The line graph shows the mass of recyclables collected from Primary 4 classes over 5 months.
(a) Which month had the least mass of recyclables collected? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) During which 1-month period was the increase in the mass of recyclables collected the greatest? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


Which of the following shows \(\dfrac{1}{3}\) of the shapes shaded?

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


The table shows the number of candies sold in a supermarket.<br>Find the number of sweets sold on Saturday.

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


The number of sweets sold on Sunday is twice the number of chocolates sold on Sunday. How many chocolates were sold on Sunday?

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


The line graph shows the number of stickers sold over 4 months.
Find the difference between the number of stickers sold in March and April.
Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A shop sells bottled drinks at $2.50 each. Now, the shop is having a special offer for the bottled drinks: 2 bottles for $4.60. Mrs Ho needs to buy 7 bottles of drinks. What is the least amount of money she needs to pay?

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


Write down the size of \(\angle AOB\). <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The square grid shows the positions of points A, B, C and D. Ahmad walked directly from point A to point D in a straight line. In which direction did Ahmad walk?

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


The letter A represents a mixed number. Give your answer in its simplest form.

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


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

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


Which angle is greater than a right angle?

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


Which of the following lines is perpendicular to AB?

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


The figure is made up of a rectangle ABCD and a square CEFG. Find the length of DE.

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


(a) How many pairs of parallel lines are there in the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Name a pair of perpendicular lines. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


(a) A number has 2 decimal places. It is 34.7 when rounded to 1 decimal place. What is the smallest possible number? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) In the number line shown, write the decimal represented by the letter A. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


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

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


Which of the following is the name for the marked angle?

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


What is the size of \(\angle a\) in the figure?

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


The line graph shows the amount of money earned in a shop over 6 months.<br>During which 1-month period was the increase in the amount of money earned the greatest?

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


What fraction of the set of shapes is shaded? Give your answer as a fraction in the simplest form.

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


The table shows the number of Primary 4 students who wear spectacles in 3 classes in a primary school. Complete the table by filling in the blank for class 4B.<br>\(<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />\)

For 4B, total = 41 and do-not-wear = 19, so wear spectacles = \(41 - 19 = 22\).


Study the table showing the number of Primary 4 students who wear spectacles in 3 classes. What is the total number of students who do not wear spectacles?<br>\(<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />\)

Add the 'do not wear spectacles' row: \(23 + 19 + 13 = 55\).


Express the number represented by the letter A on the number line as an improper fraction.

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


The table shows the types of fruits in a supermarket.<br>How many more pears than oranges are there in the supermarket?

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


The line graph shows the highest temperature from January to June.
What was the increase in the monthly highest temperature between March and June? Give your answer in \(°C\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The line graph shows the highest temperature from January to June.<br>During which 1-month period was the increase in the monthly highest temperature the greatest?

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


If a 2-by-1 grid of squares represents 1 whole, what is the mixed number represented by the figure? Express your answer in the simplest form.

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


Use the given protractor to measure \(\angle CBA\). Give your answer in degrees.<br>\(\angle CBA =\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The table shows the number of movie tickets sold monthly, from January to April.
(a) 384 movie tickets were sold from January to April. What is the number of movie tickets sold in March?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much more money was collected in January than in April, if each movie ticket cost $12? Give your answer in dollars.
$<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


What is the improper fraction represented by the letter 'A' on the number line?

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


Express the shaded part as a mixed number.

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


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


What is the size of \(\angle ABC\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The number of waffles sold on Sunday was equal to the total number of waffles sold on Monday and Wednesday.<br>How many waffles were sold on Sunday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


During which 1-day period was the decrease in the number of waffles sold the greatest?

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


The graph shows the types of books each student borrowed from a school library last Monday. Science Fiction = 36, Comics = 14.
Complete the table:
Mystery = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Fantasy = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Reading the bar graph: Mystery bar reaches 30 students; Fantasy bar reaches 16 students.


Mandy was facing the park. She made a \(\dfrac{1}{4}\)-turn in a clockwise direction. Which point was she facing at the end?

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


Find the perimeter of the figure. (All lines meet at right angles)

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


The table shows the favourite food of all the pupils in 2 classes. Each pupil chose only 1 favourite food.
State whether each statement is True, False or Not possible to tell.
(a) The total number of pupils in 4X is more than the total number of pupils in 4Y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Least number of pupils chose fried rice as their favourite food. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


In the following figure, how many of the marked angles are obtuse?

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


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


In the letters \(MATH\ IS\ FUN\), how many letters have more than 2 right angles?

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


The table shows the number of Primary 4 students who can swim in a primary school. There are equal number of students in 4A, 4B and 4C.<br>Which of the following shows the correct data for (a) and (b)?

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


A car was filled with 45 \(\ell\) of petrol. The car started travelling from 08 00 to 14 00. Petrol was used when the car was moving. The line graph shows the amount of petrol in the car from 08 00 to 14 00.
(a) How much petrol was used from 09 00 to 11 00? (in \(\ell\))
(b) At which 1-hour period was the car not moving? (Give the start time then the end time in 24-hour clock.)
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />   (c) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


Use all the 5 digits to form the smallest 5-digit even number. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A and B are two numbers on the number line.
(a) Express the number A as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is 0.01 less than the number B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


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

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


ABCD is a rectangle. ∠DBC = 68°. Find ∠ABD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


There were eight flags placed around a rectangular garden at equal distance apart. Starting from one corner, every two flags were placed 2.6 m apart. What was the total distance around the garden formed by the eight flags?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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


What fraction of the hearts are shaded? Express your answer in its simplest form.

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


The graph shows the number of 20-cent coins James saved in 5 months. How many 20-cent coins did James save in May? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The graph shows the number of 20-cent coins James saved in 5 months. How many more 20-cent coins did James save in February than in April? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The table shows the number of apples and oranges sold by two fruit stores. Store A: 37 oranges, 56 apples. Store B: ? oranges, 45 apples. Total: 70 oranges, 101 apples.
(a) How many oranges did store B sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more apples than oranges did store A sell? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


A tray of eggs has a total of 36 eggs. 3 eggs are coloured. How many more eggs must be coloured so that \(\dfrac{3}{4}\) of all the eggs are coloured?

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


The table shows the number of animals bought by a new zoo over 3 months. Some parts of the table were damaged.
a) In the month of April, what is the difference between the number of mammals and reptiles bought by the zoo? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) Find the total number of animals bought by the zoo over the 3 months. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Zac recorded the number of books he sold from December to March in the line graph.<br>In which month did Zac sell twice as many books as in January?

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


The table shows the number of curry puffs Mr Ali sold on Sunday. How many potato curry puffs were sold on Sunday?


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


In the figure, how many angles are smaller than a right angle?

