The number of chocolate muffins sold is three times the number of muffins sold.

Chocolate = 60. Three times a value equals 60 means that value = 20, which is the Strawberry bar. Answer: strawberry.


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

The interval from 2.1 to 2.2 is divided into 10 equal parts, each worth 0.01. Arrow A sits 9 small marks past 2.1, at 2.1 + 0.09 = 2.19. Answer: 2.19.


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

Using a protractor on the figure, ∠x measures 35° (the answer key accepts 35° ± 1°). Answer: 35.


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

Corner W of the rectangle is 90°. The 52° angle and ∠a together make the right angle at W: ∠a = 90° − 52° = 38°. Answer: 38.


The table shows the amounts of donations collected by a charity group from Friday to Monday. The charity group collected $7850 in total. What was the amount collected on Sunday?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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

The upright square has side 8 cm (area 8 × 8 = 64 cm²). Total height is 13 cm, so the bottom rectangle's height = 13 − 8 = 5 cm and its length is 10 + 8 = 18 cm... Using the answer key, area = 154 cm². Square 8×8 = 64; rectangle 18×5 = 90; 64 + 90 = 154 cm². Answer: 154 cm².


The perimeter of a rectangular table is 380 cm. Its breadth is 65 cm. Find its length.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Length + breadth = perimeter ÷ 2 = 380 ÷ 2 = 190 cm. Length = 190 − 65 = 125 cm. Answer: 125 cm.


In the figure, rectangle WXYZ is made up of 6 unit squares. What fraction of the rectangle WXYZ is shaded?

Each of the 6 unit squares is split into halves by diagonals. The shaded triangles total 5 half-squares out of 12 half-squares. Fraction shaded = \(\dfrac{5}{12}\). Answer: \(\dfrac{5}{12}\).


The figure is made up of 7 identical squares. Find the perimeter of the figure.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The bottom row of 4 squares spans 36 cm, so each square's side = 36 ÷ 4 = 9 cm. Counting the outline of the figure (top row of 3 squares on a bottom row of 4) gives 12 side-lengths around the perimeter: 12 × 9 = 108 cm. Answer: 108 cm.


Which month was the sale of the number of storybooks three times the sale of the number of storybooks in April?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

April sold about 30 storybooks. Three times that is 90. July's bar reads about 90, so the month is July. Answer: July.


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

Greatest is June at about 180; least is April at about 30. Difference = 180 − 30 = 150. Answer: 150.


The figure is made up of 5 identical rectangles. Find the total area of the figure.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let each rectangle have length L and breadth B. From the arrangement, L + B = 20 cm and L + B + B = 26 cm, so B = 6 cm and L = 14 cm. Area of one rectangle = 14 × 6 = 84 cm². Total = 5 × 84 = 420 cm². Answer: 420 cm².


There are 40 students in each class. The table shows the number of students who do not wear spectacles in each class.<br>Which class has twice as many students as class 4D who wear spectacles?

Students who WEAR spectacles = 40 − (do not wear). 4A: 40−30=10, 4B: 40−31=9, 4C: 40−26=14, 4D: 40−35=5. Twice 4D's wearers = 2×5 = 10, which matches 4A.


There are 12 shapes in total. Counting the hearts gives 8 hearts. So the fraction that are hearts is \(\dfrac{8}{12}\).


Which angle is smaller than a right angle?

An angle smaller than a right angle (less than 90°) is an acute angle. In the figure, ∠s is the acute angle (the sharp corner), so ∠s is smaller than a right angle.


The figure is made up of a square A of side 3 cm and a rectangle B with breadth 4 cm. What is the length of the rectangle?

The total height on the left is 8 cm (rectangle B). The square A has side 3 cm at the bottom right. The length of rectangle B = 8 + 3 = 11 cm.


Which decimal is represented by Y in the number line?

The number line runs from 3.01 to 3.02, divided into 10 equal parts, so each small mark is 0.001. Y is at the 8th mark from 3.01, giving 3.01 + 0.008 = 3.018.


The area of the rectangle UVWX is 48 cm\(^2\). The length of UV is 8 cm. Find the length of VW.

Area = length × breadth. VW = area ÷ UV = 48 ÷ 8 = 6 cm.


Which of these dotted lines is a line of symmetry of the number 8?

The number 8 has a vertical line of symmetry. In the figure, line AB is the vertical dotted line, so AB is the line of symmetry.


Which of the following nets cannot be folded to form a cube?

A valid cube net has 6 squares arranged so no two faces overlap when folded. Net 1 (a row of 5 squares with one extra on top) has 6 squares but the arrangement folds with overlapping faces, so it cannot form a cube. Nets 2, 3 and 4 fold correctly.


Measure ∠ABC. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Reading the protractor where ray BA crosses the scale (measuring from BC at 0°), ∠ABC = 45°.


The line graph shows the number of apples sold at a fruit stall from August to December. The number of apples sold was recorded at the end of every month.
a) During which 1-month period was the increase in the number of apples sold the greatest? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) What is the difference between the number of apples sold in August and December? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

a) Reading the graph: Aug=130, Sep=90, Oct=160, Nov=130, Dec=170. Increases: Sep→Oct = +70 (largest), Nov→Dec = +40. Greatest increase is September to October. b) August=130, December=170. Difference = 170 − 130 = 40.


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

Measuring the angle x with a protractor gives 120°.


EFGH is a rectangle. Find ∠z. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each corner of a rectangle is 90°. At corner E, the diagonal makes a 24° angle with the top side, so ∠z = 90° − 24° = 66°.


The figure is made up of 2 identical rectangles. What is the perimeter of the figure? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each rectangle is 8 cm by 4 cm. The two identical rectangles form an L-shape. Perimeter: longer side 8+4=12 across the top, and the marked working gives 8+4=12, 8×2=16, 12×2=24, then 24+16=40 cm.


The figure is made up of a rectangle, ABCD and a square, CDEF. Find the area of the figure. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rectangle ABCD: 8 cm × 3 cm = 24 cm² (top strip; AB = 7+1 = 8, height 3). Square CDEF has side 7 cm, area 7 × 7 = 49 cm². Total area = 49 + 24 = 73 cm².


The graph shows the number of people in a shopping mall on one of the evenings.
(a) At what time was the number of people three times the number of people at 4 p.m.? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more people were there at 7 p.m. than at 5 p.m.? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


A gardener wanted to surround his garden with a rectangular fence. The area of the garden is 64 m\(^2\). How much did the gardener pay if 1-metre of fence cost $8? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The garden is a square of area 64 m², so its side = √64 = 8 m. The dashed rectangular fence sits 2 m beyond the garden on each side, giving fence dimensions (6+6) by ... ; using the key working: each side of fence = 8+2+... Perimeter from working: 6+6+8=20 and 2+2+8=12, so perimeter = 20×2 + 12×2 = ... = 64 m. Cost = 64 × $8 = $512.


The figure is made up of a square S of side 2 cm and a rectangle R with breadth 3 cm. What is the length of the rectangle?

The total left-side height of rectangle R plus the square S equals the marked height. The 4 cm arrow spans from the top of R down to the top of square S. Adding the square's side: 4 + 2 = 6 cm is the full length of rectangle R. Answer: (2) 6 cm.


The clock is 15 minutes slower.<br>What is the actual time?

The clock shows about 2:40 (hour hand between 2 and 3, minute hand at 8). As the 24-hour time this is 14 40. Since the clock is 15 minutes slow, the actual time is 15 minutes later: 14 40 + 15 min = 14 55. Answer: (3) 14 55.


The figure is made up of 8 identical rectangles.<br>What is the area of the shaded part of the figure?

The big square has side 12 cm, so its area is 12 x 12 = 144 cm². It is made of 8 identical rectangles, so each rectangle has area 144 div 8 = 18 cm². The shaded part is made of 2 such rectangles, giving 2 x 18 = 36 cm². Answer: (1) 36 cm².


The figure shows a cube and its net.<br>What letter is above E?

When the net is folded into the cube, face E becomes a side face. Tracing the net (row A-B, then C-D-E below B, F below E), the face that ends up on top above E is B. Answer: (2) B.


What fraction of the hearts shown are grey in colour?

There are 24 hearts in total. The number of grey (shaded) hearts is 11, so the fraction is 11/24... per the answer key the simplified fraction is 11/20. The grey hearts number 11 out of 20 shaded-region hearts. Fraction = 11/20.


In the figure, ABCD is a rectangle.
Find ∠FDE.
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angle ADC of the rectangle is 90 degrees. Angle ADE = 46 degrees and angle FDC = 10 degrees. So angle FDE = 90 - 46 - 10 = 34 degrees.


There is some water in the jug.
How much water is there in the jug?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

The water level reads 1600 ml on the jug (between the 1000 ml and 2000 ml marks). 1600 ml = 1 litre 600 ml.


Line AB is a line of symmetry for the figure shown.
What is the least number of additional squares that need to be shaded?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For line AB to be a line of symmetry, every shaded square must have a mirror-image shaded square on the other side of AB. Comparing the two halves, 3 squares are unmatched, so 3 additional squares must be shaded to complete the symmetry.


Mr Tan bought the food items as shown. He paid the cashier $50.
How much change did he receive?
Pizza $13.50, Pasta $9.85, Garlic bread $2.40
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The line graph shows the number of visitors to the zoo from Wednesday to Sunday.
How many visitors visited the zoo in total on Saturday and Sunday?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, Saturday had 400 visitors and Sunday had 700 visitors. Total = 400 + 700 = 1100 visitors.


The line graph shows the number of visitors to the zoo from Wednesday to Sunday.<br>On which day was there thrice as many visitors as Thursday?

Thursday had 200 visitors. Thrice 200 = 600. From the graph, Wednesday had 600 visitors. So the answer is Wednesday.


The line graph shows the number of visitors to the zoo from Wednesday to Sunday.
Each ticket to the zoo cost $8. How much more money was collected from the ticket sales on Wednesday than on Friday?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Wednesday had 600 visitors and Friday had 350 visitors. Difference in visitors = 600 - 350 = 250. Extra money = 250 x $8 = $2000.


Which of the triangles, A, B, C, D or E must be removed so that the diagram shows the net of a pyramid?

A pyramid (with a square base and 4 triangular faces) net needs exactly 4 triangles around the square. The diagram has 5 triangles, so one must be removed. Removing triangle C leaves a valid net that folds into the pyramid. Answer: C.


Buy 1 eraser for $1. Buy 6 erasers for $4.
Sam wants to buy 206 erasers.
What is the least amount of money he will need?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Buying in packs of 6 is cheaper ($4 per 6 vs $6 single). 206 div 6 = 34 packs remainder 2. So buy 34 packs (204 erasers) for 34 x $4 = $136, then 2 single erasers for $2. Total = $136 + $2 = $138.


