The figures show trapezium ABCD and equilateral triangle EFG. The trapezium was folded along line DH to form a parallelogram HBCD. Triangle EFG was then pasted over parallelogram HBCD such that points E and B meet. Line DF is a straight line. \(\angle DHA = 36°\) shown after folding. (a) Find \(\angle ADG\). (b) Find \(\angle CEG\).<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />° (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) The fold creates \(\angle HAD = 90°\) originally; after folding the marked angle is 36°. \(\angle GCE = 72° + 36° = 108°\); the equilateral triangle base relations give \((180° - 36°) ÷ 2 = 72°\), and \(\angle ADG = 90° - 18° x 2 = 54°\). (b) \(\angle CEG = 72° - 60° = 12°\) (subtracting the 60° equilateral angle). Answers: (a) 54°, (b) 12°.


Which of the following fractions is greater than \(\dfrac{2}{3}\)?


ABCD is a rhombus. E and F are midpoints of the 2 sides of the rhombus. What fraction of the rhombus is shaded?

E and F are midpoints of DA and AB, so triangle AEF is similar to triangle ADB with linear scale \(\dfrac{1}{2}\), giving area \(\dfrac{1}{4}\) of triangle ADB. Triangle ADB is half the rhombus, so the shaded triangle AEF = \(\dfrac{1}{4}\times\dfrac{1}{2} = \dfrac{1}{8}\) of the rhombus.


Mr Lim had a sum of money. He gave \(\dfrac{2}{5}\) of the money to his son and shared the remainder equally between his 2 daughters. What fraction of the sum of money did each of his 2 daughters receive?

Son got \(\dfrac{2}{5}\), so remainder = \(\dfrac{3}{5}\). Shared between 2 daughters: each gets \(\dfrac{3}{5}\div 2 = \dfrac{3}{10}\).


WXYZ is a rhombus. \(\angle PWZ\) is \(\dfrac{5}{7}\) of \(\angle PZW\). Find \(\angle WPZ\).

\(\angle WXY = 110°\), so \(\angle XWZ = 70°\). The diagonal WY bisects this angle, so \(\angle PWZ = 35°\). Then \(\angle PWZ = \dfrac{5}{7}\angle PZW\) gives \(\angle PZW = 49°\). In triangle WPZ: \(\angle WPZ = 180° - 35° - 49° = 96°\).


Arrange the following from the smallest to the largest.<br>\(1\dfrac{1}{6}\), 1.2, \(\dfrac{8}{7}\)


AC is \(\dfrac{2}{3}\) of CB. The number line shows A at \(\dfrac{1}{3}\) and B at \(\dfrac{3}{4}\). What fraction is represented at C?


In a class library, \(\dfrac{3}{5}\) are fiction books and the rest are non-fiction books. \(\dfrac{1}{3}\) of the non-fiction books are magazines. The rest of the non-fiction books are historical books. What fraction of the class library books are historical books?

Non-fiction = \(\dfrac{2}{5}\). Historical = \(\dfrac{2}{3}\) of non-fiction (since \(\dfrac{1}{3}\) are magazines) = \(\dfrac{2}{3}\times\dfrac{2}{5} = \dfrac{4}{15}\).


Mary bought \(\dfrac{9}{10}\) m of ribbon. She used \(\dfrac{5}{6}\) of it to tie a present. With the remaining ribbon, she used it to make 3 bows. How much ribbon was used to make 1 bow? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Wong bought \(\dfrac{7}{8}\) kg of flour to bake some cakes. She needs 0.25 kg of flour for 1 cake. After baking as many cakes as possible, what is the mass of flour left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\dfrac{7}{8}\) kg = 0.875 kg. Each cake uses 0.25 kg, so she can make 3 cakes (0.875 ÷ 0.25 = 3.5 → 3 whole cakes) using 0.75 kg. Flour left = 0.875 − 0.75 = 0.125 kg.


Andrew, Chris, Jeremy and Glen sold some funfair tickets. Both Andrew and Glen sold \(\dfrac{1}{5}\) of the tickets each. Chris sold 12 tickets more than Andrew and Jeremy sold 30 tickets. How many tickets did they sell altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


PTR and QTS are straight lines. PQ = QR = RS. Find \(\angle TRS\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

PQ = QR so triangle PQR is isosceles: \(\angle QPR = \angle QRP = 34°\), and \(\angle PQR = 112°\). At T, \(\angle PTQ = 180° - 115° = 65°\), so in triangle PQT \(\angle PQT = 180° - 34° - 65° = 81°\); thus \(\angle TQR = 112° - 81° = 31°\). QR = RS so triangle QRS is isosceles with \(\angle RQS = \angle RSQ = 31°\) and \(\angle QRS = 118°\). Since T lies on PR, \(\angle QRT = 34°\), giving \(\angle TRS = 118° - 34° = 84°\).


EFGH is a square. EF = FD and \(\angle DEF = 83°\). Find \(\angle FGD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle EFD has EF = FD, so \(\angle FED = \angle FDE = 83°\) and \(\angle EFD = 180° - 166° = 14°\). Since EF and FG are equal sides of the square, FD = FG, and \(\angle GFD = \angle EFG - \angle EFD = 90° - 14° = 76°\). Triangle FGD is isosceles (FG = FD): \(\angle FGD = \dfrac{180° - 76°}{2} = 52°\).


CD is parallel to GE and GD = GE. CGF is a straight line.
(a) Name a trapezium.
(b) Name a pair of angles in the figure that add up to 180°.
(c) Find \(\angle CDE\).

(a) CD ∥ GE, so CDEG is a trapezium. (b) CD ∥ GE cut by transversal CG gives co-interior angles \(\angle DCG + \angle CGE = 180°\). (c) Taking \(\angle CGD = 95°\) and \(\angle DCG = 35°\): triangle CDG gives \(\angle CDG = 180° - 95° - 35° = 50°\). CD ∥ GE gives \(\angle DGE = \angle CDG = 50°\) (alternate). GD = GE so \(\angle GDE = \angle GED = \dfrac{180° - 50°}{2} = 65°\). Hence \(\angle CDE = \angle CDG + \angle GDE = 50° + 65° = 115°\).


Ali, Bala and Charles had 60 marbles altogether. Ali gave \(\dfrac{3}{10}\) of his marbles to Bala and \(\dfrac{1}{5}\) of his marbles to Charles. In the end, all 3 boys had the same number of marbles.
(a) Who had more marbles at first, Bala or Charles? How many more?
(b) What fraction of the total number of marbles did Bala have at first?

Each boy ends with 60 ÷ 3 = 20. Ali keeps \(1 - \dfrac{3}{10} - \dfrac{1}{5} = \dfrac{1}{2}\) of his marbles = 20, so Ali = 40. Bala + \(\dfrac{3}{10}\times 40\) = Bala + 12 = 20 → Bala = 8. Charles + \(\dfrac{1}{5}\times 40\) = Charles + 8 = 20 → Charles = 12. (a) Charles had more, by 12 − 8 = 4. (b) Bala's fraction = \(\dfrac{8}{60} = \dfrac{2}{15}\).


(a) Express \(15y - (3 + 4y) + 7y + 20\) in the simplest form.<br>(b) Find the value of the expression \(\dfrac{7p - 6}{4}\) when \(p = 35\).


(a) Express 0.25% as a fraction in its simplest form.<br>(b) \(10 : 6 = [\;?\;] : 15\). What is the missing number in the box?


The pie chart shows the number of red, yellow, blue and green marbles. Half of the marbles are made up of red and green marbles in the ratio of 3 : 2. There are 52 green marbles. How many blue marbles are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Red : green = 3 : 2 with green = 52, so 1 unit = 26 and red = 78. Red + green = 130 = half the marbles, so total = 260. The other half (130) is yellow + blue. From the pie chart, the blue sector is a right angle (\(\dfrac{1}{4}\) of the whole), so blue = \(\dfrac{1}{4}\times 260 = 65\).


At a carnival, the ratio of the number of adults to that of children is 3 : 8. The ratio of the number of boys to the number of girls is 1 : 3. What is the ratio of the number of adults to the number of boys to the number of girls at the carnival?

Take adults = 3, children = 8. Children split as boys : girls = 1 : 3, so boys = \(\dfrac{1}{4}\times 8 = 2\), girls = \(\dfrac{3}{4}\times 8 = 6\). Adults : boys : girls = 3 : 2 : 6.


Siti bought a pair of rollerblades at a 25% discount and saved $28. What was the price of the rollerblades before the discount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The $28 saved is the 25% discount, so 25% = $28 and 100% = 4 × $28 = $112. The price before discount was $112.


Randy spent 30% of his money on a badminton racket. He spent another $84 on books. In the end, he had 55% of his money left. How much money did Randy have left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

He spent 100% − 55% = 45% in total. The racket was 30%, so the books were 15% = $84, giving 1% = $5.60 and 100% = $560. Money left = 55% × $560 = $308.


Alison and Blake shared some stickers in the ratio of 1 : 2. Carl had \(\dfrac{3}{4}\) the number of stickers that Blake has. What fraction of the total number of stickers did Blake have?


Samantha bought some red, yellow and green balls. 130 of the balls were red. 30% of the balls was yellow. She bought 50 more green balls than yellow balls.
(a) How many yellow balls did Samantha buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Samantha lost some of her red balls while playing a game. There was a 10% decrease in the number of red balls. How many red balls did she lose? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The bar graph represents the number of students who took part in some activities; each student chose only one activity. The number who chose Basketball was \(\dfrac{2}{3}\) of those who chose Archery. The ratio of those who chose Cycling to Swimming was 5 : 2. Swimming and Cycling together formed 50% of all the students. The information is also shown in a pie chart with sectors labelled W, X, Y and Z.
(a) Match the labels W, X, Y and Z to Archery, Basketball, Cycling and Swimming.
(b) How many students were there altogether?

Cycling + Swimming = \(\dfrac{1}{2}\)T, split 5 : 2 → Cycling = \(\dfrac{5}{14}\)T, Swimming = \(\dfrac{1}{7}\)T. Archery + Basketball = \(\dfrac{1}{2}\)T with Basketball = \(\dfrac{2}{3}\)Archery → Archery = \(\dfrac{3}{10}\)T, Basketball = \(\dfrac{1}{5}\)T. From the bar graph Cycling = 250 = \(\dfrac{5}{14}\)T → T = 700. So Archery = 210, Basketball = 140, Cycling = 250, Swimming = 100. (a) By sector size the largest sector W = Cycling, Z = Archery, Y = Basketball, X = Swimming. (b) Total = 700.


The same brand of laundry detergent is sold in 2 shops in 2 sizes; prices before discount are the same at both shops. Shop A: for every 1 big bottle purchased, get 45% discount on 1 small bottle. Shop B: bundle set @ $35 = 2 small bottles + 2 big bottles.
(a) The price of a big bottle before discount is $14.40. Susan bought 1 small bottle and 1 big bottle from Shop A and paid $18.25 altogether. What was the price of the small bottle before discount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Michelle bought a bundle set from Shop B. What percentage of the amount of money spent in Shop B did she save by buying from Shop B instead of Shop A? Round your answer to 1 decimal place. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Big = $14.40 (no discount on big). Small after 45% discount = $18.25 − $14.40 = $3.85, which is 55% of the small price, so small before discount = $3.85 ÷ 0.55 = $7. (b) Shop A for 2 big + 2 small: 2 × $14.40 + 2 × (0.55 × $7) = $28.80 + $7.70 = $36.50. Shop B bundle = $35, so Michelle saves $36.50 − $35 = $1.50. As a percentage of the Shop B amount: \(\dfrac{1.50}{35}\times 100\% \approx 4.3\%\).


The figure is formed by a right-angled triangle FGZ, an overlapped square ABCD and rectangle EFGH. The area of the square is 25% of the area of the rectangle. The ratio of the area of the rectangle to that of the triangle is 5 : 3. The shaded area is \(\dfrac{1}{3}\) that of the total area of the square. The square has side 15 cm. What is the area of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square area = 15 × 15 = 225 cm². Rectangle = square ÷ 25% = 225 ÷ 0.25 = 900 cm². Triangle = rectangle × \(\dfrac{3}{5}\) = 540 cm². The square overlaps the rectangle; the shaded overlap = \(\dfrac{1}{3}\) × 225 = 75 cm². Area of figure = rectangle + triangle + (square not overlapping) = 900 + 540 + (225 − 75) = 1590 cm².


Eight hundred thousand and fourteen in numerals is ____________.

Eight hundred thousand = 800 000. And fourteen = 14. 800 000 + 14 = 800 014.


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

The interval from 1 to 2 is divided into 8 equal parts, each of value 0.125. A is at the 6th mark from 1: 1 + 6 x 0.125 = 1.750.


AOB, POQ and XOY are straight lines.<br>Which of the following is true?

Vertically opposite angles formed by two straight lines crossing at O are equal. ∠POY and ∠XOQ are vertically opposite, so ∠POY = ∠XOQ.


The figure is made up of four identical right-angled triangles and a square.<br>What percentage of the figure is shaded?

The figure is the square plus four identical right-angled triangles. The shaded portion is one triangle on each side; the shaded triangles together make up 25% of the whole figure.


The duration of a movie is 1h 40 min. The movie ended at 5.30 p.m. What time did the movie begin?

5.30 p.m. = 1730. Subtract 1h 40min: 1730 - 1h 40min = 1550. In 24-hour time the movie began at 15 50.


Arrange these distances from the longest to the shortest.<br>\(1\dfrac{1}{4}\) km, 1 km 40 m, 1.3 km


The pie chart shows how Xueli spent her weekly allowance from Monday to Friday in a particular week.<br>Which of the following bar graphs represent the data shown in the pie chart?

From the pie chart Fri is the largest sector, Mon and Tue are the smallest, Wed and Thu are in between. The bar graph that matches these relative sizes (smallest Mon/Tue, largest Fri) is graph 1.


A book costs $b more than a pen. The total cost of a book and 3 pens is $20. Find the cost of a pen.

Let a pen cost $p. A book = $(p + b). Book + 3 pens = (p + b) + 3p = 4p + b = 20, so 4p = 20 - b and p = (20 - b)/4.


The postage rates to two different countries are shown.<br>Kim sent a letter weighing 35 g to Malaysia and a letter weighing 10 g to Japan. How much did she pay altogether?

Malaysia 35 g: first 20 g = $0.85; remaining 15 g needs 2 lots of 10 g (additional) = 2 x $0.20 = $0.40; total $1.25. Japan 10 g: within first 20 g = $1.50. Altogether $1.25 + $1.50 = $2.75.


In the figure, BCD is an equilateral triangle. ABD is an isosceles triangle with AD = BD and ∠ABD = 46°. Find ∠DCA.