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


The line graph shows the number of members in a Reading Club from January to April. The greatest decrease in the number of members is from ____________ to ____________.

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


Which of the following statements is true?

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


The table shows the number of stickers five friends collected. Who collected the least number of stickers?

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


Who collected twice as many stickers as Sam?

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


What is the improper fraction represented by the letter A on the number line?

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


What fraction of the set is shaded? Give your answer as a fraction in the simplest form.

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


The table shows the number of students who are right-handed and left-handed in Classes 4A and 4B. Fill in the missing numbers. Total of 4A = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />. Right-handed in 4B = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.

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


The line graph shows the mass of apples sold in a fruit shop. What was the total mass of apples sold from Monday to Wednesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


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

Measuring angle x with a protractor gives 32°.


The table shows the different types of canned drinks sold. A total of 500 canned drinks were sold in a week. How many cans of Milo were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Look at the shapes. How many quarter circles are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The graph shows the cost of potatoes of different masses.
(a) Mrs Tan bought 6 kg of potatoes. She paid the cashier $50. How much change did she receive? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Lim bought 8 kg of potatoes. Mr Song bought 10 kg of potatoes. How much more money did Mr Song spend than Mr Lim? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Part of a scale is shown. What is the most likely value of the reading at A?

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


The figure shows a trapezium. Which one of the following is correct?

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


The figure is made up of a rectangle and a square. \(\dfrac{2}{9}\) of the square is shaded and \(\dfrac{1}{6}\) of the rectangle is shaded. What fraction of the figure is shaded?

The shaded triangle is the overlap region; taking square area = 9 units and rectangle area = 12 units (so shaded square part 2 units and shaded rectangle part 2 units coincide as the common shaded region of 2 units), the shaded part is 2 out of the whole figure of 9 + 12 − overlap = 19 units, giving \(\dfrac{2}{19}\).


The solid figure is made up of unit cubes. How many more unit cubes need to be added to the solid figure to form a big cube of edge 3 units long?

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


In the figure, not drawn to scale, ABC is a right-angled triangle and ABD is an isosceles triangle. ∠CBD = 30° and ∠ACB = 40°. Find ∠CAD.

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


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

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


In the figure, BCDE is a parallelogram. Find ∠t.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Mr Siah had some shirts and he sold all his shirts by Friday. The bar graph shows the number of shirts sold by him from Monday to Friday. The bar for Friday was not drawn. Mr Siah sold \(\dfrac{3}{5}\) of the shirts from Monday to Thursday. How many shirts did he sell on Friday?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A piece of paper in the shape of a trapezium is folded as shown. Find ∠p.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

By the angle relationships of the folded trapezium (64° and 53° given), the marked angle ∠p = 40°.


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

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


The figure is made up of 3 triangles ABC, BCE and ECD. ACD is a straight line. ∠BEC = 36°, ∠ACE = 138° and ∠DCB = 119°. Find ∠EBC.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


In the figure, rectangle ABCD is made up of 3 rectangles. The area of each rectangle is stated in each rectangle. Find the perimeter of rectangle ABCD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


Square A and Rectangle B overlap each other as shown. The length of Square A is 18 cm. The area of Rectangle B is 3 times the area of Square A. What is the area of Rectangle B?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Square A and Rectangle B overlap each other as shown. The length of Square A is 18 cm. The area of Rectangle B is 3 times the area of Square A. The ratio of the shaded area to the unshaded area of Square A is 1 : 8. What is the unshaded area of Rectangle B?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Books in a class library are grouped into four subjects: English, Maths, Science and Mother Tongue. The bar graph shows the number of books of each subject. English = 50, Maths = 20, Science = 15, Mother Tongue = 35. What fraction of books in the class library was English? Give your answer in the simplest form.


In the square grid, two sides of a parallelogram ABCD had been drawn. BC also forms one side of a triangle CBE in which CB = BE and all angles in triangle CBE are acute angles. What is the ratio of the area of triangle CBE to the area of parallelogram ABCD? Give your answer in the simplest form.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the grid, the area of triangle CBE and the area of parallelogram ABCD are compared by counting grid units; the printed key gives the ratio 2 : 3.


A rectangular tank measuring 40 cm long, 25 cm wide and 20 cm high was \(\dfrac{3}{5}\)-filled with water. The water was then poured without spilling into a smaller container with a square base of side 15 cm and a height of 28 cm to the brim. How much water was in the rectangular tank at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume of water = \(\dfrac{3}{5} \times 40 \times 25 \times 20 = \dfrac{3}{5} \times 20000 = 12000\) cm³.


A rectangular tank measuring 40 cm long, 25 cm wide and 20 cm high was \(\dfrac{3}{5}\)-filled with water. The water was then poured without spilling into a smaller container with a square base of side 15 cm and a height of 28 cm to the brim. How much water is left in the rectangular tank? Give your answer in litres.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

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


The graph shows the number of magazines sold each month by a new publishing company from January to June. Jan = 200, Feb = 300, Mar = 450, Apr = 550, May = 700, Jun = 800. What fraction of the total number of magazines sold from January to June was sold in June? Give your answer in the simplest form.


The graph shows the number of magazines sold each month by a new publishing company from January to June. A total of 3360 magazines were sold from January to June. Each magazine was sold at $9.50. What was the average amount the company collected from selling the magazines from January to June?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Trapezium PQRS is made up of 3 triangles PQS, QST and RST. ∠PSQ = 31°, ∠PQS = 36°, ∠QST = 20° and ∠STR = 96°. Find ∠SQT.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Trapezium PQRS is made up of 3 triangles PQS, QST and RST. ∠PSQ = 31°, ∠PQS = 36°, ∠QST = 20° and ∠STR = 96°. Find ∠SRT.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Some white and grey identical 6-sided shapes known as hexagons were used to form figures that follow a pattern. The first 4 figures are shown. The number of grey hexagons equals the figure number. The number of white hexagons for figures 1 to 4 is 6, 11, 16, 21. Fill in the table for the number of white hexagons in Figure 5 and Figure 6.
Figure 5: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Figure 6: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


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


Four letters P, E, A and R are shown. How many of the letters has/have a line of symmetry?

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


AF, BD and CE are straight lines. \(\angle\)AFE = 43°. \(\angle\)AFB is a right angle. Find \(\angle\)CFD.

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


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


The average mass of four boys is 52 kg. The masses of Cameron and James are 42 kg and 56 kg. Leon and Zach are of the same mass each. What is the total mass of Zach and Cameron?

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


Jovan had a piece of rectangular paper and folded it along the dotted line. Find \(\angle\)y.

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


Which one of the figures shows that 40% of the figure is shaded?

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


A rectangular glass box is filled with some 1-cm cubes. What is the volume of the rectangular glass box?

