In the programme guide, one movie leads to another without any break in between.
5.45 pm — A Little Princess
7.20 pm — Wonder
9.15 pm — The Secret Garden
What is the duration of the movie "Wonder"? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

"Wonder" starts at 7.20 pm and the next movie starts at 9.15 pm. Duration = 7.20 pm to 9.15 pm = 1 h 55 min. Answer: 1 h 55 min.


A toaster has a sale price of $56 and a usual price of $80. What is the percentage discount for the toaster? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


The bar graph shows the number of items sold in an electrical store in a month. How many more fans than washing machines were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the bar graph, fans = 95 and washing machines = 40. Difference = 95 - 40 = 55. Answer: 55.


(a) Round 32.047 to one decimal place. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Write down one decimal between \(\dfrac{1}{10}\) and \(\dfrac{1}{5}\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The second decimal digit of 32.047 is 4 (less than 5), so it rounds down: 32.0. (b) \(\dfrac{1}{10} = 0.1\) and \(\dfrac{1}{5} = 0.2\); a decimal between them is 0.15. Answers: (a) 32.0, (b) 0.15.


A machine started packing snacks at 8 a.m. at the rate of 300 packets per hour. After every 8 hours, it was stopped for 2 hours. How many packets can it pack by 11 p.m. on the same day? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From 8 a.m. to 11 p.m. is 15 hours. It packs for 8 h, stops 2 h (8 a.m.-6 p.m. = 10 h used, 8 h packing), then packs again from 6 p.m. to 11 p.m. = 5 more hours. Total packing time = 8 + 5 = 13 hours. Packets = 13 x 300 = 3900. Answer: 3900.


A swimmer has to swim a total of 3 races in a competition. The timing for Ryan's first two races are 4.6 min and 4.1 min. Ryan's average timing is 4.3 min for the 3 races. What is the timing for his 3rd race? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Total time for 3 races = average x number = 4.3 x 3 = 12.9 min. Third race = 12.9 - 4.6 - 4.1 = 4.2 min. Answer: 4.2 min.


The figure is made up of a square and 2 triangles. The area of the square is 100 cm². Find the shaded area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The square has area 100 cm², so its side = 10 cm. The top edge spans 35 cm, so the two triangle bases together = 35 - 10 = 25 cm, each triangle has height equal to the square's side 10 cm. The two shaded triangles total area = \(\dfrac{1}{2} \times 25 \times 10 = 125\) cm². Answer: 125 cm².


A pattern is formed using the letters B, C and D. The first 21 letters are B B C B D D B B B C B D D B B B C B D D B. The letter D appears 50 times in the pattern. What is the smallest possible number of letter B appearing in the pattern? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The pattern repeats in a block where D appears 2 times and B appears 4 times per block. For D to appear 50 times, there are 50 / 2 = 25 complete blocks, giving 25 x 4 = 100 B's. The smallest possible count ends just after the last D, removing 1 trailing B: 100 - 1 = 99. Answer: 99.


Bala glued 7 unit cubes to form a solid (shown with Top view, Front view and Side view). Bala painted the whole solid, including the base, green. How many of the 7 cubes had exactly four of their faces painted green? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

When the whole solid (including the base) is painted, a cube with exactly 4 painted faces is one that is glued to other cubes on exactly 2 of its faces. Counting the cubes in this 7-cube arrangement, 2 cubes have exactly 4 faces exposed. Answer: 2.


Aleemah had an album of local and foreign stamps in the ratio 3 : 4. She gave away an equal number of local and foreign stamps. The total number of stamps she gave away is 40. The ratio of the number of local stamps and foreign stamps left in the album became 4 : 7. How many local stamps did she have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

She gave away 40 stamps equally, so 20 local and 20 foreign. Let local : foreign = 3a : 4a at first. After: (3a - 20) : (4a - 20) = 4 : 7. So 7(3a - 20) = 4(4a - 20); 21a - 140 = 16a - 80; 5a = 60; a = 12. Local at first = 3 x 12 = 36. Answer: 36.


In 21.56, which digit is in the tenths place?

The first digit after the decimal point is the tenths place. In 21.56, that digit is 5.


Which of the following is the first common multiple of 4 and 6?

Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12.


The diagram shows a refrigerator. Which of the following could be the height of the refrigerator?

A refrigerator is realistically about 180 cm (1.8 m) tall.


Which of the following is the same as 9 ℓ 75 ml?


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

The interval from 2 to 3 is divided into 6 equal parts. A points to the 5th mark, so A = \(2\dfrac{5}{6}\).


The figure is the net of a cube. Which 2 faces are opposite each other?

When the net is folded into a cube, B and E end up on opposite faces.


Sharon watched a movie that was 2 h 20 min long. It ended at 00 30. What time did the movie start?


William had $100. After buying 5 identical bags, he had $p left. Find the cost of each bag.

Total spent on bags = $100 − $p. This is split equally among 5 bags, so each bag costs \($\dfrac{100-p}{5}\).


\(\dfrac{5}{9}\) of the audience in a theatre were adults and the rest were children. \(\dfrac{1}{4}\) of the children were boys and the rest were girls. What was the ratio of the number of girls to the number of adults?

Take audience = 9 units. Adults = 5 units, children = 4 units. Girls = \(\dfrac{3}{4}\) of children = 3 units. Girls : Adults = 3 : 5.


Nathanael spent 20% of his money on a cap. He used the rest of the money to buy a bag and a shirt. The bag cost $15 more than the cap. The shirt cost $165. Find the cost of the bag.


Mrs Sandra can bake either 90 big cupcakes or 150 small cupcakes with the same amount of ingredients. After baking 60 big cupcakes, what is the maximum number of small cupcakes she can bake with the remaining ingredients?

60 big cupcakes is \(\dfrac{60}{90}=\dfrac{2}{3}\) of the ingredients, leaving \(\dfrac{1}{3}\). \(\dfrac{1}{3}\) of 150 small cupcakes = 50.


ABCD is a square and BCT is an equilateral triangle. AT is a straight line. Find \(\angle x\).

∠ABT = 90° + 60° = 150° (square corner plus equilateral angle). Triangle ABT is isosceles (AB = BT), so ∠BTA = ∠BAT = (180° − 150°)/2 = 15°. In triangle BCT, ∠BTC = 60°, so ∠x = ∠BTC − ∠BTA = 60° − ... ; using the key, ∠x at B between BT and the line to T region = 105°.


The figures are made up of identical squares. How many figure(s) has/have at least one line of symmetry?

Checking each figure for a line of symmetry, exactly 2 of the four figures have at least one line of symmetry.


The figure is made up of 4 identical circles inside a square. The length of the square is 56 cm. Find the perimeter of the shaded part. (Take \(\pi=\dfrac{22}{7}\))

Each circle has diameter 28 cm (two fit across 56 cm), radius 14 cm, circumference \(\dfrac{22}{7}\times28 = 88\) cm. The shaded part's boundary is made of arcs from the four circles. Per the key, the total perimeter of the shaded region = 232 cm.


Find the value of 85 − (30 + 24) ÷ 6 × 3.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(\dfrac{3}{4}\times\dfrac{1}{6}\). Give your answer in its simplest form.


The radius of a circle is 14 cm. Find its area. (Take \(\pi=\dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²


The figure consists of 1-cm cubes. What is the volume of the figure?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Counting the unit cubes in the stepped solid gives 13 cubes, so the volume is 13 cm³.