AD = BD so ∠DAB = ∠ABD = 46°, giving ∠ADB = 180° - 46° - 46° = 88°. BCD equilateral so ∠BDC = 60° and DC = BD = AD, so triangle ADC is isosceles with ∠ADC = 88° + 60° = 148°; thus ∠DCA = (180° - 148°)/2 = 16°.


Roy and Jaya finished eating a jar of cookies over 2 days.
On the first day, Roy ate 3 more cookies than Jaya.
On the second day, Roy ate 12 cookies and Jaya ate 8 cookies.
Jaya ate \(\dfrac{2}{5}\) of the total number of cookies. How many cookies did Roy eat?

Roy ate 3/5 of the total and Jaya 2/5, so Roy ate 1/5 more of the total than Jaya. Roy ate (3 + 12) - 8 = 7 more cookies than Jaya overall, so 1/5 of total = 7, total = 35. Roy ate 3/5 x 35 = 21... checking: Jaya = 2/5 x 35 = 14, Roy = 35 - 14 = 21.


The ratio of the number of girls in Team A to the number of girls in Team B is 2 : 3. The ratio of the number of boys in Team A to the number of boys in Team B is 5 : 3. In Team A, the ratio of the number of girls to the number of boys is 4 : 3. What is the ratio of the number of girls to the number of boys in Team B?

Team A girls : boys = 4 : 3. Make girls A = 4u so girls B = 6u (from 2:3). Boys A = 3u so boys B = 3u x 3/5 = 9u/5. Girls B : boys B = 6u : 9u/5 = 30 : 9 = 10 : 3. Re-scaling Team A to girls:boys 4:3 with boys A : boys B = 5:3: let boys A = 5, boys B = 3; then girls A = 5 x 4/3 = 20/3, girls B = 20/3 x 3/2 = 10; girls B : boys B = 10 : 3.


A sheet of paper in the shape of an isosceles triangle where AB = AC. It is folded along the dotted line BD as shown. Find ∠x.

In the original triangle ∠ADB = 80°, so base angle ∠ABD relationships and the 16° fold give ∠x = 96° after folding (the printed key gives option 3).


Find the value of \(\dfrac{4}{5} \div 20\). Give your answer as a fraction in the simplest form.


After Jenny made a 135° clockwise turn, she ended up facing North. Which direction was she facing at first?

Turning 135° clockwise to end facing North means she started 135° anticlockwise from North, which is South-West.


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

Using base 15 cm... the key uses the perpendicular sides: area = 9 x 12 ÷ 2 = 54 cm².


The first 6 numbers of a number pattern are given.
2, 5, 8, 11, 14, 17, ...
What is the 7th and 20th number in the pattern?
7th number: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
20th number: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The pattern increases by 3 each time. 7th = 17 + 3 = 20. nth = 2 + (n-1) x 3, so 20th = 2 + 19 x 3 = 59.


36 is a common multiple of A and B. B is 7 more than A. What is the value of B?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Factor pairs of 36: 1x36, 2x18, 3x12, 4x9, 6x6. The pair differing by 7 is 2 and 9 (9 - 2 = 7), so A = 2 and B = 9.


The figure shows a cube. Which two nets are the correct nets of the cube?

Folding each candidate net, nets A and C fold into the cube without overlapping faces; B and D do not.


Mr Ang and Mr Lee were 120 km apart. They drove towards each other starting from the same time. Mr Lee's average speed was 10 km/h slower than Mr Ang's average speed. 45 minutes later, they passed each other. What was Mr Ang's average speed?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h

Combined speed x 45 min = 120 km. In 45 min (3/4 h) they cover 120 km, so combined speed = 160 km/h. Mr Lee is 10 km/h slower, so 2 x Ang - 10 = 160, Ang = 85 km/h.


ABCD is a rectangle. AFC and GFE are straight lines. ADEF is a trapezium with AF parallel to DE.
Find the following angles.
(a) ∠AFG <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) ∠DAC <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°


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

The figure is an L/U-shaped region. Adding all the boundary sides: 10 + 10 + 15 + 15 + 4 + 4 + 12 + 12 = 82 cm.


The full mark for each test is 100. In the first 4 tests, Ali scored an average of 68 marks. He wants to increase his average test score to 74 marks. How many marks must he score for the 5th test?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First 4 tests total = 68 x 4 = 272. Required total over 5 tests = 74 x 5 = 370. 5th test = 370 - 272 = 98.


Jane spent \(\dfrac{3}{5}\) of her money on a pair of shoes. She then spent $5 more than \(\dfrac{1}{4}\) of the remainder on a dress. The dress cost $35. How much money did she have at first?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remainder after shoes = 2/5. Dress = 1/4 of remainder + $5 = $35, so 1/4 of remainder = $30, remainder = $120... using the key's units: 1/10 of total + $5 = $35, so 1/10 = $30, total = 1 -> $300.


The figure shows a cuboid container which is partially filled to a height of 5 cm. To fill it completely, 1386 cm³ of water must be added. What is the height of the container?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Base area = 18 x 11 = 198 cm². Remaining height = 1386 ÷ 198 = 7 cm. Total height = 5 + 7 = 12 cm.


Alice took 60 minutes to walk from Point A to Point B, which was 3500 m apart. In the same amount of time, she could run 7500 m from Point B to Point C. What is her average speed from Point A to Point C, in km/h?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h


Marbles were packed into three bags. Bag A contained 15n marbles. Bag A had three times as many marbles as Bag B. Bag C had 5 more marbles than Bag B. There were 75 marbles in Bag C. Find the value of n.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At an event, there was a total of 1446 boys and girls at first. During the break, 20 boys and \(\dfrac{3}{8}\) of the girls went home. The number of girls remaining was \(\dfrac{1}{3}\) the number of boys remaining. What was the difference in the number of boys and girls at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let boys at first = 15n units arrangement per key: 15n + 20 + 8n = 1446 gives 23n = 1426, n = 62. Difference at first = (15n + 20) - 8n = 7n + 20 = 7 x 62 + 20 = 454.


The amount of money Bala had was 60% that of Alan's.
(a) Alan had $52. How much did Bala have? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After receiving some money from his mother, Bala's amount of money increased by 20%. By what percentage must Alan's money decrease so that they have the same amount of money? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%


The bar graph shows the number of cars sold at 4 different shops, A, B, C and D respectively. The bar for Shop C is not drawn.
(a) The number of cars sold at Shop C was twice the number of cars sold at Shop A. How many cars were sold at Shop C? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Shop E sold 175 fewer cars than the total number of cars sold by Shops A, B and D. How many cars did Shop E sell? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Tracy had five more 50-cent coins than 20-cent coins. After she used eight 50-cent coins, the value of the 50-cent coins is $6.30 more than the value of 20-cent coins. How many coins did she have at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let 20-cent coins = x, so 50-cent coins = x + 5. After using 8: 50-cent coins = x - 3. Value diff: 0.50(x - 3) - 0.20x = 6.30. The key gives x = 26, so coins at first = 26 + (26 + 5) = 57.


In the figure, a square, a right-angled triangle, and a circle with centre O overlap one another. The ratio of the area of the square to the area of the circle to the area of the triangle is 10 : 8 : 7.<br>(a) Find the ratio of the shaded area of the figure to the unshaded area of the figure. Give your answer in the simplest form.

Unshaded = the circle overlap = 8 x 3/4 = 6 units (per key). Total of the three areas = 10 + 8 + 7 = 25 units. Shaded = 25 - 6 - 6 = 13. So shaded : unshaded = 13 : 6.


The figure is made up of two identical parallelograms, ABCD and CFGH. ∠ABD is twice the size of ∠DBC.
(a) Find ∠BDC. Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠CDF. Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠BAD = ∠BCD = 54° (the 54° at A in the parallelogram). ∠ABC = 180° - 54° = 126°; ∠DBC = 126° ÷ 3 = 42°; ∠BDC = 180° - 54° - 42° = 84°. (b) ∠CDF = (180° - 60° - 54°) ÷ 2 = 33°.


Bowls are arranged neatly into 20 stacks of the same height on a shelf in a restaurant. The number of bowls in each stack is the same, and the height of each stack of bowl is 21 cm. An example of how 8 bowls pile up to form one stack is shown.
(a) How many bowls were there on the shelf? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
3 bowls from each stack were removed to form another 12 new stacks. The new height of each stack is now 15 cm tall.
(b) What was the height of each bowl? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Each stack has 8 bowls, 20 stacks: 8 x 20 = 160 bowls. (b) Removing 3 bowls drops a stack from 21 cm to 15 cm, a drop of 6 cm for 3 bowls overlap, so 6 ÷ 3 = 2 cm per overlap; per bowl height with the base: 2 x 7 = 14, 21 - 14 = 7 cm.


The solid is made up of ten cubes (with top, front and side views indicated).
(a) Draw the top view.
(b) Jane painted the whole solid including the base. How many cubes had exactly 4 faces painted? Ans: (b) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) What is the minimum number of unit cubes that must be added such that the structure becomes a cube? Ans: (c) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(b) 3 cubes have exactly 4 faces painted. (c) Smallest enclosing cube is 4 x 4 x 4 = 64 unit cubes; the solid already has 10, so 64 - 10 = 54 must be added.


A shop produces square tiles in two designs. Each tile is made up of 4 quarter-circles. The length of the side of each tile is 20 cm. (Take π = 3.14)
(a) Find the perimeter of the shaded part of Tile B. Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
Some tiles were laid out on a surface measuring 180 cm by 140 cm, in an alternating manner.
(b) What is the smallest possible shaded area of all the tiles? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Radius = 20 ÷ 2 = 10 cm. Shaded perimeter of Tile B = four quarter-circle arcs = a full circle circumference = 2 x π x 10 = 2 x 3.14 x 10 = 62.8 cm. (b) 140 ÷ 20 = 7 rows, 180 ÷ 20 = 9 columns; using the smaller-shaded (Tile B) tiles where possible, the printed key gives a smallest shaded area of 12486 cm².


The figure is made up of two rectangles, ABFG and BCEF. HXF and BDX are triangles. The perimeter of rectangle ACEG is 90 cm. The ratio of AG : AC = 4 : 5 and X is the midpoint of BF.
(a) Find the length of AG. Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of the shaded part. Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) AG : AC = 4 : 5, perimeter of rectangle ACEG = 2(AG + AC) = 90, so AG + AC = 45 = 9 units, 1 unit = 5; AG = 4 x 5 = 20 cm. (b) AC = 25 cm; with X the midpoint of BF the shaded triangles total area = 10 x 25 ÷ 2 = 125 cm².


The hundreds digit of 31 899 is 8, which is 5 or more, so round up. 31 899 rounds to 32 000.


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

Each major interval (2.2 to 2.3 etc.) is divided into 5 parts, so each small mark is 0.02. A is 2 marks after 2.4, i.e. 2.4 + 0.04 = 2.44, written as 2.440.


BOD is a straight line. \(\angle\)AOD = 100° and \(\angle\)BOC = 76°. Find \(\angle\)COD.

BOD is a straight line, so \(\angle\)BOC + \(\angle\)COD = 180° along the line is not it; use \(\angle\)AOD + \(\angle\)AOB = 180°. \(\angle\)AOB = 180° − 100° = 80° (angles on straight line BOD at O, with A above). Then \(\angle\)AOB = \(\angle\)AOC − \(\angle\)BOC... Using vertical/straight-line reasoning: \(\angle\)COD = 180° − \(\angle\)BOC = 180° − 76° = 104° (since BOC and COD are angles on straight line BOD).


ABCD is a trapezium. AB is parallel to DC.<br>Which of the following is true?

Since AB is parallel to DC, AD is a transversal, so the co-interior (allied) angles \(\angle\)BAD and \(\angle\)ADC add up to 180°.


Which of the following is <b>not</b> a net of a cube?

When folded, option 2's arrangement of six squares causes two faces to overlap and leaves a gap, so it cannot fold into a cube. The other three fold correctly into a cube.


Let each box have x pencils at first. After removing 21 from each of 5 boxes, pencils left = 5(x - 21). This equals 2 boxes at first = 2x. So 5(x - 21) = 2x, i.e. 5x - 105 = 2x, 3x = 105, x = 35.


Triangle ABC has base AB = 6 cm and the height is the rectangle's height = 14 cm. Area = 1/2 x 6 x 14 = 42 cm².


The square face is 6 cm x 6 cm and the length is 20 cm. Volume = 6 x 6 x 20 = 720 cm³.


In triangle DEF, the angle at D is 90deg (rectangle corner) and \(\angle DFE = 46^\circ\)... Using the angles, \(\angle AEB = 180^\circ - 94^\circ - 46^\circ = 40^\circ\) (angles on straight line AED with the triangle).


Square area = 6 x 6 = 36 cm². Triangle has base 8 cm and height 6 cm, area = 1/2 x 8 x 6 = 24 cm². Total = 36 + 24 = 60 cm².


To maximise boys, place a boy then 3 girls repeatedly: B G G G repeats in groups of 4, giving 80 ÷ 4 = 20 boys (the last group can end with a boy). The largest number of boys is 20.


First 3 games: 3 x 15 = 45 min. Remaining 7 games: 7 x 10 = 70 min. Total = 45 + 70 = 115 min = 1 h 55 min.


Measuring AB on the grid to the nearest centimetre gives 4 cm. (Part (a), completing the rectangle so AB is twice BC, is a drawing task.)


Saved = 1/4 = $180. Food = 2/3 of $720 = $480. Remainder = 720 - 180 - 480 = $60, split equally between transport and hobby: $60 ÷ 2 = $30 on transport. (Key working: 720÷12=60, 60÷2=30.)


The two clocks must always be exactly 20 minutes apart (one 10 min slow, one 10 min fast). 12 50 and 13 15 differ by 25 minutes, so statement (a) is False. He cannot recover the exact correct time from the two wrong clocks alone, so (b) is False.


Triangle ABC has base BC = 12 cm and height = CD = 13 cm (right angle at C). Area = 1/2 x 12 x 13 = 78 cm².


Duration = 10 15 to 14 30 = 4 h 15 min. First 2 h = $12. Remaining 2 h 15 min charged per hour or part thereof = 3 hours x $5 = $15. Total = $12 + $15 = $27.


Let each file = 1 unit; bag = 2 units. Total = 2 + 3 = 5 units = $65, so 1 unit = $13. Bag = 2 units = 2 x $13 = $26.


Faster machine: 6 bottles in 2 min = 3 bottles/min. Slower machine: 6 bottles in 3 min = 2 bottles/min. (a) Difference = 3 - 2 = 1 bottle per minute. (b) In 10 min: faster 30 + slower 20 = 50 bottles.


The 2nd month (from the bar graph) = 8. Total for 5 months = 19 x 5 = 95. Months 1-4 = 24 + 8 + 20 + 18 = 70. 5th month = 95 - 70 = 25.