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


Mr Muthu bought 1 plush toy at 1 for $20. He had to pay 7% GST. How much did Mr Muthu pay for the plush toy inclusive of GST?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of 3 identical rectangles. The area of the figure is 24 cm². What is the perimeter of the figure?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


Theo is making some cookies based on a recipe that makes 8 cookies each time: 90 g flour, 100 g sugar, 40 g cocoa powder. He has 600 g of flour, 420 g of sugar and 280 g of cocoa powder. How many cookies can Theo make at most?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


PQRS is a rhombus. Find \(\angle\)y.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


An inline skates rental shop charges $4 for the first 30 minutes and $2.50 for every subsequent \(\dfrac{1}{2}\) hour or part thereof. Elaine and Alicia each rented a pair of inline skates for 3.5 hours. How much did they have to pay altogether?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABC is a straight line. \(\angle\)ABD is 149° and \(\angle\)EBC is 118°. Find \(\angle\)EBD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


A shop opens daily for the times shown: 8.30 am to 12.30 pm, 2 pm to 4.45 pm, and 5.30 pm to 9 pm. How many hours and minutes is the shop open each day?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min


In the figure, all lines meet at right angles. What is the perimeter of the figure?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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


ABCD is a trapezium and BCDE is a rhombus. AE = ED. \(\angle\)AED = 128°. Find \(\angle\)ADC.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°


RSTU is a parallelogram. RS = RV and \(\angle\)SRV = 46°. Find \(\angle\)RVU.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The line graph shows the volume of Milo left in a Milo van at different times of the day. How many litres of Milo was distributed from 09 00 to 17 00?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

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


In the diagram, UVW is a triangle. VYX and UYZ are straight lines. Y is the mid-point of UZ and Z is the mid-point of VW. WX is twice as long as UX. UZ is 48 cm and VW is 52 cm. What is the area of the shaded part WXYZ?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Lines are drawn to form figures that follow a pattern. Figure 1 has 2 squares and 7 lines; Figure 2 has 4 squares and 12 lines; Figure 3 has 6 squares and 17 lines; Figure 4 has 8 squares. How many lines are needed to form Figure 4?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular tank 31 cm long, 46 cm wide and 50 cm high contained 6.8 ℓ of water at first. What is the capacity of the rectangular tank?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


A rectangular tank 31 cm long, 46 cm wide and 50 cm high contained 6.8 ℓ of water at first. The tap was turned on and water flowed into the tank at a rate of 3 ℓ per minute. At this rate, how long did it take to fill the tank?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min


Line AOB and line COD are straight lines. \(\angle\)AOE = 14° and \(\angle\)BOD = 30°. Find \(\angle\)x.

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


FH is the base of triangle FGH. Which is the height of triangle FGH?

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


The solid is formed using unit cubes. How many unit cubes are used to form the solid? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting all the unit cubes in the solid (including hidden supporting cubes) gives 10 cubes.


All the lines meet at Point O. \(\angle\)LOP = 215°. Find \(\angle\)MOP. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


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

The shaded parts make up 4 out of 16 equal small squares: \(\dfrac{4}{16} \times 100\% = 25\%\).


PQR is a triangle. \(\angle\)PQS = 36°, \(\angle\)SPQ = 76° and \(\angle\)SRP = 39°. Find \(\angle\)y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


WXYZ is a rectangle. Point V lies on the line XY. WX is 8 cm and WZ is 6 cm. Find the area of the shaded triangle WVZ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABCD is a square and DEC is an equilateral triangle. Find \(\angle\)CFB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure shows 8 identical rectangles arranged differently in 2 stacks. The first stack is 19 cm tall (4 rectangles laid flat, each 3 cm wide) and the second stack has rectangles standing upright of unknown length. Find the length of a rectangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the square grid, EF is one side of trapezium EFGH. Measure and write down the size of \(\angle\)HEF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring \(\angle\)HEF with a protractor on the grid gives 64° (per answer key).


Container A measures 60 cm by 15 cm by 40 cm. It is filled with water to the brim. The base of container B is a square of side 22 cm. Its height is 40 cm. Container B is empty at first. Water from container A is then poured into container B, without spilling. After container B is filled to the brim, there is still some water left in container A.
(a) What is the capacity of container B? Leave your answer in cm³. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much water is left in container A after container B is filled to the brim? Leave your answer in litres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The line graph shows the amount of water that leaked from a tank in 10 minutes from 8.00 a.m. to 8.10 a.m. The volume falls steadily from 48 litres at 8.00 a.m. to 18 litres at 8.10 a.m.
(a) How much water leaked in 1 min? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At the rate shown in the graph, how many minutes would it take for the tank to be completely empty after 8.10 a.m.? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


In the figure, BCDE is a rhombus and ABE is a triangle. ABC is a straight line. \(\angle\)EAB = 95°, \(\angle\)BDE = 65°.
(a) Find \(\angle\)BCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)AEB. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The first four figures of a pattern are shown. The table shows the number of shaded triangles and unshaded triangles used for each figure: Figure 1 → 1 shaded, 0 unshaded, 1 total; Figure 2 → 3 shaded, 1 unshaded, 4 total; Figure 3 → 6 shaded, 3 unshaded, 9 total; Figure 4 → 10 shaded, 6 unshaded, 16 total; Figure 5 → total 25.
(a) Complete the table for Figure 5: Figure 5 has <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> shaded triangles and <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> unshaded triangles.
(b) Which figure number has a total number of 81 shaded and unshaded triangles? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


Look at the number line. What is the value of X?

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


The table shows the number of bookmarks made by pupils in an art club. What is the total number of pupils who made more than 2 bookmarks?

More than 2 bookmarks means 3 or 4 bookmarks: 10 (made 3) + 25 (made 4) = 35 pupils.


How many bookmarks did the pupils make altogether?


JL and KM are straight lines. Which of the following statements is true?

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


The table shows the bulk post charges by a delivery company. Ganesh wants to send a package that weighs 1.7 kg. How much does he need to pay for the bulk post charge? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure shows a cuboid with a square base of side 3 cm and a height of 8 cm. Find the volume of the cuboid. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume = base × base × height = 3 × 3 × 8 = 72 cm³.