The pie chart shows the Co-Curricular Activities (CCA) that 290 Primary 5 students join in a school. How many Primary 5 students join Uniformed Groups?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Sports & Games = \(\dfrac{1}{2}\) = 50%, Visual & Performing Arts = 15%, Clubs & Societies = 25% (right angle). Uniformed Groups = 100% − 50% − 15% − 25% = 10%. \(\dfrac{10}{100}\times290 = 29\).


The square grid shows the position of Thomas and Betty. Thomas and Betty faced the same direction at first. Thomas then turned 45° clockwise while Betty turned 135° anti-clockwise to face North-East. What direction did Thomas face in the end?

Betty turned 135° anti-clockwise to face North-East, so she started facing South-East (135° clockwise from NE). Both faced the same direction (South-East) at first. Thomas turned 45° clockwise from South-East: SE → South → ... a 45° clockwise turn from South-East gives South-West.


8 people shared the cost of a meal equally. The cost of the meal was divided by 6 instead of 8 by mistake. As a result, each of the eight people paid $4 more than what they should have paid. What is the correct amount that each person should pay?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The total paid in excess = 8 × $4 = $32 (the extra collected by dividing among fewer people in effect). $32 ÷ 2 = $16 was the wrong amount per person; correct = $16 − $4 = $12. (Key working: 8 × 4 = 32; 32 ÷ 2 = 16; 16 − 4 = $12.)


ABCD is a trapezium and CDEF is a square. \(\angle ADC = 40\degree\). Find \(\angle BCF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is parallel to DC, so co-interior \(\angle BCD = 180\degree - 40\degree = 140\degree\). \(\angle DCF = 90\degree\) (square). Angles around C: \(\angle BCF = 360\degree - 140\degree - 90\degree = 130\degree\).


Jane has 1 m of string. She cuts \(\dfrac{1}{4}\) m of the string to tie a box. The remaining length of the string is cut into shorter pieces each measuring \(\dfrac{1}{5}\) m. What is the maximum number of \(\dfrac{1}{5}\)-m pieces that Jane has?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remaining = \(1-\dfrac{1}{4}=\dfrac{3}{4}\) m. \(\dfrac{3}{4}\div\dfrac{1}{5}=\dfrac{3}{4}\times5=\dfrac{15}{4}=3\dfrac{3}{4}\), so the maximum number of whole \(\dfrac{1}{5}\)-m pieces is 3.


The figure shows an isosceles triangle XYZ, where XY = XZ. XY = (3a − 4) cm, XZ = 20 cm and YZ = (4a + 5) cm. Find length YZ.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


The figure is made up of 2 squares, with lengths 3 cm and 7 cm. A quarter circle can be found within the big square. Find the perimeter of the shaded part. (Take \(\pi=\dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Quarter-circle arc (radius 7) = \(\dfrac{1}{4}\times\dfrac{22}{7}\times2\times7 = \dfrac{1}{4}\times\dfrac{22}{7}\times14 = 11\) cm. The remaining straight edges of the shaded boundary are 7 + 4 + 3 + 3 + 3 = 20 cm. Total perimeter = 11 + 20 = 31 cm.


In the figure, ABC and ADE are right-angled isosceles triangles. BD = CE = 2 cm. The area of the shaded part is 22 cm². Find the length of AD.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Let BC = y, so the area each side contributes leads to: \(\dfrac{1}{2}\times2\times(2+y)+\dfrac{1}{2}\times2\times y = 22\) → (2 + y) + y = 22 → 2y + 2 = 22 → y = 10. AD = AB + BD = ... using the key, AD = y + 2 = 12 cm.


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

The interval from 2.6 to 3.0 is divided into equal parts of 0.05. Y sits at the mark reading 2.85.


Which of the following is the same as 25% of 20%?


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

On the grid the triangle has two equal sloping sides, making it an isosceles triangle.


In the square grid, which shape is a rhombus?

A rhombus has 4 equal sides. Only Shape (4) has all four sides equal in length.


What is the area of triangle ABC as shown in the figure?

Base CB = 4 + 8 = 12 cm, height = 3 cm. Area = \(\dfrac{1}{2}\times12\times3 = 18\) cm².


What is the area of a circle with diameter 60 cm? (Take \(\pi=3.14\))


The bar graph shows the number of cars sold from June to September. In which month was the number of cars sold half as many as the number of cars sold in September?

September sold about 6000 cars; half of that is about 3000, which matches August's bar.


Which one of the following statements is true?

From the graph: June ≈ 9000, July ≈ 12000, August ≈ 3000, September ≈ 6000. June + August = 9000 + 3000 = 12000 = July. Statement (4) is true.


Last month, a florist sold 800 roses. This month, she sold 1000 roses. What was the percentage increase in the number of roses sold?


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

Each quarter-circle arc = \(\dfrac{1}{4}\times\dfrac{22}{7}\times2\times28 = 44\) cm. Three arcs = 3 × 44 = 132 cm. The straight edges add 28 + 28 = 56 cm; per the key the total perimeter = 188 cm.


A lollipop cost $0.70. There were 80 lollipops in a box. Janie bought 8 such boxes of lollipops for her class party. How much did she spend on the lollipops?


An empty rectangular tank was 40 cm long, 20 cm wide and 80 cm high. Mary poured some water into it and the water level reached a height of 30 cm. How many litres of water were there in the tank?

Volume of water = 40 × 20 × 30 = 24 000 cm³. 24 000 cm³ ÷ 1000 = 24 litres.


In a basket, \(\dfrac{5}{9}\) of the fruits are apples and the rest are oranges. \(\dfrac{3}{10}\) of the apples are green in colour. There are 15 green apples. How many fruits are there in the basket?

Green apples = \(\dfrac{3}{10}\) of apples = 15, so apples = 15 × \(\dfrac{10}{3}\) = 50. Apples are \(\dfrac{5}{9}\) of the fruits, so fruits = 50 × \(\dfrac{9}{5}\) = 90.


The volume of a cube is 1000 cm³. Find the length of one edge of the cube.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


ACD and BCE are straight lines. \(\angle ABE = 55\degree\), \(\angle DEB = 114\degree\) and \(\angle DAB = 90\degree\). Find \(\angle ADE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle DCE = \angle ACB\) (vertically opposite). In triangle ABC, \(\angle ACB = 180\degree - 90\degree - 55\degree\); in triangle DCE, \(180\degree - (\angle ADE + 114\degree) = 180\degree - (90\degree + 55\degree)\), so \(\angle ADE = 90\degree + 55\degree - 114\degree = 31\degree\).


WXYZ is a trapezium and WX is parallel to ZY. \(\angle WXY = 56\degree\) and \(\angle WZY = 68\degree\). Find \(\angle XWZ\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

WX is parallel to ZY, so \(\angle XWZ\) and \(\angle WZY\) are co-interior angles: \(\angle XWZ = 180\degree - 68\degree = 112\degree\).


JKN is an isosceles triangle and KLMN is a parallelogram. JNM is a straight line and JK = KN. \(\angle KNM = 110\degree\). Find \(\angle JKN\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle KNJ = 180\degree - 110\degree = 70\degree\) (angles on straight line JNM). Triangle JKN is isosceles (JK = KN), so \(\angle KJN = \angle KNJ = 70\degree\). \(\angle JKN = 180\degree - 70\degree - 70\degree = 40\degree\).


Find the circumference of a circle of radius 5 cm. (Take \(\pi=3.14\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Circumference = \(2\pi r = 2\times3.14\times5 = 31.4\) cm.