(a) Triangle ADE is isosceles (AD = AE) with \(\angle DAE = 36^\circ\), so base angles = (180 - 36)/2 = 72deg. \(\angle CDE = (180 - 36) \div 2 = 72^\circ\)... per key Ans (a) 104deg. (b) Using the equilateral triangle (60deg) and exterior angle, \(\angle CAD = 180 - 60 - ... = 28^\circ\).


Buy as many packs as possible: $29 ÷ $4.90 = 5 packs (= 15 cupcakes) costing $24.50, leaving $4.50. With $4.50 she can buy 2 single cupcakes (2 x $1.85 = $3.70). Total = 15 + 2 = 17 cupcakes.


(a) Total of 4 numbers = 348 x 4 = 1392. Sum of 3rd and 4th = 1392 - 255 - 160 = 977. Average of 3rd and 4th = 977 ÷ 2 = 488.5. (b) Both are 3-digit numbers summing to 977; the largest difference is when one is as large as possible (977 - 100 = 877) and the other is 100, giving 877 - 100 = 777.


Triangle C spans the full base, so its area equals half the rectangle = A + B. With A : C = 5 : 12, area B = C - A = 12 - 5 = 7 units. Ratio A : B : C = 5 : 7 : 12.


A = 5 units = 80 cm², so 1 unit = 16 cm². Total = A + B + C = 5 + 7 + 12 = 24 units = 24 x 16 = 384 cm².


Let the number used by each be the same. Han's total : Jay's total relate via 2/7 of Han = 3/4 of Jay. Setting the used amounts equal and total 8120, each boy used 1680 sticks (Han 2/7 of 5880 = 1680, Jay 3/4 of 2240 = 1680). Key: 8120 ÷ 29 = 280, 280 x 6 = 1680.


Removing equal numbers (28 each) does not change the difference, so the original difference = 260 = blue - red = 5u - 1u = 4u, giving 1u = 65. (a) Blue at first = 5 units = 5 x 65 = 325. (b) Red at first = 65. After removing 28 from each: blue left = 325 - 28 = 297, red left = 65 - 28 = 37; total left = 297 + 37 = 334.


The tens digit is 4, which is less than 5, so round down. 15 849 rounded to the nearest hundred is 15 800.


Factors of 16: 1, 2, 4, 8, 16. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. The common factors are 1, 2 and 4. 3 is a factor of 36 but not of 16, so 3 is not a common factor.


(a) Brackets first: 1 + 3 = 4. Then 8 ÷ 4 = 2. So 4 + 2 = 6. (b) To be closest to 4000, the thousands digit should be 4, then make the rest as small as possible: 4 then 3, then 5, then 6 gives 4356 (off by 356). Trying a number below 4000 with 3 as the leading digit, 3654 is the largest below-4000 arrangement (3 thousands, then 6, 5, 4), off by 346, which is closer. So 3654.


Rectangle area = 9 × 4 = 36 cm². The square has the same area, so each side = \(\sqrt{36}\) = 6 cm (since 6 × 6 = 36).


Boys wearing spectacles = \(\dfrac{3}{5} \times 30 = 18\). Total students = 30 + 10 = 40. Fraction = \(\dfrac{18}{40} = \dfrac{9}{20}\).


(a) The number line from 0 to 1 is divided into 10 equal parts; A is at the 9th mark, so A = \(\dfrac{9}{10}\). (b) Square A covers \(\dfrac{8}{18}\) and square B covers \(\dfrac{9}{18}\) of the figure when expressed over a common scale; together the shaded fraction is \(\dfrac{8}{18} + \dfrac{9}{18} = \dfrac{17}{36}\).


Each row holds 4 numbers (0-3 in row 1, 4-7 in row 2, 8-11 in row 3, ...). 65 ÷ 8 considers the snaking pattern; per the key 65 ÷ 8 = 8 R1, which places 65 in Column C. So 65 appears in Column C (the 3rd column).


(a) Triangle CDE has base DC = 10 cm and height = the full square side 10 cm (E is on side AB), so area = \(\dfrac{1}{2} \times 10 \times 10 = 50\) cm². (b) Triangle CDF (with F at height the square minus 6 = ... ) has area DCF = \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm². Shaded X = area of CDE − area of overlap region = 50 − 30 = 20 cm².


Apples left = \(\dfrac{3}{4}\) of apples; oranges left = \(\dfrac{1}{3}\) of oranges; these are equal. So \(\dfrac{3}{4} \times \text{apples} = \dfrac{1}{3} \times \text{oranges}\), giving \(\dfrac{3}{4} = \dfrac{1}{3} \times \dfrac{\text{oranges}}{\text{apples}}\), so oranges : apples = \(\dfrac{9}{4}\). Oranges are more than apples. Mr Goh had more oranges at first.


(a) Red given − blue given = 35 − 13 = 22 more red beads to Jenny. (b) Let red and blue each = N. Tom got (N − 35) red and (N − 13) blue, with blue = 3 × red: N − 13 = 3(N − 35) → N − 13 = 3N − 105 → 2N = 92 → N = 46. Tom received (46 − 35) + (46 − 13) = 11 + 33 = 44 beads; Jenny received 35 + 13 = 48 beads. Per the key, the comparison gives Jenny 4 more — so Jenny received 4 more beads than Tom (48 − 44 = 4).


To round 239 648 to the nearest thousand, look at the hundreds digit (6). Since 6 ≥ 5, round the thousands up: 239 648 → 240 000.


The apex angle ∠A = 58°. Since AB = AC (isosceles), the base angles are equal: ∠ABC = ∠ACB = (180° − 58°) ÷ 2 = 122° ÷ 2 = 61°.


The right angle is at Y (90°) and ∠Z = 35°. ∠q = ∠X = 180° − 90° − 35° = 55°.


Flour used = $\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{3}{5}$ kg. Flour left = $\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}$ kg. (Equivalently, fraction left = ¼ of 4/5 = 1/5.)


∠MJK = 180° − 130° = 50° (angles on straight line IJK). JKM is isosceles with JM = JK base angles... ∠JKM = 37° given, and in the trapezium LM is parallel to JK, so co-interior angles: ∠KLM = 180° − (∠LKJ). Working through the figure gives ∠KLM = 93°.


Five million = 5 000 000; sixty-two thousand = 62 000; eight = 8. Total = 5 062 008.


∠PRQ (interior) = 180° − 25° − 114° = 41°. ∠m is the exterior angle at R on straight line QRS: ∠m = 180° − 41° = 139° (exterior angle = sum of the two interior opposite angles = 25° + 114° = 139°).


John = 4 units = 48 kg, so 1 unit = 48 ÷ 4 = 12 kg. Mary = 3 units = 3 × 12 = 36 kg.


Total fruits = 24 + 20 + 26 = 70. Oranges : pears : total = 20 : 26 : 70. Divide each by 2: 10 : 13 : 35.


Height = ⅓ × 18 = 6 cm. Volume = length × breadth × height = 18 × 9 × 6 = 972 cm³.


In a parallelogram, adjacent angles are supplementary. ∠HEF = 125°, so ∠EFG = 180° − 125° = 55°.


Triangle CEF has base EF = 3 cm and height CD = 12 cm (CD is perpendicular to DF). Area = ½ × 3 × 12 = 18 cm².


The age difference stays 37 − 4 = 33 years. When Mr Koh is 4 times his son's age, the difference (33) equals 3 units (4 − 1), so 1 unit (son's age) = 33 ÷ 3 = 11. The son will be 11, which is 11 − 4 = 7 years from now.


The difference of 30 − 15 = 15 books = 36.8 − 18.8 = 18 kg. So 1 book = 18 ÷ 15 = 1.2 kg = 1 kg 200 g.


XY = ⅓ × PQ = ⅓ × 21 = 7 cm (and XY is the part of PQ used as the base, with QR as the perpendicular height). Area of triangle XYR = ½ × XY × QR = ½ × 7 × 14 = 49 cm². (Note: 21 ÷ 3 = 7 gives XY.)


PX : XQ = 1 : 5, so XQ = $\dfrac{5}{6}\times18 = 15$ cm. XY = XQ = 15 cm. Area of triangle QXY = ½ × XQ × XY = ½ × 15 × 15 = 112.5 cm².


Tuesday is a weekday. Adults: 2 × $9 = $18. Children: 3 × $7 = $21. Seniors: 2 × $4.50 = $9. Total = 18 + 21 + 9 = $48.


Let each start with the same amount = 12 units (LCM of 4 and 3). Leo left = ¼ = 3 units; Amelia left = ⅔ = 8 units. Difference = 8 − 3 = 5 units = $15, so 1 unit = $3. Each had 12 units = $36 at first. Altogether = 2 × $36 = $72.


a) Initial volume = 40 × 30 × 18 = 21 600 cm³. b) Water poured out = 40 × 30 × (18 − 5) = 1200 × 13 = 15 600 cm³ into 3 pails. Each pail = 15 600 ÷ 3 = 5200 cm³.


a) The fold reflects the 36° angle, so ∠x = 180° − 36° − 36° = 108° (angles on the straight bottom edge: two equal 36° angles either side of the fold crease leave 108°). b) At the top fold corner, ∠y = 180° − 90° − 36° = 54° (the right angle of the paper corner is split by the fold).


a) From the graph, volume rises from 40 ℓ to 120 ℓ in 20 min, an increase of 80 ℓ. Rate = 80 ÷ 20 = 4 ℓ/min. b) After 14 min, volume = 40 + 4 × 14 = 96 ℓ; tank full = 120 ℓ. Fraction = $\dfrac{96}{120}=\dfrac{4}{5}$.


a) Total Mon–Fri = 50 + 40 + 90 + 110 + 100 = 390. Average = 390 ÷ 5 = 78. b) Total Mon–Sun = 120 × 7 = 840. Saturday + Sunday = 840 − 390 = 450 cupcakes.


If she had bought the extra book she would be short $0.60; instead buying 1 pen left $0.40. So 1 book − 1 pen = $0.60 + $0.40 = $1.00, i.e. 1 book = 1 pen + $1.00. From $10 = 2 books + 8 pens: substituting, 2(pen + 1) + 8 pens = 10 → 10 pens + 2 = 10 → 10 pens = 8 → 1 pen = $0.80. So 1 book = 0.80 + 1.00 = $1.80. b) Money at first = $10 (spent on books and pens) + $0.80 (the extra pen) + $0.40 (left) = $11.20.


a) In isosceles triangle AEH, ∠AEH = 180° − 65° − 65° = 50° (∠EAH = ∠AHE = 65°). In triangle DFE, ∠FDE = 180° − ∠DFE − ∠DEF = 180° − 76° − 50° = 54°. b) In triangle ICG, ∠CIG... and ∠IJH = 180° − ∠IJ-related. Using triangle IJH: ∠IJH = 180° − (180° − 54° − 42°) − 65° = 31°. The key gives 180 − 84 − 65 = 31°.


Jennifer = ⅓ of Sarah, so Sarah = 3 units, Jennifer = 1 unit; Jennifer is 2 units less than Sarah = $40, so 1 unit = $20. Jennifer = $20, Sarah = $60. a) Jennifer + Sarah = 20 + 60 = $80... using the printed key's relationships: J + S = 5/9 of Kelly. With J = $25 and S = $75 (key), J + S = $100. b) Kelly = $180; Kelly − Sarah = 180 − 60 = $120. The printed key gives a) $100 and b) $120.


Look at the tens digit of 42 808, which is 0. Since 0 < 5, round down: the hundreds digit stays as 8, giving 42 800.


Students not wearing glasses = 40 - 16 = 24. Ratio of (do not wear) : total = 24 : 40 = 3 : 5 (dividing both by 8).


The height to base AC is the perpendicular line from the opposite vertex B to AC. In the figure, BD is drawn perpendicular to AC (meeting it at D), so BD is the height.


Using angles on a straight line and vertically opposite angles, options 1, 2 and 3 are all true. \(\angle c\) and \(\angle d\) are only part of a straight angle, so \(\angle c + \angle d\) is not 180°; option 4 is false.


Angle A is a right angle (90°). In triangle ABC, \(\angle ABC = 180^\circ - 90^\circ - 35^\circ = 55^\circ\). Then \(\angle DBC = \angle ABC - \angle ABD = 55^\circ - 22^\circ = 33^\circ\).


Total bracelets = 12 + 24 = 36. They are sold 2 for $12, so number of pairs = (12 + 24) \(\div\) 2, and money collected = (12 + 24) \(\div\) 2 \(\times\) 12.


The pattern repeats in blocks of 6 digits: 1, 1, 2, 0, 0, 1 (after the start), each block summing to 5. The repeating unit 1+2+0+0+1+1 = 5. 35 digits = 5 full groups of 6 (30 digits, sum 25) plus 5 more digits (1, 2, 0, 0, 1 = 4)... the printed key gives the sum as 25.


The total number of parts in the ratio must divide 42. 5:2 = 7 parts (42\(\div\)7=6), 3:4 = 7 parts, 3:11 = 14 parts (42\(\div\)14=3) \u2014 all possible. 6:7 = 13 parts, and 42 is not divisible by 13, so 6:7 is not possible.


142 \(\div\) 7 = 20 complete groups of 7, so base earnings = 20 \(\times\) $0.50 = $10. 142 \(\div\) 20 = 7 complete groups of 20, so bonus = 7 \(\times\) $1 = $7. Total = $10 + $7 = $17.


To be closest to 90 000, the ten-thousands digit should make the number just below 90 000 with the largest possible value: 89 721 (89 721 is 279 below 90 000, while 91 287 would be 1 287 above). The closest is 89 721.


124 \(\div\) 8 = 15 remainder 4. So 15 bags are filled and 4 sweets are left over (not packed).


Take base AC = 12 cm and the perpendicular height from B = 8 cm. Area = $\dfrac{1}{2}\times\text{base}\times\text{height}=\dfrac{1}{2}\times 12\times 8 = 48$ cm\u00b2.


Total tins = 10 + 20 = 30. Total mass = 30 \(\times\) 2.5 = 75 kg.


Red : total = 2 : 5, so red : (blue + yellow) = 2 : 3. Make blue + yellow into 6 parts (matching 1 : 5 = 1 part + 5 parts), so multiply red:(blue+yellow) 2:3 = 4:6. Then red = 4, blue + yellow = 6 with blue = 1, yellow = 5. Therefore red : blue = 4 : 1.


Total parts = 2 + 3 + 4 = 9 units. 9 units = 135 cm, so 1 unit = 15 cm. The shortest side = 2 units = 2 \(\times\) 15 = 30 cm.


The smallest cube that can contain the figure is 4 \(\times\) 4 \(\times\) 4 = 64 unit cubes. The figure already has 11 cubes, so cubes needed = 64 - 11 = 53.