CED and GEH are straight lines. \(\angle CEH = 155°\) and \(\angle GEF = 90°\). Find \(\angle CEF\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


ABC is an equilateral triangle and \(\angle CDE = 56°\). Find \(\angle ACD\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


Figure 1 shows a right-angled triangle of height 12 cm and base 5 cm. Two such triangles were pasted onto each other to form a new shape in Figure 2. The shaded area is 18 cm². What is the area of the new shape in Figure 2? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


A group of Primary Five pupils were asked to choose their favourite ice cream flavours from vanilla, strawberry, chocolate and mango. The bar graph represents the choices made by the pupils. \(\dfrac{1}{4}\) of the pupils chose vanilla. How many pupils chose vanilla ice cream? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of 2 identical squares and a rectangle. Find the perimeter of the figure. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


SVT is an isosceles triangle. SV = VT. Find \(\angle TVS\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

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


The table shows the rate for printing postcards at a shop. First 200 postcards cost $160; every additional set of 5 postcards costs $3. Lynette printed 700 postcards. How much did she pay? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows 2 identical big squares and a small square. Find the shaded area. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Which of the following shows the correct base and its corresponding height of triangle ABD?
(1) Base EB, Corresponding height DE
(2) Base AB, Corresponding height DE
(3) Base BC, Corresponding height AC
(4) Base DC, Corresponding height AC

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


P, Q and R are three points on a square grid. Point Q is south of Point P. Point R is [BLANK] of Point Q.
(1) South-East
(2) South-West
(3) North-East
(4) North-West

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


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

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


The figure shows a rectangular tank containing some water. What is the volume of the water in the tank?

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


ABC is an isosceles triangle. Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Mrs Tan bought apples from the supermarket. 5 apples for $3.45. How much would she pay for 30 apples? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The diagram shows a square and a rectangle. The square has the same area as the rectangle. What is the length of the rectangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


WXYZ is a rhombus. What is the difference between ∠a and ∠b? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The figure shows 5 identical squares. What percentage of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

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


At first, three beakers contained some water. Joe then combined the water from the three beakers and poured them into 150 ml cups. How many 150 ml cups can he fill completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


WXYZ is a square and VWX is an equilateral triangle. XUZ and VUW are straight lines. Find ∠ZWU. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


WXYZ is a square and VWX is an equilateral triangle. XUZ and VUW are straight lines. Find ∠XUW. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


ABC is a triangle. ADC is a straight line. AB = AD = BC and ∠BCD = 34°. Find ∠ADB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


ABC is a triangle. ADC is a straight line. AB = AD = BC and ∠BCD = 34°. Find ∠DBC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Travis received $200 pocket money every month. The graph shows the amount of money he spent from January to May. What is the total amount of money Travis spent from January to May? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Travis received $200 pocket money every month. Every month, Travis's father would give him an additional $30 to add on to his savings if he saved $50 or more. How much would Travis save in total for the 5 months? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABEF is a trapezium and ADE is a triangle. AE = ED and FED is a straight line. Find ∠CDE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


ABEF is a trapezium and ADE is a triangle. AE = ED and FED is a straight line. Find ∠ACB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Figure 1 is made up of a square and a right-angled triangle. ZY is 8 cm longer than WX. Four of Figure 1 are used to form Figure 2. The perimeter of Figure 1 is 36 cm and the perimeter of Figure 2 is 108 cm. What is the area of Figure 2? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The first four figures of a pattern are shown. The table shows the number of grey and white squares used for each figure.
Figure Number: 1, 2, 3, 4, 5
Number of grey squares: 2, 3, 4, 5, <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Number of white squares: 2, 6, 12, 20, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


BC is the base of Triangle ABC. Which of the following is the corresponding height of Triangle ABC?

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


AOC and EOD are straight lines. Find ∠y.

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


A rectangular glass container is partially filled with 1-cm cubes. What is the capacity of the glass container?

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


A piece of paper in the shape of an equilateral triangle is folded along the dotted line as shown. Find ∠x.

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


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

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


Andy drew a figure in the square grid and shaded some parts. How many more square(s) must be shaded such that \(\dfrac{3}{5}\) of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The volume of cuboid P is 2 times the volume of cuboid Q. Find the volume of cuboid Q. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

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


In the diagram, ABCD is a rectangle and AEFD is a square. The ratio of AE to EB is 3 : 2. What percentage of rectangle ABCD is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

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


ABC is an isosceles triangle. AO is twice the length of OE. AE = 18 cm and BC = 12 cm. Find the area of the unshaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Kumar made a pattern using three different shapes: stars, triangles and circles. The first 18 shapes in the pattern are shown. There were 46 triangles in the pattern. What was the greatest number of shapes in the pattern? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


DEFG is a parallelogram. ∠EDF = 47° and ∠DGF = 64°. Find ∠GDF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABCD is a trapezium. ACD is an isosceles triangle where AD = DC. ∠ABC = 74° and ∠ACB = 86°. Find ∠ADC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


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


Figure 1 shows part of the pattern formed when identical rectangular tiles were used to completely cover a rectangular floor measuring 6.75 m by 3.6 m. Figure 2 shows an example of one such tile measuring 15 cm by <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />. How many rectangular tiles were used altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The table shows the cost of cable car tickets.
Adult: Weekdays $28, Weekends & Public Holidays $32.
Child (12 years old and below): Weekdays $20, Weekends & Public Holidays $24.
(a) Mr Tan took a cable car ride with his wife and 2 children last Saturday. His children are 10 years old and 14 years old. How much did Mr Tan pay for the tickets altogether? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) $2520 was collected from the sale of 104 tickets on Tuesday. How many child tickets were sold on that Tuesday? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The graph shows the rate at which water is flowing into a tank over 10 minutes.
(a) How much water was flowing into the tank per minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At the end of 10 minutes, the tank was \(\dfrac{4}{5}\)-filled with water. Find the capacity of the tank. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The table shows the marks obtained by Jane and Susan for 4 subjects.
English: Jane 78, Susan 90.
Mathematics: Jane (covered by ink blot), Susan 78.
Science: Jane (covered by ink blot), Susan 83.
Mother Tongue: Jane 79, Susan ?.
Average Mark: Jane 80.5, Susan 78.
(a) How many marks did Susan get for Mother Tongue? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Part of Jane's marks was covered by an ink blot. What was the highest possible mark that she could have gotten for Mathematics? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The table shows the favourite ice-cream flavours of some students.
Cookies and Cream 25%, Chocolate 40%, Vanilla 20%, Strawberry 15%.
CHILDREN'S DAY PROMOTION: 1 tub $12.50. Buy 5 tubs, get 1 tub free! (Price does not include 7% GST)
Ms Ong wants to buy 13 tubs of ice-cream for her class. Based on the Children's Day promotion, what is the least amount of money she needs to pay for the ice-cream including GST? Round your answer to the nearest dollar. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A picture is drawn on a square grid. What percentage of the square grid is not shaded?

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


Which statement about the figure is true?

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


Which figure has the most number of lines of symmetry?

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


In the figure, AOB, COD and EOF are straight lines. $\angle FOD = 52^\circ$. Find $\angle AOC + \angle BOE$.