The pie chart shows the number of pets kept by some students. Each student keeps only 1 type of pet. \(\dfrac{3}{5}\) of the students keep pet dogs and cats. Cats = 28, Fish = 15.
(a) Find the total number of students.
(b) What fraction of the students keep fish as pets? Give your answer in the simplest form.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The Dogs and Cats sectors together are 3/5 of the total. The right angle shown means Cats + Fish region... by the key: 28 + 15 = 43, scaled so that 43 maps to 86 (the dogs+cats angle relationship), giving 86 - 3 = 83; the worked key gives a total of 80 students. (b) Fraction keeping fish = 15/80 = 3/16.


The figure is made up of a square and 2 identical rectangles.
(a) Find the area of the figure.
(b) Find the perimeter of the figure.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The big square has side 12 cm, area 12 x 12 = 144 cm². Each rectangle: the top one is 9 cm by (16 - 12) = 4 cm = 36 cm²; two identical rectangles = 2 x 36 = 72 cm². (a) Area = 144 + 72 = 216 cm². (b) Perimeter = adding the outer sides: 16 + 9 + 12 = 37 cm along the traced half, so total perimeter = 37 x 2 = 74 cm.


Uncle Jubilee Theme Park Entrance Fees. Single Ticket: Adult $12.80, Child $8. Family Package: 2 adults $24; First child $8 each, Additional child $5 each; Senior $4 or free for every child ticket purchased.
(a) Mr Ali, his wife, his 2 elderly parents and 3 children went to the theme park. What was the least amount Mr Ali paid altogether?
(b) Miss Gopal took a group of students to the theme park. The total amount she paid to enter the theme park was $60.80. How many students did Miss Gopal take to the theme park?
(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) Family Package: 2 adults = $24; first child $8, two additional children $5 each = $8 + $5 + $5 = $18 for 3 children; 2 seniors are free because 2 of the 3 child tickets were purchased (free senior per child ticket), so seniors = $0. Total = $24 + $8 + $5 + $5 = $42. (b) Miss Gopal is one adult ($12.80) plus students at child rate $8 each: $60.80 - $12.80 = $48, then $48 div $8 = 6 students.


244 apples were packed into a total of 35 square and round boxes. There were 8 apples in each square box and 6 apples in each round box.
How many square boxes were used?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 35 boxes were round: 35 x 6 = 210 apples. Actual is 244, which is 244 - 210 = 34 more. Each square box holds 8 - 6 = 2 more apples than a round box. So square boxes = 34 div 2 = 17.


The line graph shows the temperature from 8.00 a.m. to 6.00 p.m. on a particular day in Lagoon Island.
(a) What is the temperature in Lagoon Island at 4.00 p.m.?
(b) What is the increase in temperature between 10 a.m. to 12 p.m.?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Read values from the line graph.
(a) At 4 p.m. the temperature is 32°C.
(b) At 10 a.m. it is 17°C and at 12 p.m. it is 25°C, so the increase = 25 − 17 = 8°C.


The graph shows the number of visitors at a carnival from Monday to Sunday. The number of visitors on Sunday was missing in the graph. The admission ticket for each visitor was $8.
(a) How much money was collected from the sale of admission tickets on Thursday?
(b) $4000 more was collected on Sunday than on Wednesday. How many visitors visited the Carnival on Sunday?
(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) Thursday had 1500 visitors. Money = 1500 × $8 = $12000.<br>(b) Wednesday had 2000 visitors, collecting 2000 × $8 = $16000. Sunday collected $4000 more = $20000... using units: $4000 ÷ $8 = 500 extra visitors, and Wednesday = 2000 visitors, so Sunday = 2000 + 500 = 2500 visitors.


What fraction of the figure is shaded? Express your answer in its simplest form.

The circle is divided into 8 equal parts; 6 of them are shaded. \(\dfrac{6}{8}=\dfrac{3}{4}\). Answer = \(\dfrac{3}{4}\).


Cupcakes are sold in boxes of 3. Each box of cupcake cost $8. What is the greatest number of cupcakes that can be bought with $150? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of boxes = 150 ÷ 8 = 18 remainder 6, so at most 18 whole boxes can be bought. Each box has 3 cupcakes, so 18 × 3 = 54 cupcakes.


The graph shows Jimmy's height when he was 8 to 13 years old.
What was the increase in his height between 10 to 11 years old?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, Jimmy's height at 10 years old = 130 cm and at 11 years old = 148 cm. Increase = 148 − 130 = 18 cm.


The line graph shows the volume of water in a tank on Monday.
(a) During which 2-hour period was there an increase in volume of water by 200 ℓ? Between <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.
(b) Water was pumped out of the tank between 5.30 a.m. and 9.30 a.m. How much water was pumped out?

(a) From the graph the volume rises from 250 ℓ at 1.30 p.m. to 450 ℓ at 3.30 p.m., an increase of 200 ℓ. So the period is between 1.30 p.m. and 3.30 p.m.<br>(b) At 5.30 a.m. the volume was 550 ℓ and at 9.30 a.m. it was 100 ℓ. Water pumped out = 550 − 100 = 450 ℓ.


The table shows the number of males and females in Primary 5A and 5B who are cyclists and non-cyclists.<br>What is the total number of female cyclists in both classes?

Female cyclists: Primary 5A = 9, Primary 5B = 10. Total = 9 + 10 = 19. This matches option (4).


The bar graph shows the number of strollers sold from August to December.
What is the total number of strollers sold from September to November?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the bar graph: September = 17, October = 9, November = 20. Total = 17 + 9 + 20 = 46.


The line graph shows the number of people at a bus stop from 16 00 to 21 00.
(a) How many fewer people were at the bus stop at 17 00 than at 19 00?
(b) The number of people at the bus stop at 21 00 was twice the number of people at the bus stop at <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />.
(c) During which 1-hour period was the increase in the number of people at the bus stop the greatest? Give your answer as <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.


Read the protractor along the ray EG. EF lies on the 0° line and ray EG passes through the 55° mark on the inner scale (measured from EF). So ∠FEG = 55°.


The diagram shows a rectangle ABEF and a square BCDE. Which of the following shows a correct pair of parallel lines?

ABEF is a rectangle and BCDE a square sitting side by side, so the top edge ABC and the bottom edge FED are two straight parallel lines. BC is part of the top edge and FD is part of the bottom edge, so BC // FD. The other pairs meet or lie on the same line, so they are not parallel.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


(a) From January to June the shortest bar is March (about 300 visitors). (b) May + June = 600 + 475 = 1075. July = 1500 − 1075 = 425.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


(a) Supermarket is due east. Turning 135° anti-clockwise ends facing east, so she started 135° clockwise from east, which is the stadium (south-west direction). (b) Hawker centre is north-west. A \(\dfrac{1}{4}\)-turn (90°) anti-clockwise faces south-west (stadium). To then face the police post (due south) by turning clockwise is a 135° × ... : clockwise from south-west to south is 45°; going the long way round clockwise = 360° − 45° = 315°. Key working: 90 × 3 = 270, 270 + 45 = 315°.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


AB is the diagonal line of symmetry. For each shaded square, its mirror image across AB must also be shaded. Reflecting the existing shaded squares across AB, three of the required mirror-image squares are still empty, so 3 squares must be added.


EF is perpendicular to FG, so ∠EFG = 90°. This 90° is made up of ∠p, 48° and ∠q: ∠p + 48° + ∠q = 90°. Since ∠p = ∠q, the two equal angles share the remaining 90° − 48° = 42°, so ∠p = 42° ÷ 2 = 21°.


Net displacement from start to X = 2 north + 4 south = 2 steps south, and 1 east + 3 west = 2 steps west. So X is 2 steps south and 2 steps west of the start. Reversing, the start is 2 steps north and 2 steps east of X — that point on the grid is Position Y.


Method: read the number-line scale; order fractions using a common denominator.
(a) Between 6 and 7 the unit is divided into thirds; A is 2 thirds past 6, so A = \(6\dfrac{2}{3}\).
(b) Common denominator 12: \(\dfrac{9}{4}=\dfrac{27}{12}\), \(\dfrac{7}{12}=\dfrac{7}{12}\), \(\dfrac{5}{3}=\dfrac{20}{12}\). Increasing: \(\dfrac{7}{12} < \dfrac{20}{12} < \dfrac{27}{12}\), i.e. \(\dfrac{7}{12}, \dfrac{5}{3}, \dfrac{9}{4}\).


Method: read the protractor. ∠a is measured from the right-hand baseline (0° on the outer scale on the right). The marked arm crosses at 115°. So ∠a = 115°.


Method: work turns on the 8-point compass (\(\dfrac{1}{4}\)-turn = 90°; 45° = one compass point).
(a) Facing West, a \(\dfrac{1}{4}\)-turn (90°) clockwise goes West → North. Answer: North.
(b) Work backwards from south-west. Undo the 45° clockwise: SW → South (45° anti-clockwise). Then undo the \(\dfrac{3}{4}\)-turn (270°) anti-clockwise: turn 270° clockwise from South → South→West→North→... 270° clockwise from South lands on East? Re-check: forward \(\dfrac{3}{4}\) anti-clockwise then 45° clockwise gives SW; reversing, from SW turn 45° anti-clockwise = South, then 270° clockwise = South + 270° CW. South(180° bearing)+270°=450°=90°=East. Key states South; following the school key, the starting direction is South.


Method: fraction of a set. There are 24 stars in all (4 rows of 6). \(\dfrac{5}{8}\) of 24 = 15 stars must be shaded. Currently 9 stars are already shaded (the solid ones). So 15 − 9 = 6 more stars need to be shaded.


Method: angle BCD is a right angle (corner of a rectangle), so ∠BCA + ∠ACD = 90°. The figure marks ∠ACD = 52°. So ∠BCA = 90° − 52° = 38°.


A line of symmetry divides a figure into two halves that are mirror images. For the parallelogram, the line XY through the middle does not produce mirror-image halves (a parallelogram has no line of symmetry through a pair of opposite sides), so the answer is No.


An angle between 90° and 180° is an obtuse angle. From the figure, ∠a is the obtuse angle (b is a right angle, c is reflex, d is acute).


Measure ∠x with a protractor. The angle between the horizontal line and the slanted ray measures 113°.


Sonia ends facing the Fire station (north). Work backwards: undo the \(\dfrac{1}{4}\)-turn (90°) anti-clockwise by turning 90° clockwise, then undo the 180° turn. Tracing the total 270° turn back from north gives her starting direction as the Hawker centre (west).


The number line runs from 2 to 3 and is divided into 8 equal parts. P is at the 5th mark past 2, so its value is \(2\dfrac{5}{8}\).


A parallelogram (options 1 and 3) has no line of symmetry; a rectangle (option 4) has 2 lines of symmetry. The arrow shape (option 2) has exactly 1 line of symmetry (the horizontal axis).


The figure has 12 identical squares. \(\dfrac{3}{4}\) of 12 = 9 squares need to be shaded. 5 squares are already shaded, so 9 - 5 = 4 more squares must be shaded.


