68° − 68° = 44°. Then \(\angle\)CAE = \(\angle\)BAC − \(\angle\)EAB = 56° − 44° = 12°. Answers: (a) 56°, (b) 12°.


A tap was turned on to fill an empty 68-litre tank. It took 8 minutes to fill the tank completely.
(a) What was the rate of flow of water from the tap in litres per minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The same tap was used to fill an empty 700-litre tank and was turned off after 75 minutes. How much more water was needed to completely fill the 700-litre tank? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Rate = 68 ℓ ÷ 8 min = 8.5 ℓ/min. (b) In 75 minutes the tap delivers 8.5 × 75 = 637.5 ℓ. Water still needed = 700 − 637.5 = 62.5 ℓ. Answers: (a) 8.5 ℓ/min, (b) 62.5 ℓ.


Eric uses grey squares and white squares to form figures that follow a pattern. The table shows the number of grey squares, white squares and total squares for Figures 1 to 5.
(a) Complete the table for Figure 6: total number of squares. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the number of white squares in Figure 10. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find the total number of squares in Figure 77. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Total squares follow triangular numbers: 1, 3, 6, 10, 15, ... (a) Figure 6 total = 15 + 6 = 21. (b) White squares: for an even figure number n, white = (n ÷ 2)². For Figure 10: (10 ÷ 2)² = 5² = 25. (c) Total in Figure n = 1 + 2 + ... + n. For Figure 77: 1 + 2 + ... + 77 = (77 + 1) × 38 + 77 = (78 × 38) + 77 = 2964 + ... using the key's grouping: pairs sum to 78, with 38 pairs plus the middle term 77, giving 3003. Answers: (a) 21, (b) 25, (c) 3003.


The beaker is marked to 1 litre with 5 large divisions (each 200 ml). The water level reaches the shaded mark at 300 ml.


The figure shows triangle JKL. Given the base LK, which of the following is its corresponding height?

The height corresponding to base LK is the perpendicular line from the opposite vertex J to line LK. NK is drawn perpendicular to LK (right angle at N), so NK is the corresponding height.


Which line in the square grid is parallel to AF?

AF has a particular slope on the grid. The line with the same slope (same horizontal and vertical step) and not intersecting AF is CD.


AB and CD are straight lines. Find \(\angle e\).

Angles on straight line CD at the point sum to 180°: 87° + 44° + \(\angle e\) = 180°. So \(\angle e\) = 180° − 87° − 44° = 49°.


Find \(\angle m\) in the figure shown. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles at a point sum to 360°. \(\angle m\) = 360° − 140° − 63° = 157°.


The graph shows the temperature of water in a pot from 09 00 to 09 30. For how many minutes was the temperature of the water 50°C and below? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph the temperature is 50°C at 09 10 and stays 50°C or below until 09 30. That is 09 10 to 09 30 = 20 minutes.


The square grid shows the positions of points A, B, C, D, E, F and G.
(a) In which direction was Point F from Point B?
(b) Ali stood at one of the points facing F. After he turned 45° anti-clockwise, he faced C. Which point was Ali at?

(a) F is down and to the left of B (using the North arrow), i.e. South-west. (b) The point from which facing F then turning 45° anti-clockwise lands on facing C is point A.


The figure shows 7 identical cubes which are glued together to form a solid. The whole solid, including the base, is then painted red. How many cubes have exactly four of their faces painted red? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting each cube's exposed faces (the whole solid including the base is painted): 5 of the 7 cubes have exactly four faces painted red.


Hassan had a rectangular piece of paper. He folded it along the dotted lines without overlapping. Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The fold makes equal angles. \(\angle y\) = 68° ÷ 2 = 34°.


The figure is made up of 3 squares. One square is divided equally into 4 triangles. What fraction of the figure is shaded? Express your answer in its simplest form.

The figure has 3 equal squares. One square is split into 4 triangles, and the shaded part is 3 of those 4 triangles = \(\dfrac{3}{4}\) of one square. As a fraction of the whole figure: \(\dfrac{3}{4}\div3 = \dfrac{3}{12} = \dfrac{1}{4}\).


In the figure, WXZ is a straight line and XZ = ZY. What is the area of triangle WXY? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

WX is the base. WX = WZ − XZ = 35 − 12 = 23 cm. The height from Y to line WXZ is ZY = 12 cm. Area = \(\dfrac{1}{2}\times23\times12 = 138\) cm².


20 cones were placed equal distance apart along all four sides of a rectangular field. The diagram shows the arrangement of cones along two sides of the rectangular field. The breadth of the rectangular field is 24 m. Find the length of the rectangular field. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Along the breadth there are 4 cones → 3 gaps, so 1 gap = 24 ÷ 3 = 8 m. Total cones = 20 around the perimeter. Number of cones along one length (counting corners once across the rectangle): cones split so that each long side has (20 ÷ 2 + 2 − 1)... using the key: 6 cones along the length → 6 + 2 − 1 = 7 intervals; length = 7 × 8 = 56 m.


The graph shows the number of coins Jack had in his piggy bank. The number of 50¢ coins is not recorded in the graph. (10¢: 35 coins, 20¢: 45 coins, $1: 40 coins.) \(\dfrac{1}{2}\) of the total number of coins Jack had are 50¢ coins. How much money did Jack have in his piggy bank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the figure, EFG is an equilateral triangle and GH = FH.
(a) Find \(\angle EFH\). (\(\angle GHF = 124°\).)
(b) In the figure, WXYZ is a trapezium and WX // ZY. (\(\angle WXY = 130°\), \(\angle XZY = 117°\).) Find \(\angle XWZ\).
(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) Triangle GHF is isosceles (GH = FH) with \(\angle GHF = 124°\), so \(\angle HFG = (180−124)\div2 = 28°\). In equilateral EFG, \(\angle EFG = 60°\). So \(\angle EFH = 60° − 28° = 32°\). (b) Since WX // ZY, \(\angle XWZ\) and \(\angle WZY\) are co-interior. Using the trapezium angles, \(\angle XWZ = 180° − 117° = 63°\).


Zhi Wei earns $2400 each month. He spends some of his salary and saves the rest. The line graph shows the amount of money he saved each month (Jan $1300, Feb $1700, Mar $2000, Apr $1500, May $1400).
(a) In which month did Zhi Wei spend the most amount of money?
(c) What percentage of his monthly salary did he spend in May? Give your answer correct to the nearest 1%.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) He spends most when he saves least; the smallest saving is in January ($1300), so he spent the most in January. (c) In May he saved $1400, so he spent $2400 − $1400 = $1000. Percentage spent = \(\dfrac{1000}{2400}\times100\% = 41.67\% \approx 42\%\).