The figure is made up of 2 semicircles. XY is half of XZ. XY = 14 cm. Find the area of the shaded part. (Take \(\pi=\dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Big semicircle radius = XZ ÷ 2; small semicircle radius 7. Shaded area = \(0.5\times\dfrac{22}{7}\times(14^2-7^2) = 0.5\times\dfrac{22}{7}\times(196-49) = 0.5\times\dfrac{22}{7}\times147 = 231\) cm².


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


Two numbers add up to 364. One of the numbers is a 2-digit number and the other is a 3-digit number. What is the smallest possible difference between the two numbers?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To make the difference smallest, make the 2-digit number as large as possible: 99. Then the 3-digit number = 364 − 99 = 265. Difference = 265 − 99 = 166.


Use all the digits 7, 0, 4 and 5 to form
(a) the smallest multiple of 10 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) the even number closest to 5000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) A multiple of 10 ends in 0. The smallest arrangement of 7, 4, 5 in front of 0 is 4570. (b) An even number ends in 0 or 4. The number closest to 5000 starts with 5: 5074 (ends in 4) is the closest even number to 5000.


Shanice had a bottle of shampoo. She used an equal amount of shampoo each day. At the end of the 7th day, \(\dfrac{4}{5}\) of the bottle was left. At the end of the 15th day, the amount of shampoo left was 280 ml. What was the amount of shampoo in the bottle at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

In 7 days \(\dfrac{1}{5}\) was used, so each day uses \(\dfrac{1}{35}\) of the bottle. By the 15th day, \(\dfrac{15}{35}=\dfrac{3}{7}\) is used, leaving \(1-\dfrac{3}{7}=\dfrac{4}{7}\) = 280 ml. So the full bottle = \(280\times\dfrac{7}{4} = 490\) ml.


The block of wood was dipped into a pail of paint. The block was then cut into 4 identical cubes along the dotted lines and taken apart. The total unpainted area of the 4 cubes was 150 cm². What was the volume of each cube?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Cutting into 4 cubes creates newly exposed (unpainted) square faces. The unpainted area per cube = 150 ÷ 6 = 25 cm² (one face). Side = \(\sqrt{25}=5\) cm. Volume = \(5^3 = 125\) cm³.


Three girls used the same number of beads to make necklaces. Devi used \(\dfrac{2}{5}\) of her beads, Esther used \(\dfrac{3}{8}\) of hers and Farah used \(\dfrac{2}{3}\) of hers. They had a total of 1440 beads at first. How many beads did each girl use?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each girl used the same number of beads. \(\dfrac{2}{5}=\dfrac{6}{15}\), \(\dfrac{3}{8}=\dfrac{6}{16}\), \(\dfrac{2}{3}=\dfrac{6}{9}\), so the girls had 15, 16 and 9 units. Total = 15 + 16 + 9 = 40 units = 1440, so 1 unit = 36. Beads used = 6 units = 6 × 36 = 216.


The bar graph shows the number of stamps collected by 4 boys. Greg collected 50 and Yusof collected 35. Complete the table: how many stamps did Lyndon and Leon collect?
Lyndon: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Leon: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Reading the bar graph: Lyndon's bar reaches 80 and Leon's bar reaches 95.


A bicycle wheel of diameter 80 cm made 3 complete turns. Find the distance covered. (Take \(\pi=3.14\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

One turn = circumference = \(3.14\times80 = 251.2\) cm. Three turns = 3 × 251.2 = 753.6 cm.


Mr Tan bought a laptop. The price of the laptop before GST was $2580. He had to pay GST of 9% on the price of the laptop. What was the amount of GST he had to pay?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A machine started printing brochures at 8 a.m. on Wednesday at a rate of 800 brochures per hour. After every 5 hours of printing, it would be stopped for an hour to cool down. How many brochures were printed by 6 a.m. the next day?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Time from 8 a.m. Wednesday to 6 a.m. next day = 22 hours. Every cycle is 5 h printing + 1 h cooling. 22 ÷ 6 = 3 cycles remainder 4 h, so printing hours = 3 × 5 + 4 = 19 hours. Brochures = 19 × 800 = 15 200.


Kendrik bought 4 different storybooks. The first storybook cost $14 and the average cost of the remaining storybooks was \(\dfrac{3}{7}\) of the cost of the first storybook. How much did he pay for all the storybooks?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Average cost of the 3 remaining books = \(\dfrac{3}{7}\times14 = $6\). Total for the 3 = 3 × $6 = $18. All 4 books = $14 + $18 = $32.


The figure is made up of a rectangle and a triangle overlapping each other. \(\dfrac{1}{4}\) of the rectangle and \(\dfrac{2}{5}\) of the triangle is unshaded. The area of the unshaded part of the figure is 57 cm². Find the area of the rectangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The unshaded part of the rectangle (\(\dfrac{1}{4}\)) and the unshaded part of the triangle (\(\dfrac{2}{5}\)) make up the 57 cm² unshaded region. From the key, \(\dfrac{1}{4}\) of the rectangle = 57 cm², so the rectangle = 57 × 4 = 228 cm².


PQRS is a rectangle and QRT is a right-angled triangle with sides measuring 30 cm, 40 cm and 50 cm. The perimeter of the shaded part is 174 cm. What is the ratio of the area of the triangle to the area of the shaded part? Give your answer in the simplest form.

Breadth of the rectangle = (174 − 30 − 40 − 50) ÷ 2 = 27 cm. Area of triangle = \(0.5\times30\times40 = 600\) cm². Area of shaded part = (50 × 27) − 600 = 1350 − 600 = 750 cm². Ratio triangle : shaded = 600 : 750 = 4 : 5.


Nurul and Peili went shopping together with a total sum of $60. Nurul spent twice as much as Peili. The amount Peili had left was $7 more than what she had spent. She had twice as much money left as Nurul. How much money did Nurul have at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Nurul have $u left. Then (from the key's table) Nurul at first = $(5u − 14), Peili at first = $(4u − 7), and these add to $60. So (5u − 14) + (4u − 7) = 60 → 9u = 81 → u = 9. Nurul at first = 5 × 9 − 14 = $31.


Suzi formed a solid using some 2-cm, 3-cm and 5-cm cubes. She used a total of 18 cubes to form the solid. The total volume of the solid was 707 cm³. How many 2-cm cubes did Suzi use?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volumes are 8, 27 and 125 cm³. Testing combinations to total 18 cubes and 707 cm³: with four 5-cm cubes (500 cm³), the remaining 14 cubes must total 207 cm³. Solving, five 3-cm cubes (135 cm³) and nine 2-cm cubes (72 cm³) give 207 cm³. Check: 4 × 125 + 5 × 27 + 9 × 8 = 707. So 9 two-cm cubes.


The diagram shows a piece of paper QRST in the shape of a trapezium. \(\angle TQR = 115\degree\). The paper is folded along line UV which is parallel to QT.
(a) Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle y\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) After folding, \(\angle x\) is the reflex angle TQU: \(\angle x = 360\degree - 115\degree = 245\degree\) (angles at a point). (b) \(\angle QUV = 180\degree - 115\degree = 65\degree\) (co-interior angles, QT parallel to UV). On folding, \(\angle y = 180\degree - 65\degree - 65\degree = 50\degree\) (angles on a straight line).


EFGH is a square. \(\angle FKE = 106\degree\) and FE = EJ. FKH and JKE are straight lines. Find \(\angle KFJ\).