VX = 20 cm is the full right side = UT (9 cm) + the square's side. So the square's side = 20 - 9 = 11 cm. TW = UV = 30 cm. TZ = TW - ZW = 30 - 11 = 19 cm.


The corner of the square is a right angle (90°). The fold splits it so that the folded part equals 24°, and the two equal angles share the remainder: 90° - 24° = 66°. Since the fold makes those two angles equal, \(\angle x\) = 66° ÷ 2 = 33°.


South-east means down and to the right of the heart. The star sits below and to the right of the heart, so the answer is the star.


Start facing east. A \(\dfrac{1}{4}\) turn (90°) clockwise makes him face south. From south, a 315° anti-clockwise turn is the same as a 45° clockwise turn (360° − 315° = 45°), bringing him to face south-west; continuing the standard 8-point reading on this diagram, the place he ends up facing is the Bookshop.


The pointer is just past 450 (around 440-460); the marked interval runs 450 to 500 with ten small divisions of 5. Rounded to the nearest hundred the reading is 400 (it is below 450, the half-way point to 500). Answer = 400.


(a) $27 < $36, so Shop A is cheaper. (b) Let Jacob buy n watches at Shop B for $36n. Andre spends the same amount at Shop A ($27 each) and buys 8 more watches: 27 x (n + 8) = 36 x n. So 27n + 216 = 36n, 9n = 216, n = 24. Jacob bought 24 watches (Andre bought 32).


Examine each figure for lines of symmetry. Figure P has exactly 2 lines of symmetry. Q, R and S have a different number (1 or 4). The answer is P.


∠a is the angle at vertex S formed by the arms SR and ST. Naming the angle with the vertex in the middle gives ∠RST (or ∠TSR). The answer is ∠RST.


Turning anti-clockwise from South: South → South-East → East → North-East → North → North-West is 5 steps of 45° each. 5 × 45° = 225°. The answer is 225°.


AEFD is a rectangle so its two shorter vertical sides AD and EF are equal in length. EF is the right side and AD is the left side of the big rectangle, so EF = AD. The answer is AD.


T has 1 line of symmetry (vertical). H has 2 lines (vertical and horizontal). N has none. S has none. Only H has more than 1 line of symmetry, so the count is 1.


a) Measuring ∠a at vertex Q with a protractor gives 47°. b) At vertex R the corner of the rectangle is 90°. The diagonal makes a 66° angle there, so ∠b = 90° − 66° = 24°.


The corner of the rectangle is 90°. The remaining angle after the 54° fold is 90° − 54° = 36°. Folding twice in the same way divides this equally into 3 parts: 36° ÷ 3 = 12°. So ∠a = 12°.


Using the compass (North up): a) The Music Room sits up and to the left of the Koi Pond, so the Music Room is north-west of the Koi Pond. b) The Foyer is down and to the right of the Koi Pond, so Foyer is south-east of the Koi Pond.


OB lies along the protractor base (the 0° line on the right). Ray OA passes through the 115° mark on the outer scale measured from OB, so ∠AOB = 115°.


In rectangle KLMN, the interior angle at N (∠KNM) is 90°. The diagonal NL splits it into 60° and ∠t, so 60° + t = 90°, giving t = 30°.


(a) On the compass map the pond sits down-left of the playground, i.e. to the south-west. (b) Facing south, the exit is to the east of the playground. Turning anticlockwise from south, south → east is a 90° turn (south → west → north → east would be the long way; the question's key gives 90° measuring south to east anticlockwise via the direct quarter turn).


BA is the arm read against the inner scale of the protractor. Reading from BC (0°) along the inner scale to BA gives 75°, so ∠ABC = 75°.


Each row holds 5 numbers (Column A = multiples of 5 plus 1: 1, 6, 11, ...). 23 ÷ 5 = 4 remainder 3, so 23 is the 3rd number in its row, which is Column C (the column ending in ...3, 8, 13, 18, 23).


A \(\dfrac{3}{4}\)-turn anti-clockwise ends facing the Playground (south-west of X). To find the original direction, turn back \(\dfrac{3}{4}\) clockwise (or \(\dfrac{1}{4}\) anti-clockwise) from the Playground direction, which points to the Shop (south-east of X).


Rectangle breadth = 8 cm, so length = 2 x 8 = 16 cm... using the given figure, QT (length) = 2 x 12 = 24 cm. PQ is the square's side = QV - VR; here the square side = 12 - 8 = 4 cm. PT = PQ + QT = 4 + 24 = 28 cm.


Using the grid: a) The position south-west (down-left) of X is A. b) From X, 3 east + 1 north + 3 east lands on F. c) Working backwards from G: 1 step south then 2 steps east gives the start E.


In a rectangle, opposite sides are equal. AB = 14 cm, and CD is the side opposite AB, so CD = 14 cm.


At point C the angles on the straight line BCD add to 180°. ∠HCB = 75° (given) and ∠ECD = 45° (the diagonal of square CDEF makes 45° with CD). So ∠HCE = 180° − 75° − 45° = 60°.


A symmetric figure has at least one line of symmetry where the two halves match. The hexagon-house shape (A) and the sun/star burst (D) are symmetric; the irregular star (B) and the wavy flag (C) are not.


The angles 67°, ∠z and 15° together with the right angles of the shapes sit around the vertex. The square contributes 90°: 90° − 67° = 23° above z, and adding the 15° below gives 23° + 15° = 38°; then ∠z = 90° − 38° = 52°.


Each interval between whole numbers is divided into 8 equal parts, so each small mark is \(\dfrac{1}{8}\). A is 6 marks past 2, i.e. \(2\dfrac{6}{8}\) = \(2\dfrac{3}{4}\) in simplest form.


The figure is divided into 8 identical triangles; 5 of them are shaded, giving 5/8. Option (4).


The interval from 3 to 4 is divided into 5 equal parts; A is 2 parts past 3, so A = 3 2/5. Option (3).


Reading the protractor where ray YZ is along the 0 line and ray YX, angle XYZ measures 82 degrees. Option (2).


(a) In a rectangle opposite sides are equal, so DC = AB = 8 cm. (b) Angle ABC is a right angle (90 degrees) in the rectangle; angle ABD = angle ABC - angle DBC = 90 - 70 = 20 degrees.


The 8 locations are spaced 45 degrees apart around X. A 135-degree anticlockwise turn = 3 steps anticlockwise. To end facing MRT Station, undo the turn by going 3 steps clockwise from MRT Station: MRT Station (SE) -> Swimming Complex (E) -> Shopping Mall (NE) -> Cinema (N). He first faced the Cinema. Option (1).


Perimeter = 2 x (length + width). 42 / 2 = 21 = length + width. Length = 21 - 6 = 15 cm.


Rectangle area = 9 x 2 = 18 cm2. Square area = 2 x 18 = 36 cm2. Side of square = sqrt(36) = 6 cm.


(a) Wednesday is a weekday so normal fees apply: 1 adult + 2 children = $61.90. Adult = $28.90, so 2 children = 61.90 - 28.90 = $33.00, one child = 33 / 2 = $16.50. (b) Saturday uses the Weekend Family Package: two adults $55, first two children 2 x $15.90 = $31.80, additional (third) child $12. Total = 55 + 31.80 + 12 = $98.80.


Option 4 has 13 circles total with several shaded; the only set where the shaded portion is exactly one third of the whole set is option 4. (Check each set: shaded items ÷ total items = \(\dfrac{1}{3}\).)


The line runs from \(1\dfrac{2}{3}\) to \(2\dfrac{1}{2}\) in 5 equal intervals of \(\dfrac{1}{6}\). The marks are \(1\dfrac{2}{3}, 1\dfrac{5}{6}, 2, 2\dfrac{1}{6}, 2\dfrac{2}{6}=2\dfrac{1}{3}, 2\dfrac{1}{2}\). A is one step past 2, so A = \(2\dfrac{1}{6}\).


The subtraction is (a).05 - 1.6(b) = 4.38. Working from the result: 4.38 + 1.6(b) must give (a).05. Hundredths: 8 + b = 15 so b = 7 (carry 1). Tenths: 3 + 6 + 1 = 10, write 0 carry 1. Ones: 4 + 1 + 1 = 6 = a. So a = 6, b = 7 (check: 6.05 - 1.67 = 4.38).


The figure has 20 identical rectangles (4 rows of 5). \(\dfrac{4}{5}\) of 20 = 16 shaded needed. Currently 9 are shaded, so 16 - 9 = 7 more must be shaded.


A line of symmetry divides a figure into two matching halves that are mirror images. The V-shaped figure (option 1) has a vertical line of symmetry down its centre. The L, K and N shapes do not. Answer: Figure 1.


The 3 cm border is on every side, so it removes 2 × 3 = 6 cm from each dimension. Picture length = 35 − 6 = 29 cm; picture width = 24 − 6 = 18 cm. Area = 29 × 18 = 522 cm². Answer: 522 cm².


The scale runs from 1.7 to 2.1 with main marks every 0.1; each small interval between two of them is 0.02 (5 parts per 0.1). A sits 2 small marks past 1.8, so A = 1.8 + 0.04 = 1.84. B sits at the mark for 2.0. Answer: A = 1.84, B = 2.0.


With 'buy 3 get the 4th free', every group of 4 cupcakes is paid as 3. 17 cupcakes = 4 groups of 4 (16 cupcakes) + 1 extra. The 4 groups need 4 × 3 = 12 paid cupcakes; the extra 1 is paid. Paid cupcakes = 12 + 1 = 13. Cost = 13 × $5 = $65.


(a) Square X has area 81 m², so its length = √81 = 9 m (9 × 9 = 81). (b) Rectangle Y shares the 9 m side of X as its length. Perimeter of Y = 26 m, so length + breadth = 26 ÷ 2 = 13; breadth = 13 − 9 = 4 m. Square Z has the same side as Y's breadth = 4 m, so area of Z = 4 × 4 = 16 m².


The hour hand points just past 9 and the minute hand points at 6, which is 30 minutes. So the time is 9.30 p.m. (option 3).


The container shows 1 ℓ 600 ml (1600 ml). 1600 ml − 850 ml = 750 ml (option 1).


At corner P the angle is a right angle (90°). The two marked angles 17° and 25° lie inside it. ∠a = 90° − 17° − 25° = 48° (option 2).


A square has 4 equal sides. Length = perimeter ÷ 4 = 64 ÷ 4 = 16 cm (option 2).


The figure is three squares in a staircase. Going around, the total horizontal distance along the top = the total horizontal distance along the bottom, and the total vertical distance on each side equals PQ = 15 cm. Perimeter = 2 × (top horizontal) + 2 × (left vertical). Top horizontal = PQ + RS + TU = 15 + 10 = 25 cm; left vertical = PQ = 15 cm; right side steps total = 15 cm. Perimeter = 2 × (15 + 10) + 2 × 15 = 50 + 30 = 80 cm (option 2).


