The solid is made up of nine 1-cm cubes. Find the greatest number of cubes that can be added to the solid without changing the top view and the side view.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The maximum cubes that can be added while keeping the same top view and side view footprint is 4 (per the printed answer key).


A strip of paper 60 cm long is folded to form the shape shown. The difference in area between the strip of paper and the folded strip is 18 cm². Find the area of the shaded folded strip.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The figure is made up of a trapezium and 2 parallel lines, with angles c, b, a, d, e and given angles 240°, 75°, 50° marked. Which 2 angles add up to 75°?

At the bottom-left vertex, angles a, the 75° and angle b lie on the parallel line; reflex/straight-line relationships give \(\angle a + \angle c = 75°\) (the printed key answers angles A and C).


There was 21.6 litres of water altogether in container A and container B. The water level in container A was the same as container B. The base area of container A was 700 cm² and the base of container B has dimensions 40 cm by 50 cm.
(a) What was the height of water in container B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) From which container should water be poured out so both containers would have the same amount of water, and how much water should be poured? Pour out <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³ from Container B.


ABCD is a rhombus and CDEF is a parallelogram. \(\angle ABC = 46°\) (at B) and \(\angle DEF\)/\(\angle EFG = 128°\) (at F). Find \(\angle ACG\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The figure is formed by a square ABCD, equilateral triangle ABE and triangles BDG and BEG. AFE and BFD are straight lines, with \(\angle G = 18°\).
(a) Find \(\angle CBE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle BHG\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°


Shop A, Shop B and Shop C sell an identical pen at different prices: Shop A $1.20, Shop B $1.50, Shop C $1.00. The bar graphs show the number of pens sold by the 3 shops on Monday and Tuesday. Shop A sold 95 pens on Monday and 100 pens on Tuesday.
(a) How much money was collected altogether by Shop A from the sale of the pens on Monday and Tuesday?
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The graph shows the total number of cars painted by two robots, Robot A and Robot B, over 10 hours at a constant rate. Robot B stopped working after 5 hours while Robot A continued painting at the same constant rate.
(a) How many cars did Robot A and B paint altogether before Robot B stopped working? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many cars did Robot A paint over the 10 hours? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The figure is made up of rectangle ABCD, a circle with FG as the diameter, 2 identical small semi-circles with diameters EF and GH, and 2 larger semi-circles with diameters EG and FH. The radius of the circle FG is 18 cm. EH is a straight line. The length of BC is 60 cm.
(a) Find the length of EF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the total area of all the shaded parts (Take \(\pi = 3.14\)). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


A spiral number pattern begins with the number 8. 9 is the second number which happens at the first corner. 10 is the third number at the second corner. 12 is the fifth number at the third corner, and the spiral pattern continues.
(a) What is the 103rd number of the pattern? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the number at the 20th corner <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> and the 21st corner <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.

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


The diagram shows a cup of milo sold in a school canteen.<br>Which of the following could be the volume of the milo sold in the cup?

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


The table shows the height of different plants, W, X, Y and Z grown in a room over a month.<br>Which plant has the greatest increase in height?


Kamie is standing at K facing north. Kamie turns through an angle of 135° in an anti-clockwise direction, followed by \(\dfrac{1}{4}\)-turn in the clockwise direction. Which point will Kamie be facing?

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


Figure 1 is formed using four identical squares. The perimeter of Figure 1 is 80 cm. Another three identical squares were added to Figure 1 to form Figure 2.<br>Find the perimeter of Figure 2.

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


Xavier planned to save $21 using his pocket money. The table shows his saving plan for his weekly pocket money from Monday to Friday. Xavier started to save on a Friday. On which day of the week did Xavier manage to save exactly $21?

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


Five letters, T, E, A, M and S are drawn on a square grid.<br>(a) How many of the letters have a line of symmetry? (b) Which letter(s) have both parallel and perpendicular lines?

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


The table shows the timing of the movie XYZ shown in a cinema: 11.15 a.m., 2.30 p.m., 4.30 p.m., 7.15 p.m., 10.30 p.m.
(a) The duration of the movie is 2 h 10 min. What time will the last screening of movie XYZ end? Express your answer using the 24-hour clock. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Wendy arrived at the cinema at 11.45 a.m. How long must she wait for the next screening? Give your answer in h and min. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> min

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


The pie chart shows the different ways a group of students go to school. The number of students who travel by car and MRT are shown in the bar graph. How many students walk to school? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The pie chart shows the different ways a group of students go to school. The number of students who travel by car and MRT are shown in the bar graph. How many students travel to school by bus? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure ABCD is formed by two different squares and two identical rectangles. The area of the small square is \(\dfrac{1}{4}\) the area of the big square.<br>(a) What fraction of EBCF is shaded? (b) What fraction of ABCD is shaded?

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


In the circle, O is the centre and AOB is a straight line.<br>(a) Measure and write down the diameter of the circle in cm. (b) Using your answer in (a), find the circumference of the circle. Leave your answer in terms of \(\pi\).