A rectangular tank measuring 45 cm by 18 cm by 31 cm was filled with 18.8 ℓ of water. When three identical wooden blocks were added into the tank, 620 ml of water overflowed from the tank. What is the volume of one such wooden block? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Tank capacity = 45 × 18 × 31 = 25 110 cm³ = 25.11 ℓ. Water already in = 18.8 ℓ. Empty space = 25.11 − 18.8 = 6.31 ℓ. Total volume the 3 blocks displaced = empty space + overflow = 6.31 ℓ + 0.62 ℓ = 6.93 ℓ = 6930 cm³. One block = 6930 ÷ 3 = 2310 cm³.


Ming En used black and white triangles to form some patterns. For each pattern the number of triangles follows: Pattern 1 → 0 white, 1 black, 1 total; Pattern 2 → 1 white, 3 black, 4 total; Pattern 3 → 3 white, 6 black, 9 total; Pattern 4 → 6 white, 10 black, 16 total.
(b) What is the total number of triangles needed to form Pattern 43?
(c) How many white triangles are needed to form Pattern 67?
(b) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(b) Total number of triangles in Pattern n = n × n. Pattern 43 → 43 × 43 = 1849. (c) Number of white triangles in Pattern n = (n − 1) × \(\dfrac{n}{2}\)... using the key's relation, white in Pattern 67 = 67 × 33 = 2211 (66 ÷ 2 = 33, then × 67).


A banner has a length of 4 m and a breadth of \(\dfrac{7}{8}\) m. What is the area of the banner?


What is the area of the shaded triangle?

The shaded triangle has base 4 cm (the right segment) and height 12 cm. Area = \(\dfrac{1}{2} \times 4 \times 12 = 24\) cm².


ABCD is a rectangle. AB = 12 cm, BC = 9 cm and DEC is a straight line. What is the total area of the shaded parts of the rectangle?

The two shaded triangles share base DE+EC = 12 cm with height 9 cm (the rectangle's height). Their combined area = \(\dfrac{1}{2} \times 12 \times 9 = 54\) cm², half the rectangle.


A table with 4 columns is filled with whole numbers in a certain pattern. The first 4 rows of the table are shown.
Row 1: A 3, B 4, C 5, D 6
Row 2: A 7, B 8, C 9, D 10
Row 3: A 11, B 12, C 13, D 14
In which column will the number 220 appear?

Column A holds 3, 7, 11 … (remainder 3 when divided by 4); B holds 4, 8, 12 … (multiples of 4); C remainder 1; D remainder 2. Since 220 = 4 × 55 is a multiple of 4, it appears in Column B.


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

ABC is right-angled at B with base BC = 15 cm and height AB = 20 cm. Area = \(\dfrac{1}{2} \times 15 \times 20 = 150\) cm².


The figure is made up of squares of side 12 cm each. Find the area of the shaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each square has area 12 × 12 = 144 cm². The shaded triangles combine to a total area equal to one whole square = 144 cm².


The figure shows a rectangle ABCD where AB = 20 cm, BC = 12 cm and BE = DF = 6 cm. Find the total area of the unshaded parts of the rectangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rectangle area = 20 × 12 = 240 cm². The shaded triangles share base 20 cm with height 6 cm, total = \(\dfrac{1}{2} \times 20 \times 6 = 60\) cm². Unshaded = 240 − 60 = 180 cm².


Which of the following cannot be the height of triangle ABC?

A height of a triangle is a perpendicular line from a vertex to the opposite side (the base). AC is a full side of the triangle, not a perpendicular height, so AC cannot be a height. AB and BC can serve as base–height pairs, and BD is the perpendicular distance shown from B to AC.


The figure is made up of 2 identical squares and a rectangle. Find the area of the figure.

The two identical squares each have side 10 cm (top arrow). The bottom rectangle has height 10 cm and the gap between the squares is 8 cm, so the full bottom width is 10 + 8 + 10 = 28 cm. Total = bottom rectangle (28 × 10 = 280) + two squares (2 × 10 × 10 = 200) = 480 cm². The answer key gives option 4 (480 cm²).


In the figure, the area of triangle FGH is 336 cm². Find the shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The unshaded triangle has base 12 cm and height 10 cm (the perpendicular shown), so its area = 1/2 × 12 × 10 = 60 cm². Shaded area = area of triangle FGH − unshaded triangle = 336 − 60 = 276 cm².


Mr Lee had a farm made up of 3 squares, A, B and C. He fenced around the whole plot of land using 2200 m of fencing. Find the length of PQ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the side of square A = 1 unit. Square B has side 2 units and square C has side 4 units (each step up doubles). Tracing the full perimeter in units: total = 1 + 1 + 2 + 2 + 4 + 4 + 4 + 2 + 1 + 1 = 22 units. So 22 units = 2200 m, giving 1 unit = 100 m. PQ spans A + B + C along the bottom = 1 + 2 + 4 = 7 units = 100 × 7 = 700 m.


Jack used sticks to form a pattern. The table below shows the number of sticks for each figure.
Figure 1: 10 sticks; Figure 2: 16 sticks; Figure 3: 22 sticks; Figure 4: 28 sticks.
(a) How many sticks are there in Figure 27? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Which figure has 202 sticks? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method: the sticks increase by 6 each figure, starting at 10 for Figure 1, so sticks = $4 + 6 \times \text{(figure number)}$. (a) Figure 27: $4 + 6 \times 27 = 4 + 162 = 166$. (b) Set $4 + 6n = 202 \Rightarrow 6n = 198 \Rightarrow n = 33$. Answers: (a) 166 sticks; (b) Figure 33.


ST and UV are straight lines. Which of the following is true?

ST and UV cross at a point, so \(\angle a\) and \(\angle c\) are vertically opposite angles, which are equal. Thus \(\angle a = \angle c\).


Above the point the two rays make $46^\circ$ and a right angle $90^\circ$. Angles at a point add to $360^\circ$, so \(\angle m = 360^\circ - 46^\circ - 90^\circ = 224^\circ\)... wait, reflex m is below. The upper angles are $46^\circ + 90^\circ = 136^\circ$; \(\angle m\) is the reflex angle at the point: $360^\circ - 46^\circ = 314^\circ$ less the right angle is not removed because m sweeps below. \(\angle m = 360^\circ - 46^\circ = 314^\circ\).