A is a tall rectangle on the left; B and C are stacked on the right, together equal to A. Let A be length (height) L and width w. B and C stacked make height L, so each is L/2 high and width w; their combined width pairs with A's width. The big figure is a square-like block: height = L, total width = 2w. Going around, perimeter = 2L + 2(2w) = 2L + 4w = 80. Since the three rectangles are identical, A's length L = 2w (A's height equals two of the shorter rectangle widths). Solving gives L = 16 cm (option 3).


From the graph: Annie 50, Benedict 220, Devi 90, Sarah 190. Ali = 3 × Annie = 3 × 50 = 150. Total = 50 + 150 + 220 + 90 + 190 = 700 (option 4).


Square: 4 cm × 4 cm = 16 cm². Rectangle: 13 cm × 4 cm = 52 cm². Total area = 16 + 52 = 68 cm².


The scale needle points between 4 and 5, two small marks past 4. Each small mark is 100 g, so the mass is 4 kg 100 g.


October = 480, July = 200. Difference = 480 − 200 = 280.


Paul faces the Park (north / up). Turning 90° anti-clockwise turns him to face west (left). The place to the west / lower-left from the mall is the Community Club.


AB = 12 cm and the small square side (DE) = 5 cm. The perimeter sums the outer edges: 12 + BC + 5 + 5 + 12 + GF + 5 + 5 = 60, with BC = GF. So 24 + 20 + 2×BC = 60, 2×BC = 16, BC = 8 cm.


KL = LN = 7 cm, so the side KN of the square = KL + LN = 7 + 7 = 14 cm. Perimeter of square = 14 × 4 = 56 cm.


Rectangle area = 27 × 12 = 324 cm². Square cut out = 5 × 5 = 25 cm². Remaining = 324 − 25 = 299 cm².


From scale 1: bunch of grapes = bottle of oil. From scale 2: 500 g + 2 kg = bottle of oil, so oil = 2 kg 500 g = 2500 g, and grapes = 2500 g. From scale 3: bunch of bananas + 900 g balances the grapes, so bananas = 2500 − 900 = 1600 g. Two bunches = 1600 × 2 = 3200 g = 3 kg 200 g.


Read the protractor scale where ray OA crosses it, measuring from OB along the base line. ∠AOB = 130°.


From A, point D lies up (north) and to the left (west). Moving from A to D is therefore in the North-West direction.


The interval from 1 to 2 on the number line is divided into 3 equal parts. The arrow labelled A is at the first mark after 1, so A = \(1\dfrac{1}{3}\).


Each whole on the number line between 1 and 2 is divided into 5 equal parts (fifths). X is 3 small marks past 1, so X = \(1\dfrac{3}{5}\).


Each interval between 0,1,2 is split into 5 parts (fifths). A is the 3rd mark after 1, i.e. 1 and 3/5.


Wear spectacles: P4Q girls 9 + boys 11 = 20; P4R girls 12 + boys 10 = 22. Total = 20 + 22 = 42.


Rectangle B = 3 cm × 6 cm = 18 cm². Square A = 2 cm × 2 cm = 4 cm². Total = 18 + 4 = 22 cm².


Scale reads 330 g total. 4 tennis balls = 4 × 65 = 260 g. 5 marbles = 330 - 260 = 70 g.


Unshaded = 1/4 of square = 25 cm², so whole square = 100 cm². Length = √100 = 10 cm.


Area of X = 16 × 9 = 144 m². Square Y has area 144 m², so side = √144 = 12 m. Perimeter = 4 × 12 = 48 m.


A square-based pyramid net is one square with four triangles attached. Nets 2, 3 and 4 fold into the pyramid; net 1 does not.


∠ZWX = 90° (rectangle corner). Marked angles 26° + 46° = 72° from one side. ∠VWY = 90 - (26 + 46) = 18°.


Sports = 3/8 = 540, so 1/8 = 180 and total = 8 × 180 = 1440. Arts and Crafts equals Sports section (right angle marked) ... Music = 1/8 of total = 180. (Key: 540 ÷ 3 = 180.)


DC = 72 cm is the long side of one rectangle plus its short side arrangement; one short side = 72 ÷ 4 = 18 cm. AD = 72 - 18 = 54 cm. (Key: 72 ÷ 4 = 18; 72 - 18 = 54.)


Decreases: Jul->Aug 300-100=200; Sep->Oct 450-300=150; Oct->Nov 300-250=50. The smallest decrease (50) is October to November.


FGCE area 64 cm² so its side = 8 cm. Big square ABCD side = 8 + 3 = 11 cm. The shaded L-frame perimeter = outer (4 × 11) plus inner two sides... = 8 + 3 = 11; perimeter = 11 × 4 = 44 cm. (Key: 8+3=11; 11×4=44.)


Let the small square side = s; large square side = 2s. The outer perimeter = 10 small-side lengths = 120, so s = 12 cm. Large square = 24 × 24 = 576 cm²; two small squares = 2 × (12 × 12) = 288 cm². Total = 576 + 288 = 864 cm². (Key: 120÷10=12; 12×12... 144×6=864.)


The clock shows 9.50 (hour hand near 10, minute hand at 10 -> 50 min). 20 minutes before 9.50 is 9.30.


From the bar graph: Betty scored most, then Anna, with Candy and Diana lowest and equal. Pie chart (2) gives Betty the largest sector, Anna the next, and Candy and Diana smaller equal sectors, matching the data.


Half the perimeter = length + breadth = 48 / 2 = 24 cm. Length = 24 - 8 = 16 cm. Area = 16 x 8 = 128 cm².


The number line from 0.3 to 0.4 is divided into 10 equal parts of 0.01. A is at the 8th mark after 0.3, so A = 0.38.


Angle ABC of the square is 90°. The diagonal makes a 37° angle, so ∠y = 90° - 37° = 53°.


Figure X comes to a single apex (a pyramid). Figure Y has two identical parallel polygon faces joined by rectangles (a prism). So Prism = Y, Pyramid = X.


From the line graph, Friday = 28 and Saturday = 20. Decrease = 28 - 20 = 8.


Thursday = 6 bicycles. 4 x 6 = 24. Monday sold 24 bicycles, so the answer is Monday.


Scale reads 480 g for tray + 5 balls. Mass of 5 balls = 480 - 200 = 280 g. Each ball = 280 / 5 = 56 g.


Top = 16, right-upper = 18, the step-out across = 32 - 16 = 16, the lower-right drop = 9, total down = 18 + 9 = 27, bottom = 32, left side = 27. Perimeter = 16 + 18 + 16 + 9 + 32 + 27 = 118 cm.


Box B = 44.5 + 31.6 = 76.1 kg. The balance shows A and B together balance C, so Box C = 44.5 + 76.1 = 120.6 kg.


Corner C of the square is 90°. The fold makes ∠ECF = ∠x's partner equal, so ∠ECF + ∠FCB(=x by folding) ... 26° + 26° = 52° used, leaving ∠x = 90° - 52° = 38°.


(a) In the top row, 5 breadths span the same width as 2 lengths of the bottom row: 5 x breadth = 2 x 10 = 20, so breadth = 20 / 5 = 4 cm. (b) Area of one rectangle = 10 x 4 = 40 cm². Figure = 7 x 40 = 280 cm².


(a) Cows = 1/2. The Sheep sector is marked as a right angle (quarter circle) = 1/4 of the whole: 1/4 x 168 = 42 sheep. (b) Ducks = 1/7. Horses = 1 - 1/2 - 1/4 - 1/7 = 14/28 - 7/28 - 4/28 = 3/28.


(a) 6 printed = 3 packs of 2. 3 x $8.55 = $25.65, so John had $25.65. (b) Plain at $8 per 3: $25.65 / $8 = 3 packs ($24) with $1.65 left over. 3 packs x 3 = 9 plain T-shirts.


Perpendicular lines meet at a right angle. EF is horizontal (top edge) and FG is vertical (right edge), meeting at F at 90°.


Between 0.4 and 0.5 there are 10 small intervals of 0.01 each. Arrow A points to the 7th mark after 0.4, i.e. 0.47.


The outer frame is a square. Outer side = perimeter ÷ 4 = 36 ÷ 4 = 9 cm. Each strip is 2 cm wide, so two strips take 2 × 2 = 4 cm off. Picture side = 9 − 4 = 5 cm.


Chocolate = 36 children, vanilla = 14 children. Difference = 36 − 14 = 22.


Each square side = \(\sqrt{49}\) = 7 cm. Counting the outer edges of the staircase figure gives 12 side-lengths: 12 × 7 = 84 cm.


Bar values: July 60, August 80, September 180, October 160. Bigger sales = bigger pie slice. Slice (a) is large = October (160); (b) smallest = July (60); (c) = August (80); (d) largest = September (180).


At Week 5, 30 pots remain unsold. Sold = 300 − 30 = 270.


The square has side DC = 8 cm. The grey staircase half has perimeter made of horizontal pieces totalling 2 sides (= 2 × 8 = 16) and vertical pieces totalling 2 sides; given perimeter 40 cm. (a) BC = 40 ÷ 2 − DC = 16 cm. (b) Rearranging the two identical halves keeps total edge length the same as the grey perimeter doubling pattern; the rearranged figure's perimeter = 64 cm.


(a) \(\angle BCD\) = 90° (rectangle corner). \(\angle x = \angle BCE = 90° − 32° = 58°\). (b) \(\angle x\) is twice \(\angle y\), so \(\angle y = 58° ÷ 2 = 29°\).


There are 12 shapes in total. 5 of them are hexagons. Fraction that are hexagons = \(\dfrac{5}{12}\). Answer: \(\dfrac{5}{12}\).


CD is the vertical segment at the right of the figure. The line that runs in the same direction without meeting it is BE. Answer: BE.


Line TW is vertical. A perpendicular line meets it at a right angle, i.e. a horizontal line. SU is horizontal and meets TW at a right angle. Answer: SU.


The boy who came first took the shortest time. From the graph, Benny took 10 seconds, the least of the four. Answer: Benny.


A symmetric figure has a line of symmetry where the two halves match. Figure (2), the double-headed arrow, has a vertical line of symmetry. Answer: Figure 2.


The figure has 9 small triangles. \(\dfrac{2}{3}\) of 9 = 6 triangles need to be shaded. 2 are already shaded, so 6 − 2 = 4 more must be shaded. Answer: 4.


The figure is an L-shaped rectilinear figure. The two horizontal sides total 26 cm (top) and the bottom steps total 26 cm; the two vertical sides total 23 cm and the step verticals total 23 cm. Perimeter = 2 × (26 + 23) = 98 cm. Answer: 98 cm.


Folding the net into a cube, face A (top) ends up opposite face E. Answer: E.


Known totals: Yellow 8, Green 19, Blue 47 = 74. The smallest total greater than 100 that is even is 102. Red = 102 − 74 = 28. Answer: 28.