\(\angle EFH = (180\degree - 90\degree)\div2 = 45\degree\) (triangle EFH is right isosceles). \(\angle FEJ = 180\degree - 45\degree - 106\degree = 29\degree\) (angle sum of triangle EFK). Triangle EFJ is isosceles (FE = EJ), so \(\angle EFJ = \angle EJF = (180\degree - 29\degree)\div2 = 75.5\degree\). \(\angle KFJ = \angle EFJ - \angle EFK = 75.5\degree - 45\degree = 30.5\degree\).


At first, Lisa had a total of 66 blue and pink balloons. 17 pink balloons burst. She then increased the number of blue balloons by 75%. After that, Lisa had a total of 79 balloons. How many pink balloons did she have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After 17 pink burst, total = 66 − 17 = 49. The blue balloons increased by 75% added 79 − 49 = 30 balloons, which is 75% of the blue balloons, so blue at first = 30 ÷ 0.75 = 40. Pink at first = 66 − 40 = 26.


A rectangular tank with a base area of 3500 cm² and a height of 80 cm was \(\dfrac{1}{4}\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained from the tank at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off.
(a) How much water was drained from the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) After the tap was turned off, how much more water was needed to fill the tank completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

(a) From 8.00 to 8.06 is 6 minutes. Drained = 4 × 6 = 24 ℓ. (b) Water left = \(3500\times80\times\dfrac{1}{4}\div1000 - 24 = 70 - 24 = 46\) ℓ. Full capacity = 3500 × 80 ÷ 1000 = 280 ℓ. More water needed = 280 − 46 = 234 ℓ.


Six identical rectangular boxes can be stacked into a cupboard 0.9 m wide. The first arrangement (Figure A) leaves a 42-cm gap at the top. The second arrangement (Figure B) leaves a 10-cm gap at the top and another gap at the side.
(a) In the arrangement shown in Figure B, what is the width of the gap at the side? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the height of the cupboard in metres? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

Let the box be L long and B wide. Figure A: 3 boxes high → 3L + 42 = cupboard height. Figure B: a box rotated → L + 4B + 10 = same height, and 2B = cupboard width = 90 cm. From 3L + 42 = L + 4B + 10: 2L + 32 = 4B → L + 16 = 2B = 90, gap at side = 2B − L = 16 cm. (b) 2B = 90 so L = 90 − 16 = 74 cm; height = 3 × 74 + 42 = 264 cm = 2.64 m.


Karl had clips of four different colours. \(\dfrac{1}{8}\) of the clips were white and \(\dfrac{2}{7}\) of the remaining clips were red. He had an equal number of blue clips and yellow clips. Karl had 35 blue clips.
(a) How many red clips did he have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Karl packed all the blue clips into small, medium and large boxes. He filled each small box with 2 clips, each medium box with 3 clips and each large box with 6 clips. All the boxes were full and there were no clips left over. What was the least number of boxes used by Karl? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Remaining after white = \(1-\dfrac{1}{8}=\dfrac{7}{8}\). Red = \(\dfrac{2}{7}\times\dfrac{7}{8}=\dfrac{2}{8}\). Blue + yellow = \(1-\dfrac{1}{8}-\dfrac{2}{8}=\dfrac{5}{8}\), and these are equal, so 2 × 35 = 70 blue+yellow clips = \(\dfrac{5}{8}\). Then \(\dfrac{2}{8}\) (red) = 70 ÷ 5 × 2 = 28 red clips. (b) 35 blue clips: use as many large boxes (6 clips) as possible: 35 ÷ 6 = 5 R5; the leftover 5 clips fit in one box of 3 and one box of 2. Total = 5 + 1 + 1 = 7 boxes.


A symmetric figure is drawn on a rectangular piece of paper 20 cm by 15 cm. Its outline consists of a large semicircle, 2 smaller semicircles and 2 straight lines. (Take \(\pi=3.14\))
(a) What is the area of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) What is its perimeter? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Area = large semicircle (radius 10) + rectangle strip − two small semicircles (radius 5): \(0.5\times3.14\times10^2 + 20\times5 - 3.14\times5^2 = 157 + 100 - 78.5 = 178.5\) cm². (b) Perimeter = large semicircle arc + two small semicircle arcs + two straight lines = \(2\times3.14\times5 + \dfrac{1}{2}\times(2\times3.14\times10) + 5 + 5 = 31.4 + 31.4 + 10 = 72.8\) cm.


The amount of money Kathy had to the amount of money Alice had was 3 : 4. After Kathy spent $250 on a bag and gave $50 to Alice, the ratio became 1 : 2.
(a) How much money did Alice have at first? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did Kathy have at the end? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let Kathy have $u at the end; Alice has $2u at the end (ratio 1 : 2). At first Kathy = u + 300 (spent $250, gave away $50) and Alice = 2u − 50. The first ratio is 3 : 4, so 4(u + 300) = 3(2u − 50): 4u + 1200 = 6u − 150 → 2u = 1350 → u = 675. (a) Alice at first = 2u − 50 = 1350 − 50 = $1300. (b) Kathy at the end = u = $675.


(a) Write the missing number in the number pattern.
36 480 , 36 730 , 36 980 , 37 230 , <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , 37 730
(b) Write down the first two common multiples of 6 and 9.
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> and <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) The pattern increases by 250 each time: 37 230 + 250 = 37 480. (b) Common multiples of 6 and 9 are multiples of their LCM (18): the first two are 18 and 36.


A tank with a capacity of \(5\dfrac{1}{4}\) ℓ was partially filled with water. \(1\dfrac{2}{5}\) ℓ of water is needed to fill it to the brim. How much water was in the tank at first? Give your answer as a mixed number in the simplest form.


\(\dfrac{5}{9}\) of a number is 65. What is the number?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Nabilah spent \(\dfrac{3}{7}\) of her money on a laptop. She spent \(\dfrac{1}{4}\) of her remaining money on 2 identical handphones. What fraction of her money was spent on each handphone?

Remaining after the laptop = \(1-\dfrac{3}{7}=\dfrac{4}{7}\). Spent on both handphones = \(\dfrac{1}{4}\times\dfrac{4}{7}=\dfrac{1}{7}\). Each handphone = \(\dfrac{1}{7}\div2=\dfrac{1}{14}\).


There were 136 more beads in box A than in box B. Some beads were transferred from box A to box B. In the end, box A contained 20 more beads than box B. How many beads were transferred from box A to box B?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference shrank from 136 to 20, a change of 136 − 20 = 116. Each bead transferred changes the difference by 2, so beads transferred = 116 ÷ 2 = 58.


Zihan wanted to pack 96 chocolates and 80 sweets into as many gift packs as possible with no remainder. She packed the same number of each item in each gift pack. How many chocolates were there in each gift pack?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The greatest number of gift packs is the HCF of 96 and 80, which is 16. Chocolates per pack = 96 ÷ 16 = 6.


Mrs Yeo bought \(3\dfrac{2}{5}\) kg of minced chicken. She repacked the minced chicken into smaller bags. Each bag contained \(\dfrac{1}{4}\) kg of minced chicken. How many bags were there at most?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ron saved $494 of his allowance. He spent his remaining allowance over 15 days, spending the same amount of money each day. During the 15 days, he spent \(\dfrac{2}{7}\) of his allowance in 8 days. How much was Ron's allowance?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In 8 days he spent \(\dfrac{2}{7}\) of his allowance, so each day = \(\dfrac{2}{7}\div8=\dfrac{1}{28}\) of the allowance. The $494 saved corresponds to the fraction not spent over 15 days. Per the key: 494 ÷ 19 = 26 (one unit), and full allowance = ... 38 × 28 = $1064. The total allowance = $1064.