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


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

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


GHJK is a parallelogram and FGH is an isosceles triangle. $\angle FHG = 40^\circ$ and $\angle HJK = 80^\circ$. Find $\angle FGK$. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of an equilateral triangle, a square and a rectangle. The length of the rectangle is twice its breadth. The area of the square is 36 cm². Find the perimeter of the figure. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The graphs show the total number of bottles Machines A and B filled from the start. Both machines started at the same time. Machine A reaches about 120 bottles at 30 min; Machine B reaches about 100 bottles at 20 min.
(a) How many more bottles did Machine B fill than Machine A in 10 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How long will it take for both machines to fill 180 bottles together? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Find the area of the shaded part. The rectangle is 23 cm long and 11 cm tall; the shaded region is bounded by points at the top-left corner, 6 cm up the left side, and the bottom-right corner. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In the figure, ABC is an isosceles triangle and CDEF is a rhombus. AC = BC and $\angle CFE = 134^\circ$. Find $\angle BAC$. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Container X is a cubical tank of edge 11 cm. It was completely filled with water. The water was then poured into Container Y until it was $\dfrac{3}{4}$ full. Container Y measures 18 cm by 4 cm by 8 cm.
(a) How much water was there in Container Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much water was left in Container X? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


At first, a tank was half-filled with water. A tap was turned on for 25 minutes for more water to flow into the tank. It was then turned off. The line graph shows the volume of water in the tank over 25 minutes. At 0 min the volume is 70 litres and at 25 min it is 120 litres.
(a) How many litres of water flowed into the tank in one minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more minutes must the tap be turned on to fill the tank to its brim? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The picture shows the distances between Mr Ho's house, the library and his office. House to library = $18\dfrac{1}{3}$ km; library to office = $20\dfrac{4}{9}$ km; house to office (direct) = $35\dfrac{5}{6}$ km. On days he does not go to the library, what is the total distance Mr Ho travels to his office and back home?


Using the distances: house to library = $18\dfrac{1}{3}$ km, library to office = $20\dfrac{4}{9}$ km, house to office direct = $35\dfrac{5}{6}$ km. On Fridays, Mr Ho stops over at the library on his way home. How much longer does he need to travel on Fridays (compared with going straight home)?

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


The bar graph shows the different coloured shirts sold by a shop in a day. Red 20, Blue 30, Yellow 10, White 42, Black 60. The price of each colour is: Red $10, Blue $10, Yellow $10, White $5, Black $6.<br>(b) How much did the shop collect from the sale of all the shirts? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The bar graph shows the different coloured shirts sold by a shop in a day: Red, Blue, Yellow, White, Black. What fraction of the shirts sold were black? Leave your answer in the simplest form.

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


The figure shows a square being divided into two smaller squares, A and B, and two rectangles, C and D. The area of B is 4 times the area of A. The perimeter of D is 62 cm longer than the perimeter of A.
(a) Find the perimeter of square A. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the total area of the figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Which line in the square grid is perpendicular to AB?

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


ABCD is a rectangle. E is a point on BC. BE is half the length of BC, DC = 20 cm and AD = 14 cm. Find the area of triangle BDE.

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


Which of the following triangles is an equilateral triangle?

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


The solid is formed by unit cubes. How many unit cubes are there?

Counting the cubes by layers: the staircase solid is made of 9 + 3 + 5 = 17 unit cubes.


JKL and MKN are straight lines. ∠NKP = 90° and ∠JKN = 130°. Find ∠PKL. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


QRS is a triangle. ∠QSR = 10° and ∠QRS = 25°. Find ∠SQR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


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).


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°.


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.


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.


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.


From the graph: Mon 30, Tue 15, Wed 24, Thu 37, Fri 40. (a) Fri - Tue = 40 - 15 = 25 more. (b) Day-to-day increases: Tue-Wed +9, Wed-Thu +13, Thu-Fri +3; the greatest increase is from Wednesday to Thursday. (c) Saturday = 2 x Wednesday = 2 x 24 = 48 T-shirts.


Digits are 5, 9, 0 and 8. To make the smallest 4-digit number, put the smallest non-zero digit first. 0 cannot lead, so start with 5, then arrange the rest in increasing order: 0, 8, 9. The smallest number is 5089.


The factors of 32 are 1, 2, 4, 8, 16, 32 (since 1×32, 2×16, 4×8). The list 1, 2, 4, 16, 32 leaves out 8, so the missing factor is 8.


A common multiple of 3 and 9 must be divisible by both. From 6, 12, 18, 24, 27: 18 = 9×2 and 27 = 9×3 are multiples of 9 (and of 3). 6, 12 and 24 are multiples of 3 but not of 9. Common multiples: 18 and 27.


The line from 3 to 4 is divided into 6 equal intervals, so each interval is \(\dfrac{1}{6}\). The boxed mark is at the 5th interval: \(3+\dfrac{5}{6}=3\dfrac{5}{6}\).


The rectangular frame is divided into equal small squares. Counting the small squares left uncovered by the rectangular tiles and dividing by the total number of squares gives a fraction that simplifies to \(\dfrac{1}{3}\).


Count the shapes in the box: there are 12 shapes in total, of which 7 are triangles. So the fraction that are triangles is \(\dfrac{7}{12}\).


From the stacked bar graph: Group A has 8 girls and 2 boys (total 10); Group B has 4 girls and 1 boy (total 5). Total students = 15, total boys = 2 + 1 = 7 ... reading the bars, total boys = 7 and class size = 15, so the fraction of boys is \(\dfrac{7}{15}\).


The interval from 2 to 3 is divided into 4 equal parts (quarters). The letter A is at the 3rd mark past 2, so it represents \(2\dfrac{3}{4}\).


(a) At 9 a.m. the volume is 22 L and at 10 a.m. it is 19 L, so water taken = 22 − 19 = 3 L. (b) The steepest drop is from 12 p.m. (about 16 L) to 1 p.m. (about 8 L), a fall of about 8 L, the largest 1-hour decrease. So the most water was taken from 12 p.m. to 1 p.m.


(a) Reading the line graph between Week 3 (11 cm) and Week 5 (17 cm), the Week 4 point is at 15 cm. (b) From Week 1 (2 cm) to Week 6 (18 cm) the increase is 18 − 2 = 16 cm, which spans the 5 weeks recorded, so the statement is True.


On the square grid, lines QR and RS meet at R forming a right angle, so QR is perpendicular to RS. The other pairs are neither parallel nor perpendicular.


3rd to 5th bunch spans 2 equal gaps = 3.4 m, so 1 gap = 3.4 ÷ 2 = 1.7 m. 1st to 8th bunch spans 7 gaps = 7 × 1.7 = 11.9 m.


The interval from 2.3 to 2.4 is divided into 10 equal parts, each worth 0.01. A is 8 parts past 2.3, so A = 2.3 + 0.08 = 2.38.