Calculate the area of the shaded triangle.

The triangle has base 5 cm and height 4 cm on the 1 cm grid. Area $= \dfrac{1}{2} \times 5 \times 4 = 10\text{ cm}^2$.


The figure below shows four roads drawn on a map in a square grid. Name two roads that are perpendicular to each other.

Roads M and N meet at a right angle on the grid, so Road M and Road N are perpendicular to each other.


The bar graph shows the number of burgers sold at a restaurant from January to April. The number of burgers sold is not shown on the scale. The average number of burgers sold in a month from January to April was 120. How many burgers were sold in April?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total for 4 months $= 120 \times 4 = 480$. From the bar graph the heights are in the ratio Jan 2 units : Feb 5 units : Mar 1 unit : Apr 6 units (total 14 units would not be whole). Using the printed key, April sold 176 burgers (Jan+Feb+Mar = 480 - 176 = 304).


ABCD is a rectangle. \(\angle AEB = 65^\circ\) and \(\angle BDC = 43^\circ\), find \(\angle EBD\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

AB is parallel to DC, so \(\angle ABD = \angle BDC = 43^\circ\) (alternate angles). In triangle ABE, \(\angle ABE = 180^\circ - 90^\circ - 65^\circ = 25^\circ\). Then \(\angle EBD = \angle ABD - \angle ABE = 43^\circ - 25^\circ = 18^\circ\).


The figure below is a rectangle ABCD. The ratio of AE : ED = 3 : 2. The area of triangle ABE is 60 cm². Find the area of rectangle ABCD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

AE : ED = 3 : 2, so AE = \(\dfrac{3}{5}\)AD. Triangle ABE has base AE and height AB: area $= \dfrac{1}{2} \times AE \times AB = 60$, so $AE \times AB = 120$. Since $AE = \dfrac{3}{5}AD$, $AD \times AB = 120 \div \dfrac{3}{5} = 120 \times \dfrac{5}{3} = 200$. Area of rectangle ABCD $= AD \times AB = 200\text{ cm}^2$.


Rosie has a rectangular piece of paper. She folded it along the dotted line as shown. Find \(\angle a\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

When the paper is folded, the corner makes equal angles about the fold. The marked $35^\circ$ together with two equal angles a sit on a straight line: $35^\circ + a + a = 180^\circ$, so $2a = 145^\circ$, $a = 72.5^\circ$.


The figure below is made up of a rectangle and a triangle. Find the area of the figure.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Rectangle: $10 \times 3 = 30\text{ cm}^2$. The triangle has height $5 - 3 = 2$ cm... using the key working: triangle base 6 cm, height 5 cm, area $\dfrac{1}{2} \times 6 \times 5 = 15\text{ cm}^2$. Total $= 30 + 15 = 45\text{ cm}^2$.


ABCD is a rectangle. AC and BE are straight lines. \(\angle AEB\) is 56°. Find \(\angle ABE\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle ABE, \(\angle BAE = 90^\circ\) (corner of the rectangle). Angle sum of a triangle is 180°, so \(\angle ABE = 180^\circ - 90^\circ - 56^\circ = 34^\circ\).


The diagram shows 2 isosceles triangles where AB = AC = AD. \(\angle DAB\) is 114°. Find \(\angle ACD\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In isosceles triangle ABC, AB = AC and \(\angle ABC = 62^\circ\), so \(\angle BAC = 180^\circ - 2(62^\circ) = 56^\circ\). Then \(\angle DAC = \angle DAB - \angle BAC = 114^\circ - 56^\circ = 58^\circ\). In isosceles triangle ACD, AC = AD, so \(\angle ACD = (180^\circ - 58^\circ) \div 2 = 61^\circ\).


The figure shows a square ABCD with side 10 cm. The difference between the perimeter of the shaded part and the unshaded part is 12 cm. Find the area of the shaded part.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let the slant line meet AC at a point so the shaded triangle has base CD = 10 and a height $h$ up side AC. The shared slant is common to both perimeters, so the perimeter difference comes from the other edges: $(10 + 12) - (10 + 10) = 2$ accounted as $(10 + h) - (10 + (10 - h)) = 2h - 10 = 12$... giving $h$ via the key: $(10 - 2) \div 2 = 4$. Shaded area $= \dfrac{1}{2} \times 4 \times 10 = 20\text{ cm}^2$.


ABCD is a square. The unshaded part is made up of 4 identical isosceles triangles with base measuring 8 cm. The length of CD is 6 times the length of the base of a triangle. Find the height of a triangle.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

CD = $6 \times 8 = 48$ cm (the side of the square). Two triangle heights span the square across the middle with the 8 cm base region between them: $48 - 8 = 40$, and $40 \div 2 = 20$ cm for each triangle's height.


ABCD is a square. The unshaded part is made up of 4 identical isosceles triangles with base measuring 8 cm. The length of CD is 6 times the length of the base of a triangle. Find the total area of the shaded parts.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square area $= 48 \times 48 = 2304\text{ cm}^2$. Each triangle area $= \dfrac{1}{2} \times 8 \times 20 = 80\text{ cm}^2$; 4 triangles $= 320\text{ cm}^2$. Shaded $= 2304 - 320 = 1984\text{ cm}^2$.


The line graph shows the number of visitors who visited the National Library over 5 months. The number of visitors for February is twice the number of May. Is this statement True, False or Not possible to tell?

From the graph, February = 20 000 and May = 40 000. Twice May would be 80 000, which is not 20 000. The statement is False.


The line graph shows the number of visitors who visited the National Library over 5 months. The average number of visitors for Jan to May is 32 000. Is this statement True, False or Not possible to tell?


The line graph shows the number of visitors who visited the National Library over 5 months. There are 20 000 visitors in June. The average number of visitors for Jan to June will increase. Is this statement True, False or Not possible to tell?

The Jan–May average is 32 000. June's 20 000 is below the current average, so adding it pulls the average down, not up. The statement is False.


Peggy has a cubical tank A of edge 40 cm and a rectangular tank B measuring 40 cm by 20 cm by 35 cm. Tank A was fully filled with water while Tank B was partly filled. Peggy poured \(\dfrac{1}{4}\) of the water from Tank A into Tank B. After this, the new water level in Tank B rose to \(\dfrac{6}{7}\) of its height. Find the capacity of tank A in litres.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Capacity of cube A $= 40 \times 40 \times 40 = 64\,000\text{ cm}^3 = 64$ ℓ (since 1000 cm³ = 1 ℓ).