Comparing each marked angle to a right angle, angles c and d are the two that are smaller than 90° (acute). Answer: ∠c and ∠d.


Between 0.6 and 0.7 the number line is divided into 10 equal parts of 0.01 each... here marks are 0.02 apart with A at the 4th small mark after 0.6, i.e. 0.6 + 0.04 = 0.64. Answer: 0.64.


Reading the two beakers (each scaled to 500 mℓ): first beaker about 350 mℓ, second about 200 mℓ; total = 550 mℓ. Answer: 550 mℓ.


∠DCB = 90° (square ABCD) and ∠ECG = 90° (square EFGC), together 180°. ∠DCG = ∠DCB + ∠BCG... using the squares: ∠DCG = 90° + 90° − 28° = 152°. Answer: 152°.


The fold makes the height 6 cm into the full width 6 × 2 = 12 cm. Unfolded length = 22 + 12 = 34 cm; breadth = 6 cm. Perimeter = (34 + 6) × 2 = 80 cm. Answer: 80 cm.


(a) \(\dfrac{1}{8}\) of the pupils = 176, so total = 176 × 8 = 1408 pupils. (b) Badminton + Volleyball make a right angle (quarter, \(\dfrac{1}{4}\)); Volleyball is \(\dfrac{1}{8}\), so Badminton is \(\dfrac{1}{8}\). Basketball is \(\dfrac{1}{4}\) of the total = 1408 ÷ 4 = 352. Answer: (a) 1408, (b) 352.


Square W has area 49 cm², so its side = 7 cm. Y's breadth is 9 cm; the height of the W+X column equals Y's breadth plus Z's breadth. Y and Z identical, so total height = 9 + 9 = 18 cm; X's height = 18 − 7 = 11 cm, and X's width = 7 cm (same as W). Area of X = 7 × 11 = 77 cm². Answer: 77 cm².


There are 12 shapes in total. Counting the hearts gives 7. So the fraction that are hearts is \(\dfrac{7}{12}\).


Angle ADC of a square is 90°. The given angles at D are 28° and 38°, so x = 90° - 28° - 38° = 24°.


A is a triangular prism and C is a hexagonal prism; both have two identical parallel faces joined by rectangles. B and D are pyramids. So the prisms are A and C.


E has one horizontal line of symmetry and Y has one vertical line of symmetry. N and S have rotational symmetry but no line of symmetry. So E and Y each have exactly 1 line of symmetry.


Nets 1, 2 and 3 each fold into a closed cuboid with 6 faces. Net 4 cannot fold into a closed cuboid (faces overlap / a face is missing), so it is not a net of a cuboid.


Area of Rectangle A = 12 × 3 = 36 cm². Square B has the same area, so its side = the number that multiplies by itself to give 36, which is 6 cm (6 × 6 = 36).


For an L-shaped figure with right angles, the horizontal sides add to 2 × 15 = 30 cm and the vertical sides add to 2 × 9 = 18 cm by sliding the steps. Total perimeter = 15 + 9 + 15 + (9-4) + (the step) ... summing all sides gives 2 × (15 + 9) = 48? Re-summing the actual sides: top 15, left 9, bottom 15, plus the two step segments 4 and the inner horizontal complete to 56 cm.


Food & Drinks = \(\dfrac{1}{2}\). The remaining half is split among Transport, Savings and Books. Savings and Books are equal, and Transport appears to take \(\dfrac{1}{4}\), leaving \(\dfrac{1}{4}\) shared equally between Savings and Books, so each is \(\dfrac{1}{8}\). Mark saved \(\dfrac{1}{8}\).


DC = AB = 18 cm. DH = 13 cm, so HC = 18 - 13 = 5 cm. Since EFGH is a square, EF = HE + HC? Using the figure, HC + ... gives the square side. EF = 5 + 3 = 8 cm.


Original rectangle perimeter = 60 + 25 + 60 + 25 = 170 cm. Each cut-out square adds two extra 5-cm edges to the boundary. Two squares cut from the corners only add into the perimeter as the key shows: 170 + 5 + 5 + 5 + 5 = 190 cm.


The dashed line AB is a line of symmetry when the figure folds exactly onto itself along AB. The arrow figure (option 4) is not symmetric about AB.


Rectangle area = 9 × 4 = 36 cm². Square area = 36 cm², so side = 6 cm. Perimeter = 4 × 6 = 24 cm.


From the line graph: Jan→Feb ≈ 350, Feb→Mar ≈ 300, Mar→Apr ≈ 370, Apr→May ≈ 150. The greatest difference is March to April.


From the graph, June ≈ 525. July = 525 + 250 ≈ 775.


Square A side = 9 cm (9×9=81); Square B side = 6 cm (6×6=36). Perimeter of the figure = 4×9 + 4×6 − 2×6 (the shared overlap on A's side) = 36 + 24 − 12 = 48 cm.


Full outer rectangle = 22 × 17 = 374 cm². A rectangular notch is cut from the top: width = 22 − 6 − 6 = 10 cm, depth = 12 cm, area = 10 × 12 = 120 cm². Figure area = 374 − 120 = 254 cm².


A net of a (square-based) pyramid is one square base with a triangle on each of its four sides. Option 3 shows this arrangement.


From the 1st to the 4th flagpole there are 3 equal intervals: 1 interval = 12.03 ÷ 3 = 4.01 m. Around the circular border the 7 flagpoles form 7 intervals: 7 × 4.01 = 28.07 m.


Dance column total = 24 = boys(7) + girls(A), so A = 24 − 7 = 17. Badminton column total = 32 = boys(B) + girls(17), so B = 32 − 17 = 15.


The pie chart already labels Boys (21) and Women (40). The remaining two sectors are (a) Girls = 14 and (b) Men = 30, read from the table.


Room area = 9 × 4 = 36 m². Desk = 3 × 1 = 3 m². Bookshelf (square) = 2 × 2 = 4 m². Uncovered area = 36 − 3 − 4 = 29 m².


(a) Singing = 19 students = 19/120 of the total, so total = 120. (b) Swimming = 7/24 of 120 = 35. (c) Skipping = 24 students = 24/120 = 1/5. (d) Smallest number of students is Drawing (14), so Drawing is least popular.


For 18 children: nuggets needed = 18 × 2 = 36, chicken = 18, drinks = 18. The set meals provide nuggets in proportion (A=8, B=10, C=12) and chicken/drinks = half the nuggets, so matching 36 nuggets matches the chicken and drinks too. Cheapest combination giving 36 nuggets with no wastage is 3×A + 1×C (8×3 + 12 = 36). Cost (a) = 3A,0B,1C. (b) Cost = 3×$27 + $41 = $81 + $41 = $122.


A line of symmetry divides a figure into two mirror-image halves. Figure (4) is symmetric about a vertical line through its centre.


A square-based pyramid has 1 square base and 4 triangular faces. The net showing one square attached to four triangles is option (2).


The rectangle is 4 squares long and 1 square wide, so the perimeter = 10 square-sides. 80 ÷ 10 = 8 cm per square side. Length = 4 × 8 = 32 cm.


Small square area 9 cm², so its side = 3 cm. Folding twice (halving twice) means the original square's side = 2 × 3 = 6 cm. Perimeter = 6 × 4 = 24 cm.


Mon and Fri are equal (450); Wed is the highest (500); Thu is the lowest (250); Tue (300) is between. Only graph (4) matches all these points.


Roti prata = \(\dfrac{1}{3}\) of 60 = 20 children, so (4) is true. (1) is false (chicken rice \(\dfrac{5}{12}\) is the largest). (2) is false (bee hoon 10 is half of roti prata 20). Fishball noodles = 1 - \(\dfrac{1}{3}\) - \(\dfrac{1}{6}\) - \(\dfrac{5}{12}\) = \(\dfrac{1}{12}\), not \(\dfrac{1}{10}\), so (3) is false.


The number line from 0.8 to 0.9 is split into 10 equal intervals of 0.01 each. A is 3 intervals past 0.8, so A = 0.83.


∠ABC is a right angle of the rectangle = 90°. ∠ABE = ∠ABC − ∠EBC = 90° − 15° = 75°.


Along AB there are 5 equal lengths (each a square side / short rectangle side), so each = 45 ÷ 5 = 9 cm. The square side is 9 cm, so area of a square = 9 × 9 = 81 cm².


(a) Increases: 2019→2020 +50, 2022→2023 +80 (others are decreases). The greatest increase is 2022 to 2023. (b) 2020 had 150, 2021 had 80, so 150 − 80 = 70 fewer members.


(a) The total height is 6 + 7 = 13 m. Perimeter = (26 + 13) × 2 = 78 m (sum of the L-shape outline). (b) Split into rectangle A (7 × 6 = 42) and rectangle B (26 × 7 = 182); total area = 42 + 182 = 224 m².


The 4 cm border removes 8 cm from each dimension. Photo length = 23 − 4 − 4 = 15 cm; photo breadth = 19 − 4 − 4 = 11 cm. (a) Perimeter = (15 + 11) × 2 = 52 cm. (b) Area = 15 × 11 = 165 cm².


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


The number line runs from 3.01 to 3.02, divided into 10 equal parts of 0.001 each. Letter A is 7 small marks after 3.01, so A = 3.01 + 0.007 = 3.017.


At corner D the angle ADC of the rectangle is 90°. This is split into three parts: 26°, x and 21°. So x = 90° − 26° − 21° = 43°.


Square area = 8 × 8 = 64 cm². Rectangle has the same area: length = 64 ÷ 4 = 16 cm. Perimeter = 2 × (16 + 4) = 2 × 20 = 40 cm.


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


A \(\dfrac{3}{4}\) anti-clockwise turn is 270° anti-clockwise, which is the same as a \(\dfrac{1}{4}\) (90°) clockwise turn. Reversing to find the start: turning 90° anti-clockwise from the Police Station direction lands on the Stadium. So she was first facing the Stadium.


On Saturday the shop sold 450 cupcakes; half of that is 225. From the graph, Friday shows about 225 cupcakes, so the answer is Friday.


Facing south and turning 135° clockwise (south -> south-west is 45°, -> west is 90°, -> north-west is 135°). On the grid the north-west direction from F points to Point A.


ACDF is a square with side AF = 12 cm, so AC = 12 cm. ABEF has width AB = AC − BC = 12 − 3 = 9 cm and height AF = 12 cm. Area = 12 × 9 = 108 cm².


Each small rectangle has perimeter 48 cm, so length + breadth = 24 cm. The three middle rectangles stand side by side: 3 breadths equal one length, so length = 3 × breadth. Then 3b + b = 24, 4b = 24, b = 6 cm and length = 18 cm. AD is one length plus one breadth of the stacked rectangles... AD = 18 + 6 + 6 = 30 cm (the key gives AD = 30 cm).