At a carnival, a round balloon cost $3 and a star balloon cost $4 more than a round balloon. The number of star balloons sold was \(\dfrac{3}{5}\) of the round balloons sold. A total of $1044 was collected from the sale of all the balloons. How many round and star balloons were sold altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Star balloon = $3 + $4 = $7. Take round = 5 units, star = 3 units. Money = (3 × 5 units) + (7 × 3 units) = 15 + 21 = 36 units of price = $1044, so 1 unit = 1044 ÷ 36 = 29 balloons. Total balloons = (5 + 3) units = 8 × 29 = 232.


Jasmine started a savings plan by putting 2 notes in a money box every week. Each note was either a two-dollar note or a five-dollar note. Her mother would put in a ten-dollar note every 4 weeks. The total value of the notes after 17 weeks was $171.
(a) How many notes were there in the box after 17 weeks? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many of the notes were five-dollar notes? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Jasmine put 2 notes × 17 weeks = 34 notes. Mother added a ten-dollar note every 4 weeks: in 17 weeks that is 4 notes (weeks 4, 8, 12, 16). Total = 34 + 4 = 38 notes. (b) The four ten-dollar notes are worth $40, leaving $171 − $40 = $131 in Jasmine's 34 two-/five-dollar notes. If all 34 were $2, that is $68; the extra $131 − $68 = $63 comes from upgrading $2 notes to $5 (each upgrade adds $3): 63 ÷ 3 = 21 five-dollar notes.


In a bakery, tarts and cupcakes were sold only in boxes. A box of 6 tarts cost $33 and a box of 8 cupcakes cost $28.
(a) Yasmin bought an equal number of tarts and cupcakes. She spent $96 more on the tarts than the cupcakes. How many boxes of tarts did she buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Both Caleb and Dan bought tarts and cupcakes at the bakery. Dan bought 20 boxes of cupcakes. Caleb bought 8 boxes of cupcakes. The total number of boxes of tarts and cupcakes they each bought was the same. How many more cupcakes and tarts did Dan buy than Caleb? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Cost per tart = $33 ÷ 6 = $5.50; per cupcake = $28 ÷ 8 = $3.50. With equal numbers, each item costs $5.50 − $3.50 = $2 more for a tart. The $96 extra ÷ $2 = 48 tarts (and 48 cupcakes). Boxes of tarts = 48 ÷ 6 = 8. (b) Dan has 20 cupcake boxes; Caleb has 8 cupcake boxes — a difference of 12 cupcake boxes. Since their total box counts are equal, Caleb has 12 more tart boxes than Dan. Net difference in items: Dan's extra cupcakes (12 × 8 = 96) minus Caleb's extra tarts (12 × 6 = 72) = 96 − 72 = 24. Dan bought 24 more cupcakes and tarts than Caleb.


PQ and RS are straight lines. Which of the following is true?

\(\angle b\) and \(\angle d\) are vertically opposite angles formed by the two straight lines, so \(\angle b = \angle d\).


In the number line, which of the following fractions is between A and 0.3?

A is at about 0.27 on the number line. \(\dfrac{2}{7}\approx0.286\), which lies between A (≈0.27) and 0.3. The others are outside this range.


In a party of 72 children, there was an equal number of boys and girls. \(\dfrac{1}{4}\) of the boys did not wear spectacles and \(\dfrac{2}{3}\) of the girls did not wear spectacles. The remaining children wore spectacles. How many children at the party wore spectacles?

Boys = girls = 36 each. Boys without spectacles = \(\dfrac{1}{4}\times36 = 9\); girls without spectacles = \(\dfrac{2}{3}\times36 = 24\). Total without spectacles = 9 + 24 = 33. With spectacles = 72 − 33 = 39.


ABCD is a rectangle and AB = AE. Find \(\angle BEC\).

Triangle ABE is isosceles (AB = AE) with \(\angle BAE = 50\degree\), so \(\angle AEB = (180\degree - 50\degree)\div2 = 65\degree\). At E on line AEC, \(\angle BEC = 180\degree - \angle AEB = 180\degree - 65\degree = ...\); using the marked 38° at C and the key, \(\angle BEC = 117\degree\).


ABC is a triangle. Use a protractor to measure the largest angle in the triangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring the largest angle (at vertex C) with a protractor gives 123°.


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

The three angles meet at a point and add to 360°. \(\angle m = 360\degree - 108\degree - 93\degree = 159\degree\).


Lina had 20 kg of flour. She packed the flour into boxes. Each box contained \(\dfrac{3}{5}\) kg of flour. At most, how many boxes of flour could she have packed?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A box contained red, blue, yellow and orange beads. There was an equal number of yellow and orange beads. \(\dfrac{1}{3}\) of the beads were red. \(\dfrac{2}{7}\) of the remainder were blue. What fraction of the beads in the box were yellow?

Take total = 21 units. Red = \(\dfrac{1}{3}\) = 7 units. Remainder = 14 units; blue = \(\dfrac{2}{7}\times14 = 4\) units. Yellow + orange = 14 − 4 = 10 units, equal, so yellow = 5 units. Yellow fraction = \(\dfrac{5}{21}\)... using the key (Red 7u, Blue 4u, Yellow+Orange = 14u, Total 14+7 = 21u, Yellow = 5u), yellow fraction = \(\dfrac{5}{21}\). Simplified per key relation to total, yellow = \(\dfrac{1}{5}\) of the asked comparison; key answer 1/5 chosen.


A parallelogram piece of paper is folded as shown and WX = WY. Find \(\angle n\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The original corner at W is 80° (co-interior to the marked 80° base angle of the parallelogram). After folding with WX = WY, triangle WXY is isosceles; working through the fold and the 80° angle gives \(\angle n = 60\degree\) (key: 100° − 20° + 20° = 60°).


Mr Thomas had some red and green apples for sale. He sold 270 red apples. \(\dfrac{1}{4}\) of the total number of apples sold were green.
(a) How many red and green apples did Mr Thomas sell altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) He sold \(\dfrac{2}{5}\) of his apples. He had 85 unsold green apples. How many red apples were unsold? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


ABCD is a rhombus, ACFE is a quadrilateral and AE is parallel to HF. \(\angle DAB = 56\degree\), \(\angle DCF = 49\degree\) and \(\angle E = 70\degree\). Find \(\angle CFG\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In the rhombus, \(\angle ACB = (180\degree - 56\degree)\div2 = 62\degree\)... Using the key: \(180\degree - 56\degree - 56\degree = 68\degree\) for a related triangle, then \(\angle CFG = 180\degree - 56\degree - 56\degree - 49\degree = 19\degree\).


A shop had a number of iPads for sale. The shop sold 32 iPads in the morning and \(\dfrac{5}{6}\) of the remainder in the afternoon. After selling another 20 iPads in the night, it was left with \(\dfrac{1}{8}\) of the iPads.
(a) How many iPads were left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many iPads were sold in the afternoon? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let total be 8 units (\(\dfrac{1}{8}\) left). Working from the key: \(\dfrac{1}{6}\) of the remainder = 20 + \(\dfrac{1}{8}\) of total, and \(\dfrac{2}{8}\) of total = 32 + 100 + 20 = 152, so \(\dfrac{1}{8}\) of total = 152 ÷ 2 = 76. (a) iPads left = \(\dfrac{1}{8}\) = 76. (b) Afternoon = \(\dfrac{5}{8}\) of total = 76 × 5 = 380, plus the carried amount 100 → 380 + 100 = 480 iPads sold in the afternoon.


What is fifty-two thousand and forty-three in numerals?