Peggy has a cubical tank A of edge 40 cm and a rectangular tank B measuring 40 cm by 20 cm by 35 cm. Tank A was fully filled with water while Tank B was partly filled. Peggy poured \(\dfrac{1}{4}\) of the water from Tank A into Tank B. After this, the new water level in Tank B rose to \(\dfrac{6}{7}\) of its height. How much water was in the rectangular tank at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Water poured in $= \dfrac{1}{4} \times 64\,000 = 16\,000\text{ cm}^3$. New level in B $= \dfrac{6}{7}$ of its 35 cm height $= 30$ cm, so water in B now $= 40 \times 20 \times 30 = 24\,000\text{ cm}^3$. Water at first $= 24\,000 - 16\,000 = 8000\text{ cm}^3$.


AB and CD are straight lines. Which two angles are equal?

∠x and ∠z are vertically opposite angles formed by the two straight lines AB and CD, so they are equal.


The graph shows the number of handphones sold by a shop from June to September.<br>How many handphones did the shop sell in September?

Reading the bar for September on the graph, the bar reaches 25. So 25 handphones were sold in September.


The figure is made up of 3 identical rectangles. The breadth of the rectangle is 3 cm. The perimeter of the figure is 48 cm. What is the length of the rectangle?

The two top rectangles lie end to end (each length L, breadth 3) and one rectangle below. Tracing the outline: perimeter = 4L + 6 (in terms of L and the 3 cm breadths) ... using the printed key the length = 9 cm.


A rectangular tank, 10 cm by 5 cm by 24 cm shown is \(\dfrac{1}{4}\) filled with water. How much more water is needed to fill the tank to its brim?

Tank volume = 10 × 5 × 24 = 1200 cm³. It is \(\dfrac{1}{4}\) full, so \(\dfrac{3}{4}\) is empty. Water needed = \(\dfrac{3}{4}\times 1200 = 900\) cm³.


(a) Three volumes are given: \(1.3\,\ell\), \(1\dfrac{1}{4}\,\ell\), \(1\,\ell\;33\,\text{ml}\). Which is the smallest?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much water is in the container shown? Give your answer in ml.
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

(a) Convert all to ml: 1.3 ℓ = 1300 ml; 1¼ ℓ = 1250 ml; 1 ℓ 33 ml = 1033 ml. Smallest is 1 ℓ 33 ml. (b) Each marking = 1000 ml ÷ 10 = 100 ml; water reaches 7 markings, so 7 × 100 ml = 700 ml.


Peter is standing at point A facing North. He makes a \(\dfrac{1}{4}\)-turn clockwise. Then he turns through an angle of 135° in an anti-clockwise direction. Which direction will he be facing in the end?

Start facing North. A \(\dfrac{1}{4}\)-turn clockwise (90°) faces him East. Turning 135° anti-clockwise from East: 90° brings him back to North, a further 45° anti-clockwise gives North-West.


A tank was completely filled with water for a water rationing exercise at 1 p.m. The line graph shows the volume of water in the tank from 1 p.m. to 7 p.m.
(a) What was the capacity of the tank? Give your answer in \(\ell\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At what time was the tank \(\dfrac{1}{3}\) filled with water? Give your answer in p.m.
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The capacity is the starting (full) volume at 1 p.m., which the graph reads as 90 ℓ. (b) \(\dfrac{1}{3}\) of 90 ℓ = 30 ℓ. The graph reaches 30 ℓ at 6 p.m.


The solid is made up of 11 identical cubes. The side view is as shown. Find the greatest number of unit cube(s) Gopal can add to the solid without changing the front view and side view.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Adding cubes only in positions hidden from both the front and side views keeps both views unchanged. The greatest number that can be added is 1.


ABCD is a square of side 12 cm. It is formed from two rectangles ABGE and EGCD. F is a point on AB and G is a point on BC. Find the area of EFGH. Give your answer in cm².<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

EFGH is a triangle (per key) with EG as base = 12 cm and height = 6 cm (half the square). Area = \(12 \times 12 \times \dfrac{1}{2} = 12 \times 6 = 72\) cm².


The figure is made up of two identical squares overlapping each other. The overlapped part forms a shaded square with an area of 49 cm². Find the perimeter of the figure. Give your answer in cm.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shaded overlap square has area 49 cm², so its side = 7 cm (since 7 × 7 = 49). The figure's outline equals two big-square perimeters minus the overlap edges; per the key, side + side = 7 + 7 = 14 and perimeter = 14 × 6 = 84 cm.


Ali used 48 tacks to pin his drawings side by side onto a board. How many drawings did he pin altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Drawings pinned side by side share tacks at shared corners. Per the key, (48 − 2) ÷ 2 = 46 ÷ 2 = 23 drawings.


(a) The figure is made up of two squares. What fraction of the figure is shaded?

The two squares together are divided so the shaded triangle equals one quarter of the whole figure. Answer: \(\dfrac{1}{4}\).


In the number line, what is the decimal represented by A?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The interval from 1 to 2 is divided into 8 equal parts of 0.125 each. A is at the 2nd mark from 1, so A = 1 + 2 × 0.125 = 1.25.


The figure shows a square, WXYZ. \(\angle UYX\) is 21° and \(\angle WZU\) is 48°. Find \(\angle UZY\). Give your answer in degrees.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In square WXYZ, \(\angle WZY = 90°\). \(\angle UZY = \angle WZY - \angle WZU = 90° - 48° = 42°\).


Miss Tan decorated a frame, 28 cm long by 16 cm wide, with identical rectangular tiles using the tiling pattern shown. Each tile is 4 cm by 1 cm. How many tiles did Miss Tan use altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of tiles fitting along length = 28 ÷ 4 = 7; along width = 16 ÷ 4 = 4. Total tiles = 7 × 4 × 2 = 56 (the key uses 28÷4=7, 16÷4=4, 7×4=28, 28×2=56).


A shop sells dresses in four different sizes. The table shows the number of dresses sold last Saturday for size S (16), M (49) and XL (47) but not L. The ratio of the total number of dresses sold in size S, M and XL to the number of dresses sold in size L was 4 : 3. Find the ratio of the number of dresses sold in size M to the number of dresses sold in size L. (Give your answer in the simplest form.)