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


Vinesh has a rectangular piece of paper. He folded it along the dotted lines without overlapping. In the folded figure one angle is 62°. Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the figure, ABCD is a rhombus and AB = AE. \(\angle\)EBC = 29° and \(\angle\)BCE = 80°.
(a) Find \(\angle\)BAE in degrees. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)ADE in degrees. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Jill glued 12 cubes of side 1 cm to form a solid. (a) Draw the top view of the solid on the square grid. (b) Jill painted the whole solid, including the base. How many cubes would have only two painted faces? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (c) What was the total painted area of the solid in cm²? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

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


ABC is a triangle, BFC is a straight line and \(\angle\)DEF is a right angle. \(\angle\)BAC = 61°, \(\angle\)EDC = 122° (exterior), \(\angle\)ABC = 58°. Find \(\angle\)BEF in degrees. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At first, 80% of a tank was filled with water. Chris turned on Tap A to let water out. 4 minutes later he turned on Tap B to let water in. After a while both taps were turned off. The graph shows the amount of water in the tank over 12 minutes.
(a) Find the capacity of the tank in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> l
(b) How much water flowed out of the tank from Tap A in 1 minute? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> l
(c) How much water flowed into the tank from Tap B in 1 minute? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> l

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


A market and a stadium are 240 m apart, located between Paul's and Rani's houses. The market is exactly half-way between the two houses. One morning Paul and Rani started cycling from their houses at the same time and arrived at the stadium together. Rani cycled at 100 m/min while Paul cycled 15 m/min faster than Rani.
(a) How much further did Paul cycle than Rani? Give your answer in m. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m
(b) How far was Rani's house from the stadium? Give your answer in m. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

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


Shermaine used rectangular blocks (length 2 cm, breadth 1 cm) to form figures following a pattern. The table shows blocks used per figure: Fig1=4, Fig2=7, Fig3=13, Fig4=16, Fig5=19, Fig6=23.
(a) Complete the table for Figure 7: number of blocks used. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many rectangular blocks were used in Figure 123? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Leon designed a logo made up of two identical small semi-circles, a circle and a big quadrant. YZ is the diameter of the circle and O is its centre. X is the centre of the big quadrant and XY = XZ. The radius of the small semi-circle is 7 cm. (Take \(\pi = \dfrac{22}{7}\).)
(a) Find the length of YZ in cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of triangle XYZ in cm². <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²
(c) Find the total area of the shaded parts of Leon's logo in cm². <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> cm²

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


In a bakery, donuts and cupcakes are sold only in boxes. A box of 8 donuts costs $15 and a box of 5 cupcakes costs $7. Alice and Ben were given a sum of money. Alice needed 20 donuts and 3 cupcakes for her party. After buying, Alice passed the rest of the money to Ben.
(a) What was the least amount of money Alice spent? Give your answer in dollars. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) Ben spent \(\dfrac{4}{7}\) of the rest of the money and had \(\dfrac{1}{3}\) of the sum left. How many donuts and how many cupcakes did Ben buy? Donuts: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, Cupcakes: <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


In the given figure, how many pairs of lines are perpendicular to each other?

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


In the number line, which mixed number is represented by X?

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


The table shows the number of pupils with the following Math scores.<br>Prizes were given to the top 8 pupils. Gwen won a prize. What was the lowest she could have scored?

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


The figure is not drawn to scale. PQRT is a parallelogram. ∠TPQ = 115° and ∠SQP = 43°. Find ∠RQS.

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


The figure shows a prism.<br>Which one of the following is a net of the prism?

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


The pie chart shows the mode of transport taken by Primary Six students to school. \(\dfrac{1}{2}\) of the students take the train to school.<br>Which of the following bar graphs best represents the information in the pie chart?

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


The figure shows 4 points (A, B, C and D) in a square grid.
Which of the following statements is false?

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


Study the table. Which algebraic expression is for Row \(n\), Column C?

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


The figure shows two rectangular tanks, X and Y. 540 cm³ of water was poured into X and Y until the heights of the water levels were the same. What was the height of the water level in each tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


A rectangular piece of paper was folded to form the figure shown. Find the perimeter of the rectangular piece of paper when it was <b>unfolded</b>. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


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

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


The table shows the cost of parking in SC Mall. Monday to Thursday: 1st hour $2.70, subsequent 30 minutes or part thereof $1.30. Friday to Sunday: 1st hour $2.60, subsequent 30 minutes or part thereof $1.89.
(a) On Tuesday, Mr Salim parked his car for 3 hours in the carpark starting from 8.30 a.m. How much did he pay for carpark charges? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Jamie was charged $12.05 for parking her car in the mall from 10 a.m. on a Saturday. What is a possible time she left the mall (in 24-hour or a.m./p.m. clock)? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The bar graph shows the number of cars sold from January to March. The number of cars sold in March is \(\dfrac{1}{4}\) of the number of cars sold in April. (a) Draw and shade the bar representing the number of cars sold in April. (b) What is the average number of cars sold from January to April? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The radius of Circle A is 3 times the radius of Circle B. Circle A makes 12 revolutions to roll from X to Y. (Take π = 3.14)
(a) How many revolutions will Circle B take to roll from X to Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The radius of Circle A is 12 cm. What is the length of XY? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

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


In the figure, not drawn to scale, EFGH is a square overlapping a trapezium ABCD. AB // DC, ∠CHG = 25° and ∠IFC = 33°.
(a) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠EBF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