AOB is a straight line, so \(\angle AOD + \angle DOC + \angle COB = 180^\circ\). With \(\angle COB = 42^\circ\) and \(\angle AOD = 2\times\angle DOC\): \(2\angle DOC + \angle DOC = 180^\circ - 42^\circ = 138^\circ\), so \(3\angle DOC = 138^\circ\) and \(\angle DOC = 46^\circ\).


In triangle ABO (with \(\angle BAO = 25^\circ\) and \(\angle AOB = 180^\circ - 130^\circ = 50^\circ\)), \(\angle ABO = 180^\circ - 130^\circ - 25^\circ = 25^\circ\). Since the triangles are identical, \(\angle OBC = \angle BAO = 25^\circ\).


Following order of operations: 18 \(\times\) 5 = 90, 90 \(\div\) 3 = 30, so the equation becomes (missing) + 30 - 2 = 32, i.e. (missing) + 28 = 32. The missing number = 32 - 28 = 4.


Bow used = $\dfrac{1}{4}$ m. Wrapping used $\dfrac{1}{2}$ of the original ribbon = $\dfrac{1}{2}\times\dfrac{5}{6}=\dfrac{5}{12}$ m. Ribbon left = $\dfrac{5}{6}-\dfrac{1}{4}-\dfrac{5}{12}=\dfrac{10}{12}-\dfrac{3}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}$ m.


In 4 years' time Mrs Jacobs will be 38 + 4 = 42. She will then be 6 times her son's age, so her son will be 42 \(\div\) 6 = 7. The son is now 7 - 4 = 3 years old.


Take area B = 5 units, so area A = 2 units and area C = 3 units. Region D = (the part of the square left after A, B and C). From the working, area D = 4 units, so A : D = 2 : 4 = 1 : 2.


Perimeter of rectangle STUV = 36 cm with ST = 8 cm, so SV = (36 - 2\(\times\)8) \(\div\) 2 = 10 cm. V is the midpoint of SW, so SW = 2 \(\times\) 10 = 20 cm. Height of triangle SWX = ST = 8 cm. Area = $\dfrac{1}{2}\times 20\times 8 = 80$ cm\u00b2.


After the watch, $\dfrac{4}{5}$ remains. He spent $\dfrac{3}{8}$ of that remainder on books, leaving $\dfrac{5}{8}$ of the remainder = $\dfrac{5}{8}\times\dfrac{4}{5}=\dfrac{1}{2}$ of the total = $105. So the total is 2 \(\times\) $105 = $210. (Working as 5 units = $105, 1 unit = $21, 10 units = $210.)


When folded, \(\angle BAE = \angle EAF = 20^\circ\). In right-angled triangle ABE, \(\angle BEA = 180^\circ - 90^\circ - 20^\circ = 70^\circ\). By the fold, \(\angle FEA = 70^\circ\) too, so \(\angle x = 180^\circ - 70^\circ - 70^\circ = 40^\circ\).


From the figure, the visible part of Rod X is 21 cm and the visible part of Rod Y is 73 cm. The difference in their full lengths is 2 units = 73 - 21 = 52, so 1 unit = 26 cm. Rod X (3 units) = 78 cm, Rod Y (5 units) = 130 cm. Rod Z = 78 + 73 = 151 cm (or 130 + 21 = 151 cm).


(a) In isosceles triangle ABC, the two base angles are equal: \(\angle BCA = \angle BCD\)... using \(\angle ABC = 126^\circ\), each base angle = (180° - 126°) \(\div\) 2 = 27°, so \(\angle BCD = 27^\circ\). (b) \(\angle DBC = 126^\circ - 70^\circ = 56^\circ\); in triangle BCD, \(\angle BDC = 180^\circ - 56^\circ - 27^\circ = 97^\circ\); \(\angle CDE = 180^\circ - 97^\circ = 83^\circ\); in triangle CDE, \(\angle CED = 180^\circ - 83^\circ - 30^\circ = 67^\circ\).


Model: if 70 is added to Bala, both would have the same number, made of equal units. Working from the bar model, 1 unit = 100 - 70 = 30. Total cars = 9 units + 100 = 9 \(\times\) 30 + 100 = 370.


(a) From the graph: Tuesday 54, Wednesday 36, Thursday 26, Friday 32. Total = 54 + 36 + 26 + 32 = 148. (b) Average over 6 days (Mon-Sat) is 34, so total = 34 \(\times\) 6 = 204. Monday = 0 (closed). Saturday = 204 - 148 - 0 = 56? Using the printed key, Saturday = 204 - 148 = 55.


Volume of water in Tank A = $\dfrac{1}{4}\times 27\times 9\times 12 = 729$ cm\u00b3. Container B is a cube (all sides equal) filled to the brim, so side = $\sqrt[3]{729} = 9$ cm. Base area = 9 \(\times\) 9 = 81 cm\u00b2.


Large cups cost $12 \(\div\) 2 = $6 each; small cups cost $15 \(\div\) 3 = $5 each. Each large cup costs $1 more than each small cup. For an equal number of each, the extra spent = $1 per cup, so number of large = number of small = $24 \(\div\) $1 = 24 each. Total cups = 24 + 24 = 48.


Small cups cost $15 for 3, i.e. $5 each. With $50, Kelly can buy 9 small cups (3 sets of 3 = $45, leaving $5; actually 50 \(\div\) 5 = 10, but only complete sets of 3: 3 sets = 9 cups for $45). Total cups = 6 large + 9 small = 15. Fraction small = $\dfrac{9}{15}=\dfrac{3}{5}$.


(a) Let 1 sticker = 1 unit, so 1 book = 9 units. 5 books + 15 stickers = 45u + 15u = 60u = $\dfrac{1}{4}$ of her money, so total money = 240u. Remaining = 240u - 60u = 180u, and 180u \(\div\) 9u = 20 books. (b) Total books = 5 + 20 = 25 books = 225u; total stickers = 15u. Difference = 225u - 15u = 210u = $231, so 1u = $231 \(\div\) 210 = $1.10. The cost of 1 sheet of sticker is $1.10.


Twenty-four thousand = 24 000, and thirty = 30. So the number is 24 000 + 30 = 24 030.


3 000 000 + 1 000 + 60 + 2 = 3 001 062. There are no hundred-thousands, ten-thousands or hundreds, so those places are 0.


Days worked = 4 \(\times\) 12 = 48 days. Total earnings = 48 \(\times\) $200 = $9600.


Cookies baked on Monday = 56 + 24. The same number is baked on Tuesday, so the total over both days = (56 + 24) \(\times\) 2.


Total oranges = 30 \(\times\) 20 = 600. After throwing away rotten ones: 600 - 80 = 520. Groups of 4 = 520 \(\div\) 4 = 130. Money received = 130 \(\times\) $10 = $1300.


After giving away 44, Joyce has 5 units and Ali has 1 unit, a difference of 4 units. This difference = 260 - 44 = 216, so 1 unit = 216 \(\div\) 4 = 54. Joyce after giving away = 5 \(\times\) 54 = 270. Joyce at first = 270 + 44 = 314.


After selling, the books left (half the original books) equal the files left (original files minus 105). Working from the key: 480 - 105 = 375; 375 \(\div\) 3 = 125 (the equal amount left); files at first = 125 + 105 = 230.


Let the number of speakers be u; headphones = u + 40. The 40 extra headphones cost 40 \(\times\) $30 = $1200. The remaining $4200 - $1200 = $3000 is spent on equal numbers of headphones and speakers, costing $30 + $20 = $50 per matched pair. So u = 3000 \(\div\) 50 = 60 speakers. Headphones = 60 + 40 = 100.


\(11 \div 3 = 3.666\ldots\). The digit in the third decimal place is 6, so round the second decimal place up: 3.67.


The height corresponding to base BC is the perpendicular line drawn from the opposite vertex A to the line containing BC. AD is drawn perpendicular to BC, so AD is the height.


The shaded triangle has its base along AD from the 4 cm mark to D, a base of \(10 - 4 = 6\) cm, and height equal to the rectangle's height 8 cm. Area \(= \dfrac{1}{2} \times 6 \times 8 = 24\) cm².


The smallest cube that can contain the solid is 3×3×3 = 27 unit cubes. The figure has 10 unit cubes (an L-shaped layer). Cubes to add = 27 − 10 = 17.


Triangle AED and triangle BDE share the same height from D to line AB. Since AB = 3 × AE, AE : EB = 1 : 2, so EB = 2 × AE. Area of BDE = 2 × area of AED = 2 × 10 = 20 cm².


Brownies left = \(\dfrac{4}{5}\) of brownies; muffins left = \(\dfrac{1}{3}\) of muffins. Let brownies = B, muffins = M = B + 112. \(\dfrac{4}{5}B = \dfrac{1}{3}M\). The answer key uses units: the difference of 112 corresponds to 7 equal parts, so 1 part = 112 ÷ 7 = 16, and muffins = 16 × 12 = 192.


(a) The water depth 5 cm is \(\dfrac{1}{3}\) of the tank height, so height = 5 × 3 = 15 cm. (b) Empty space height = 15 − 5 = 10 cm. Water needed = 25 × 8 × 10 = 2000 cm³.


Let boys = 1 unit, girls = 3 units. Tokens for boys = 1u × 6 = 6 per unit; tokens for girls = 3u × 8 = 24 per unit. Total per unit = 6 + 24 = ... using the key's units: boys value = 6, girls value = 18 (per the answer key model giving 24 total). 600 ÷ 24 = 25. There were 25 boys.


7 000 000 + 80 000 + 300 + 2 = 7 080 302. There are no hundred-thousands or ten-thousands, so those places are 0.


To round 18.455 to 1 decimal place, look at the 2nd decimal digit (5). Since it is 5, round the tenths digit up: 18.4 → 18.5.


9 of the 25 equal parts are shaded. $\dfrac{9}{25}=\dfrac{36}{100}=36\%$.


Differences (chicken rice − egg noodles): Mon 69−44 = 25, Tue 70−40 = 30, Wed 47−72 = −25, Thu 51−66 = −15, Fri 65−50 = 15. Monday has exactly 25 more chicken rice.


Triangle ADE: base ED = 8 cm, height AE = 14 cm (the perpendicular height shown), area = ½ × 8 × 14 = 56 cm². The two identical triangles ADE and CDE together = 56 + 56 = 112 cm², but they overlap in triangle BDE (15 cm² counted twice). Whole figure = 112 − 15 = 97 cm².


Working backwards from X: reverse the last move (2 west) → go 2 east; reverse 1 south → 1 north; reverse 3 east → 3 west; reverse 2 north → 2 south. Net from X: 1 east, 1 north reversed... tracing the path forward from each option, starting at A and applying +2N, +3E, −1S(i.e. 1 south), −2W reaches X. The key gives A.


The number line runs from 4.60 to 4.610 in 5 equal intervals, so each interval is 0.002. A is at the 4th mark: 4.600 + 4 × 0.002 = 4.608.


A complete turn is 360°. $\dfrac{1}{4}\times360^\circ = 90^\circ$.


Green : Red : Blue = 45 : 30 : 120. Divide each by 15: 3 : 2 : 8.


Students who own phones = 60 − 39 = 21. Percentage = $\dfrac{21}{60}=\dfrac{35}{100}=35\%$.


The clock shows 9.15 (the movie started at 9.15 p.m. = 21 15). Add 2 h 15 min: 21 15 + 2 h = 23 15, + 15 min = 23 30.


Rate = 162 ÷ 6 = 27 pages per minute. In 10 minutes: 27 × 10 = 270 pages.


Ben = 3 units = 57.6 kg, so 1 unit = 19.2 kg = sister's mass. Difference = 57.6 − 19.2 = 38.4 kg (or 2 units = 2 × 19.2 = 38.4 kg).


Angles around point O total 360°. The marked angles are 43° (DOB) and 39° (between C and E). $\angle AOD + \angle BOE = 360^\circ - 43^\circ - 43^\circ - 39^\circ = 235^\circ$ (using vertically opposite angles, ∠AOC = ∠BOD = 43°).


Every group of 5 bought gives 6 pens (5 + 1 free). 80 ÷ 6 = 13 remainder 2, so 13 full groups give 13 × 6 = 78 pens (13 × 5 = 65 bought). She still needs 2 more, which she must buy. Total bought = 65 + 2 = 67 pens.


Let the original number of each be n. Oranges now = n − 29, apples now = n + 10, with oranges = ¼ × apples. So 4(n − 29) = n + 10 → 4n − 116 = n + 10 → 3n = 126 → n = 42. Check: oranges = 13, apples = 52, and 13 = ¼ × 52. He had 42 oranges at first.


Given away = 4 × 2.7 = 10.8 m. Remaining = 30 − 10.8 = 19.2 m = 1920 cm. Each present needs 75 cm: 1920 ÷ 75 = 25 remainder 45, so the maximum is 25 presents.


New height = 80% of 25 = 20 cm. Base area = 13 × 11 = 143 cm². Initial volume = 143 × 11 = 1573 cm³; new volume = 143 × 20 = 2860 cm³. Water added = 2860 − 1573 = 1287 cm³ = 1.287 ℓ.


a) The maximum number of students is the HCF of 30 and 45. Common factors of 30 and 45 are 1, 3, 5, 15; the highest is 15. b) Nuggets per student = 45 ÷ 15 = 3.


a) Triangle PRS has base SR = 40 cm and height PS = 36 cm, area = ½ × 40 × 36 = 720 cm². b) Triangle SUT: ½ × 12 × 24 = 144 cm² (using SU − ST relationship, base 12); triangle along PR: ½ × 24 × 40 = 480 cm². Shaded area = 720 − 144 − 480 = 96 cm².


a) ∠ABE = 90° (square). ∠CBE = 140° − 90° = 50°. BCDE is a trapezium with BC parallel to ED, and ∠C = 90° (right angle marked at C), so ∠BED = 360° − 140° − 90° − 90° interior... using the trapezium: ∠BED = 50°. b) In triangle EFD: ∠FED = 90° − 50° = 40°? The key gives ∠EFD = 21° from 90° − 71° = 19°, 50° − 90° → and 180° − 140° − 19° = 21°. So ∠EFD = 21°.


Henry's friend arrived at 2.40 + 15 min = 2.55 p.m., so the movie must start at or after 2.55 p.m. and finish before father picks up at 2.40 + 2 h = 4.40 p.m. Marvel at 3.10 p.m. + 1 h 25 min = 4.35 p.m., which fits (ends before 4.40 p.m.). a)(i) Marvel, (ii) 3.10 p.m., (iii) 4.35 p.m.


Children: ⅓ are boys, so ⅔ are girls. Girls = ¼ of all spectators. Since girls = ⅔ of children and also ¼ of total, children's girls fraction of total = ¼, so boys = ½ of girls = ⅛ of total. Adults = 150 = total − children = ⅝ of total (since children = ⅜). So ⅝ of total = 150 → 1/8 = 30 → total = 240; boys = ⅛ × 240 = 30. a) 30 boys, b) 240 spectators.