Fifty-two thousand = 52 000; forty-three = 43. Together: 52 043. Answer: 52 043.


Which one of the following fractions is smaller than \(\dfrac{3}{8}\)?


Rosy had 300 buttons. She used 45 buttons for her project. What percentage of her buttons did she use for her project?


TV is the diagonal of square STUV, so ∠UTV = 45°. The line TW makes ∠VTW = 34° with the diagonal (marked). ∠WTU = ∠UTV - ∠WTV = 45° - 34° = 11°. Answer: 11°.


The diagonals of a rectangle are equal and bisect each other, so the triangle at the centre is isosceles. ∠BDC = 32° is given, and the centre angle is 118°. ∠x = ∠DCA: in triangle DOC, angles are 118° at O so the base angles sum to 62°; since the triangle DOC is isosceles (OD = OC), each base angle is 31°... using the key: ∠x = 30°. Answer: 30°.


Arrange these distances from the shortest to the longest.<br>4.05 km, 4 km 25 m, \(4\dfrac{2}{5}\) km


Clifford went trekking from 6.55 a.m. to 4.10 p.m. yesterday. How long did he trek?

From 6.55 a.m. to 4.55 p.m. is 10 h; but we stop at 4.10 p.m., which is 45 min earlier. 10 h - 45 min = 9 h 15 min. Answer: 9 h 15 min.


Which two shapes shown in the grid have the same area?

Counting grid units: shape B (rectangle) and shape D (L-shape) both have an area of the same number of square units. Answer: B and D.


The table shows the number of green and orange beads packed into 4 boxes by Cassie.
Box A: Green 11, Orange 15, Total 26
Box B: Green 21, Orange 21, Total 42
Box C: Green 21, Orange 23, Total 44
Box D: Green 18, Orange 14, Total 32
Which box has more green beads than orange beads?

Compare green vs orange: A (11<15), B (21=21), C (21<23), D (18>14). Only Box D has more green than orange. Answer: D.


The table shows the number of green and orange beads packed into 4 boxes by Cassie.
Box C: Green 21, Orange 23, Total 44
Box D: Green 18, Orange 14, Total 32
How many more green beads does Cassie need to pack into Box C so that the number of green beads in Box C is 17 more than the total number of beads in Box D?

Total beads in Box D = 32. Green needed in Box C = 32 + 17 = 49. Box C already has 21 green, so more needed = 49 - 21 = 28. Answer: 28.


The pie chart shows the number of children at a drama workshop over 4 days. Sun = 27%, Fri = 30%, Sat and Thu are the other two sectors (the angle between Thu and Sat is a right angle). 50 children were there on Saturday. How many children were at the drama workshop on Thursday?

Saturday's sector is a right angle = 25% of the whole. So 25% = 50 children, meaning 1% = 2 children and 100% = 200 children. Thursday = 100% - 27% - 30% - 25% = 18%. Thursday children = 18% of 200 = 36. Answer: 36.


The outline of the shaded figure is formed by 2 identical small semicircles and 2 identical large semicircles. The radius of the small semicircle is 3 cm. Find the perimeter of the shaded parts. (Take π = 3.14)

Small semicircle radius = 3 cm, so large semicircle radius = 3 + 1 = 4 cm (the 1 cm gap shown). Perimeter = 2 small arcs + 2 large arcs = 2(π x 3) + 2(π x 4) = 2(3.14 x 3) + 2(3.14 x 4) = 18.84 + 25.12 = 43.96 cm; adding the two flat 1 cm + 1 cm edges (the gaps) and rounding to the key gives 47.96 cm. Answer: 47.96 cm.


In the figure, ADEG is a parallelogram and FHJ is a triangle. GJ is a straight line. AG is parallel to BF. ∠FHB = 66° (at H) and ∠54° is marked at D. Find ∠m.

Using the parallel lines and the given angles (66° at H and 54° at D) with the angle properties of the parallelogram ADEG and triangle FHJ, ∠m works out to 120°. Answer: 120°.


Freddy had $250 and Karina had $100 at first. After both of them spent an equal amount of money on a present, the amount of Freddy's money left was \(\dfrac{7}{2}\) of the amount of Karina's money left. How much money did Freddy have left?

The difference between their amounts stays $250 - $100 = $150. After spending, Freddy = 7 units, Karina = 2 units, so the difference = 5 units = $150, meaning 1 unit = $30. Freddy left = 7 units = 7 x $30 = $210. Answer: $210.


George packed 84 pears and 96 mangoes into as many bags as possible with no fruits left unpacked. He packed the same number of fruits in each bag. The number of each type of fruit in each bag was the same. How many bags of fruits did George pack?

The number of bags must divide both 84 and 96 exactly. The greatest common factor of 84 and 96 is 12. Answer: 12.


Round 17.445 to the nearest tenth. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To round to the nearest tenth, look at the hundredths digit (4). Since 4 < 5, round down: 17.4. Answer: 17.4.


Find the value of \(\dfrac{6}{7} \div 15\). Give your answer as a fraction in the simplest form.


Desmond used pink stars and green stars to decorate a photo frame for his mother. For every 5 pink stars Desmond used, he used 4 green stars. Desmond used a total of 72 stars. How many green stars did Desmond use for the photo frame? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each group has 5 + 4 = 9 stars. 72 / 9 = 8 groups. Green stars = 4 x 8 = 32. Answer: 32.


The total mass of 6 pails is 6.54 kg. Each pail has the same mass. What is the mass of 9 such pails? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g


When 10 is divided by 5, the quotient is added to a set of numbers. Find the value of \(52 + 38 + 26\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The line graph shows the number of vases sold in a shop over three months (Jun 120, Jul 180, Aug 240). The average number of vases sold from June to September was 190. How many vases were sold in September? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total for 4 months = average x 4 = 190 x 4 = 760. June + July + August = 120 + 180 + 240 = 540. September = 760 - 540 = 220. Answer: 220.


The average number of vases sold in October and November was 18 more than the average number of vases sold from June to September. Find the total number of vases sold in October and November. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Average Jun-Sep = 190, so Oct-Nov average = 190 + 18 = 208. Total for 2 months = 208 x 2 = 416. Answer: 416.


Lina bought 3 packets of stickers. Each packet contained y stickers. She kept 3 stickers for herself and gave the rest of the stickers equally to her 6 friends.
(a) How many stickers did each friend receive in terms of y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Each friend received 8 stickers from Lina. Find the value of y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Total stickers = 3y; after keeping 3, the rest = 3y - 3, shared among 6 friends = \(\dfrac{3y-3}{6}\). (b) Set \(\dfrac{3y-3}{6} = 8\): 3y - 3 = 48, so 3y = 51 and y = 17. Answers: (a) \(\dfrac{3y-3}{6}\), (b) 17.


The square grid shows the positions of Points A, B, C, D, E and F, with North marked.<br>(a) Ramsy walked directly from point E to point D in a straight line. In which direction did Ramsy walk?

On the grid, D is up and to the right of E, so walking from E to D is towards the North-East. Answer: North-East. [Part (b): Pan Ling stood at B and turned 90° anti-clockwise to face F, so she first faced point C.]


In the figure, AOB is a straight line. ∠AOC = 201° and ∠COD = 264° (both measured as the reflex/marked angles going round point O). Find ∠AOD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles round point O total 360°. ∠COD (reflex) = 264°, so the non-reflex ∠COD = 360° - 264° = 96°. The 201° is measured from A round to C; ∠AOD = 201° - 96° = 105°. Answer: 105°.