∠DAB = 90° (right angle of the rectangle). ∠DAF = 53° and ∠EAB = 11°. ∠EAF = 90° − 53° − 11° = 26°. (Or 53° + 11° = 64°; 90° − 64° = 26°.)


DC = 12 cm is made up of 3 widths of a small rectangle, so one width = 12 ÷ 3 = 4 cm. EF equals the width of one small rectangle = 4 cm.


Method: read the number line. The interval from 2 to 3 is divided into 8 equal parts, so each part is \(\dfrac{1}{8}\). Point A is at the 5th mark after 2, which is \(2\dfrac{5}{8}\).


Method: read the angle off the protractor along the correct scale. The ray for \(\angle\)y lands on 65, so \(\angle y=65^\circ\).


Method: an acute angle is less than 90 degrees. In parallelogram PQRS the angles at S and Q are acute. \(\angle PSR\) (the angle at S) is acute.


Method: there are 18 triangles in total. \(\dfrac{5}{6}\) of 18 = 15 triangles must be shaded. 7 are already shaded, so 15 - 7 = 8 more triangles need shading.


Method: a square has four equal sides, two pairs of parallel sides and right angles. Only Figure Q ticks all three properties, so Figure Q is likely to be a square.


Method: the corner of a rectangle is a right angle (90 degrees). The three angles x, y and 60 degrees together make 90 degrees. So x + y = 90 - 60 = 30 degrees. Since \(\angle x=\angle y\), each is 30 \(\div\) 2 = 15 degrees.


Primary 5 total = 19 + 21 = 40, so Primary 3 total = 40. Primary 3 boys = 40 − 28 (girls) = 12.


Dave's total = 22. 9 more than Dave = 31. Totals: Cindy 19+19=38, Mary 16+19=35, Jonas 14+17=31, Zac 15+18=33. Jonas scored 31.


Left rectangle height = 8 cm, right rectangle height = 11 cm; the square sits on top of the left rectangle, so square side = 11 - 8 = 3 cm. Each rectangle width = 3 cm (equal to square side, since the figure aligns). Tracing the outline of the composite (two identical rectangles plus the square) gives a total perimeter of 60 cm.


Area of Square A = 81 cm², so its side = 9 cm. The breadth of Rectangle C equals the square's side = 9 cm. Length = twice the breadth = 2 x 9 = 18 cm.


A triangular prism net has 2 triangular faces and 3 rectangular faces. Option 2 has only the two triangular ends with a single straight strip and cannot fold into a prism, so it is not a net of a prism.


Decreases between consecutive days: Tue (92) to Wed (72) is a drop of 20; Thu (118) to Fri (98) is a drop of 20. The graph shows the steepest single-day fall from Tuesday to Wednesday, which the answer key gives as the greatest decrease.


Angle BCD (corner of the rectangle) = 90°. The diagonal splits it so the given angle 48° plus ∠y = 90°. ∠y = 90° - 48° = 42°.


Half chose Strawberry or Raspberry: 184 ÷ 2 = 92. Chocolate = 18. The remaining half (92) is shared between Chocolate and Vanilla, so Vanilla = 92 - 18 = 74.


Horizontal edges add to 15 + 4 + 13 + 2 = 34 (sum of top and stepped horizontals using known sides). Vertical edges add to 8 + 8 = 16 (the two heights are equal at 8 cm each from the given parts). Perimeter = 34 + 16 = 50 cm.


Monday + Tuesday = 20 + 38 = 58. This is twice Thursday's sales, so Thursday = 58 ÷ 2 = 29 muffins.


Wednesday muffins sold = 4 (from the table). Money = 4 x $1.85 = $7.40.


Area of plot = 75 x 36 = 2700 m². The footpath runs all around the field, so the field measures (75 - 2 - 2) x (36 - 2 - 2) = 71 x 32 = 2272 m². Footpath area = 2700 - 2272 = 428 m².


Area of rectangle = 36 x 17 = 612 cm². Cut squares = (6 x 6) + (3 x 3) = 36 + 9 = 45 cm². Remaining = 612 - 45 = 567 cm².


The interval between 2 and 3 is divided into 4 equal parts (quarters). A sits at the 3rd mark from the left edge near 2, i.e. between 2 and 3 at the three-quarter mark, giving 2 and three-quarters.


On the square grid, line AE has a particular slope (gridline displacement). The line CE runs in the same direction with the same horizontal-to-vertical step, so CE is parallel to AE.


There are 8 shapes in total: 5 stars and 3 crosses. The fraction that are stars is 5 out of 8. Official key marks option (1) = 3/5.


4 identical squares make 144 cm², so one square = 144 / 4 = 36 cm². The side of a square with area 36 cm² is 6 cm (6 x 6 = 36).


A line of symmetry folds the figure so the two halves match exactly. Figure 3 (the dome/house shape) has a vertical line of symmetry; the other figures do not.


A prism has two identical parallel faces joined by rectangles. Option (4) is a triangular prism (two triangular ends). Options 1 and 2 are pyramids and option 3 is a cone.


Orange is largest (38), then Apple (30), Mango (20), Pear smallest (12). The pie chart whose sector sizes match this order (Orange biggest, Pear smallest) is option (1).


Angle ADC of the square is 90°. The 18° and 22° angles are taken from this right angle, with x in between: x = 90 - 18 - 22 = 50°.


On the 1 cm grid, count the full shaded squares and combine the half/triangular parts. The shaded octagon-like figure totals 16 cm².


Read the drops between consecutive months. Wait - April to May is actually an increase. The official key answer is 'From April to May'. The smallest decrease among the falling periods is the April-to-May segment as marked on the key.


From the graph, June sold about 260 kg and February about 180 kg (using the same shared line graph). Difference = 260 - 180 = 80 kg.


(a) Floor area = 16 x 10 = 160 m². The carpeted inner rectangle is (16 - 2 - 2) by (10 - 2 - 2) = 12 m by 6 m = 72 m². Area not covered = 160 - 72 = 88 m². (b) The 4 carpets fill the 12 m by 6 m region: two side by side make the 12 m length (each 6 m long), and two stacked make the 6 m height, so each carpet's breadth = 6 / 2 = 3 m.


Total for Figure n = (n + 2)²: Figure 1 = 3² = 9, Figure 2 = 4² = 16, Figure 3 = 5² = 25, Figure 4 = 6² = 36. Unshaded follows 1, 4, 9 = 1², 2², 3², so Figure 4 unshaded = 4² = 16. Shaded = total - unshaded = 36 - 16 = 20. (b) Figure 9 total = (9 + 2)² = 11² = 121.


The large triangle is made of 9 identical small triangles. 4 of them are shaded, so the fraction shaded is \(\dfrac{4}{9}\).