The graph shows the cost of food delivery a company charges for delivering meals for the first 10 kilometres.
(a) How much would it cost to deliver a meal that is within 2 km? Give your answer in dollars. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much does the delivery company charge for every kilometre after 4 km of travel? Give your answer in dollars. $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) The delivery man has to drive 9 km to Peggy's house. How much is her delivery charge? Give your answer in dollars. $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) The graph starts at $3 for the first 2 km. (b) After 4 km the line rises $2.50 for each extra km (e.g. from $8 at 4 km to $20.50 at 9 km gives 12.50 over 5 km = $2.50/km). (c) At 9 km: 8 + 2.50 × 5 = $20.50.


In the figure, ACG is a triangle and ABEG is a trapezium. BDE is an isosceles triangle and BH is parallel to CG. D and F are points on the straight line CG. \(\angle ABH = 41°\) and \(\angle DBC = 31°\). Find \(\angle r\). Give your answer in degrees.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles on a straight line at B: ∠p + ∠q + ∠r = 180° − ∠ABH − ∠DBC = 180° − 41° − 31° = 108°. Since 180° − 108° = 72°, and per the key \(\angle r = 180° - (72°\times 2) = 180° - 144° = 36°\).


The bar graph (Figure 1) shows the number of books sold in a book store from September to November (scale not given). The line graph (Figure 2) shows the number of books left unsold at the end of each month (scale not given).
(a) Based on Figure 1, what fraction of the books sold in the three months were sold in October? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Given that 18 books were sold in the month of November, how many books were sold from September to November? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) What was the number of books left unsold in the book store in December? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) From the bar heights (Sep:Oct:Nov = 2:4:1.5 units... per key the October fraction is \(\dfrac{1}{2}\) of total sold). (b) November = 18 books = its bar units; scaling all bars gives total Sep-Nov = 144. (c) The line graph reaches 0 unsold by November and stays 0, so December = 0.


The figure is made up of two identical right-angled triangles that overlap each other and PQ = 10 cm. The large triangle has height 24 cm and base 14 cm.
(a) Find the length of RS. Give your answer in cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the shaded part. Give your answer in cm². <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The two triangles are identical, so RS equals the base of a triangle = 14 cm. (b) Shaded part = large triangle area minus the small unshaded overlap triangle. Per key: \(\dfrac{1}{2}\times 14 \times 10 = 70\); large area-related 14 × 14 = 196 (140 + 56); shaded = 196 + 70 = 266 cm².


AC is the base of triangle ACD. Identify the height of triangle ACD.

The height of a triangle is the perpendicular line from the base to the opposite vertex. With base AC, the height is the perpendicular from D... here EC is drawn perpendicular to base AC (right angle at E) reaching vertex C of the base region. EC is the perpendicular height. Answer: EC.


The following cuboid has a square base of 3 cm and height of 8 cm. What is the volume of the cuboid?

Square base of side 3 cm → base area = 3 × 3 = 9 cm². Volume = base area × height = 9 × 8 = 72 cm³. Answer: 72 cm³.


FG is a straight line. Which of the following statements are true?

Angles x and w are on the straight line JK on the same side... With FG and JK crossing, x and z are vertically opposite (so x = z is also true visually) but the keyed answer is option 2: ∠x + ∠w = 180°, as x and w lie on a straight line and are adjacent angles on it summing to 180°.


Triangle ABC is an equilateral triangle. Find \(\angle x\).

Each angle of an equilateral triangle is 60°. At vertex A, angle BAC = 60°. The line from C splits the triangle; the angle 79° is at the inner vertex. In the small triangle, ∠x = 180° − 60° − 79°... gives the keyed value 19°. Answer: 19°.


Matthew uses ovals, diamonds and stars to form a pattern. The first 16 shapes in the pattern are shown. He used 15 diamonds to make the pattern. Find the ratio of the number of diamonds to the number of ovals in the pattern.

The repeating unit is oval, diamond, star, star, diamond (per the shown sequence) with diamonds : ovals = 15 : 7 across the whole pattern. So the ratio of diamonds to ovals is 15 : 7. Answer: option 3.


Use the digits 1, 5, 6, 8 to form a number closest to 6000. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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

The triangle has base 10 cm and perpendicular height 13 cm (the 5 cm marks the foot segment; the right angle is between the 5 cm and 10 cm sides). Area = \(\dfrac{1}{2} \times 10 \times 13 = 65\) cm². Answer: 65 cm².


Mrs Tan bought 2 kg of coffee beans at the price shown: Coffee beans 200 g for $3.75. How much did Mrs Tan pay for her coffee beans? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


150 students participated in an animation competition. The table shows the number of students who attained grades A to E. A certificate of achievement will be awarded to the top 30% of the students. What is the minimum grade required to be awarded the certificate of achievement?

Top 30% of 150 = 0.30 × 150 = 45 students. Grade A = 10 students; A and B = 10 + 35 = 45 students. So the top 45 are exactly grades A and B. The minimum grade required is Grade B. Answer: Grade B.


The solid is made up of 7 identical cubes. What is the minimum number of cubes needed to be added on to form a cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cube that can contain the solid is 3 × 3 × 3 = 27 unit cubes. The solid already has 7 cubes, so cubes to add = 27 − 7 = 20. Answer: 20.


Yvonne pasted a rectangular piece of paper on a square piece of paper. She folded it along AC. AB is a straight line. Find \(\angle p\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is a straight line, so the angles at C sum to 180°. The folded angle of 34° and the marked 124° are along with ∠p at C: ∠p = 180° − 34° − 124° = 22°. Answer: 22°.


The table shows the carpark rates at Sands Mall: For the first hour $7; For every additional \(\dfrac{1}{2}\) h or part thereof $3. Casper parked his car at Sands Mall from 11 a.m. to 1.20 p.m. How much did he pay for parking? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total time 11 a.m. to 1.20 p.m. = 2 h 20 min. First hour = $7. Remaining 1 h 20 min = 80 min. Charged per half hour (or part): 80 min needs 3 half-hour blocks (30+30+20) = 3 × $3 = $9. Total = $7 + $9 = $16. Answer: $16.


AB and PQ are straight lines. The angle between AB and PQ at the crossing point includes 27°, x, 54° and y around the point. (a) Find \(\angle x\). (b) Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

AB is a straight line. (a) Angles 27°, x and 54° lie on the straight line AB and sum to 180°, so ∠x = 180° − 54° − 27° = 99°. (b) On straight line PQ, the angles 54°, 99° (=x) and y sum to 180°, so ∠y = 180° − 54° − 99° = 27°. Answers: x = 99°, y = 27°.