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


In the figure, ACE is a triangle, BDEF is a parallelogram, BFG is an isosceles triangle and AE is parallel to BD. ∠ABG... the marked angles are 22° (at B, angle GBF region) and 58° (∠DBC).
(a) Find ∠GBD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠GBF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°
(c) Find ∠BDC. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />°

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


Freddy had a rectangular block measuring 10 cm by 3 cm by 6 cm. He painted all the faces of the block.
(a) Calculate the total painted area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Freddy wants to cut the largest possible cube from the rectangular block. What is the greatest number of such cubes he can cut from the rectangular block? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Tank A and Tank B shown are empty at first. Tank A has a base area of 441 cm². When water was poured into Tank A, some of the water flowed into Tank B through the hole which was just above Tank B. Tank B was \(\dfrac{4}{7}\) filled and the total amount of water in both tanks became 5324 cm³. The height of Tank B is 4 cm.
(a) Find the amount of water in Tank A. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) What is the capacity of Tank B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

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


The figure is formed by 2 quadrants and a rectangle overlapping one another. The radius of the smaller quadrant is 10 cm and the rectangle width is 4 cm. Find the area of the shaded part of the figure. (Take π = 3.14) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The table shows the time taken by 4 swimmers in a competition. Who is the fastest swimmer?

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


Identical cubes were glued together to form the 4 solids, A, B, C and D. Which 2 solids could be joined to form the solid shown?

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


The figure shows a cube. Which one of the following faces, P, Q, R or S, is <u>not</u> part of the net of the cube?

Folding the net, faces P, Q and R form valid faces of the cube but face S overlaps an existing face, so S cannot be part of the net.


Gayatri is standing at point Q facing the market. Which place will she be facing if she turns 225° anti-clockwise?

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


A group of workers was asked to vote for their favourite food, carrot cake, mee siam and nasi lemak. The pie chart shows the workers' choices. The workers' choices were also represented by 4 graphs. Which of the following graphs best represents the information in the pie chart?

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


The table shows the number of families with different number of children in a block of flats. There are 300 children altogether. How many families are there with 3 children each?

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


The figure is formed by 3 big identical equilateral triangles and 3 small identical equilateral triangles. The length of GH is 54 cm. What is the perimeter of the figure?

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


Figure X is a trapezium and Figure Y is a parallelogram. Which pair of the following statements is true?

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


What is the length of the arrow? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


The table shows the number of Primary 6 pupils who had their weight measured. Some information was smudged with ink. What is the difference between the number of girls who are underweight and overweight? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The pie chart represents the number of the different flavours of cakes in a bakery. What is the total number of blueberry cakes and mango cakes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, PQ = PS and QR is a straight line. Find \(\angle RQS\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The line graph shows the number of text messages Marco received each day from Tuesday to Friday. On which day was the number of messages Marco received the closest to the average number of messages he received from Tuesday to Friday?

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


In the figure, ACEG is a rectangle and ABJH is a square. The area of the shaded rectangle JDEF is 36 cm² and the length of HG is 9 cm. Find the length of AC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


The figure shows 4 identical triangles inside a square. The perimeter of the square is 280 cm. What is the length of \(y\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


The figure shows 4 identical triangles inside a square. The perimeter of the square is 280 cm. The 4 triangles are then cut from the square. What is the area of the remaining figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


In the figure, KN and JL are straight lines. LPK is an equilateral triangle and JK ∥ ML. The sum of \(\angle MJL\) and \(\angle LMJ\) is 138°. Find \(\angle QLP\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A tank was \(\dfrac{1}{3}\) filled with water at 4 p.m. Water flowed out of the tank from 4 p.m. to 9 p.m. The amount of water that flowed out of the tank is not shown on the scale. From 5 p.m. to 9 p.m., 125 ℓ of water flowed out of the tank. What is the capacity of the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

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


A tank was \(\dfrac{1}{3}\) filled with water at 4 p.m. Water flowed out from 4 p.m. to 9 p.m. From 5 p.m. to 9 p.m., 125 ℓ of water flowed out. At what time was 100 ℓ of water left in the tank? It was <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> p.m.

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


The table shows the prices of 5 items: Cap $4.00, Towel $0.90, Haversack $8.90, Water bottle $17.50, Sleeping bag $16.50. Of the 5 items, Glynis bought 3 of them and Levivia bought the remaining items. The items each of them bought were different. Glynis spent $11 more than Levivia. How much money did Glynis spend on the 3 items? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the prices of 5 items: Cap $4.00, Towel $0.90, Haversack $8.90, Water bottle $17.50, Sleeping bag $16.50. Of the 5 items, Glynis bought 3 (spending $29.40) and Levivia bought the remaining 2; the items each bought were different. What was the cheapest item each of them bought?

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


The figure is made up of a semicircle, a large circle and a small circle. O is the centre of the semicircle and the large circle. X is the centre of the small circle. OP = 56 cm. The small circle has a radius of 14 cm. Use the calculator value of \(\pi\) to find the perimeter of the unshaded parts, correct to 2 decimal places. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Using OP = 56 cm (semicircle radius) and the small circle radius 14 cm with the calculator value of \(\pi\), the perimeter of the unshaded parts works out to 467.82 cm (correct to 2 decimal places).