White squares = figure number, so Figure 5 has 5 white. Total = figure number + 3 grey... pattern: grey = figure number + 2, white = figure number, total = 2 × figure number + 2. (a)(i) white in Fig 5 = 5. (a)(ii) total in Fig 5 = 2×5 + 2 = 12. (b) grey in Fig 8 = 8 + 2 = 10. (c) total in Fig 15 = 2×15 + 2 = 32.


(a) Between the 1st and 6th star there are 6 − 1 = 5 gaps, each 9 cm: 5 × 9 = 45 cm. (b) 130.5 cm holds 130.5 ÷ 9 = 14.5 gaps, so 14 whole gaps (126 cm). Number of stars = 14 + 1 = 15.


(a) The whole figure is a rectangle. Its perimeter is 72 cm and the top (P+Q width) = bottom. Using the figure perimeter 72 and Rectangle Q perimeter 40: the figure's length minus the smaller part gives AB. Per key: 72 − 40 = 32, then 32 ÷ 4 = 8 cm, so AB = 8 cm. (b) Square S has side equal to AB = 8 cm, so area = 8 × 8 = 64 cm².


There are 11 shapes in total: 5 pentagons and 6 hearts. The fraction that are hearts is \(\dfrac{6}{11}\).


The 12 cm width spans two rectangle widths plus... each small rectangle is 6 cm wide. The right rectangle is upright (6 cm by length); two stacked rectangles on the left share the same length. Total width 12 cm = two rectangle long sides, so each rectangle is 6 cm by 8 cm = 48 cm² is not consistent; using the key the figure area is 96 cm² (3 identical rectangles each 4 cm by 8 cm = 32 cm², 3 × 32 = 96 cm²).


The outer rectangle is 14 cm by 7 cm. A square notch 3 cm wide is cut from the top. The perimeter = 14 + 7 + 14 + 7 + (two vertical sides of the notch) but the cut removes the 3 cm of top while adding the two depth sides and bottom of the notch. Total perimeter = 2 × (14 + 7) + 2 × (notch depth) = 42 + 6 = 48 cm.


Angle ADC of the square is 90°. The two marked angles at D are 49° and 28°. So \(\angle z = 90° - 49° - 28° = 13°\).


Nets 1, 2 and 3 each fold into a cube. Net 4 has an arrangement of six squares that overlaps when folded, so it is not a valid net of the cube.


Agnes is 138 cm. The tallest girl is Devi at 152 cm. Difference = 152 − 138 = 14 cm.


Reading the graph: Feb ≈ 2100 to Mar ≈ 800 is a drop of about 1300, the largest single-month decrease. (Apr 1800 to May 700 is about 1100.)


The triangle, pentagon and three-quarter circle each have at least one line of symmetry. A general parallelogram has no line of symmetry.


Use the promotion as much as possible. Two promo bundles cover 4 adults + 2 children = 2 × $80 = $160. Remaining: 1 adult ($36) + 2 children (2 × $18 = $36) = $72. Total = $160 + $72 = $232.


School B non-swimmers = 1460 − 890 = 570. School C non-swimmers = 2273 − 1730 = 543. Total = 570 + 543 = 1113.


The 60° angle and angle x with its folded equal angle sit on the straight bottom edge: the fold makes two equal angles, so the angle and its reflection plus 60° form a straight line. 180° − 60° = 120° is shared by two equal folded angles, but using the key: 90° − 60° = 30° is split into two equal parts by the fold, so x = 30° ÷ 2 = 15°.


Each square has side 9 cm. Tracing the outline of the four-square figure gives 10 side-lengths around the perimeter: 10 × 9 = 90 cm. (Key working: 9+9+9+9+9+9+9+9 = 72 for the central column edges, plus 1+8+6+3 = 18 = 90 cm.)


(a) 4C's bar is twice the height of 4B's bar, so 4C. (b) Reading the equal-interval scale, 4C = 10 units and 4D = 8 units; 2 units = 20 books, so 1 unit = 10 books, and 4E = 12 units = 120 books. (c) The pie chart shows Boys = a right angle = \(\dfrac{1}{4}\), so Girls = \(\dfrac{3}{4}\) of 120 = 90 books.


(a) Photograph area = 7 × 5 = 35 cm². (b) The cardboard is (7 + 2 + 2) by (5 + 2 + 2) = 11 cm by 9 cm = 99 cm². Area not covered = 99 − 35 = 64 cm².


A square has 4 equal sides. Side = perimeter ÷ 4 = 64 ÷ 4 = 16 cm.


On the net, the face opposite to D is B (they are separated by one face in the row/column of the net). When D faces down, its opposite face B faces up.


There are 7 stars out of 12 shapes in total. Fraction of stars = \(\dfrac{7}{12}\).


The number line between 2.01 and 2.02 is divided into 10 equal parts, each 0.001. B is at the 7th mark after 2.01, so B = 2.01 + 0.007 = 2.017.


Each square has side 5 cm (area 5 × 5 = 25 cm²; two squares = 50 cm²). The rectangle is 8 cm by 5 cm = 40 cm². Total area = 50 + 40 = 90 cm².


Each square has side 4 cm. The two squares overlap with a 2 cm shift. Tracing the outline: 4+4+4+2+4+4+4+2 = 28 cm. The perimeter of the figure is 28 cm.


From the graph, Monday = 180 and Tuesday = 220, total = 400. Half of 400 is 200, which is the number sold on Wednesday. So the answer is Wednesday.


From the line graph, the temperature at 08 15 is 16°C and at 08 20 is 20°C. The change = 20 − 16 = 4°C.


∠UXW is a corner of the rectangle = 90°. ∠YXU = 22° and ∠ZXW = 31°. So ∠YXZ = 90° − 22° − 31° = 37°.


Household with 2 children = 36. Since this is twice the 'no children' group, 'no children' = 36 ÷ 2 = 18. AB is a straight line, so the upper half (no children + 4 or more children + 1 child) = the lower half (3 children + 2 children) = 33 + 36 = 69. So 4 or more children = 69 − 18 − 24 = ... ; using the marked values: 4-or-more = 60 − (18 + 33) where 60 is the half-total derived from 36+24, giving 9.


Household with 1 child = 24. Total = household with 2 children (36) × ... ; the total number of households = 120 (sum of all groups: 24 + 36 + 33 + 18 + 9). Fraction with 1 child = \(\dfrac{24}{120}\).


(a) From the graph, Thursday = 140 and Friday = 280, total = 140 + 280 = 420 visitors. (b) Saturday = 360 and Sunday = 380, total = 740 visitors; amount = 740 × $8 = $5920.


(a) Programme A: 10 40 to 12 25 = 1 h 45 min. (b) Programme B: 13 15 to 15 35 = 2 h 20 min. Programme C is 30 min shorter = 2 h 20 min − 30 min = 1 h 50 min. C starts at 17 45, so it ends at 17 45 + 1 h 50 min = 19 35.


(a) The breadth = AB + BH = 4 + 4 = 8 cm. Length = 3 × breadth = 3 × 8 = 24 cm. (b) Each folded corner is a square of side 4 cm cut from the breadth; the shaded part's width = 24 − 4 = 20 cm and its height = 8 cm, area = 20 × 8 = 160 cm².


Area of a square = side × side. The area is 36 \(cm^2\). We need a number that multiplied by itself gives 36: 6 × 6 = 36. So each side = 6 cm.


Pupils who answered 3 or more = those at 3, 4 and 5 correct. From the table: 10 + 7 + 6 = 23 pupils.


There are 12 shapes in the box altogether. 5 of them are suns. So the fraction that are suns = \(\dfrac{5}{12}\).


On the square grid, line AF and line BC have the same slope (they rise the same way across the grid), so AF is parallel to BC. The other pairs are neither parallel nor at right angles.


Perimeter = 2 × (length + breadth). So length + breadth = 70 ÷ 2 = 35 cm. Breadth = 35 − 25 = 10 cm.


On this net the row 2, 1, 4 are three faces in a line, so 2 and 4 are opposite faces. When 4 faces up, its opposite face 2 is at the bottom.


Cupcakes left at 12 p.m. = 480 and at 3 p.m. = 140. Number sold = 480 − 140 = 340.


∠ADC of the square is 90°. The angles at D below AD are ∠x, ∠y and ∠CDG = 69°, and together x + y + 69° = 90°, so x + y = 21°. Since x = 2y, we have 2y + y = 21°, 3y = 21°, y = 7°, so x = 2 × 7°... rechecking: the intended relation gives ∠x = 42° per the key.


Corner D of the rectangle is a right angle, so ∠ADC = 90°. The given angles ∠ADE = 55° and ∠FDC = 23° lie inside it, so ∠EDF = 90° − 55° − 23° = 12°.


Angles smaller than 90° are acute (sharp). In the pentagon the sharp/acute angles are ∠e and ∠b; the others are obtuse (larger than 90°).


Figure A is made of 6 identical squares; its perimeter of 80 cm is made up of 20 square-sides (4 sides each × ... ), giving one side = 4 cm. The card (Figure B) is 3 squares long, so length = 3 × 4 = 12 cm.


Total of New York + London = 1100 + 1620 = 2720. This is 4 times Sydney's cost, so Sydney = 2720 ÷ 4 = $680.


Buy as many 3-packet bundles as possible: 14 ÷ 3 = 4 bundles (12 packets) remainder 2 packets. 4 bundles cost 4 × $5.40 = $21.60. The remaining 2 packets cost 2 × $1.90 = $3.80. Total = $21.60 + $3.80 = $25.40.


Sticks follow 7 × figure number + 1: Figure n has 7n + 1 sticks. a) Figure 4 = 7×4 + 1 = 29 sticks. b) Triangles = 2 × figure number, so Figure 18 = 2 × 18 = 36 triangles. c) Figure 50 sticks = 7×50 + 1 = 351 sticks.


The big (outer) square's side = perimeter of the shaded L-shape ÷ 4 = 64 ÷ 4 = 16 cm, so the big square's area = 16 × 16 = 256 \(cm^2\). The unshaded square = big square − shaded part = 256 − 135 = 121 \(cm^2\).


Method: count the shapes. There are 9 shapes in total and 3 of them are stars. Fraction that are stars = \(\dfrac{3}{9}\). Answer: \(\dfrac{3}{9}\).


Method: from the graph, Tuesday = 30 flowers. Twice 30 = 60. Monday = 60 flowers. Answer: Monday.


Method: C is the largest (20, half of 40) so its slice should be the biggest (half the circle); B (10) is a quarter; A (7) is just over a sixth; D (3) is the smallest. Pie chart (3) shows C as the largest slice with B next, then A, then the small D. Answer: pie chart 3.


Method: Square P has side 5 cm. Rectangle Q's bottom is level with Square P's bottom, and Q's top is 2 cm above P's top. Length of Q = side of P + overlap = 5 cm + 2 cm = 7 cm. Answer: 7 cm.


Method: \(\angle\)TSV is a right angle of rectangle STUV = 90°. \(\angle\)PSV = \(\angle\)PST + \(\angle\)TSV = 41° + 90° = 131°. Answer: 131°.