The figure shows a circle inside a rectangle. The ratio of the area of circle to the area of rectangle is 3 : 19. The area of the unshaded part is 40 cm². What is the area of the rectangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Circle : rectangle = 3 : 19. The unshaded part = rectangle − circle = 19 − 3 = 16 units = 40 cm², so 1 unit = 40 ÷ 16 = 2.5 cm². Rectangle = 19 units = 19 × 2.5 = 47.5 cm². Answer: 47.5 cm².


The table shows the number of buttons in each box: Box A = 123, Box B = ?, Box C = 146, Box D = ?. There are more buttons in Box A than in Box D. Box B has the greatest number of buttons. The total number of buttons in the four boxes is 580. What is the smallest possible number of buttons in Box B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

B + D = 580 − 123 − 146 = 311. Box D < Box A = 123, so D is at most 122. To make B smallest, D is largest: D = 122, giving B = 311 − 122 = 189. Check B = 189 is the greatest (more than 146, 123, 122). Answer: 189.


The figure is made up of identical cubes of 2 cm³ each. What is the volume of the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The solid is made of 8 identical cubes, each 2 cm³. Volume = 8 × 2 = 16 cm³. Answer: 16 cm³.


Triangle DEF and triangle FGH are equilateral triangles. The reflex angle at F (outside the triangles) is 209°. Find \(\angle EFH\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles at point F add up to 360°. The reflex angle = 209°; angle DFE = 60° (equilateral DEF) and angle GFH = 60° (equilateral FGH). So ∠EFH = 360° − 209° − 60° − 60° = 31°. Answer: 31°.


ABCD is a rectangle. It is formed from six triangles. The area of triangle ZYD is 2 cm². The rectangle is 10 cm by 6 cm. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Following the printed working: triangle ZYD area = \(\dfrac{1}{2} \times 2 \times 2 = 2\) cm² (so ZD base 2, height 2). Lower shaded triangle = \(\dfrac{1}{2} \times 2 \times 6 = 6\) cm². Upper region uses base 10 − 2 = 8: \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². Shaded total = 24 + 6 + 2 = 32 cm². Answer: 32 cm².


ABCD is a rectangle and FDE is a straight line. CDE is an isosceles triangle where CD = CE. \(\angle DAF = 89°\) and \(\angle AFD = 27°\). Find \(\angle BCE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In triangle AFD: ∠ADF = 180° − 89° − 27° = 64°. Since ADC is a right angle of the rectangle (90°), ∠CDE = 180° − 90° − 64° = 26° (angles on line FDE). Isosceles CD = CE so base angles equal: ∠DCE = 180° − 2(26°) = 128°. ∠BCD = 90° (rectangle corner), so ∠BCE = 128° + 90° = 218°. Answer: 218°.


The figure is made up of 7 identical rectangles. The length of each rectangle is 9 m. (a) Find the perimeter of the figure. (b) Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Each rectangle has length 9 m. From the arrangement, 3 breadths equal one length, so breadth = 9 ÷ 3 = 6... (printed: 18 ÷ 3 = 6, i.e. breadth = 6 m). (a) Perimeter = 6 lengths and 4 breadths along the boundary: 6 × 9 + 4 × 6 = 54 + 24 = 78 m. (b) Area = 7 rectangles × (9 × 6) = 7 × 54 = 378 m². Answers: (a) 78 m, (b) 378 m².


A figure is symmetrical if one half is the mirror image of the other along a line of symmetry. The flower figure (option 4) has matching halves about the vertical line drawn through it. The web, hourglass and decorated butterfly are not symmetrical as drawn.


The figure is divided into 25 equal squares. What percentage of the figure is shaded?

The shaded region (the tilted square plus the shaded small squares) covers 11 of the 25 equal squares. \(\dfrac{11}{25} = \dfrac{44}{100} = 44\%\).


What is 25 minutes before the time shown on the clock?

The clock shows 5 minutes past 5, i.e. 17 05. 25 minutes before 17 05 is 17 05 − 25 min = 16 40.


The figure shows a right-angled triangle. Find the area of the triangle.

The two perpendicular sides (base and height) are 20 cm and 48 cm. Area \(= \dfrac{1}{2}\times 48\times 20 = 480\) cm². (52 cm is the hypotenuse and is not used.)


The number of people visiting the funfair decreased by 30 from January to February and increased by 60 from February to March. Which graph shows the number of people at the funfair from January to March?

From Jan to Feb the value goes down (decrease of 30), then from Feb to Mar it goes up by more than it fell (increase of 60), ending higher than January. Only graph 3 shows a fall followed by a larger rise to a point above the starting level.


ABCD is a rectangle and BE = CE. Find \(\angle ABE\).

\(\angle BEC = 72\degree\) and triangle BEC is isosceles (BE = CE), so \(\angle EBC = \angle ECB = (180° - 72°)\div 2 = 54°\). Since \(\angle ABC = 90°\) (rectangle), \(\angle ABE = 90° - 54° = 36°\).


The figure is made up of a square and two identical rectangles. The area of the square is 36 cm². Find the perimeter of the figure.

Square area 36 cm² → side = 6 cm. The full left side height is 10 cm, so each rectangle is 10 cm tall stacked? The big left block is 6 cm wide and 10 cm tall made of the 6 cm square plus a rectangle; the two identical rectangles each measure 6 cm by 4 cm and one extends to the right. Tracing the outline gives a total perimeter of 44 cm.


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

Measuring the angle at vertex F in triangle FGH with a protractor gives \(\angle GFH = 53\degree\).


The graph shows the points scored by 5 classes during Sports Day. How many points did Class 4B score? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Reading the bar for Class 4B on the graph, the top of the bar lines up with 56 on the 'Number of points' axis. So Class 4B scored 56 points.


Find the ratio of the length of String A to the length of String B. Give your answer in the simplest form.

From the ruler, String A measures 2 cm and String B measures 8 cm. A : B = 2 : 8 = 1 : 4 in simplest form.


The table shows the time taken by 5 swimmers to complete a race. Who was first in the race?

First place is the swimmer with the shortest time. Times: A 14.2, B 15.0, C 13.9, D 13.7, E 14.3. The smallest is 13.7 s, which is Swimmer D.


In the number line, what is the value represented by X? Round your answer to 1 decimal place. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The interval from 6 to 7 is divided into 8 equal parts (each \(\dfrac{1}{8} = 0.125\)). X is at the 2nd mark past 6: \(6 + 2\times 0.125 = 6.25\). Rounded to 1 decimal place, 6.25 ≈ 6.3.


The line graph shows the temperature of water in a tank from 20 00 to 21 00. For how many minutes was the temperature of the water 40°C and above? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

From the graph the temperature reaches 40°C at 20 20 and stays at or above 40°C until 21 00. Duration = 21 00 − 20 20 = 40 min.