ABCD is a rectangle. AE and DG are straight lines. \(\angle ADF = \angle FDE\). Find \(\angle FDE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABCD is a rectangle. AE and DG are straight lines. \(\angle ADF = \angle FDE = 39^\circ\). Find \(\angle BGD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABCD is a rectangle. AE and DG are straight lines. \(\angle ADF = \angle FDE = 39^\circ\). Find \(\angle FAG\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the angle facts in the figure (the 43° at E and the rectangle's right angles), \(\angle FAG\) works out to 31° as given in the key.


On Saturday, Mr Zusin sold five different types of items. He prepared 66 of each item at the start of Saturday. He sold every item at the same price. The bar graph shows the number of items sold on Saturday. Which item had the most number left unsold at the end of Saturday?

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


On Saturday, Mr Zusin sold five different types of items (66 of each prepared). He sold the pens at $1.95 each. From the bar graph, 48 pens were sold. How much money did he collect from the pens? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


On Saturday, Mr Zusin prepared 66 of each of five items and sold some (bar graph). On Sunday, he continued to sell the remaining unsold items from Saturday at any 3 for $5. He sold all the items. How much money did he collect on Sunday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular piece of paper was folded at two of its corners F and H as shown. The angle at L (DLK region) is 157° and the angle at K is 67°. Find \(\angle HKJ\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular piece of paper was folded at two of its corners F and H as shown. The angle at L is 157°. Find \(\angle DLF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the fold and the given angles, \(\angle DLF\) works out to 44° (the printed key value).


A rectangular piece of paper was folded at two of its corners F and H as shown. The angle at K is 67°. Find \(\angle GFH\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the fold and the angle facts in the figure, \(\angle GFH = 44^\circ\) as given in the key.


An empty rectangular tank with a base area of 1250 cm² was to be filled with water from 2 taps, X and Y. Tap X filled the tank at 2.7 ℓ per minute; tap Y filled it at 1.5 ℓ per minute. Tap Z drained water out at 1.8 ℓ per minute. At 12.41 p.m., all 3 taps were turned on. How much water was there in the tank at 12.47 p.m.? Leave your answer in ℓ and ml. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml


An empty rectangular tank with base area 1250 cm² is filled by tap X (2.7 ℓ/min) and tap Y (1.5 ℓ/min); tap Z drains at 1.8 ℓ/min. All 3 taps were on from 12.41 p.m. At 12.47 p.m. only tap X was turned off. At 12.55 p.m., \(\dfrac{1}{3}\) of the tank was filled with water. What is the height of the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


WXYZ is a rhombus. Find \(\angle\)XYW. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Which of the following solids has the greatest number of faces?

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


The bar graph shows the range of marks scored by a group of students in a Maths test.<br>What fraction of the pupils obtained at least 45 marks in the test? Give your answer in the simplest form.


The figure is made up of 2 identical quadrants of radius 7 cm that overlapped each other. The shaded square has sides of 3 cm. What is the area of the whole figure? (Take \(\pi=\dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, how many more smiley faces must be added to make the ratio of the number of smiley faces to the total number of shapes become 2 : 3? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure shows a \(\dfrac{3}{4}\) circle of radius 14 cm. Find the perimeter of the figure. (Take \(\pi=\dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


In the figure, ABCD is a parallelogram. ECG is a straight line and BCE is an isosceles triangle. BC = CE and \(\angle\)BCG = 100°. Find \(\angle\)AEC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a square and 4 semicircles. The perimeter of the figure is 20\(\pi\) cm. What is the perimeter of the square? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


The figure is formed by five straight lines, AC, AD, BD, BE and CE. Find the sum of \(\angle\)p, \(\angle\)q, \(\angle\)r and \(\angle\)s. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


At first, 15 lamp posts were arranged in a row at an equal distance from point P to Q. The distance between 2 lamp posts is 26.6 m. Then, 5 more lamp posts were added. The lamp posts were rearranged at a new equal distance from point P to Q. Find the new distance between two lamp posts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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


The figure shows taps A and B and an empty tank with dimensions 25 cm, 30 cm and 60 cm. At 1.30 p.m., only tap A was turned on. Water flowed into the tank from tap A at a rate of 1.8 litres per minute. After 20 minutes, tap B was turned on. At 2 p.m., the tank was half-filled with water. How much water flowed out from tap B in one minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\)

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


The figure is made up of three rectangles. A straight line drawn across the rectangles divides the figure into two parts: shaded and unshaded. Heights are 2 cm, 7 cm, 3 cm, 2 cm and base widths are 9 cm, 6 cm and x cm. (a) Find the total area of the figure in terms of x.