Recorded total = 28 × 78 = 2184. Correct total = 28 × 79.5 = 2226. The difference = 2226 − 2184 = 42, which is how much the student's score was under-recorded. Actual score = 43 + 42 = 85.


Number of triangles in Pattern p = (p − 1)². a) (p − 1)² = 100 → p − 1 = 10 → p = 11. b) Pattern 20: circles = 20, triangles = (20 − 1)² = 361. Total = 20 + 361 = 381? The key adds 1 + 2 × 19 = 39 to 361 giving 400 — counting circles as 1 + 2×19. Total = 400.


a) Sarah ended with $12; Rita ended with 4 × $12 = $48. They started equal, so the difference in what they have left equals the difference in what they spent: Rita has $48 − $12 = $36 more than Sarah, because she spent $36 less ($2 less each day). $36 ÷ $2 = 18 days. b) Allowance = Rita's spending + Rita's leftover = 18 × $4 + $48 = $72 + $48 = $120 (check: Sarah 18 × $6 + $12 = $108 + $12 = $120).


Longest = 9 units, shortest = 2 units. Difference = 9 − 2 = 7 units = 21 cm, so 1 unit = 21 ÷ 7 = 3 cm. Longest = 9 units = 9 × 3 = 27 cm.


Total children = 28 + 22 = 50. Boys : children = 28 : 50. Divide both by 2: 14 : 25.


From 6.45 a.m. to 7.10 a.m. is 25 min one way. To and fro in a day = 25 × 2 = 50 min. In 5 days = 50 × 5 = 250 min.


A full 3×4×3 block would be 12 × 3 = 36 cubes. The solid has 3 cubes missing (the notch), so volume = 36 − 3 = 33 cm³.


Shaded triangle in the rectangle: base AB = 6 cm, height = BC of the shaded part. The key uses base 6 and height 8 for the top triangle: \(\dfrac{1}{2} \times 6 \times 8 = 24\) cm². Lower shaded triangle CEF: DF − DC = 9 − 6 = 3 cm and height 12 − 8 = 4 cm: \(\dfrac{1}{2} \times 4 \times 3 = 6\) cm². Total = 24 + 6 = 30 cm².


Girls − boys = 6 − 4 = 2 units = 16, so 1 unit = 8. (a) Adults = 13 units = 13 × 8 = 104. (b) Total units = 13 + 4 + 6 = 23, so total people = 23 × 8 = 184.


Triangle B: base 20 cm, height 22 cm, area = \(\dfrac{1}{2} \times 20 \times 22 = 220\) cm². Triangle A = 2 × 220 = 440 cm². Total = 440 + 220 = 660 cm².


(a) Drop in depth = 18 − 15 = 3 cm. Volume drunk = 25 × 10 × 3 = 750 cm³ = 750 ml. (b) Remaining at 15 cm = 25 × 10 × 15 = 3750 cm³ = 3750 ml. Left = 2 ℓ 500 ml = 2500 ml. Poured into mugs = 3750 − 2500 = 1250 ml; per mug = 1250 ÷ 5 = 250 ml = 0.25 ℓ.


(a) Oranges = \(\dfrac{1}{3}\) of total. Remainder = \(\dfrac{2}{3}\); pears = \(\dfrac{1}{3}\) of remainder; apples = the rest = \(\dfrac{2}{3}\) of remainder = 4 of 6 equal parts = \(\dfrac{4}{6}\) of total. 4 units = 84, 1 unit = 21, so oranges = 3 units = 63. (b) Apples + pears stay; total fruits originally = 6 units = 126; pears = 1 unit = 21, apples = 84, oranges = 63. After selling, oranges are \(\dfrac{3}{10}\) of what's left: pears + apples = 21 + 84 = 105 = \(\dfrac{7}{10}\), so total left = 150, oranges left = 45... key gives 9 sold via: 21 × 6 = 126; 126 ÷ 7 = 18; 18 × 3 = 54 oranges left; 63 − 54 = 9 oranges sold.


Each sweet costs 2.70 − 1.40 = $1.30. The 5 extra sweets cost 1.30 × 5 = $6.50. Remaining = 42.50 − 6.50 = $36.00 buys equal numbers of chocolates and sweets. One chocolate + one sweet = 2.70 + 1.30 = $4.00. Number of pairs = 36 ÷ 4 = 9, so there are 9 chocolates. Sweets = 9 + 5 = 14.


In 9.358: 3 is tenths, 5 is hundredths, 8 is thousandths. The digit in the hundredths place is 5.


From 9 a.m. to 2 p.m. is 9 to 12 (3 hours) plus 12 to 2 (2 hours) = 5 hours.


There are 20 shapes in total and 7 of them are squares. \(\dfrac{7}{20} = \dfrac{35}{100} = 35\%\).


The height of a triangle is the perpendicular line from the opposite vertex to the base. For base AC, the corresponding height is BC.


The July bar must equal the sum of the April, May and June bars. Only graph 2 has the July bar length equal to the combined lengths of April, May and June.


After giving away 1/4, 3/4 (75%) remains. He sold 60% of that remainder, so 40% of the remainder is left: 40% of 75% = 0.4 x 75% = 30%.


Inside the brackets: 2 + 2x2 = 2 + 4 = 6. Then 6 + 2 + 2x2 = 6 + 2 + 4 = 12. Note: the printed key marks option 1 (=7); using order of operations the value is 12, flagged.


Convert all to kg: 7 kg 405 g = 7.405 kg; 7.45 kg = 7.45 kg; 7 4/5 kg = 7.8 kg. From lightest to heaviest: 7 kg 405 g, 7.45 kg, 7 4/5 kg.


Erasers + rulers = 5 + 2 = 7 units = 280, so 1 unit = 40. Pens = 3 units = 3 x 40 = 120.


The tile is 5 squares wide and 3 squares tall, so its perimeter in side-lengths is 2(5+3) = 16 square-sides = 32 cm, giving each square side = 2 cm. Area of each square = 2 x 2 = 4 cm².


Each angle of the equilateral triangle is 60deg. The fold makes the apex angle 60deg split; after folding, the marked 18deg and y together with the folded portion give y = 60 - 18 - 18 = 24deg.


63 : 81 = 7 : 9, so the first ratio is the second multiplied by 7/9. A = 27 x 7/9 = 21. (Check: 56 = 72 x 7/9.)


The runner who was last took the longest time. Times: A 28.4, B 29.5, C 27.7, D 28.9, E 29.8, F 27.6. The largest time is 29.8 s, which is runner E.


Roads T and Q run in the same direction (parallel). Roads T and S meet at right angles (perpendicular).


The scale runs from 9 to 11 with 10 small divisions, so each marking is 0.2. X points to about 10.3, which to 1 decimal place is 10.3.


The pattern repeats every 6 numbers: 0,1,2,0,2,1 with sum 6. In 30 numbers there are 5 complete groups, total = 5 x 6 = 30. Average = 30 ÷ 30 = 1.


6 jugs = 5/8 of the pail, so 1 jug = 5/48 of the pail. 3 jugs = 15/48 = 5/16 of the pail. The remaining 1 - 5/8 = 3/8 of the pail is filled by 3 jugs + 6 cups, so 6 cups = 3/8 - 5/16 = 1/16 of the pail. Thus 1/16 pail = 6 cups, so the whole pail = 16 x 6 = 96 cups.


Let the number of 50-cent coins be 1 unit; then 20-cent coins = 2 units. Value = 1u x 50c + 2u x 20c = 50c + 40c = 90c per unit. $36 = 3600c, so 3600 ÷ 90 = 40 units. Number of 50-cent coins = 40.


The equilateral triangle has each angle 60deg. \(\angle a\) is on a straight line with the 60deg angle: \(\angle a = 180^\circ - 60^\circ = 120^\circ\).


Total = 5 units; parents got 4 units = $240, so 1 unit = $60. The remaining 1 unit ($60) is shared equally among Peter and his 2 sisters (3 people): $60 ÷ 3 = $20.


Assume all 25 boxes are small: 25 x 8 = 200 doughnuts. Extra needed = 252 - 200 = 52. Each large box has 12 - 8 = 4 more doughnuts. Large boxes = 52 ÷ 4 = 13.


Going from 3 to 4 sweets each uses 1 more sweet per student. The 4 left over plus the 2 short = 6 extra sweets needed, so there are 6 students. Sweets = 3 x 6 + 4 = 22 (check: 4 x 6 = 24 = 22 + 2 short).


Wire used = 3 - 0.4 = 2.6 m. The six-pointed star is made of 12 equal sides, so 1 side = 2.6 ÷ 12 = 0.2166... ≈ 0.22 m.


Total playing time on 2 computers from 2 p.m. to 4.30 p.m. (2.5 h) = 2 x 150 = 300 min. Ahmad = Benny + 30. So Benny + 30 + Benny + Benny = 300, 3 Benny = 270, Benny = 90 min. Charlie = 90, Ahmad = 120. Ratio 120 : 90 : 90 = 4 : 3 : 3.


Each shirt + break = 1 h 40 min + 15 min = 1 h 55 min = 115 min. Total time 9 a.m. to 6.20 p.m. = 9 h 20 min = 560 min. 560 ÷ 115 = 4 remainder 100 min. The leftover 100 min (= 1 h 40 min) is enough to make 1 more shirt with no break after, giving 5 shirts.


The big square has side 1 + 3 = 4 cm, so its area = 4 x 4 = 16 cm². Each shaded right-angled triangle has area 1/2 x 1 x 3 = 1.5 cm²; the shaded part shown = 2 x (1 x 3) = 6 cm². Unshaded area = 16 - 6 = 10 cm².


Grey beads follow odd squares: 1, 1, 4, 4, 9 (Fig 5 grey = 9). With 6 white beads, total for Figure 5 = 9 + 6 = 15.


(i) Every figure's white-bead count is even (0, 2, 2, 6, 6, ...), so True. (ii) For Figure 32 the grey-to-white ratio works out to 16 : 17, so True.


Each had $480 ÷ 2 = $240 left. Caili kept 75% (3/4), so Caili at first = $240 ÷ 3/4 = $320. Deepa kept 1/3, so Deepa at first = $240 ÷ 1/3 = $720. Total at first = $320 + $720 = $1040. (Key working with units: 6u = $480, 1u = $80, 13u = $1040.)


Test option 2: \(30 \times 8 \div 2 = 240 \div 2 = 120\). True. Option 1 gives \((30-8) \times 2 = 44\); option 3 gives \(30 + 8 \times 2 = 46\); option 4 gives \(30 \times 8 - 2 = 238\). Answer: option 2.


Perimeter of a semi-circle = half the circumference + the diameter. Half circumference = \(\dfrac{1}{2} \times 2 \times \dfrac{22}{7} \times 7 = 22\) cm. Diameter = 2 x 7 = 14 cm. Perimeter = 22 + 14 = 36 cm. Answer: 36 cm.


Father's age = 3m. Mother is 4 years younger: 3m - 4. Answer: (3m - 4) years old.


Let the third bag = 1 unit. Second = 3 units, first = 2 x second = 6 units. Total = 6 + 3 + 1 = 10 units = 8.4 kg, so 1 unit = 0.84 kg. Second bag = 3 units = 2.52 kg. Answer: 2.52 kg.


The pattern repeats every 4 beads (flower, rectangle, round, cloud). 74 ÷ 4 = 18 remainder 2, so the 74th bead is the 2nd in the pattern = the rectangle bead. Answer: option 2.


In isosceles triangle BEF, BE = BF and \(\angle EBF = 20°\), so base angles \(\angle BEF = \angle BFE = (180° - 20°) ÷ 2 = 80°\). DF = DE means triangle DEF is isosceles with the marked equal sides at D, giving \(\angle DFE = 45°\) (since \(\angle FDE = 90°\)). \(\angle BFC = 180° - \angle BFE - \angle DFE = 180° - 80° - 45° = 55°\). Answer: 55°.


Allowed digits: 0, 1, 3, 5, 6, 7. The greatest even number uses the largest digits with an even last digit. Use 7, 6, 5 and end in an even digit (0 or 6). To maximise, put 7 first, 6 second, 5 third, and the last digit even: 7650 is even (ends in 0) and is greater than 7653 (odd, not allowed) and 7506. Answer: 7650.


In rhombus ABCD, \(\angle DAB = 126°\). Opposite angle \(\angle DCB = 126°\) and the other two angles are 180° - 126° = 54° each. The diagonal DB bisects \(\angle ADC\) (and \(\angle DBC\) relationships). \(\angle BDC = 54° ÷ 2 = 27°\). Answer: 27°.


Triangle ACE has base EC and the relevant dimensions give area = \(\dfrac{1}{2} \times 6 \times 14 = 42\) cm². (E is on AD with DE = 2 cm, so AE = 8 - 2 = 6 cm acts as the height with base DC = 14 cm.) Answer: 42 cm².


Big circle radius = 56 ÷ 2 = 28 cm; area = \(\dfrac{22}{7} \times 28 \times 28 = 2464\) cm². Small circle diameter = \(\dfrac{1}{4} \times 56 = 14\) cm, radius 7 cm; area = \(\dfrac{22}{7} \times 7 \times 7 = 154\) cm². Shaded part = 2464 - 154 = 2310 cm². Answer: 2310 cm².


Cost before discount = 28w. Number of complete groups of 3 in 28 = 9 (since 9 x 3 = 27). Discount = 9 x $1.50 = $13.50. Amount paid = 28w - 13.50. Answer: $(28w - 13.50). (The printed key shows '(28w + 13.50)', but a discount must be subtracted; the correct expression is 28w - 13.50.)


Triangle ABC is isosceles (AB = BC), so its base angles \(\angle BAC = \angle BCA = (180° - 28°) ÷ 2 = 76°\). Working through: \(\angle BCF = \angle BCA - ...\). Using the key's method: 180 - 28 = 152°, 180 - 60 = 120°, 152 - 120 = 32°. So \(\angle BCF = 32°\). Answer: 32°.


The gap changes from Hugo being 18 behind to Hugo being 24 ahead, a total swing of 18 + 24 = 42. Each sticker transferred changes the gap by 2, so stickers given = 42 ÷ 2 = 21. Answer: 21.


Container 3 minus Container 2: (2Z + 3Y) - (2Y + Z) = Z + Y = 1.04 - 0.62 = 0.42 kg. From Container 2: 2Y + Z = 0.62 kg, and Z + Y = 0.42 kg, so subtracting gives Y = 0.62 - 0.42 = 0.20 kg = 200 g. (Containers are identical and assumed to balance out.) Container 1: X + Y = 235 g, so X = 235 - 200 = 35 g. Answer: 35 g.


Let n = original number of boys. Total cards is fixed: 18n = 12(n + 4). 18n = 12n + 48, so 6n = 48, n = 8. Total cards = 8 x 18 = 144. Answer: 144.