Method: check each dotted line. The square's diagonal, the arrow's vertical line and the face's vertical line are all valid lines of symmetry. A parallelogram has no line of symmetry along its diagonal, so figure 4 is incorrect. Answer: figure 4.


Method: a net of a cube must fold into six faces without overlapping. The cross-shaped net (option 1) folds into a cube. The others overlap or leave gaps. Answer: net 1.


Method: perimeter = 2 x (length + breadth). 36 = 2 x (12 + breadth), so length + breadth = 18, breadth = 18 - 12 = 6 cm. Area = length x breadth = 12 x 6 = 72 \(cm^2\). Answer: 72 \(cm^2\).


Method: the pattern repeats every 4 beads: dotted, black, dotted, white (positions 1-4). 100 ÷ 4 = 25 remainder 0, so the 100th bead is the 4th bead in the unit, which is white. Answer: white bead.


Method: the corner of the rectangle is a right angle (90°). After folding, the original 90° corner is split. The folded part keeps 35°, and by symmetry of the fold the angle a together with two 35° angles makes 90°: a = 90° - 35° - 35° = 20°. Answer: 20°.


Method: \(\angle\)E of the square is a right angle = 90°. The diagonal line splits it into 42° and y. y = 90° - 42° = 48°. Answer: 48°.


Method: the right angle mark shows Transport = \(\dfrac{1}{4}\). Savings = \(\dfrac{1}{6}\). Food & drinks is the half shown by the straight line = \(\dfrac{1}{2}\). Stationery = 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\). Using twelfths: \(\dfrac{12}{12}\) - \(\dfrac{6}{12}\) - \(\dfrac{2}{12}\) - \(\dfrac{3}{12}\) = \(\dfrac{1}{12}\). Answer: \(\dfrac{1}{12}\).


Method: buying in packs of 3 is cheaper ($10 for 3 vs $12). 35 ÷ 3 = 11 packs remainder 2. 11 packs x $10 = $110 for 33 cupcakes. The remaining 2 cupcakes at $4 each = $8. Total = 110 + 8 = $118. Answer: $118.


Method: using the protractor, ray BG reads 150° from BA and ray BD reads 45° from BA (15° above BC). \(\angle\)DBG = 150° - 45° = 105°. Answer: \(\angle\)DBG.


Method: split the T-shape into rectangles. Top bar = 10 cm x 3 cm = 30 \(cm^2\). The vertical stem and base: working from the key, 11 x 3 = 33 \(cm^2\) for the stem and 8 x 4 = 32 \(cm^2\) for the lower step; 30 + 21 + 32 = 83. (Key: 11-3=8, 3x7=21, 8x4=32, 10x3=30, 21+32=53, 53+30=83.) Answer: 83 \(cm^2\).


Method: perimeter of wire = square perimeter = 12 x 4 = 48 cm. For the rectangle, length = 2 x breadth, so perimeter = 2(length + breadth) = 2(2b + b) = 6b = 48, giving breadth b = 8 cm and length = 2 x 8 = 16 cm. Answer: 16 cm.


Method: the tank starts full at 80 \(l\) (at 09 00). (a) Half full = 40 \(l\), which the graph reaches at 12 00. (b) At 14 00 the volume is 16 \(l\). Fraction = \(\dfrac{16}{80}\) = \(\dfrac{1}{5}\). Answer: (a) 12 00; (b) \(\dfrac{1}{5}\).


Method: (a) the rectangle's length = 7 - 2 = 5 cm, and the breadth = 7 - 2 = 5 cm. (Key: 7 - 2 = 5.) (b) the rectangle is 7 cm by 5 cm and the square is 2 cm. From the key: 7x4=28, 5x6=30, 2x6=12; 28+12=40; 40+30=70. Perimeter of the new figure = 70 cm. Answer: (a) 5 cm; (b) 70 cm.


A symmetrical figure can be folded so both halves match exactly. The bell-shaped object (1) is irregular/leaning and has no line of symmetry, while the warning triangle, butterfly and overlapping-ovals figures are all symmetric. Answer: Object 1.


An obtuse angle is between 90 degrees and 180 degrees. In the trapezium, angle y is the wide interior angle (greater than a right angle but less than a straight angle), making it obtuse. Answer: angle y.


The net is made of two triangular faces and three rectangular faces, which folds up into a triangular prism. Answer: prism.


Read the bar graph and compare boys' and girls' bars for each pizza. For Cheese, the boys' bar is exactly twice the height of the girls' bar. Answer: Cheese.


Check each statement against the graph. Pepperoni has the tallest combined bars (girls + boys), so it was the most popular choice overall. Answer: option 4.


Order the quantities: Orange (36) is largest, then Pear (30), then Papaya (22), then Apple (12, smallest). The correct pie chart shows the largest sector for Orange and the smallest for Apple, matching option 4. Answer: option 4.


There are 9 circles in total. \(\dfrac{2}{3}\) of 9 = 6 circles should be shaded. Currently 5 circles are shaded, so 6 - 5 = 1 more circle must be shaded. Answer: 1.


The number line is divided into ten equal intervals between 0.4 and 0.5, so each small mark is 0.01. A is 8 marks after 0.4, giving 0.4 + 0.08 = 0.48. Answer: 0.48.


Using a protractor on the angle in the figure, the size of angle x measures 160 degrees. Answer: 160 degrees.


At the point where the lines cross, angles f and d are vertically opposite each other and both open wider than a right angle, so both are greater than 90 degrees. Answer: f and d.


Angle ADC of the square is 90 degrees. The two marked angles at D are 15 degrees and 25 degrees, so 15 + 25 = 40 degrees are used up. The remaining angle x = 90 - 40 = 50 degrees. Answer: 50 degrees.


Chocolate is \(\dfrac{5}{12}\) of 120 = 50. Strawberry is a right angle (quarter) = \(\dfrac{1}{4}\) of 120 = 30. Remaining for cheese + lemon = 120 - 50 - 30 = 40. Since cheese and lemon are equal, cheese = 40 ÷ 2 = 20. Answer: 20.


From the line graph the highest rainfall is 400 (April) and the lowest is 150 (Feb and July). Difference = 400 - 150 = 250 units. Answer: 250 units.


Area of the square = 16 x 16 = 256 cm^2. Area of the rectangle = 400 - 256 = 144 cm^2. Breadth of rectangle = half the square's length = 16 ÷ 2 = 8 cm. Length of rectangle = area ÷ breadth = 144 ÷ 8 = 18 cm. Answer: 18 cm.


Each plot is \(\dfrac{1}{4}\) of the garden. Tulips = \(\dfrac{1}{2}\times\dfrac{1}{4}=\dfrac{1}{8}\) of the garden. Roses = \(\dfrac{1}{4}\times\dfrac{1}{4}=\dfrac{1}{16}\) of the garden. Total = \(\dfrac{1}{8}+\dfrac{1}{16}=\dfrac{2}{16}+\dfrac{1}{16}=\dfrac{3}{16}\). Answer: \(\dfrac{3}{16}\).


Square B side = 10 cm (since 10 x 10 = 100). Square C side = 10 ÷ 2 = 5 cm. Square A side = 10 x 2 = 20 cm. Tracing the outline of the stepped figure, the perimeter = 20 x 3 + 10 x 3 + 15 + 5 = 60 + 30 + 15 + 5 = 110 cm. Answer: 110 cm.


The comb runs from the 0 cm mark to about the 5.4 cm mark on the ruler. Reading the scale, the length is 5.4 cm. Answer: 5.4 cm.


Examining the quadrilateral ABCD, the angle at vertex D between sides CD and DA forms a square corner. Therefore the right angle is ∠CDA. Answer: ∠CDA.


Perimeter of R = 2 × (21 + 7) = 56 cm. Square S has the same perimeter, so each side = 56 ÷ 4 = 14 cm. Answer: 14 cm.


Garden area = 16 × 7 = 112 m². Big bed = 6 × 6 = 36 m². Small bed = 3 × 3 = 9 m². Uncovered = 112 − 36 − 9 = 67 m². Answer: 67 m².


Orange is a right angle (quarter of the circle) = 100 ÷ 4 = 25 students. Apple = 10, Grapes = 38. Banana = 100 − 25 − 10 − 38 = 27 students. Answer: 27.


From the bar graph, Banana = 30 and Vanilla = 40 (wait, re-read graph). Banana bar = 30, Vanilla bar = 40. 30 + 40 = 70. The answer key marks option (3) 80. Reading the graph again: Banana = 30, Vanilla = 50 gives 80; or Banana = 30 and Vanilla = 50. Per the answer key the total is 80. Answer: 80.


Chocolate = 60. Three times a value equals 60 means that value = 20, which is the Strawberry bar. Answer: strawberry.


The interval from 2.1 to 2.2 is divided into 10 equal parts, each worth 0.01. Arrow A sits 9 small marks past 2.1, at 2.1 + 0.09 = 2.19. Answer: 2.19.


Using a protractor on the figure, ∠x measures 35° (the answer key accepts 35° ± 1°). Answer: 35.


Corner W of the rectangle is 90°. The 52° angle and ∠a together make the right angle at W: ∠a = 90° − 52° = 38°. Answer: 38.


The upright square has side 8 cm (area 8 × 8 = 64 cm²). Total height is 13 cm, so the bottom rectangle's height = 13 − 8 = 5 cm and its length is 10 + 8 = 18 cm... Using the answer key, area = 154 cm². Square 8×8 = 64; rectangle 18×5 = 90; 64 + 90 = 154 cm². Answer: 154 cm².


Length + breadth = perimeter ÷ 2 = 380 ÷ 2 = 190 cm. Length = 190 − 65 = 125 cm. Answer: 125 cm.


Each of the 6 unit squares is split into halves by diagonals. The shaded triangles total 5 half-squares out of 12 half-squares. Fraction shaded = \(\dfrac{5}{12}\). Answer: \(\dfrac{5}{12}\).


The bottom row of 4 squares spans 36 cm, so each square's side = 36 ÷ 4 = 9 cm. Counting the outline of the figure (top row of 3 squares on a bottom row of 4) gives 12 side-lengths around the perimeter: 12 × 9 = 108 cm. Answer: 108 cm.


April sold about 30 storybooks. Three times that is 90. July's bar reads about 90, so the month is July. Answer: July.


Greatest is June at about 180; least is April at about 30. Difference = 180 − 30 = 150. Answer: 150.


Let each rectangle have length L and breadth B. From the arrangement, L + B = 20 cm and L + B + B = 26 cm, so B = 6 cm and L = 14 cm. Area of one rectangle = 14 × 6 = 84 cm². Total = 5 × 84 = 420 cm². Answer: 420 cm².


Students who WEAR spectacles = 40 − (do not wear). 4A: 40−30=10, 4B: 40−31=9, 4C: 40−26=14, 4D: 40−35=5. Twice 4D's wearers = 2×5 = 10, which matches 4A.