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


The table shows the number of books read by 42 students. Two of the numbers in the table are accidentally painted over. The number of students reading 0, 1, 2 books are 0, 10, 9 respectively; the numbers for 3 and 4 books are hidden.
(a) On average, each student read 2.5 books. What was the total number of books read by all the students? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many students read 4 books each? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


In the figure, a piece of paper PQRS in the shape of a parallelogram is folded along 2 dotted lines, TS and UR. In Figure 2, \(\angle\)PST = 32° and \(\angle\)QRS = 14°.
(a) Find \(\angle\)UQR in Figure 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)QRU in Figure 2. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In the figure, EFGH is a rhombus and FGJK is a trapezium. EH and KF are straight lines. \(\angle\)KLH = 55° and \(\angle\)HEF = 110°.
(a) Find \(\angle\)KFH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)HGJ. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Containers X and Y are rectangular containers. The base area of Container X is 30 cm² and the base area of Container Y is 100 cm². Container X contained water to a height of 120 cm and Container Y is empty.
(a) Find the volume of water in Container X. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) John then poured some water from Container X into Container Y without spilling. After that, the height of water level in Container X is twice the height of water level in Container Y. How much water did John pour into Container Y? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

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


The figure shows 2 identical semicircles and a rectangle. O and N are the centres of the semicircles. The length and breadth of the rectangle are 34 cm and 24 cm respectively. (Take \(\pi=3.14\))
(a) Find the perimeter of total shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the total shaded area. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


Hasif uses white and black sticks to form figures that follow a pattern. The first five figures use white sticks 2, 2, 8, 8, 18 and black sticks 0, 4, 4, 12, 12 (totals 2, 6, 12, 20, 30).
(a) Complete the table for Figure 6: number of white sticks (i) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, number of black sticks (ii) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, total number of sticks (iii) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.
(b) Which figure number has a total of 650 sticks? Figure <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />.

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


PQRS is a rectangle and STUV is a square. Find ∠PSU.

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


The map shows the locations of four cities that are linked by railroads. Which one of the following statements is correct?

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


Which two figures have the same area? (Take $\pi = \dfrac{22}{7}$)

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


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

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


The line graph shows the number of cars sold monthly from January to June by a car dealer. What is the difference between the greatest and the least number of cars sold from January to June? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of two triangles, PRS and PQT. TQ = 4 cm and the length of TQ is half the length of SR. Find the area of TQRS.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


In a survey, a group of students were asked about their favourite pastime. The pie chart shows their choices. What fraction of the students liked to read? Express your answer in its simplest form.

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


The figure is made up of a triangle in a semicircle of radius 5 cm. The sides of the triangle measure 4 cm, 3 cm and 5 cm. Find the perimeter of the shaded part. (Take $\pi = 3.14$)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


In the figure, ABDE is a trapezium and BCDE is a rhombus. AFD is a straight line. $AE \parallel BD$, $\angle ADB = 21^\circ$ and $\angle DBC = 58^\circ$. Find $\angle EFD$.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


AGC is a right-angled triangle and GEDC is a parallelogram. ACD and BCF are straight lines. $\angle BCD = 104^\circ$ and $\angle EDC = 51^\circ$.
(a) Find $\angle FCG$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find $\angle CAG$. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °

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


Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The bar graph shows the number of fruits in each box. The bar representing Box D is not shown.<br>(a) How many percent more fruits are there in Box B than Box A? Give your answer as a mixed number in its simplest form.

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


Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The bar graph shows the number of fruits in each box. Box C contains only apples and oranges. There are 4 more apples than oranges in Box C. How many apples are there in Box C? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Uncle Lim has 4 boxes of fruits, Box A, Box B, Box C and Box D. The average number of fruits in the 4 boxes is 90. Find the number of fruits in Box D. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of a circle and a right-angled triangle XYZ. O is the centre of the circle. YOZ is a straight line. XY = YZ = 10 cm. Find the area of the shaded part. (Take $\pi = 3.14$)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


At a market, eggs were sold only in big trays of 50 eggs each and small trays of 12 eggs each. Hawker A and Hawker B bought the same number of trays of eggs. Hawker A bought 12 small trays of eggs while Hawker B bought 19 small trays of eggs. Hawker B used up all the big trays of eggs that he had bought. As a result, he had 1416 fewer eggs than Hawker A. How many trays of eggs did both hawkers buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, WXYZ is a trapezium and AXYB is a rhombus. XBC and XAZ are straight lines. $WX \parallel ZY$, BC = BY, $\angle XAB = 82^\circ$, $\angle ZYB = 19^\circ$ and $\angle WZX = 57^\circ$.
(a) Find $\angle c$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find $\angle w$. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °

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


The figure shows a small cube and a large cube. The length of each small cube is half the length of each large cube. Xavier wants to use 6 large cubes and some small cubes to build a new larger cube.
(a) What is the least number of small cubes that Xavier needs to build the new larger cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The volume of the new cube built by Xavier is 2744 cm³. Find the length of each small cube. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

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


The table shows numbers from 1 to 56. Kai and Ray are given a plastic frame that covers exactly 9 squares of the table with the centre square covered.
(a) Kai puts the frame on the 9 squares shown (with 38 covered at the centre). What is the average of the 8 numbers that is seen in the frame? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Ray puts the frame on another 9 squares. The sum of the 8 numbers that can be seen in that frame is 344. What is the sum of all the even numbers that Ray can see in that frame? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The figure is made up of Square ABCD and Rectangle EFGC. DCG is a straight line. DG = 49 cm and BE = 3 cm. The perimeters of Rectangle EFGC and Square ABCD are the same. Find the area of the figure.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The pie chart shows the CCAs taken up by a class of 40 students. How many students take up Rugby as a CCA?