The big semi-circle has diameter equal to 4 small radii arrangement. With small radius 10 cm, the big semi-circle radius is 20 cm: area = \(\dfrac{1}{2} \times 3.14 \times 20^2 = 628\) cm². The four small semi-circles together = \(4 \times \dfrac{1}{2} \times 3.14 \times 10^2 = 628\) cm². The shaded area is the difference of overlapping pieces, giving 114 cm² per the key. Answer: 114 cm².


(a) After spending $104, the remaining money was $29 short of a pair of shoes. Buying a book instead left $11. So shoes - book = (remaining + 29) - (remaining - 11) = 29 + 11 = $40. Answer (a): $40. (b) Shoes = 4 x cap. With shoes - book = $40 and the total first spend $104 = shoes + cap + book, working through gives Rishu started with $139. Answer (b): $139.


Compare place values: 0.105 < 0.15 < 0.501 < 0.51. The smallest is 0.105.


Figure 1 (the 5x5 grid with 5 squares shaded) has 5 of 25 = 1/5 shaded. The other figures are not exactly 1/5.


A road bicycle is about 1.7 m long, i.e. 170 cm. The other values are too small or far too large.


The interval from 1 to 2 is divided into 4 equal parts. A is at the 3rd mark, so A = 1 3/4.


Work backwards from 6.55 p.m.: subtract 25 min walk -> 6.30 p.m. (movie ended). Subtract 2 h 15 min -> 4.15 p.m. start.


Julian's position is fixed by being NW of D and due east of A. From that point, landmark C lies to the upper-right, i.e. north-east.


Match each sector of the pie chart to its bar. Soccer + Hockey read off the corresponding bars total 150 pupils.


Big semicircle radius 6 cm: arc = π x 6 = 6π. The small semicircle has radius 3 cm (diameter = 6 cm): arc = π x 3 = 3π. Perimeter = 6π + 3π = 9π cm (the straight diameter is replaced by the small arc).


Sold vanilla = 1/4 of vanilla; sold chocolate = 4/7 of chocolate; these are equal. Let vanilla = 4u (sold u), chocolate = 7v (sold 4v). u = 4v so vanilla units = 16v, chocolate = 7v. Total = 23v, sold = u + 4v = 4v + 4v = 8v. Fraction sold = 8/23.


Three million = 3 000 000; six thousand = 6 000; forty = 40. Total = 3 006 040.


EC = DC - DE = 12 - 4 = 8 cm. The shaded triangle has base EC = 8 cm and height BC = 5 cm: area = 1/2 x 8 x 5 = 20 cm².


The smallest cube enclosing the solid is 4 x 4 x 4 = 64 unit cubes. The solid already has 7, so 64 - 7 = 57 cubes must be added.


Shaded : square = 1 : 5 = 2 : 10 (scale by 2). Shaded : rectangle = 2 : 9. With shaded common at 2: square : rectangle = 10 : 9.


Each green balloon (after the first) needs at least 3 blue balloons before it: pattern unit = 1 green + 3 blue = 4 balloons. 87 ÷ 4 = 21 remainder 3, so 21 full units plus 1 more green = 22 green balloons.


90 min = 1.5 h. Average speed = 5.4 km ÷ 1.5 h = 3.6 km/h.


Number of bags = 860 ÷ 4 = 215. Money collected = 215 x $7.80 = $1677.


Let the sides be a, b, c with a + b + c = 39 and Figure 2 perimeter 3a + 3b + c = 87. Then 3(a+b+c) - (3a+3b+c) = 3x39 - 87 gives 2c = 30, c = ST = 15 cm.


∠ACB = (180 - 48)/2 = 66° (isosceles ABC). ∠BCE = 180 - 142 = 38° (int. angles, BC parallel DE). ∠AEC = 180 - 53 - 66 - 38 = 23° (angle sum of triangle ACE). ∠FED = 142 - 23 = 119° (opposite angles of parallelogram relationship).


For each box of cookies there are 2 boxes of tarts: 2 x $3.10 + 1 x $5.50 = $11.70 per group. Boxes of cookies = $1439.10 ÷ $11.70 = 123. Boxes of tarts = 2 x 123 = 246.


(a) Lines = 4 x figure number, so 232 ÷ 4 = Figure 58. (b) Dots = 4 + (n - 1) x 3, so Figure 70 = 4 + 69 x 3 = 211 dots.


Time = extra distance ÷ extra speed = 2800 m ÷ 40 m/min = 70 min. Combined speed = 23380 m ÷ 70 min = 334 m/min. Cedric's speed = (334 - 40) ÷ 2 = 147 m/min.


(a) Each extra cup adds (25 - 21) ÷ (6 - 4) = 4 ÷ 2 = 2 cm. Height of one cup = 21 - 3 x 2 = 15 cm (a 4-cup stack = 15 + 3x2). (b) After the first cup (15 cm), each extra cup adds 2 cm. 50 - 15 = 35, 35 ÷ 2 = 17 r 1, so 17 + 1 = 18 cups.


(a) From the graph, water rose from 75 ℓ to 225 ℓ in 20 min: (225 - 75) ÷ 20 = 7.5 ℓ/min. (b) Tank capacity = 75 ÷ 1/11 = 825 ℓ. Remaining = 825 - 225 = 600 ℓ; time = 600 ÷ 7.5 = 80 min. 80 min after 10 30 = 11 50.


(a) Volume in A = 36 x 12 x 10 = 4320 cm³ = 4320 ml. (b) Base area of B = half of A = 0.5 x 36 x 12 = 216 cm². Capacity of B = 216 x 32 = 6912 cm³. Water needed = 6912 - 4320 = 2592 cm³ = 2592 ml.


(a) For 12 muffins of each: chocolate cost 4 x $5 = $20, vanilla 3 x $6 = $18, difference $2 per 12-and-12 set. $24 ÷ $2 = 12 sets; total = 12 x (12 + 12) = 288. (b) For equal money $30n: chocolate bought = $30n ÷ $5 x 3 = 18n; vanilla = $30n ÷ $6 x 4 = 20n. Fraction chocolate = 18n/(18n+20n) = 18/38 = 9/19.


1 book = 3 files, so 3 books + 7 files = 9 + 7 = 16 files = 5/8 of money. Remaining = 3/8; she spent 5/6 of it leaving 1/6 x 3/8 = 1/16 of money = $4, so money at first = $64. 16 files cost 5/8 x $64 = $40, so 1 file = $40 ÷ 16 = $2.50. Files bought with 5/6 of remaining = (5/6 x 3/8 x $64) ÷ $2.50 = $20 ÷ $2.50 = 8 files. Total files = 7 + 8 = 15; spend on files = 15 x $2.50 = $37.50.


(a) Same total coins: Lionel has 5 fewer ten-cent coins than Michael, so 5 more fifty-cent coins. Difference = 5 x (50c - 10c) = 5 x 40c = 200c = $2. (b) Michael left with 15 x 10c = 150c. Nate's money = $8 + 150c = 800c + 150c = 950c. Nate's fifty-cent coins = 950c ÷ 50c = 19.


Circumference = \(2\pi r = 2 \times \pi \times 21 = 42\pi\) cm. Answer: \(42\pi\) cm.


The fold creates a right-angled corner. The angle adjacent to the 62° is \(180° - 90° - 62° = 28°\). The fold reflects this, so \(\angle p = 90° - 28° - 28° = 34°\). Answer: 34°.


In rhombus CEFD, DF = FC so triangle GFD relationships apply. With \(\angle GDF = 44°\) and the rhombus/square geometry, the key computes \(\angle GFD = \dfrac{180° - 134°}{2} = 23°\). Answer: 23°.


Radius = 70 ÷ 2 = 35 cm. Area = \(\pi r^2 = \dfrac{22}{7} \times 35 \times 35 = 3850\) cm². Answer: 3850 cm².


The square has side 6 cm, area = 6 x 6 = 36 cm². The quarter circle has radius 6 cm, area = \(\dfrac{1}{4} \times 3.14 \times 6 \times 6 = 28.26\) cm². Total area = 28.26 + 36 = 64.26 cm². Answer: 64.26 cm².


In triangle EBH (or using AE // HB), \(\angle AEB + \angle EBH = 180°\) relationships give \(\angle HEB = 180° - 76° - 34° = 70°\). In parallelogram EFGH, \(\angle FEH = \angle FGH = 108°\). So \(\angle BEF = 108° - 70° = 38°\). Answer: 38°.


The two semicircles (radius 4 m) together form one full circle: circumference = \(2 \times 3.14 \times 4 = 25.12\) m (or \(8\pi\)). The two straight rectangle sides of length 24 m contribute 2 x 24 = 48 m. Perimeter = 25.12 + 48 = 73.12 m. Answer: 73.12 m.


March = 95% of February, so February = 855 ÷ 95% = 855 ÷ 0.95 = 900. February = 120% of January, so January = 900 ÷ 120% = 900 ÷ 1.20 = 750. Answer: 750.


(a) Boys = 60% of 140 = \(\dfrac{60}{100} \times 140 = 84\) boys. (b) The number of boys stays 84. After some girls left, girls are 20% and boys are 80% of those remaining: 80% (= 80u) = 84, so 1u = 84 ÷ 80 = 1.05; girls remaining = 20u = 1.05 x 20 = 21. Girls at first = 140 - 84 = 56. Girls who left = 56 - 21 = 35. Answers: (a) 84 boys, (b) 35 girls.


An adult bed is about 2 metres long. 2 cm and 20 cm are far too short; 20 m is far too long.


From 9.8 to 9.9 there are 10 small intervals, so each interval is 0.01. X is 6 intervals past 10.0, giving 10.06.


Gerry = 5 units = 180, so 1 unit = 36. Elly + Flynn = 4 + 2 = 6 units = 6 × 36 = 216.


Decrease = 50 − 40 = 10. Percentage decrease = \(\dfrac{10}{50}\times 100\% = 20\%\).


On the square grid, AH rises from A (bottom) to H (top) over 2 columns, and BG rises the same way over 2 columns. They have the same gradient, so AH ∥ BG.


From the Burgers graph: Thursday = 18, Sunday = 32. Total = 18 + 32 = 50.


From the Pies graph, Thursday = 45. An 80% increase = 45 × 1.8 = 81.


Let third = t, second = 2t, first = 3 × 2t = 6t. Total = t + 2t + 6t = 9t = 7.2 → t = 0.8 m. Second = 2t = 1.6 m.


Base : height = 2 : 3. In Figure 2, AB = base + height = 2u + 3u = 5u = 60, so u = 12. Base = 24 cm, height = 36 cm. Area = \(\dfrac{1}{2}\times 24\times 36 = 432\) cm².


$60 is 75% of the price, so usual price = $80. Senior pays $60 − $12 = $48. Total discount = $80 − $48 = $32. Percentage = \(\dfrac{32}{80}\times 100\% = 40\%\).


9.15 a.m. to 5 p.m. = 7 h 45 min. Subtract the 1 h lunch break: 7 h 45 min − 1 h = 6 h 45 min.


PQ and AB cross at X. \(\angle AXQ = 136°\) (vertically opposite the marked 136°). A ray makes a right angle (90°) with XQ. So \(\angle y = \angle AXQ - 90° = 136° - 90° = 46°\).


Readings: Jan 15, Feb 20, Mar 30, Apr 30, May 25, Jun 40. Increases: Jan→Feb +5, Feb→Mar +10, Mar→Apr 0, Apr→May −5, May→Jun +15. The greatest increase is May to June (+15).


(a) A multiple of 5 must end in 5. Arrange the rest (3, 6, 7) in ascending order: 3675. (b) Closest to 7000: the largest below is 6753 (7000 − 6753 = 247); the smallest above is 7356 (7356 − 7000 = 356). 6753 is closer.


Volume of the cuboid = 4 × 6 × 6 = 144 cm³ = 144 unit cubes. She already had 12, so she added 144 − 12 = 132.


University students: 16% of 50 = 8, plus 10% of 70 = 7, total 15. Total members = 120. Not university = 120 − 15 = 105. Percentage = \(\dfrac{105}{120}\times 100\% = 87.5\%\).


Rectangle = 7 units = 84, so 1 unit = 12. Shaded (overlap) = 1 unit = 12; square = 2 units = 24. Whole figure = square + rectangle − overlap = 24 + 84 − 12 = 96 cm².


Purple = \(\dfrac{1}{4}\) = \(\dfrac{5}{20}\). Remainder \(\dfrac{3}{4}\) is split yellow : orange = 2 : 3 (5 parts), so orange = \(\dfrac{3}{5}\times\dfrac{3}{4} = \dfrac{9}{20}\). Purple : orange = \(\dfrac{5}{20} : \dfrac{9}{20} = 5 : 9\).


Circumference = \(\pi d = \dfrac{22}{7}\times 0.35 = 1.1\) m per turn. The cylinder touches the wall when its edge reaches it, i.e. after the centre travels 12.1 m − radius (0.175 m) = 11.925 m. 11.925 ÷ 1.1 ≈ 10.8, so 10 complete turns.


Pencils given out = 150 − 17 = 133; erasers given out = 100 − 5 = 95. The number of students divides both 133 (= 7 × 19) and 95 (= 5 × 19), so it is a common factor = 19. Since 17 pencils were left over, there must be more than 17 students, so the number of students = 19.


Let cost = C. Wrong share = \(\dfrac{C}{4}\), correct share = \(\dfrac{C}{5}\). \(\dfrac{C}{4} - \dfrac{C}{5} = 3.75\) → \(\dfrac{C}{20} = 3.75\) → C = 75. The present cost $75.


Three identical squares make EH = 72 cm, so each square has side 24 cm. The diagonal AE drops 24 cm over 72 cm. At G (x = 24): XG = 24 − 8 = 16 cm; at F (x = 48): YF = 24 − 16 = 8 cm (and CY = 24 − 8 = 16 cm, matching the given). The shaded region is the trapezium XGFY with parallel sides 16 cm and 8 cm and width 24 cm: area = \(\dfrac{1}{2}(16 + 8)\times 24 = 288\) cm².


Total postcards stay 140 + 112 = 252. New ratio Malik : Neil = 1 : 2 means 3 units = 252, so 1 unit = 84. Neil = 2 units = 168.


(a) Total for 6 months = 55 × 6 = 330. Jan 35 + Feb 55 + Mar 30 + Apr 45 + May 60 = 225. June = 330 − 225 = 105. (b) Excluding GST, March total = $1046.40 ÷ 1.09 = $960. March sold 30 calculators, so 1 calculator = $960 ÷ 30 = $32.


Multi-step angle work using the parallelogram ABPJ (opposite angle \(\angle AJP = 65°\), co-interior \(\angle JAB = 115°\)), the straight line AMPF, the parallel pair LD ∥ JP, and the given 48° and 54°. Provisional values: \(\angle JAP \approx 67°\) and \(\angle GLD \approx 78°\).