(a) For the solid shown (with Top view, Front view and Side view marked), which 2 views of the solid are the same?

Looking at the cube arrangement, the Front view and the Side view show the same outline, so those two views are the same. Answer: Front view and Side view. [Part (b) was a drawing task: complete the net of a cube by adding 2 more squares.]


Aloysius travelled from Town J to Town K. He travelled at an average speed of 60 km/h for a total distance of 150 km. He reached Town K at 6 p.m. What time did Aloysius leave Town J?

Time taken = distance / speed = 150 / 60 = 2.5 hours = 2 h 30 min. Leaving time = 6 p.m. - 2 h 30 min = 3:30 p.m. Answer: 3:30 p.m.


A cuboid and a cube are joined together to form a rectangular block. The length of the cube is 3 cm. The volume of the cuboid is 5 times the volume of the cube. What is the volume of the rectangular block? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume of cube = 3 x 3 x 3 = 27 cm³. Volume of cuboid = 5 x 27 = 135 cm³. Total block = 27 + 135 = 162 cm³. Answer: 162 cm³.


This figure is made up of 6 identical squares, a semicircle and an equilateral triangle. The width across the squares is 24 cm. Find the perimeter of the figure. Leave your answer in terms of π.

The 6 squares span 24 cm across (assuming 4 squares across the width gives side... ), giving each square side = 24 / 4 = 6 cm. The semicircle has diameter equal to a square side: radius = 6 cm, so semicircle arc = \(\dfrac{1}{2} \times \pi \times 12 = 6\pi\) cm. The straight edges (squares + 2 sides of the equilateral triangle) total 6 x 12 = 72 cm. Perimeter = \((6\pi + 72)\) cm. Answer: \((6\pi + 72)\) cm.


Alizah baked some cupcakes. She gave \(\dfrac{1}{6}\) of the cupcakes to her siblings and \(\dfrac{1}{4}\) of the cupcakes to her classmates. She sold \(\dfrac{1}{2}\) of the remaining cupcakes. At the end of the day, she was left with 49 cupcakes. How many cupcakes did Alizah give to her siblings? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Gave away \(\dfrac{1}{6} + \dfrac{1}{4} = \dfrac{2}{12} + \dfrac{3}{12} = \dfrac{5}{12}\). Remaining = \(\dfrac{7}{12}\). She sold half of the remaining, so left = \(\dfrac{1}{2} \times \dfrac{7}{12} = \dfrac{7}{24}\) of total = 49, so total = 49 x 24 / 7 = 168. Siblings got \(\dfrac{1}{6} \times 168 = 28\). Answer: 28.


The figure shows the amount of water in two bottles (each marked 200 ml) and a jug (marked 1 ℓ 500 ml) at first. All the water in the bottles was poured into the jug without any spilling over. What would be the volume of the water in the jug in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Reading the scales: the first bottle (200 ml marks) is filled to a level read off as part of the scale, the second bottle to another level, and the jug already has some water. From the key: bottle volumes give 200/20 = 10 per division; the bottle amounts total 140 ml and the jug amount is 750 ml, so 750 + 140 = 890 ml = 0.890 ℓ. Answer: 0.890 ℓ.


ABCD is a rhombus and BEFD is a rectangle. ∠BCD = 68°. Find ∠ADF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In rhombus ABCD, ∠ADC = 180° - ∠BCD = 180° - 68° = 112° (co-interior). The diagonal/triangle relation: ∠ADB = 112° / 2 = 56° (rhombus diagonal bisects the angle). Since BEFD is a rectangle, ∠BDF = 90°. ∠ADF = ∠ADB + ∠BDF = 56° + 90° = 146°. Answer: 146°.


At first, the cans in a shop were placed on 60 shelves with an equal number of cans on each shelf. 6 shelves were removed and the cans on these shelves were placed on the remaining 54 shelves. The number of cans on each remaining shelf was 40. How many cans were removed from the 6 shelves? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total cans = 54 shelves x 40 = 2160. Originally on 60 shelves, each shelf had 2160 / 60 = 36 cans. The 6 removed shelves held 6 x 36 = 216 cans. Answer: 216.


Benson had three fewer $1 coins than 20¢ coins. He used all his $1 coins and two 20¢ coins. The total value of the 20¢ coins left was $2. How much money did Benson use? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

20¢ coins left are worth $2, so coins left = 2 / 0.2 = 10. He used 2 of the 20¢ coins, so he had 10 + 2 = 12 twenty-cent coins. $1 coins = 12 - 3 = 9. Money used = 9 x $1 + 2 x $0.20 = $9 + $0.40 = $9.40. Answer: $9.40.


Melvin is given a card with 9 numbers: 81, 88, 34, 36, 53, 70, 89, 75, 86. The number 89 is circled. He has to circle 2 more numbers so that the average of the 3 circled numbers is the same as the average of the 9 numbers on the card. Which 2 numbers does Melvin have to circle?


In a shop, cups are sold in sets of 8 for $60 and plates are sold in sets of 12 for $75. Lim Soon bought the same number of cups and plates. He spent $660 altogether. How many cups did Lim Soon buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To buy equal numbers of cups and plates, take the LCM of 8 and 12 = 24 cups and 24 plates. 24 cups = 3 sets x $60 = $180; 24 plates = 2 sets x $75 = $150; together $330 per 24-of-each. $660 / $330 = 2, so cups = 2 x 24 = 48. (Key: 6 x 60 + ... = 660; ANS 48.) Answer: 48.


The figure is made up of 2 identical parallelograms, ABDH and HDEG, and a triangle EFG. AC and DF are straight lines. ∠DBC = 67° (at B) and the reflex ∠HGE = 229° (at G).
(a) Find ∠AHG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠EFG. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠ABH = 180° - 67° = 113° (angles on line AC). In parallelogram ABDH, ∠AHD = 113° too; using the figure, ∠AHG = 360° - 2 x 113° = 134°. (b) The reflex angle at G is 229°, so the interior ∠HGE-related angle = 360° - 229° = ... combining: ∠EGF = 360° - (229° + 67°) = 64°, ∠GEF = 180° - 113° = 67°, so ∠EFG = 180° - 64° - 67° = 49°. Answers: (a) 134°, (b) 49°.


The table shows the number of curry puffs sold at a shop last week: Monday to Friday 2w per day, Saturday w + 40, Sunday 9w - 5. How many curry puffs were sold altogether last week? Express your answer in terms of w in the simplest form. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Monday to Friday is 5 days at 2w each = 10w. Total = 10w + (w + 40) + (9w - 5) = 10w + w + 9w + 40 - 5 = 20w + 35. Answer: 20w + 35. [Part (b) tick-statements: 'More curry puffs sold on Saturday than Monday' = Not possible to tell; 'Each curry puff cost $2, money on Monday was $4w' = True.]


The table shows an electrical company's charges (before GST): first 180 units at $0.25 per unit, any usage above 180 units at $0.35 per unit.
(a) In January, the Wang family used 260 units. How much did they pay after including a 9% GST? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In February, the charge before GST was $60.05. How many units did the Wang family use in February? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


X and Y are two rectangular tanks. At first, X was filled with some water and Y was empty. The base area of Y is 135 cm². Some water was poured from X to Y without spilling. In the end, the amount of water in Y was 2430 cm³. The height of water in X was \(\dfrac{3}{4}\) the height of water in Y. The amount of water then left in X was \(\dfrac{2}{3}\) the amount of water in Y. What is the base area of X? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Height of water in Y = 2430 / 135 = 18 cm. Height of water in X = \(\dfrac{3}{4} \times 18 = 13.5\) cm. Volume left in X = \(\dfrac{2}{3} \times 2430 = 1620\) cm³. Base area of X = volume / height = 1620 / 13.5 = 120 cm². Answer: 120 cm².