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


The figure shows a semicircle. The diameter of the semicircle is 14 cm. What is the perimeter of the figure? Take \(\pi = \dfrac{22}{7}\)

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


ABCD is a trapezium. FED and AEC are straight lines.<br>Which of the following statements is not true?

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


The line graph shows the number of wallets sold in a shop from Monday to Saturday.
During which one-day interval was the increase in the number of wallets sold by the shop the greatest?
From <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The figure is made up of two squares. The length of the smaller square is 6 cm and the length of the larger square is 10 cm. Find the area of the shaded part.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The figure is made up of 5 rectangles. What fraction of the whole figure is shaded?

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


ABCD is a parallelogram. BC = BE, \(\angle ADE = 115^\circ\) and \(\angle AEB = 75^\circ\). Find \(\angle DAE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The pie chart shows the type of books in a library. The number of Tamil books is \(\dfrac{1}{2}\) the number of Malay books and there are 400 more Chinese books than Malay books. How many books are there in the library altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a square, a large semicircle, a small semicircle and 2 quarter circles. The length of each side of the square is 28 cm. Find the area of the shaded part. Take \(\pi = \dfrac{22}{7}\)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


AC and DF are straight lines. Find \(\angle CBF\).

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


A cuboid has a height of 8 cm and a square base of edge 5 cm. What is its volume? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume = base area × height = (5 × 5) × 8 = 25 × 8 = 200 cm³.


In the figure, ABCH is a square and DEGH is a rectangle. Given that the area of the shaded triangle is 35 cm², find the length of DE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, ABCD is a trapezium and BCD is an isosceles triangle. Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Meili had only the coins 5 cents, 10 cents, 20 cents, 50 cents and $1 in her wallet. She took three coins from her wallet and dropped them into a donation box. Which one of the following could be the amount she donated?

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


Given that the height of triangle ACF is AD, find the base that is related to the height AD.

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


The letters R, E, L, A and X are drawn on a square grid. Which of the letters have only 1 line of symmetry?

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


The figure shows the position of Jack and some facilities in his neighbourhood. Jack is facing the swimming pool. Where would he be facing if he made a 135° anti-clockwise turn?

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


ABCD is a parallelogram. ABF, BEC and DEF are straight lines. Find \(\angle DEB\).

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


The figure is formed by joining two quarter circles of radius 6 cm. Find the perimeter of the figure. Round your answer to the nearest whole number. Take \(\pi = 3.14\).

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


The ratio of the number of red beads to the number of blue beads is 3 : 2. The ratio of the number of red beads to the number of yellow beads is 1 : 2. Which one of the following graphs best represents the number of red beads, blue beads and yellow beads?

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


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

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


AB and CD are straight lines. Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A cuboid has a square base of side 5 cm and a height of 24 cm. Find the volume of the cuboid, in cm³. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, AEC and BED are straight lines and AB = BC = CD. Consider the statements: (a) \(\angle BEC = \angle AED\); (b) \(\angle BAC = \angle CDB\). Which of the following is correct?

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


The area of rectangle ABCD is 160 cm². The area of triangle AFB is 48 cm². Find the area of the shaded portion, in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Cheryl's Mathematics quiz results are: Fractions 88, Percentage [blank], Ratio [blank], Circles 75, with a Total score of 318. (The Percentage and Ratio scores are smudged.) Given the Percentage score is 6_ and the Ratio score is _2, find the difference between Cheryl's Percentage quiz score and her Ratio quiz score. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABCD is a rhombus. \(\angle ADC = 78°\) (at D) and \(\angle AEB = 69°\) (at E, where E lies on diagonal). Find \(\angle CAE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Container A measuring 45 cm by 22 cm by 28 cm was \(\dfrac{4}{7}\) filled with water. The water was then poured into another empty container, Container B (42 cm by 11 cm by 20 cm), until it was filled to the brim. What was the volume of water left in Container A? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a trapezium where AB is parallel to CD. CDE is an isosceles triangle where DE = DC. BEF is an equilateral triangle. \(\angle CDE = 38°\) (at D). Find \(\angle ABF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular tank contained some water at first. A tap was turned on to fill the tank completely with water. It was turned off at the end of 90 minutes. The graph shows the amount of water in the tank: at 0 min 40 L, rising steadily to 240 L at 50 min, then level at 240 L. (a) How much water flowed from the tap into the tank in 1 minute, in litres? (b) How many litres of water overflowed from the tank at the end of 90 minutes? (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) Rate = rise ÷ time. From 40 L to 240 L is 200 L over 50 min = 4 L/min. (b) The tank was full at 50 min; the tap kept running until 90 min, i.e. 90 − 50 = 40 more minutes. Overflow = 40 × 4 = 160 L. Answer: (a) 4 L, (b) 160 L.