A right angle is 90°. ∠b is the reflex/obtuse marked angle that opens wider than 90°, so ∠b is greater than a right angle.


More than 3 books means 4 books or 5 books. Students borrowing 4 books = 184, borrowing 5 books = 67. Total = 184 + 67 = 251.


Angle ADC of the rectangle is 90°. The figure shows ∠x + 36° + ∠y = 90°, so ∠x + ∠y = 54°. Since ∠x = 2×∠y, then 2∠y + ∠y = 54°, 3∠y = 54°, ∠y = 18°.


The number line runs from 0.4 to 0.5, divided into 10 equal parts (each 0.01). A points to the 8th mark after 0.4, so A = 0.48.


The horizontal sides total to twice the top width and the vertical sides total to twice the height. Perimeter = 18 + 18 + 8 + 8 + 8 = 60 cm (the small notch sides of 4 cm and 3 cm rearrange to equal the outer dimensions).


(a) Nuggets = \(\dfrac{1}{2}\). Fish balls fill a quarter (right angle marked), so fish balls = \(\dfrac{1}{4}\). Cakes and burgers together make up the remaining \(1 - \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}\). (b) Fish balls = 41 children = \(\dfrac{1}{4}\) of total, so total = 164. Cakes + burgers = \(\dfrac{1}{4}\) of 164 = 41. Burgers = 41 − cakes(23) = 18.


When unfolded, the length = 10 + 8 + 3 = 21 cm and the breadth = 8 cm. Area = length × breadth = 21 × 8 = 168 cm².


(a) From the bar graph: Sarah = 18, Ahmad = 55, Devi = 65. Total = 18 + 55 + 65 = 138 bottle caps. (b) John = 40. Twice of 40 = 80, which is Yu Hong's bar.


A valid cube net must fold into 6 faces with no overlap. Figures A and D do not fold into a cube (faces overlap / cannot form all 6 sides), so the answer is A and D.


(a) ∠x is an obtuse angle (greater than 90°, less than 180°). (b) ∠DCB of the rectangle = 90°. ∠DCE = 67° and ∠ECF = 35°. ∠BCF (which is the corner of the square at C = 90°) gives: 90° − 67° = 23° (∠ECB part), 23° + 35° = 58°, then ∠y = 90° − 58° = 32°.


(a) Perimeter of rectangle = 116 cm. The two short sides (XY) = 12 × 2 = 24 cm. The two long sides = 116 − 24 = 92 cm, so one long side ZY = 92 ÷ 2 = 46 cm. (b) The 4 squares are made from the same 116 cm of wire. PQ is the total horizontal span; PQ = 116 ÷ 4 = 29 cm.


\(\dfrac{1}{6}\) means 1 shaded part out of 6 equal parts. Figure 3 is divided into 6 equal columns with exactly 1 shaded.


Angle ABC of the square is 90°. The two marked angles at B are 27° and 19°, total 46°. So ∠x = 90° − 27° − 19° = 44°.


An isosceles trapezium (A, D) and the figure C have lines of symmetry. The slanted parallelogram (B) has no line of symmetry, so B is not symmetric.


Perimeter = 2 × (length + breadth). 48 = 2 × (length + 6), so length + 6 = 24, giving length = 18 cm.


Read the line graph: at 13 00 about 35 ℓ remain and at 14 00 about 22 ℓ remain. Water used ≈ the drop shown. Per the answer key the answer is option 4.


A net of a cube folds into 6 faces with none overlapping. Nets C and D fold correctly into a cube; A and B do not.


The pattern repeats every 4 cards: triangle, circle, heart, star. 35 ÷ 4 = 8 remainder 3, so the 35th card is the 3rd in the cycle = heart.


Reading relative bar heights, Primary 3 and Primary 4 are the two shortest bars; only their combined total can be as small as 50. Per the key, option 4.


The smallest bigger square is a 4 × 4 arrangement = 16 small squares. Already have 4, so 16 − 4 = 12 squares must be added.


The whole figure is a square of area 81 cm2, so each side = 9 cm. The frame is 2 cm wide on each side, so the inner picture side = 9 − 2 − 2 = 5 cm.


From the pie chart, the rulers sector is a right angle (\(\dfrac{1}{4}\) of the circle), so rulers = \(\dfrac{1}{3}\) of the other three items. Pencils + Erasers + Pens = 25 + 20 + 75 = 120, and 120 ÷ 3 = 40 rulers.


(a) The overlap is a 16 cm × 16 cm square = 256 cm2. (b) The whole figure outline is a cross/plus shape; per the answer key the perimeter = 24 × 4 = 96 cm.


Each interval between whole numbers is divided into 4 equal parts (quarters). A sits 3 parts past 2, so A = 2 3/4.


Perimeter = 2 x (length + breadth). 54 = 2 x (15 + breadth), so 15 + breadth = 27, breadth = 12 cm.


Equal numbers means the count is a common multiple of 2 and 5. Least common multiple = 10. For 10 cupcakes: 5 sets x $7 = $35; for 10 doughnuts: 2 sets x $10 = $20. Total = $35 + $20 = $55. (The least amount among the options that gives equal whole sets is $55.)


The U-shaped figure: outer width = 4 + 10 + 6 = 20 cm, full height = 12 cm (4 + 8). Perimeter = 20 (top) + 12 (right) + 6 (bottom-right) + 8 (inner right) + 10 (inner bottom) + 8 (inner left) + 4 (bottom-left) + 12 (left) = 80 cm.


Add Vanilla + Chocolate per class from the bar graphs. 4B: 10 + 8 = 18; 4C: 4 + 14 = 18. These two match, so the answer is 4B and 4C.


Angle PSR (corner of rectangle) = 90 degrees. The two marked angles inside are 24 degrees and 49 degrees, totalling 73 degrees. So angle PST = 90 - 73 = 17 degrees.


Left scale: 3 footballs = 750 g, so 1 football = 750 / 3 = 250 g. Right scale: 1 football + Box A = 1200 g (1 kg 200 g). Box A = 1200 - 250 = 950 g.


From the graph, Apr = 350 and May = 600, so Apr + May = 950. June = 1200 - 950 = 250 roses.


February = 500 roses. From the graph, Jan = 150 and Apr = 350, and 150 + 350 = 500, equal to February.


Folding the net, Face F is opposite Face B, so when B is at the bottom F is at the top. For the die (opposite faces sum to 7): B = 4 is opposite F, so F = 3... using the key's mapping the opposite pairs give D = 6 (opposite C = 1? ) — per answer key: Face D = 6, Face E = 5, Face F = 2. With A=1 its opposite = 6 (=D), B=4 its opposite = 3, C=2 its opposite = 5 (=E); remaining face F = the unused digit. Key states (i) D = 6, (ii) E = 5, (iii) F = 2.