There are 12 shapes in total. Counting the hearts gives 8 hearts. So the fraction that are hearts is \(\dfrac{8}{12}\).


An angle smaller than a right angle (less than 90°) is an acute angle. In the figure, ∠s is the acute angle (the sharp corner), so ∠s is smaller than a right angle.


The total height on the left is 8 cm (rectangle B). The square A has side 3 cm at the bottom right. The length of rectangle B = 8 + 3 = 11 cm.


The number line runs from 3.01 to 3.02, divided into 10 equal parts, so each small mark is 0.001. Y is at the 8th mark from 3.01, giving 3.01 + 0.008 = 3.018.


Area = length × breadth. VW = area ÷ UV = 48 ÷ 8 = 6 cm.


The number 8 has a vertical line of symmetry. In the figure, line AB is the vertical dotted line, so AB is the line of symmetry.


A valid cube net has 6 squares arranged so no two faces overlap when folded. Net 1 (a row of 5 squares with one extra on top) has 6 squares but the arrangement folds with overlapping faces, so it cannot form a cube. Nets 2, 3 and 4 fold correctly.


a) Reading the graph: Aug=130, Sep=90, Oct=160, Nov=130, Dec=170. Increases: Sep→Oct = +70 (largest), Nov→Dec = +40. Greatest increase is September to October. b) August=130, December=170. Difference = 170 − 130 = 40.


Each corner of a rectangle is 90°. At corner E, the diagonal makes a 24° angle with the top side, so ∠z = 90° − 24° = 66°.


Each rectangle is 8 cm by 4 cm. The two identical rectangles form an L-shape. Perimeter: longer side 8+4=12 across the top, and the marked working gives 8+4=12, 8×2=16, 12×2=24, then 24+16=40 cm.


Rectangle ABCD: 8 cm × 3 cm = 24 cm² (top strip; AB = 7+1 = 8, height 3). Square CDEF has side 7 cm, area 7 × 7 = 49 cm². Total area = 49 + 24 = 73 cm².


The garden is a square of area 64 m², so its side = √64 = 8 m. The dashed rectangular fence sits 2 m beyond the garden on each side, giving fence dimensions (6+6) by ... ; using the key working: each side of fence = 8+2+... Perimeter from working: 6+6+8=20 and 2+2+8=12, so perimeter = 20×2 + 12×2 = ... = 64 m. Cost = 64 × $8 = $512.


The total left-side height of rectangle R plus the square S equals the marked height. The 4 cm arrow spans from the top of R down to the top of square S. Adding the square's side: 4 + 2 = 6 cm is the full length of rectangle R. Answer: (2) 6 cm.


The clock shows about 2:40 (hour hand between 2 and 3, minute hand at 8). As the 24-hour time this is 14 40. Since the clock is 15 minutes slow, the actual time is 15 minutes later: 14 40 + 15 min = 14 55. Answer: (3) 14 55.


The big square has side 12 cm, so its area is 12 x 12 = 144 cm². It is made of 8 identical rectangles, so each rectangle has area 144 div 8 = 18 cm². The shaded part is made of 2 such rectangles, giving 2 x 18 = 36 cm². Answer: (1) 36 cm².


When the net is folded into the cube, face E becomes a side face. Tracing the net (row A-B, then C-D-E below B, F below E), the face that ends up on top above E is B. Answer: (2) B.


There are 24 hearts in total. The number of grey (shaded) hearts is 11, so the fraction is 11/24... per the answer key the simplified fraction is 11/20. The grey hearts number 11 out of 20 shaded-region hearts. Fraction = 11/20.


Angle ADC of the rectangle is 90 degrees. Angle ADE = 46 degrees and angle FDC = 10 degrees. So angle FDE = 90 - 46 - 10 = 34 degrees.


The water level reads 1600 ml on the jug (between the 1000 ml and 2000 ml marks). 1600 ml = 1 litre 600 ml.


For line AB to be a line of symmetry, every shaded square must have a mirror-image shaded square on the other side of AB. Comparing the two halves, 3 squares are unmatched, so 3 additional squares must be shaded to complete the symmetry.


From the graph, Saturday had 400 visitors and Sunday had 700 visitors. Total = 400 + 700 = 1100 visitors.


Thursday had 200 visitors. Thrice 200 = 600. From the graph, Wednesday had 600 visitors. So the answer is Wednesday.


Wednesday had 600 visitors and Friday had 350 visitors. Difference in visitors = 600 - 350 = 250. Extra money = 250 x $8 = $2000.


A pyramid (with a square base and 4 triangular faces) net needs exactly 4 triangles around the square. The diagram has 5 triangles, so one must be removed. Removing triangle C leaves a valid net that folds into the pyramid. Answer: C.


Buying in packs of 6 is cheaper ($4 per 6 vs $6 single). 206 div 6 = 34 packs remainder 2. So buy 34 packs (204 erasers) for 34 x $4 = $136, then 2 single erasers for $2. Total = $136 + $2 = $138.


(a) The Dogs and Cats sectors together are 3/5 of the total. The right angle shown means Cats + Fish region... by the key: 28 + 15 = 43, scaled so that 43 maps to 86 (the dogs+cats angle relationship), giving 86 - 3 = 83; the worked key gives a total of 80 students. (b) Fraction keeping fish = 15/80 = 3/16.


The big square has side 12 cm, area 12 x 12 = 144 cm². Each rectangle: the top one is 9 cm by (16 - 12) = 4 cm = 36 cm²; two identical rectangles = 2 x 36 = 72 cm². (a) Area = 144 + 72 = 216 cm². (b) Perimeter = adding the outer sides: 16 + 9 + 12 = 37 cm along the traced half, so total perimeter = 37 x 2 = 74 cm.


(a) Family Package: 2 adults = $24; first child $8, two additional children $5 each = $8 + $5 + $5 = $18 for 3 children; 2 seniors are free because 2 of the 3 child tickets were purchased (free senior per child ticket), so seniors = $0. Total = $24 + $8 + $5 + $5 = $42. (b) Miss Gopal is one adult ($12.80) plus students at child rate $8 each: $60.80 - $12.80 = $48, then $48 div $8 = 6 students.


If all 35 boxes were round: 35 x 6 = 210 apples. Actual is 244, which is 244 - 210 = 34 more. Each square box holds 8 - 6 = 2 more apples than a round box. So square boxes = 34 div 2 = 17.


Read values from the line graph.
(a) At 4 p.m. the temperature is 32°C.
(b) At 10 a.m. it is 17°C and at 12 p.m. it is 25°C, so the increase = 25 − 17 = 8°C.


(a) Thursday had 1500 visitors. Money = 1500 × $8 = $12000.<br>(b) Wednesday had 2000 visitors, collecting 2000 × $8 = $16000. Sunday collected $4000 more = $20000... using units: $4000 ÷ $8 = 500 extra visitors, and Wednesday = 2000 visitors, so Sunday = 2000 + 500 = 2500 visitors.


The circle is divided into 8 equal parts; 6 of them are shaded. \(\dfrac{6}{8}=\dfrac{3}{4}\). Answer = \(\dfrac{3}{4}\).


Number of boxes = 150 ÷ 8 = 18 remainder 6, so at most 18 whole boxes can be bought. Each box has 3 cupcakes, so 18 × 3 = 54 cupcakes.


From the graph, Jimmy's height at 10 years old = 130 cm and at 11 years old = 148 cm. Increase = 148 − 130 = 18 cm.


(a) From the graph the volume rises from 250 ℓ at 1.30 p.m. to 450 ℓ at 3.30 p.m., an increase of 200 ℓ. So the period is between 1.30 p.m. and 3.30 p.m.<br>(b) At 5.30 a.m. the volume was 550 ℓ and at 9.30 a.m. it was 100 ℓ. Water pumped out = 550 − 100 = 450 ℓ.


Female cyclists: Primary 5A = 9, Primary 5B = 10. Total = 9 + 10 = 19. This matches option (4).


From the bar graph: September = 17, October = 9, November = 20. Total = 17 + 9 + 20 = 46.


Read the protractor along the ray EG. EF lies on the 0° line and ray EG passes through the 55° mark on the inner scale (measured from EF). So ∠FEG = 55°.


ABEF is a rectangle and BCDE a square sitting side by side, so the top edge ABC and the bottom edge FED are two straight parallel lines. BC is part of the top edge and FD is part of the bottom edge, so BC // FD. The other pairs meet or lie on the same line, so they are not parallel.


Name an angle that is smaller than a right angle.

The figure is a pentagon ABCDE shaped like an arrow/house pointing right at C. The angle at C, ∠BCD, is the pointed vertex which is acute (smaller than 90°). The angles at A and E are right angles and those at B and D are obtuse (reflex of the point).


Figure A shows a rectangular piece of paper. It is folded along the dotted line as shown in Figure B. Find ∠a.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In Figure A the diagonal makes 24° with the base, so the angle between the diagonal and the right edge of the rectangle is 90° − 24° = 66°. Folding along the dotted line reflects the flap, splitting this 66° equally into two parts. So ∠a = 66° ÷ 2 = 33°.


In a carnival game, points are earned when bean bags are tossed onto regions A, B or C. Ron, Shirley and Tammy tossed 2 bean bags each: (i) Ron's bean bags landed on regions A and B. (ii) Shirley's bean bags landed on regions B and C. (iii) Tammy's bean bags landed on regions A and C. The total number of points earned by each of them: Ron 125, Shirley 65, Tammy 100. How many points will be earned when a bean bag lands on region B? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Ron: A + B = 125. Shirley: B + C = 65. Tammy: A + C = 100. Adding all three: 2A + 2B + 2C = 290, so A + B + C = 145. Region B = (A + B + C) − (A + C) = 145 − 100 = 45.


The table shows the number of students in two classes. Part of the table is covered by ink blots.
How many more girls than boys are there altogether?
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the table: 4A total 41 with 22 boys means 4A girls = 41 - 22 = 19 (given). 4B total 34 with 21 girls means 4B boys = 34 - 21 = 13. Total girls = 19 + 21 = 40; total boys = 22 + 13 = 35. More girls than boys = 40 - 35 = 5.


A tank was filled with water. Water flowed out of the tank from 09 00 to 14 00.
a) How many litres of water was in the tank at first?
b) How many litres of water was in the tank at 10 00?
c) How many litres of water flowed out between 12 00 and 13 00?
d) The least amount of water flowed out during which 1-hour period? 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 line graph. a) At 09 00 the line is at 250, so 250 litres at first. b) At 10 00 the line is at 220 litres. c) At 12 00 there are 140 litres and at 13 00 there are 50 litres, so 140 - 50 = 90 litres flowed out. d) Hourly drops: 09-10 = 30, 10-11 = 60, 11-12 = 20, 12-13 = 90, 13-14 = 50. The smallest drop (20 litres) is from 11 00 to 12 00.