The figure is made up of 2 quarter circles and a rectangle ABCD. AB = 6 cm and BC = 2 cm. What is the area of the shaded part, in cm²? (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In Country X, the height of six 10-cent coins is the same as that of five 20-cent coins (Diagram 1). Diagram 2 shows an unknown number of 10-cent coins stacked to the same height as another stack of 20-cent coins. The total value of the 2 stacks of coins in Diagram 2 is $8. (a) Find the number of 10-cent coins used in Diagram 2. (b) Find the value of all the 20-cent coins used in Diagram 2, in $. (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


A rectangular piece of paper has been folded from the two lower corners so the two corners meet at P. In the figure, \(\angle MNP\)-region angle is 26° (at N) and \(\angle MPN = 109°\) (at P). (a) Find \(\angle RPQ\). (b) Find \(\angle PRQ\). (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) Around point P: \(\angle RPQ = 360° − 109° − 90° − 90° = 71°\). (b) In triangle with the 26° fold: 180 − 26 − 90 = 64°; base angles 180 − 64 − 64 = 52°; \(\angle PRQ = 180 − 71 − 52 = 57°\). Answer: (a) 71°, (b) 57°.


Mrs Goh had some money. She used $53 to pay for 4 identical large potted plants and 7 identical small potted plants. If she bought another large potted plant, she would be short of $3.50. If she bought another small potted plant, she would have $1.50 left. (a) What is the difference in price between the large and the small potted plant, in $? (b) Find the price of one large potted plant, in $. (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) Buying one more large needs $3.50 more; buying one more small leaves $1.50. Difference in price = $3.50 + $1.50 = $5. (b) Money = $53. 7s + 4(1 small + $5) = 53 → 7s + 4s + 20 = 53 → 11s = 33 → s = $3. Large = small + $5 → $8. Check: 4 × 8 + 7 × 3 = 32 + 21 = 53. Answer: (a) $5, (b) $8.


The graph shows the number of people at a book fair from Monday to Friday: Mon 130, Tue 200, Wed 180, Thur unknown, Fri 160. (a) The average number of people from Monday to Friday was 174. How many people were at the book fair on Thursday? (b) The average number of people who visited on Saturday and Sunday was 206. 20 more people visited on Saturday than on Sunday. What was the percentage increase in the number of visitors from Friday to Saturday? (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In the number line, what is the value of X?

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


The solid is made up of 1-cm cubes. How many more 1-cm cubes must be added to the solid to make it a cuboid measuring 4 cm by 3 cm by 4 cm?

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


The table shows the number of bicycles sold each month by a shop. January 53, February 150, March 124, April 76, May 99. Which of the following bar graphs best represents the information in the table?

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


A wire 210 cm long was cut into 2 pieces and bent to form a rectangle and a square. The length of the rectangle is 48.5 cm and its breadth is 38.5 cm. Find the length of one side of the square.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular tank measuring 32.5 cm by 22.5 cm by 18 cm was \(\dfrac{5}{9}\)-filled with water. When Kamala poured 7 litres of water into the tank, some water overflowed. Find the volume of water that overflowed. Give your answer in litres.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


ABCD is a trapezium. APB, DQC, RCS and PQR are straight lines. APB is parallel to DQC and AD is parallel to PQR. ∠BCS = 50°, ∠BCQ = 90° and ∠QRC = 95°.
(a) Find ∠DQP. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find ∠ADQ. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The figure is made up of 2 squares, ABCD and DEFG. X is a point on GD and Y is a point on DE. BC = 7 cm, FE = 6 cm and XD = DY = YE.
(a) Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) What fraction of the figure is unshaded? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


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


The square grid shows Triangle P. What type of triangle is Triangle P?

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


WXYZ is a rhombus.<br>Which one of the following is false?

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


ABC is a triangle with AB = 10 cm and BC = 15 cm. BE = 8 cm and AD = 21 cm. Find the area of triangle ABC.

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


In which month was the number of cars sold half as many as the number of cars sold in September?

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


The bar graph shows the number of cars sold from June to September. Which one of the following statements is true?

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


The figure is made up of 3 identical quarter circles of radius 28 cm. Find its perimeter. (Take π = \(\dfrac{22}{7}\))

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


At first, a rectangular tank measuring 40 cm by 30 cm by 30 cm contained some water. After Melvin poured 2400 ml of water into the tank, the tank became \(\dfrac{2}{3}\)-filled with water. How much water was there in the tank at first?

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


ABC is an isosceles triangle. AB = AC. ∠BAC = 36°. Find ∠ABC.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


WXYZ is a trapezium and WX is parallel to ZY. ∠WXY = 56° and ∠WZY = 66°. Find ∠XWZ.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


RSVW is a square and WTUV is a parallelogram. WST is a straight line. ∠TVS = 7°. Find ∠TVU.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


The figure shows a square and a quarter circle. The length of the square is 20 cm. Find the area of the shaded part. Leave your answer in terms of π.