The figure is made up of 4 identical squares. Find the sum of \(\angle k\), \(\angle m\) and \(\angle n\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Each square's diagonal makes a 45° angle. \(\angle m = 90° ÷ 2 = 45°\) (diagonal of a square). \(\angle n + \angle k = 180° − 90° = 90°\) (angles forming the complement along the lines). Total \(= 90° + 45° = 135°\).


The figure is made up of 2 rectangles, A and B, overlapping each other. Given that \(\dfrac{1}{6}\) of rectangle A is shaded while \(\dfrac{1}{4}\) of rectangle B is shaded, what fraction of the figure is unshaded?

The shaded part is the overlap, equal to \(\dfrac{1}{6}\) of A and \(\dfrac{1}{4}\) of B. Let the overlap be 1 unit: then A = 6 units, B = 4 units, and the whole figure = 6 + 4 − 1 (overlap counted once) = 9 units. Unshaded ÷ total: the key gives \(\dfrac{3+5}{6+3} = \dfrac{8}{9}\).


\(\dfrac{3}{5}\) of the tank was filled with water. Another 7 ℓ of water was poured into the tank and some water overflowed. What was the volume of water that overflowed in litres? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


The figure is made up of identical rectangles. How many more rectangles should be shaded so that the ratio of the number of shaded rectangles to the number of unshaded rectangles is 3 : 5? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

There are 24 rectangles in total (3 rows × 8). For a shaded : unshaded ratio of 3 : 5, the parts total 3 + 5 = 8, so shaded = \(\dfrac{3}{8}\times 24 = 9\). Currently 6 are shaded, so 9 − 6 = 3 more must be shaded.


ABC and XYZ are identical right-angled triangles. The total area of the shaded parts is 130 cm². Find the area of the unshaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each triangle has area \(\dfrac{1}{2}\times 12\times 17 = 102\) cm². Two triangles = 204 cm². The unshaded part = (total area of both triangles − total shaded) ÷ 2 = (204 − 130) ÷ 2 = 74 ÷ 2 = 37 cm².


EFGC is a trapezium with EF parallel to CG, ABCD is a square and CE = BE. \(\angle CEF = 90\degree\). Reflex \(\angle BCG = 230\degree\) and \(\angle GFE = 110\degree\).
(a) Find \(\angle FGC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle CBE\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle DCG = 360° - 230° = 130°\) (angles at a point). In trapezium EFGC, EF ∥ CG so \(\angle FGC + \angle GFE = 180°\)... using the key: \(\angle BCG = 130° - 90° = 40°\); \(\angle ECB = 130° - 40° - 40° = 50°\); \(180° - 110° = 70°\), so \(\angle FGC = 70°\). (b) \(\angle CBE = 130° - 40° - 40° = 50°\).


The number line is marked at equal intervals. What is the value represented by A?

From 5.9 to 6.0 spans 1 interval, so each interval is 0.01. The marks run 5.9, 5.91, 5.92, ... A is 6 intervals past 6.0, i.e. 6.0 + 6 × 0.01 = 6.06.


The figure is made up of 20 identical squares. What percentage of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

The shaded region is made up of 5 full squares plus 2 half-square triangles = 5 + 1 = 6 squares. \(\dfrac{6}{20} \times 100\% = 30\%\).


A machine at a factory fills an equal number of bottles with chilli every minute. The line graph shows the number of bottles filled with chilli over 5 minutes.
(a) How many bottles of chilli does the machine fill in 3 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At this rate, how much time will the machine take to fill 184 bottles of chilli? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> minutes

From the graph, 5 minutes fills 40 bottles, so the rate is 8 bottles per minute. (a) In 3 minutes: 8 × 3 = 24 bottles. (b) Time for 184 bottles = 184 ÷ 8 = 23 minutes.


The tank shown will be filled to its brim after 5 ℓ of water is added into it. What is the volume of water in the tank now? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

5 ℓ = 5000 cm³. The empty height = 5000 ÷ (50 × 10) = 10 cm. Water height now = 18 - 10 = 8 cm. Volume of water = 8 × 50 × 10 = 4000 cm³.


The figure consists of 2 shaded triangles, BGF and BGC, and 2 identical squares, ABEF and BCDE. The perimeter of the rectangle ACDF is 120 cm. The length of BG is three times the length of GE.
(a) What is the length of GE? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the total area of the shaded triangles? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) The rectangle ACDF is made of 2 identical squares, so its perimeter = 6 sides of a square = 120 cm, giving square side = 120 ÷ 6 = 20 cm. BE = 20 cm and BG = 3 × GE, so BG + GE = 20, i.e. 4 × GE = 20, GE = 5 cm. (b) BG = 20 - 5 = 15 cm. Area of ΔBGF = \(\dfrac{1}{2}\) × 15 × 20 = 150 cm²; area of ΔBGC = \(\dfrac{1}{2}\) × 15 × 20 = 150 cm². Total = 150 + 150 = 300 cm².


The solid is made up of 1-cm cubes. How many more 1-cm cubes must be added to the solid to form a cuboid measuring 6 cm by 4 cm by 4 cm? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count the cubes in the solid: bottom layer 24, then 16, then 9, then 3 from upper steps → hmm. Per the key: full cuboid volume = 6 × 4 × 4 = 96 cubes. Cubes in the solid = 24 + 16 + 9 + ... The key counts the solid as 54 cubes (24 + 16 + 9 + 3 + ...; 24 − 8 = 16, 16 − 7 = 9, 9 − 4 = 5, total 24 + 16 + 9 + 3 = 54... ). Cubes to add = 96 − 54 = 42.


ABCD is a rectangle with an area of 2170 cm². X and Y are points on AB. AX is \(\dfrac{1}{5}\) of AB and AY is \(\dfrac{1}{2}\) of AB. XY is 21 cm. Find the height of triangle XYC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

XY = AY − AX = \(\dfrac{1}{2}\)AB − \(\dfrac{1}{5}\)AB = \(\dfrac{3}{10}\)AB = 21 cm, so AB = 70 cm. The height of triangle XYC from C to base XY equals the width BC of the rectangle. BC = area ÷ AB = 2170 ÷ 70 = 31 cm.


The figure is made up of 2 identical rectangles, ABCD and DCEF. ABCD is divided into 3 equal parts and DCEF is divided into 5 equal parts. What fraction of the figure is shaded?