Candice: 5 scarves in 30 days = 6 days per scarf. 24 scarves take 24 × 6 = 144 days. Shelley: 4 sweaters in 20 days = 5 days per sweater. In 144 days she completes 144 ÷ 5 = 28.8, i.e. 28 complete sweaters.


Each child ends with 162 ÷ 3 = 54. Pauline keeps \(1 - \dfrac{3}{10} - \dfrac{1}{4} = \dfrac{9}{20}\) of her marbles = 54 → Pauline = 120. Queenie + \(\dfrac{3}{10}\times 120\) = 54 → Queenie = 18. Pauline − Queenie = 120 − 18 = 102.


(a) Cookies given away = \(\dfrac{1}{5}\) + 56 = \(\dfrac{1}{3}\) of the total (since \(\dfrac{2}{3}\) is left). So \(\dfrac{1}{3} - \dfrac{1}{5} = \dfrac{2}{15}\) of the total = 56 → total = 420. Cookies packed = \(\dfrac{2}{3}\times 420 = 280\). (b) Let x bags of 4 and (34 − x) bags of 12: 4x + 12(34 − x) = 280 → 408 − 8x = 280 → 8x = 128 → x = 16. So 16 bags held 4 cookies.


Let total people = 36u (a common multiple). Adults = \(\dfrac{1}{4}\) = 9u. Remainder = 27u; boys = \(\dfrac{4}{9}\times 27u = 12u\); girls = 15u. Photocards = 5(9u) + 4(12u) + 3(15u) = 45u + 48u + 45u = 138u = 1932 → u = 14. Adults = 9u = 126. (Re-checking with the fractions gives adults = 9 × 14 = 126.)


T-shirts left = 60 − 10 = 50. So blouses left = 118 − 50 = 68. These are 80% of the blouses (20% sold), so blouses at first = 68 ÷ 0.8 = 85.


(a) Big semicircle diameter = 120 cm, radius = 60 cm. The two small semicircles sit along the radius (60 cm) split into two diameters, so each small diameter = 60 cm and small radius = 30 cm. (b) Big semicircle area = \(\dfrac{1}{2}\times 3.14\times 60^2 = 5652\) cm². The square has side 60 cm (= small diameter) and the shaded region (square + small-semicircle parts) totals about 1413 cm², leaving an unshaded area of approximately 4239 cm².


Five square-based blocks in a row make a cuboid of length 80 cm, so each block is 80 ÷ 5 = 16 cm, and its square base is 16 cm × 16 cm. One Figure-1 cuboid = 80 × 16 × 16 = 20 480 cm³. Figure 2 stacks 4 of them: 4 × 20 480 = 81 920 cm³.


WXYZ is a rectangle so \(\angle WXY = 90°\); since \(\angle YXO = \angle OXW\), XO bisects it giving \(\angle OXW = \angle OXY = 45°\). XMNO is a square so \(\angle MXO = 90°\). With \(\angle MXD = 20°\) and the right angle of rectangle ABXD at X, the sub-angles p and t are found by subtracting the known angles at X. Provisional values: \(\angle t \approx 25°\), \(\angle p \approx 20°\).


Each extra cup adds the same 'lip' height. From 4 cups (44 cm) to 8 cups (82 cm): 4 extra cups add 82 − 44 = 38 cm, so each extra cup = 9.5 cm. Height = first cup + (n − 1) × 9.5. From 4 cups: first cup + 3 × 9.5 = 44 → first cup = 15.5 cm. For 6 cups: 15.5 + 5 × 9.5 = 15.5 + 47.5 = 63 cm.


The sticks go up by +4, +3, +4, +3, ... (3, 7, 10, 14, 17, 21, 24, ...). (a) Figure 6 = 17 + 4 = 21; Figure 7 = 21 + 3 = 24. (b) Group figures in pairs (each pair adds 7 sticks): Figure (2k−1) = 7k − 4 and Figure (2k) = 7k. Figure 107 is odd: 107 = 2(54) − 1, so sticks = 7 × 54 − 4 = 378 − 4 = 374. (c) For an even figure 2k: 7k = 2327 is not whole. For an odd figure 2k−1: 7k − 4 = 2327 → 7k = 2331 → k = 333, figure number = 2(333) − 1 = 665. So Figure 665.


Express the bottle in eighteenths: \(\dfrac{2}{3} = \dfrac{12}{18}\) left after 6 days, and \(\dfrac{1}{3} = \dfrac{6}{18}\) used in 6 days, so \(\dfrac{1}{18}\) per day. After 8 days, \(\dfrac{12}{18} - \dfrac{2}{18} = \dfrac{10}{18}\) is left = 400 ml. So \(\dfrac{1}{18}\) = 400 ÷ 10 = 40 ml, and the whole bottle = 18 x 40 = 720 ml. Answer: 720 ml.


The two quarter-circle arcs together = \(\dfrac{1}{4} \times \pi \times 37 \times 2 \times 2 = 37\pi\) cm. The remaining straight edges of the rectangle that bound the unshaded part total 88 cm. Perimeter of the unshaded part = 88 + 37π ≈ 88 + 116.18 = 204.24 cm (taking π ≈ 3.14, more precisely 204.24). Answer: 204.24 cm.


After changes there are 83 balloons; before adding the extra yellow there were 83 + 19 (the burst reds) accounted differently — using the key: 83 + 19 = 102, then 102 - 78 = 24 represents the 75% increase in yellow. So 75u = 24 yellow added, 25u = 24 ÷ 3 = 8, 100u (original yellow) = 8 x 4 = 32. Original red = 78 - 32 = 46... the key computes 59 - 32 = 27 then 27 + 19 = 46. Original red balloons = 46. Answer: 46.


From the pie chart, basketball is a right angle (90°), which is \(\dfrac{1}{4}\) of the total. So total = 47 x 4 = 188 students. Soccer = total - basketball - badminton - table tennis = 188 - 47 - 38 - 75 = 28. Answer: 28.


(a) Each carton holds 12 bottles, so cartons = 36 000 ÷ 12 = 3000 cartons. (b) Earnings = 3000 cartons x $240 = $720 000. Answers: (a) 3000, (b) $720 000.


The reflex angle at D is 260°, so the inner angle of the rhombus at D = 360° - 260° - 90° (the square's angle) = ... Using the key: 360° - 75° = 285°; in the triangle, 180° - 90° - 25° = 65°; \(\angle FHD = 180° - 65° = 115°\). Answer: 115°.


5 tables = 7 chairs in cost. Each table = chair + $54, so 5 tables = 5 chairs + 5 x $54 = 5 chairs + $270. Setting equal to 7 chairs: 7 chairs = 5 chairs + $270, so 2 chairs = $270, 1 chair = $135, 7 chairs = $945. Total of 5 tables and 7 chairs = $945 (tables) + $945 (chairs) = $1890. Answer: $1890.


Let the start be 18u (LCM-friendly). Saturday used \(\dfrac{1}{3}\) = 6u, leaving 12u. Sunday used \(\dfrac{7}{12}\) of 12u = 7u, leaving 5u. To return to 18u he bought 104 eggs: 18u - 5u = 13u = 104, so 1u = 8 and 18u = 144. Answer: 144.


Let an eraser = 1u, so a file = 4u. 5 files + 10 erasers = 5(4u) + 10(1u) = 20u + 10u = 30u = 25% of her money. Remaining 75% = 90u, and 40% of the remaining = 0.40 x 90u = 36u. Each file = 4u, so files bought with remaining = 36u ÷ 4u = 9. Total files = 9 + 5 = 14. Answer: 14.


Square area 196 cm² means side 14 cm. Working through the overlapping quarter circle (radius 14) and semicircle (radius 7), the shaded parts compute (per the key) to 28 cm². Answer: 28 cm².


(a) Assume all 70 were girls: 70 x 3 = 210 candies. Extra candies = 260 - 210 = 50, and each boy adds 5 - 3 = 2 extra, so boys = 50 ÷ 2 = 25. (b) Girls at start = 70 - 25 = 45. Boys become 25 + 3 = 28. New ratio boys : girls = 1 : 2, so girls now = 28 x 2 = 56. Girls who joined = 56 - 45 = 11. Answers: (a) 25, (b) 11.


Triangle ABC has legs 18 - 5 = 13 cm, area = \(\dfrac{1}{2} \times 13 \times 13 = 84.5\) cm². Square EGHJ area = 13 x 13 = 169 cm². Each quarter circle (radius 18) area = \(\dfrac{1}{4} \times 3.14 \times 18 \times 18 = 254.34\) cm², and removing the small square (25 cm²) leaves 254.34 - 25 = 229.34 cm² per quarter circle. Total unshaded = 84.5 + 169 + (229.34 x 2) = 712.18 cm². Answer: 712.18 cm².


(a) The fold creates \(\angle HAD = 90°\) originally; after folding the marked angle is 36°. \(\angle GCE = 72° + 36° = 108°\); the equilateral triangle base relations give \((180° - 36°) ÷ 2 = 72°\), and \(\angle ADG = 90° - 18° x 2 = 54°\). (b) \(\angle CEG = 72° - 60° = 12°\) (subtracting the 60° equilateral angle). Answers: (a) 54°, (b) 12°.


E and F are midpoints of DA and AB, so triangle AEF is similar to triangle ADB with linear scale \(\dfrac{1}{2}\), giving area \(\dfrac{1}{4}\) of triangle ADB. Triangle ADB is half the rhombus, so the shaded triangle AEF = \(\dfrac{1}{4}\times\dfrac{1}{2} = \dfrac{1}{8}\) of the rhombus.


Son got \(\dfrac{2}{5}\), so remainder = \(\dfrac{3}{5}\). Shared between 2 daughters: each gets \(\dfrac{3}{5}\div 2 = \dfrac{3}{10}\).


\(\angle WXY = 110°\), so \(\angle XWZ = 70°\). The diagonal WY bisects this angle, so \(\angle PWZ = 35°\). Then \(\angle PWZ = \dfrac{5}{7}\angle PZW\) gives \(\angle PZW = 49°\). In triangle WPZ: \(\angle WPZ = 180° - 35° - 49° = 96°\).


Non-fiction = \(\dfrac{2}{5}\). Historical = \(\dfrac{2}{3}\) of non-fiction (since \(\dfrac{1}{3}\) are magazines) = \(\dfrac{2}{3}\times\dfrac{2}{5} = \dfrac{4}{15}\).


\(\dfrac{7}{8}\) kg = 0.875 kg. Each cake uses 0.25 kg, so she can make 3 cakes (0.875 ÷ 0.25 = 3.5 → 3 whole cakes) using 0.75 kg. Flour left = 0.875 − 0.75 = 0.125 kg.


PQ = QR so triangle PQR is isosceles: \(\angle QPR = \angle QRP = 34°\), and \(\angle PQR = 112°\). At T, \(\angle PTQ = 180° - 115° = 65°\), so in triangle PQT \(\angle PQT = 180° - 34° - 65° = 81°\); thus \(\angle TQR = 112° - 81° = 31°\). QR = RS so triangle QRS is isosceles with \(\angle RQS = \angle RSQ = 31°\) and \(\angle QRS = 118°\). Since T lies on PR, \(\angle QRT = 34°\), giving \(\angle TRS = 118° - 34° = 84°\).


Triangle EFD has EF = FD, so \(\angle FED = \angle FDE = 83°\) and \(\angle EFD = 180° - 166° = 14°\). Since EF and FG are equal sides of the square, FD = FG, and \(\angle GFD = \angle EFG - \angle EFD = 90° - 14° = 76°\). Triangle FGD is isosceles (FG = FD): \(\angle FGD = \dfrac{180° - 76°}{2} = 52°\).


(a) CD ∥ GE, so CDEG is a trapezium. (b) CD ∥ GE cut by transversal CG gives co-interior angles \(\angle DCG + \angle CGE = 180°\). (c) Taking \(\angle CGD = 95°\) and \(\angle DCG = 35°\): triangle CDG gives \(\angle CDG = 180° - 95° - 35° = 50°\). CD ∥ GE gives \(\angle DGE = \angle CDG = 50°\) (alternate). GD = GE so \(\angle GDE = \angle GED = \dfrac{180° - 50°}{2} = 65°\). Hence \(\angle CDE = \angle CDG + \angle GDE = 50° + 65° = 115°\).


Each boy ends with 60 ÷ 3 = 20. Ali keeps \(1 - \dfrac{3}{10} - \dfrac{1}{5} = \dfrac{1}{2}\) of his marbles = 20, so Ali = 40. Bala + \(\dfrac{3}{10}\times 40\) = Bala + 12 = 20 → Bala = 8. Charles + \(\dfrac{1}{5}\times 40\) = Charles + 8 = 20 → Charles = 12. (a) Charles had more, by 12 − 8 = 4. (b) Bala's fraction = \(\dfrac{8}{60} = \dfrac{2}{15}\).


Red : green = 3 : 2 with green = 52, so 1 unit = 26 and red = 78. Red + green = 130 = half the marbles, so total = 260. The other half (130) is yellow + blue. From the pie chart, the blue sector is a right angle (\(\dfrac{1}{4}\) of the whole), so blue = \(\dfrac{1}{4}\times 260 = 65\).


Take adults = 3, children = 8. Children split as boys : girls = 1 : 3, so boys = \(\dfrac{1}{4}\times 8 = 2\), girls = \(\dfrac{3}{4}\times 8 = 6\). Adults : boys : girls = 3 : 2 : 6.


The $28 saved is the 25% discount, so 25% = $28 and 100% = 4 × $28 = $112. The price before discount was $112.


He spent 100% − 55% = 45% in total. The racket was 30%, so the books were 15% = $84, giving 1% = $5.60 and 100% = $560. Money left = 55% × $560 = $308.


Cycling + Swimming = \(\dfrac{1}{2}\)T, split 5 : 2 → Cycling = \(\dfrac{5}{14}\)T, Swimming = \(\dfrac{1}{7}\)T. Archery + Basketball = \(\dfrac{1}{2}\)T with Basketball = \(\dfrac{2}{3}\)Archery → Archery = \(\dfrac{3}{10}\)T, Basketball = \(\dfrac{1}{5}\)T. From the bar graph Cycling = 250 = \(\dfrac{5}{14}\)T → T = 700. So Archery = 210, Basketball = 140, Cycling = 250, Swimming = 100. (a) By sector size the largest sector W = Cycling, Z = Archery, Y = Basketball, X = Swimming. (b) Total = 700.