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


A cuboid is 0.4 m long and 20 cm wide. It has a volume of 20 000 cm³. Find the height of the cuboid.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


A rectangular block P was cut along the dotted line into two smaller rectangular blocks Q and R. The volume of Q was \(\dfrac{2}{3}\) the volume of R. The difference in volume between Q and R was 12 000 cm³. Find the unknown edge of block P.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


[Paper 2 Q1] The table shows the number of storybooks read by each student in a class. Part of the table is covered by an ink blot. There were 20 students who read less than 3 storybooks. There were twice as many students who read 3 storybooks as those who read 5 storybooks. The table shows: 1 storybook → 9 students; 4 storybooks → 3 students; 5 storybooks → 4 students (the 2-storybook and 3-storybook counts are hidden).
(a) How many students read 2 storybooks? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many students were there in the class? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


[Paper 2 Q7] The figure is made up of a right-angled triangle, a rectangle and 2 semicircles. Find the total area of the shaded parts. (Take π = 3.14)<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


[Paper 2 Q10] JKLM is a rhombus. MNK is a straight line and MN = ML. ∠MNL is 24° more than ∠LMN. Find ∠MJK.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


[Paper 2 Q11] BCDE is a trapezium. BC is parallel to GED. BEH is an equilateral triangle and EFGH is a square.
(a) Find ∠DEF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find ∠EBC. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °

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


[Paper 2 Q16] The figure is drawn on a rectangular piece of paper 30 cm by 40 cm. Its outline consists of 4 identical quarter circles and 5 straight lines. (Take π = 3.14)
(a) Find the perimeter of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of the shaded figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

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


Ahmad used sticks to form the following figures. The table shows the number of triangles and sticks for the first three figures: Figure 1 has 2 triangles and 8 sticks; Figure 2 has 4 triangles and 15 sticks; Figure 3 has 6 triangles and 22 sticks. (a) Complete the table for Figure 4: number of triangles <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and number of sticks <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.<br>(b) A figure in the pattern has a total of 2507 sticks. What is the Figure Number? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


In the figure, WX is the base of the triangle WXY. Which line represents its height?

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


The figure is made up of a quadrant and 2 identical semicircles. Find the perimeter of the figure. (Take \(\pi = \dfrac{22}{7}\))

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


The line graph shows the number of cars sold in the showroom from June to November.<br>What was the percentage decrease in sale from June to July?

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


In the figure, ABCD is a trapezium. AD is parallel to BC and BC = CD. \(\angle ABD = 76^\circ\) and \(\angle BCD = 112^\circ\). Find \(\angle BAD\).

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


The figure is made up of 2 identical quadrants. Find the perimeter of the figure. (\(\pi = 3.14\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


The figure is made up of a parallelogram EFGH and a triangle EJH. \(\angle FGH = 78^\circ\), \(\angle JHE = 36^\circ\) and \(\angle JEF = 13^\circ\). Find the value of \(\angle EJH\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


In the figure, ABCD is a rhombus and AEFD is a parallelogram. \(\angle CDB = 47^\circ\). Find \(\angle x\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Five identical isosceles triangles are joined as shown in the figure. Find the area of one such triangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

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


ABCD is a rectangle made up of 8 identical right-angled triangles. The perimeter of rectangle ABCD is 364 cm, what is the area of rectangle ABCD?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


A rectangular piece of paper is folded along the dotted line such that the total area of triangles A, B and C is \(\dfrac{5}{9}\) the area of the rectangular piece of paper. The area of triangle B is 24 cm², find the area of the rectangular piece of paper.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The figure is made up of two identical squares, ABCD and CEFG, a trapezium DEKJ and a triangle DEJ. The line DE is parallel to the line JK. \(\angle CDE = 54^\circ\) and \(\angle EJK = 33^\circ\).
(a) Find \(\angle BCG\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle DHJ\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

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


The figure, not drawn to scale, shows a parallelogram ABCD and a triangle CEF. AF and DE are straight lines. Which angle is equal to \(\angle\)CAB?

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


The figure is made up of 8 unit squares. Which parts must be shaded so that the figure is \(\dfrac{3}{4}\) shaded?

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


In the figure, not drawn to scale, XZ is a straight line and \(\angle\)XOY is 3 times \(\angle\)WOZ. Given that \(\angle\)YOX is 75°, find \(\angle\)WOX. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The figure shows a rectangle ABCD. Find \(\angle\)CAD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


ACDE is a rhombus and BD is a straight line. Find \(\angle\)BAC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The figure, not drawn to scale, shows an equilateral triangle ADE overlapping with a square ABCD. DFB and AFE are straight lines. Find \(\angle\)DFE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The graph shows the number of tickets sold for a basketball match from Monday to Friday.
(a) What was the total number of tickets sold from Monday to Wednesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The usual price of a ticket was $70. On Friday, tickets were sold at a 5% discount. How much money was collected from the sale of tickets on Friday? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


ABCD is a parallelogram. CFE is an isosceles triangle with CF = EF. AFC is a straight line. \(\angle BAF = 59^\circ\) and \(\angle ABC = 71^\circ\). Find \(\angle AFE\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Figure 1 shows a rectangular tile. Figure 2 is formed using 5 such tiles. The perimeter of Figure 2 is 360 cm.
(a) Find the perimeter of Figure 1. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) The length of the rectangular tile was 3 times as long as its breadth. Find the area of the rectangular tile. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

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


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

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


The figure is divided into 20 equal parts. What percentage of the figure is shaded?

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


In the figure, ABC and DBF are straight lines. Find \(\angle FBE\).

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


At a birthday party, each child got to choose one ice-cream flavour. The table shows their choices. Which flavour was the most popular among the children?

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


In a movie marathon guide, movies are screened one after another without any break in between. Which movie has the longest duration?

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