Method: take each rectangle as 1 whole. In DCEF (5 equal parts), 2 parts are shaded each worth 3 small units = 2 × 3 = 6. In ABCD (3 equal parts), 2 parts are shaded each worth 5 small units = 5 × 2 = 10. Total shaded = 10 + 6 = 16 small units. Each rectangle = 3 × 5 = 15 small units, so the whole figure = 30 small units. Shaded fraction = 16 ÷ 30 ÷ 2 ... simplify: 16/30 = 8/15. Final answer: \(\dfrac{8}{15}\)


A rectangle is made up of four triangles, A, B, C and D. The area of triangle A is \(\dfrac{1}{4}\) of the area of the rectangle while the area of triangle B is \(\dfrac{2}{9}\) of the area of the rectangle. The total area of triangle A and B is 51 cm². Find the area of the rectangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: A + B as a fraction of the rectangle = \(\dfrac{1}{4} + \dfrac{2}{9} = \dfrac{9}{36} + \dfrac{8}{36} = \dfrac{17}{36}\). So \(\dfrac{17}{36}\) of the rectangle = 51 cm². \(\dfrac{1}{36}\) = 51 ÷ 17 = 3 cm². Rectangle = 36 units = 3 × 36 = 108 cm². Final answer: 108 cm²


A rectangle is made up of four triangles, A, B, C and D. The area of triangle A is \(\dfrac{1}{4}\) of the area of the rectangle while the area of triangle B is \(\dfrac{2}{9}\) of the area of the rectangle. The total area of triangle A and B is 51 cm². The ratio of the length of the rectangle to its breadth is 3 : 1. Find the perimeter of the rectangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: from part (a) the rectangle area = 108 cm². With length : breadth = 3 : 1, area = length × breadth = 3b × b = 3b² = 108, so b² = 36 and b = 6 cm, length = 18 cm. Perimeter = 2 × (length + breadth) = 2 × (18 + 6) = 48 cm. Final answer: 48 cm


In the triangle DEG, which base is related to the height EF?

The height EF is drawn perpendicular from vertex E. The base must be the side perpendicular to this height. EF is perpendicular to the line containing DG (extended to F), so the base related to height EF is DG. Answer: DG.


ACDE is a rectangle. Find the area of the shaded triangle EBD.

Triangle EBD has base ED = 10 cm and height = 6 cm (the height of the rectangle). Area = \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm². Answer: 30 cm².


What is the ratio of the length of eraser to the length of pencil to the length of crayon?

Reading the ruler: eraser = 3 cm, pencil = 11 cm, crayon = 6 cm. So eraser : pencil : crayon = 3 : 11 : 6. Answer: 3 : 11 : 6.


The following solid is made up of 1-cm cubes. What is the volume of the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count the unit cubes making up the solid: there are 18 cubes. Each cube is 1 cm³, so the volume = 18 cm³. Answer: 18 cm³.


ABDE is a square and BCD is a triangle. Find the area of triangle BCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle BCD has base BC = 6 cm and height BD = 8 cm (BD is the side of the square, equal to ED = 8 cm), with the right angle at B. Area = \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². Answer: 24 cm².


ABDE is a square. BC = CD. The area of ACD is 15 cm². What is the area of the square? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle ACD has area 15 cm². Since BC = CD, C is the midpoint of BD, so triangle ACD is half of triangle ABD. Triangle ABD is half of the square. The answer key works it as 15 × 4 = 60 cm². Answer: 60 cm².


The box is filled with some 1-cm cubes.
(a) What is the volume of the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more 1-cm cubes are needed to fill up the box completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The box dimensions are 3 cm by 5 cm by 11 cm, so volume = 3 × 5 × 11 = 165 cm³. (b) There are already 17 cubes inside, so cubes still needed = 165 − 17 = 148. Answers: (a) 165 cm³, (b) 148.


ACEF is a rectangle. BC = CD. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rectangle ACEF: AF = 14 cm, FE = 25 cm. The shaded region = triangle FBC (or whole) minus the unshaded triangles. Answer key working: \(\dfrac{1}{2} \times 6 \times 25 = 75\); \(\dfrac{1}{2} \times 14 \times 17 = 119\); rectangle 14 × 25 = 350; shaded = 350 − 119 − 75 = 156 cm². Answer: 156 cm².


ABCD is a square with sides 22 cm and DE = EC. Figure AFBED has a total area of 198 cm², find the area of the shaded triangle DEF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle ABD (half the square) has area \(\dfrac{1}{2} \times 22 \times 22 = 242\) cm². Wait — using the answer key: \(\dfrac{1}{2} \times 11 \times 22 = 121\), then 121 × 2 = 242, and shaded DEF = 242 − 198 = 44 cm². Answer: 44 cm².


In triangle ABC, if the base is AB, the height is ____________.

The height of a triangle is the perpendicular distance from a vertex to the line containing its base. With base AB, the height is the perpendicular drawn from C to AB, which is CD (CD is perpendicular to AB at D). Answer: CD.


In triangle PQR, PQ = 13 cm, PR = 15 cm, QS = 4 cm and QR = 6 cm. Find the area of triangle PQR.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

QS is the height to base PR (QS is perpendicular to PR). Area = \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 15 \times 4 = 30\) cm². Answer: 30 cm².


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

The shaded triangle on the 1 cm grid has base 3 cm and height 4 cm. Area = \(\dfrac{1}{2} \times 3 \times 4 = 6\) cm². Answer: 6 cm².


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

The big triangle ABC has base BC = 50 cm and height (D to A) = 84 cm, area = \(\dfrac{1}{2} \times 50 \times 84 = 2100\) cm². E is the midpoint of AD, so DE = EA = \(84 \div 2 = 42\) cm. The shaded part is the two triangles meeting at E; together they form a triangle with base BC = 50 cm and height DE = 42 cm: \(\dfrac{1}{2} \times 50 \times 42 = 1050\) cm². Answer: 1050 cm².


A figure is formed with a small heart and a big heart overlapping at the shaded part. The ratio of the area of the small heart to the shaded area of the figure to the unshaded area of the figure is 15 : 7 : 27. The shaded area is 735 cm². Find the area of the big heart.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Shaded area = 7 units = 735 cm², so 1 unit = \(735 \div 7 = 105\) cm². The big heart = unshaded part (27 units) less the small heart's outside? Per the key, big heart = unshaded (27u) minus small-heart-non-overlap (15u)... actually big heart = 19u + 7u = 26 units = \(105 \times 26 = 2730\) cm². Answer: 2730 cm².