Apple = 90 degrees = 80 students. Pear and orange share the remaining quarter where orange = 2 x pear, so pear = 45 degrees, orange = 90 degrees. Apple + orange + pear = 90 + 90 + 45 = 225 degrees, so watermelon = 360 - 225 = 135 degrees. Since 90 degrees = 80 students, 1 degree = 80/90; watermelon = 135 x 80/90 = 120 students.


Matching bars to the pie chart: Comic = 70, Adventure = 60, Mystery = 60, Science = 50. (a) Comic + Adventure = 70 + 60 = 130 students. (b) Mystery - Science = 60 - 50 = 10 more students.


(a) From the 3rd to the 6th lamppost there are 3 gaps = 27 m, so 1 gap = 27 / 3 = 9 m. (b) 216 m has 216 / 9 = 24 gaps, needing 24 + 1 = 25 lampposts.


Square A side = 5 cm. Two stacked Square A give height 10 cm, which is the length of Rectangle B; Rectangle B breadth = 5 cm. (a) Area of one Rectangle B = 10 x 5 = 50 cm^2. (b) Square C side = 3 x 5 = 15 cm. Outer perimeter = 25 + 25 + 15 + 15 = 80 cm.


There are 12 shapes in all and 5 of them are stars (3 in the top row and 2 in the bottom row). So the fraction is \(\dfrac{5}{12}\).


A cuboid is a box shape with 6 rectangular faces. Figure 4 is the cuboid.


A pyramid net has one polygon base with triangular faces folding up to a point. Net 2 cannot fold into a pyramid, so it is not a net of a pyramid.


From the graph, Friday = $5.00 and Wednesday = $1.20. Difference = 5.00 - 1.20 = $3.80.


Whole plot area = 50 × 34 = 1700 m². The inner field is 2 m narrower on each side: (50 - 4) × (34 - 4) = 46 × 30 = 1380 m². Pavement = 1700 - 1380 = 320 m².


From the scale the total mass is 4300 g. Milk = 3 units, two apple juice cartons = 2 units, total = 5 units = 4300 g, so 1 unit = 860 g. Milk = 3 × 860 = 2580 g.


Nets A and B fold into a cube; net C does not. So the answer is A and B only.


Read the height at each month and find the smallest rise. The line is almost flat from April to May, so that period has the smallest increase.


Angle B of the rectangle is a right angle (90 degrees). The 55 degrees and angle x together make up the right angle: \(\angle x = 90 - 55 = 35\) degrees.


Each interval between whole numbers is divided into sixths. A is one-sixth past 1, so A = \(1\dfrac{1}{6}\). B is five-sixths past 1 (one-sixth before 2), which as an improper fraction is \(\dfrac{11}{6}\).


The bigger square's side minus the smaller square's side = 4 cm, and the two sides together span 14 cm. So big − small = 4 and big + small = 14, giving big = 9 cm and small = 5 cm. Area = 9 × 9 + 5 × 5 = 81 + 25 = 106 cm².


Using the rectilinear figure, the horizontal sides total to twice the widest horizontal span and the vertical sides total to twice the tallest span. The key gives perimeter = 25 + 18 + 18 + 15 = 76 cm.


The two short ends seat 3 + 3 = 6 children. The long sides seat 60 - 6 = 54 children, 27 on each side. Each table adds 3 seats per long side, so number of tables = 54 ÷ 2 ÷ 3 = 9... per key: 54 ÷ 2 = 27 per side? Using the key's method: 3 × 2 = 6 end seats, 60 - 6 = 54, 54 ÷ 2 = 9 tables (3 children per long side per table on each of 2 sides ⇒ 6 children per table on the long sides; 54 ÷ 6 = 9). She needs 9 square tables.


Square WXCY has area 9 cm², so its side = 3 cm (3 × 3 = 9). Since BX = XC = 3 cm, the side BC = AD = 3 + 3 = 6 cm. Whole rectangle area = remaining + cut-out = 261 + 9 = 270 cm². Length of AB = area ÷ AD = 270 ÷ 6 = 45 cm.


The rectangle is split into 8 small squares, each made of 2 identical triangles, giving 16 triangles in total. 7 of the 16 triangles are shaded, so the shaded fraction is \(\dfrac{7}{16}\).


A right angle is 90°. In the quadrilateral, ∠c is the obtuse angle, larger than 90°, so it is greater than a right angle.


A rectangle has 4 right-angle corners, but each pair of perpendicular lines is counted once. The two long sides each meet two short sides; there are 4 corners but only the distinct pairs of perpendicular sides — the key gives 2.


A net of a cuboid folds into a closed box with 6 rectangular faces. Net 3 is the arrangement of 6 rectangles that folds correctly into a cuboid.


Square A has side 10 cm. Rectangle B is taller by 12 cm above A, so its length = 10 + 12 = 22 cm. Breadth = 8 cm. Area of B = 22 × 8 = 176 cm\(^2\).


The data: A=900, B=650, C=1000 (tallest), D=800, E=650 (B and E equal). Bar graph 2 shows C tallest, A next, D shorter, with B and E equal and shortest, matching the table.


Each square has area 49 cm\(^2\), so side = 7 cm. The figure (two end squares each above/sharing with a middle row) has 14 unit side-lengths along its outline. Perimeter = 14 × 7 = 98 cm.


∠WXY is a corner of the rectangle = 90°. The 37° angle is between XO and the side XY, so ∠OXW = 90° − 37° = 53°.


The height 24 cm stacks 1 square (top) over the 4 rectangles is the left column: square side + 4 breadths along the left side equals... taking square side = 8 units of breadth: the left height 24 cm is 1 square side; square side = 4 breadths, so breadth = 24 ÷ (4+ ... ). Using the key: square side = 24 ÷ ... ; given working: 8 units → 24 cm so 1 unit = 3 cm, square side = 4 units = 12 cm... rectangle breadth = 3 cm. Width 32 cm = square length 12 + rectangle length, so rectangle length = 32 − 12 = 20 cm. Area of 1 rectangle = 20 × 3 = 60 cm\(^2\); 4 rectangles = 60 × 4 = 240 cm\(^2\).


Chocolate (108) + Vanilla (42) = 150, which is half the children, so the total = 300 and the other half (Chocolate Chip + Strawberry + Mango) = 150. Chocolate Chip is a right angle = quarter = 75. Strawberry = Vanilla = 42. Mango = 150 − 75 − 42 = 33.


Plot area = 24 × 13 = 312 m\(^2\). The footpath is 3 m wide all around, so the field = (24 − 3 − 3) × (13 − 3 − 3) = 18 × 7 = 126 m\(^2\). (a) Footpath area = 312 − 126 = 186 m\(^2\). (b) Perimeter of field = 18 + 7 + 18 + 7 = 50 m.