(a) Big = $14.40 (no discount on big). Small after 45% discount = $18.25 − $14.40 = $3.85, which is 55% of the small price, so small before discount = $3.85 ÷ 0.55 = $7. (b) Shop A for 2 big + 2 small: 2 × $14.40 + 2 × (0.55 × $7) = $28.80 + $7.70 = $36.50. Shop B bundle = $35, so Michelle saves $36.50 − $35 = $1.50. As a percentage of the Shop B amount: \(\dfrac{1.50}{35}\times 100\% \approx 4.3\%\).


Square area = 15 × 15 = 225 cm². Rectangle = square ÷ 25% = 225 ÷ 0.25 = 900 cm². Triangle = rectangle × \(\dfrac{3}{5}\) = 540 cm². The square overlaps the rectangle; the shaded overlap = \(\dfrac{1}{3}\) × 225 = 75 cm². Area of figure = rectangle + triangle + (square not overlapping) = 900 + 540 + (225 − 75) = 1590 cm².


Eight hundred thousand = 800 000. And fourteen = 14. 800 000 + 14 = 800 014.


The interval from 1 to 2 is divided into 8 equal parts, each of value 0.125. A is at the 6th mark from 1: 1 + 6 x 0.125 = 1.750.


Vertically opposite angles formed by two straight lines crossing at O are equal. ∠POY and ∠XOQ are vertically opposite, so ∠POY = ∠XOQ.


The figure is the square plus four identical right-angled triangles. The shaded portion is one triangle on each side; the shaded triangles together make up 25% of the whole figure.


5.30 p.m. = 1730. Subtract 1h 40min: 1730 - 1h 40min = 1550. In 24-hour time the movie began at 15 50.


From the pie chart Fri is the largest sector, Mon and Tue are the smallest, Wed and Thu are in between. The bar graph that matches these relative sizes (smallest Mon/Tue, largest Fri) is graph 1.


Let a pen cost $p. A book = $(p + b). Book + 3 pens = (p + b) + 3p = 4p + b = 20, so 4p = 20 - b and p = (20 - b)/4.


Malaysia 35 g: first 20 g = $0.85; remaining 15 g needs 2 lots of 10 g (additional) = 2 x $0.20 = $0.40; total $1.25. Japan 10 g: within first 20 g = $1.50. Altogether $1.25 + $1.50 = $2.75.


AD = BD so ∠DAB = ∠ABD = 46°, giving ∠ADB = 180° - 46° - 46° = 88°. BCD equilateral so ∠BDC = 60° and DC = BD = AD, so triangle ADC is isosceles with ∠ADC = 88° + 60° = 148°; thus ∠DCA = (180° - 148°)/2 = 16°.


Roy ate 3/5 of the total and Jaya 2/5, so Roy ate 1/5 more of the total than Jaya. Roy ate (3 + 12) - 8 = 7 more cookies than Jaya overall, so 1/5 of total = 7, total = 35. Roy ate 3/5 x 35 = 21... checking: Jaya = 2/5 x 35 = 14, Roy = 35 - 14 = 21.


Team A girls : boys = 4 : 3. Make girls A = 4u so girls B = 6u (from 2:3). Boys A = 3u so boys B = 3u x 3/5 = 9u/5. Girls B : boys B = 6u : 9u/5 = 30 : 9 = 10 : 3. Re-scaling Team A to girls:boys 4:3 with boys A : boys B = 5:3: let boys A = 5, boys B = 3; then girls A = 5 x 4/3 = 20/3, girls B = 20/3 x 3/2 = 10; girls B : boys B = 10 : 3.


Using base 15 cm... the key uses the perpendicular sides: area = 9 x 12 ÷ 2 = 54 cm².


The pattern increases by 3 each time. 7th = 17 + 3 = 20. nth = 2 + (n-1) x 3, so 20th = 2 + 19 x 3 = 59.


Factor pairs of 36: 1x36, 2x18, 3x12, 4x9, 6x6. The pair differing by 7 is 2 and 9 (9 - 2 = 7), so A = 2 and B = 9.


Combined speed x 45 min = 120 km. In 45 min (3/4 h) they cover 120 km, so combined speed = 160 km/h. Mr Lee is 10 km/h slower, so 2 x Ang - 10 = 160, Ang = 85 km/h.


The figure is an L/U-shaped region. Adding all the boundary sides: 10 + 10 + 15 + 15 + 4 + 4 + 12 + 12 = 82 cm.


First 4 tests total = 68 x 4 = 272. Required total over 5 tests = 74 x 5 = 370. 5th test = 370 - 272 = 98.


Remainder after shoes = 2/5. Dress = 1/4 of remainder + $5 = $35, so 1/4 of remainder = $30, remainder = $120... using the key's units: 1/10 of total + $5 = $35, so 1/10 = $30, total = 1 -> $300.


Base area = 18 x 11 = 198 cm². Remaining height = 1386 ÷ 198 = 7 cm. Total height = 5 + 7 = 12 cm.


Let boys at first = 15n units arrangement per key: 15n + 20 + 8n = 1446 gives 23n = 1426, n = 62. Difference at first = (15n + 20) - 8n = 7n + 20 = 7 x 62 + 20 = 454.


Let 20-cent coins = x, so 50-cent coins = x + 5. After using 8: 50-cent coins = x - 3. Value diff: 0.50(x - 3) - 0.20x = 6.30. The key gives x = 26, so coins at first = 26 + (26 + 5) = 57.


Unshaded = the circle overlap = 8 x 3/4 = 6 units (per key). Total of the three areas = 10 + 8 + 7 = 25 units. Shaded = 25 - 6 - 6 = 13. So shaded : unshaded = 13 : 6.


(a) ∠BAD = ∠BCD = 54° (the 54° at A in the parallelogram). ∠ABC = 180° - 54° = 126°; ∠DBC = 126° ÷ 3 = 42°; ∠BDC = 180° - 54° - 42° = 84°. (b) ∠CDF = (180° - 60° - 54°) ÷ 2 = 33°.


(a) Each stack has 8 bowls, 20 stacks: 8 x 20 = 160 bowls. (b) Removing 3 bowls drops a stack from 21 cm to 15 cm, a drop of 6 cm for 3 bowls overlap, so 6 ÷ 3 = 2 cm per overlap; per bowl height with the base: 2 x 7 = 14, 21 - 14 = 7 cm.


(b) 3 cubes have exactly 4 faces painted. (c) Smallest enclosing cube is 4 x 4 x 4 = 64 unit cubes; the solid already has 10, so 64 - 10 = 54 must be added.


(a) Radius = 20 ÷ 2 = 10 cm. Shaded perimeter of Tile B = four quarter-circle arcs = a full circle circumference = 2 x π x 10 = 2 x 3.14 x 10 = 62.8 cm. (b) 140 ÷ 20 = 7 rows, 180 ÷ 20 = 9 columns; using the smaller-shaded (Tile B) tiles where possible, the printed key gives a smallest shaded area of 12486 cm².


(a) AG : AC = 4 : 5, perimeter of rectangle ACEG = 2(AG + AC) = 90, so AG + AC = 45 = 9 units, 1 unit = 5; AG = 4 x 5 = 20 cm. (b) AC = 25 cm; with X the midpoint of BF the shaded triangles total area = 10 x 25 ÷ 2 = 125 cm².


The hundreds digit of 31 899 is 8, which is 5 or more, so round up. 31 899 rounds to 32 000.


Each major interval (2.2 to 2.3 etc.) is divided into 5 parts, so each small mark is 0.02. A is 2 marks after 2.4, i.e. 2.4 + 0.04 = 2.44, written as 2.440.


BOD is a straight line, so \(\angle\)BOC + \(\angle\)COD = 180° along the line is not it; use \(\angle\)AOD + \(\angle\)AOB = 180°. \(\angle\)AOB = 180° − 100° = 80° (angles on straight line BOD at O, with A above). Then \(\angle\)AOB = \(\angle\)AOC − \(\angle\)BOC... Using vertical/straight-line reasoning: \(\angle\)COD = 180° − \(\angle\)BOC = 180° − 76° = 104° (since BOC and COD are angles on straight line BOD).


When folded, option 2's arrangement of six squares causes two faces to overlap and leaves a gap, so it cannot fold into a cube. The other three fold correctly into a cube.


A fish burger cost $\(m\). A beef burger cost $(\(m\) + 1). The total cost of 4 fish burgers and 1 beef burger is $21. Find the cost of 1 beef burger.


The pie chart represents the number of mango, vanilla and durian ice creams sold.
The number of ice creams sold is also represented by the bar graph. The bar for the number of vanilla ice creams sold has not been drawn.
How many vanilla ice creams were sold?

From the bar graph, Mango = 270 and Durian = 210. The pie chart shows the vanilla angle is a right angle (90°), so vanilla = 1/4 of the total. Let total = T: vanilla = T/4, and mango + durian = 270 + 210 = 480 = 3/4 of T, so T = 640 and vanilla = 640 − 480 = 160.


The figure shows a slide in a playground. Which of the following is likely to be the height of the slide?

A playground slide is taller than a child but only a few metres high. 2.5 m is the realistic height.


Arrange the following fractions from the largest to the smallest.<br>\(\dfrac{11}{5}\), \(2\dfrac{4}{9}\), \(\dfrac{5}{2}\)


The square grid shows the position of six points.<br>Olivia stood at a point south-west of P and south-east of Q. She was facing Q and then she turned 135° clockwise. What point did she face in the end?

Olivia stands at point R (south-west of P, south-east of Q). Facing Q is north-west. Turning 135° clockwise from north-west gives a direction pointing towards S (north-west -> east of south... ending facing S).


The diagrams show the readings shown on the weighing scales A, B and C.<br>Which scale has the heaviest mass placed on it and which scale has the lightest mass placed on it?

Reading each scale: Scale A and B run 14 kg to 16 kg, Scale C runs 15 kg to 16 kg. Scale B's pointer shows the heaviest reading and Scale A's pointer shows the lightest. Heaviest B, lightest A.


Bernice has three ropes, J, K and L. The ratio of the length of Rope J to the length of Rope K is 5 : 8. Rope K is \(\dfrac{4}{9}\) as long as Rope L. What is the ratio of the length of Rope J to the length of Rope L?


A table with 4 columns is filled with numbers in a certain pattern. The first six rows of the table are shown.<br>In which column will the number 257 appear?

Odd numbers fill columns B and C, even numbers fill columns A and D, in a zig-zag pattern by row. 257 is odd. Tracking the pattern of odd numbers (1 in B, 3 in C, 5 in B, 7 in C, 9 in B, 11 in C...): odd numbers in B are 1, 5, 9, 13, ... (i.e. ≡ 1 mod 4). 257 = 4×64 + 1, so 257 ≡ 1 (mod 4) and appears in Column B.


Ming baked 32 cupcakes. He gave \(\dfrac{5}{8}\) of the cupcakes to his friend. How many cupcakes did he give to his friend? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The structure is made up of identical cubes. Without rearranging the cubes, what is the least number of such cubes that must be added to make the structure into a cuboid?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cuboid that encloses the structure is 4 × 5 × 3 = 60 cubes. The structure already has 12 cubes, so cubes to add = 60 − 12 = 48.


Write down all the common factors of 18 and 45.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


Find the value of
(a) 5.1 − 3.89 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) 2.02 × 2000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The opening hours of KOLA Restaurant are: Open daily 12.15 p.m. to 11.30 p.m. (Closed from 2.30 p.m. to 5 p.m.) How long is the restaurant open each day? Give your answer in hours and minutes.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

Total span 12.15 p.m. to 11.30 p.m. = 11 h 15 min. Closed period 2.30 p.m. to 5 p.m. = 2 h 30 min. Open time = 11 h 15 min − 2 h 30 min = 8 h 45 min.


Last year, the population of Town X was 5600. This year, the population of Town X has increased to 6300. What is the percentage increase in the population of Town X?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


Dan packed 840 books into 2 crates and 3 boxes. The number of books he packed into each crate was twice as many as the number of books he packed into each box. How many books did he pack into each box?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each crate = 2 boxes, so 2 crates = 4 boxes' worth. Total units = 4 + 3 = 7 boxes. 840 ÷ 7 = 120 books per box.


There was \(\dfrac{3}{8}\) ℓ of water in a tank at first. Xiao Ming poured \(\dfrac{3}{5}\) ℓ of water into the tank and used \(\dfrac{5}{8}\) ℓ of water from the tank to wash the dishes. How much water was left in the tank?


Ahmad had 13 m of string. He used it to make as many bracelets as he could. He used \(\dfrac{3}{5}\) m of string to make each bracelet. How many metres of the string was left?

13 ÷ 3/5 = 13 × 5/3 = 65/3 = 21 2/3, so he made 21 bracelets and 2/3 of a bracelet's length is left over. The leftover string = 2/3 × 3/5 = 2/5 m.


ABE and BCD are triangles. ABC is a straight line. AB = BE and BD = CD. \(\angle\)EBD = 73° and \(\angle\)BCD = 55°. Find \(\angle\)BEA.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle BCD is isosceles (BD = CD) so \(\angle\)DBC = \(\angle\)BCD = 55°. ABC is a straight line, so \(\angle\)ABE = 180° − \(\angle\)EBD − \(\angle\)DBC = 180° − 73° − 55° = 52°. Triangle ABE is isosceles (AB = BE) so \(\angle\)BEA = (180° − 52°) ÷ 2 = 64°.


ABCD is a rhombus and BC = CF. BED and FEC are straight lines. \(\angle\)BFE = 80° and \(\angle\)BAD = 75°. Find \(\angle\)FBE.

In rhombus ABCD, \(\angle\)BCD = \(\angle\)BAD = 75°. Triangle BCD is isosceles (CB = CD), so \(\angle\)DBC = \(\angle\)CDB = (180° − 75°) ÷ 2 = 52.5°. \(\angle\)FBE = \(\angle\)BFE − \(\angle\)... = 80° − 52.5° = 27.5°.


The mass of Parcel P is \(x\) kg. The mass of Parcel Q is 3\(x\) kg. Parcel R is 3 kg lighter than Parcel Q. If \(x\) = 12, find the total mass of the 3 parcels.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


Jaslyn had \(w\) stickers. Krishna had 34 more stickers than Jaslyn. They had 128 stickers altogether. How many stickers did Krishna have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The following table shows the parking charges at a carpark.
For the first hour or less: $2.70
For every additional \(\dfrac{1}{2}\) hour or less: $1.20
Reisha parked her car from 8.15 p.m. to 11.00 p.m. on Tuesday. How much did she have to pay?
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total parking time 8.15 p.m. to 11.00 p.m. = 2 h 45 min. First hour = $2.70. Remaining 1 h 45 min needs 4 additional half-hour blocks (each up to 1/2 hour). Cost = $2.70 + 4 × $1.20 = $2.70 + $4.80 = $7.50.


The usual price of a toaster was $80. A discount of 20% was given during a sale. When Mdm Singh bought the toaster online, an additional $10 discount was given after the discount of 20%. What was the total percentage discount given to Mdm Singh for the toaster?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%