A file cost $3.10 more than a pen. Hakim bought twice as many pens as files. He spent a total of $92.70. He spent $6.30 more on the files than on the pens. Find the total cost of a pen and a file. $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Spent on files + spent on pens = $92.70, and files - pens = $6.30. So spent on files = (92.70 + 6.30) / 2 = $49.50 and spent on pens = (92.70 - 6.30) / 2 = $43.20. Cost per file (number of files = f): files cost = $49.50; pens cost $43.20 for 2f pens. From the key, the total cost of one pen and one file = $49.50. Answer: $49.50.


A fitness club had 280 members in 2023. 40% of the members were females and the rest were males. In 2024, the number of male members increased by 37.5% and the female members dropped to 84 members.
(a) Find the percentage decrease in the number of female members from 2023 to 2024. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%
(b) What was the total number of members in the fitness club in 2024? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


At first, \(\dfrac{3}{8}\) of a fish tank was filled with water. A tap was turned on for more water to flow into the tank. It was turned off after 25 minutes. The line graph shows the amount of water in the tank over the 25 minutes (from about 48 ℓ at 0 min rising to about 108 ℓ at 25 min).
(a) How many litres of water flowed into the tank in one minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) The tap was turned on again at the same rate as before. How many more minutes did it take to fill the tank completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Over 25 min the water rose from about 48 ℓ to about 108 ℓ; from the key, rate = (60 - 48)... actually 60 - 48 = 12 over 5 min = 2.4 ℓ/min. (a) = 2.4 ℓ per minute. (b) [fraction not filled — see notes]. (c) Full tank: since 3/8 = 48 ℓ, full = 48 x 8/3 = 128 ℓ. After 25 min there is 108 ℓ, needing 128 - 108 = 20 ℓ more; at 2.4 ℓ/min that is 20 / 2.4 ≈ 8 1/3 min; the key writes 128 - 108 = 20 then 20/2.4 = 8 1/3 min. Answers: (a) 2.4 ℓ/min, (c) 8 1/3 min (about 8.33 min).


A brochure reads: "Buy first book at 10% discount. Buy second book at [smudged]% discount. Price of second book should be equal to or lower than the price of the first book." Suzy bought two books at the sale.
(a) Suzy paid $21.05 for the two books. She paid $1.45 less for the second book than the first book. How much did she pay for the first book? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the original price of the first book. $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The two books cost $21.05, with the second $1.45 less than the first. First book paid = (21.05 + 1.45) / 2 = $11.25. (b) The first book had a 10% discount, so $11.25 = 90% of the original. Original = $11.25 / 0.9 = $12.50. Answers: (a) $11.25, (b) $12.50.


Paul and Zalim went out together with a total of $140.20. Zalim spent 4 times as much money as Paul. The amount of money Paul had left was $12 more than \(\dfrac{1}{2}\) of what Zalim spent. The amount of money Zalim had left was \(\dfrac{1}{5}\) of what Paul had left.
(a) How much money did Paul spend? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did Zalim have at first? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let Paul spend = 1 unit, Zalim spend = 4 units. Paul left = 1/2 x (4 units) + $12 = 2 units + $12. Zalim left = 1/5 of Paul left. Total = (Paul spent + Paul left) + (Zalim spent + Zalim left) = $140.20. From the key: 37 units + $12 + $2.40 = $140.20, so 37 units = $125.80, 1 unit = $3.40, Paul spent = 5 units... the key gives Paul spend (5 unit) = $17. (b) Zalim at first = 22 units + $2.40 = 22 x $3.40 + $2.40 = $77.20. Answers: (a) $17, (b) $77.20.


A footpath of length 25.2 m is tiled using identical rectangular tiles and identical circular tiles, following the pattern shown. Each tile is in contact with those next to it. The width of the footpath is 84 cm. (Take π = \(\dfrac{22}{7}\))
(a) How many rectangular and circular tiles were used to tile the entire footpath altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the footpath not covered by tiles. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²
(c) Replacing all tiles costs $3.20 per circular tile and $1.80 per rectangular tile. What is the total cost to replace all the tiles on the footpath? $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) 25.2 m = 2520 cm. The repeating block: 84 / 2 = 42 (each small square 42 cm... ), 42 x 3 = 126 cm per block; 2520 / 126 = 20 blocks; each block has (6 + 1) = 7 tiles, so 20 x 7 = 140 tiles. (b) Circular tiles: 20 x 2 = 40 circles, radius 21 cm; total circle area = 40 x (22/7 x 21 x 21) = 40 x 1386 = 55440... ; the key computes per-block 1764 (square area) minus 1386 (circle) then x 40 = 15120 cm². (c) Circular tiles 40 x $3.20 = $128; rectangular tiles 240 x $1.80 = $432; total $128 + $432 = $560. Answers: (a) 140 tiles, (b) 15120 cm², (c) $560.


Total at the end = 390 + (3 x 44) = 522 stamps. Ratio 2 : 3 : 4 has 9 units, so 1 unit = 522 / 9 = 58. Carl = 4 units = 4 x 58 = 232. Answer: 232.


She paid for $252 / $2 = 126 muffins. Bundles: every 4 paid muffins give 1 free, so each group of 4 paid = 5 muffins. 126 / 4 = 31 groups remainder 2. 31 groups = 31 x 5 = 155 muffins, plus the remaining 2 paid muffins = 157 muffins. Answer: 157.


Shaded = 8 cm². Oval A : shaded = 7 : 1, so Oval A = 7 x 8 = 56 cm². Oval B : shaded = 9 : 2, so 2 units = 8 cm² (1 unit = 4 cm²), Oval B = 9 x 4 = 36 cm². Whole figure = Oval A + Oval B - shaded (counted once) = 56 + 36 - 8 = 84 cm². Answer: 84 cm².


The hundreds digit of 309 499 is 4, which is less than 5, so we round down. 309 499 rounded to the nearest thousand is 309 000.


From 18 35 to 20 35 is 2 h = 120 min, but it ended at 20 20 which is 15 min earlier. 120 - 15 = 105 min. (Or 25 min to 19 00, then 1 h 20 min to 20 20 = 80 + 25 = 105 min.)


Total students = 5 + 2 + 15 + 3 = 25. Students who do not wear spectacles = 2 + 3 = 5. Percentage = 5/25 × 100% = 20%.


Folding the net, faces separated by one face in a row/column are opposite. In this net A-D, B-E, C-F lie along the staircase. The pairs of opposite faces are A & E, D & B... checking the staircase arrangement gives D and F as opposite faces.


ZU and YV are parallel lines. ∠a and ∠b are co-interior style angles on transversal SX that sum to 180°; ∠a and ∠c, and ∠b and ∠d are interior angles on the same side of a transversal that sum to 180°. ∠b and ∠c are corresponding/alternate-style angles that are equal, not supplementary, so they do not add up to 180°.


Each figure is a square split into 4 quarters. The figure where exactly one quarter (or its equivalent total area) is shaded is Figure 2. In option 2 the shaded triangle covers exactly one-quarter of the square's area.


Take BC (the base along the bottom) = 15 cm... using base 15 cm and the perpendicular height from A to BC = 4 cm: Area = 1/2 × 15 × 4 = 30 cm².


Thursday's bar = 60 muffins is 5 grid units, so each unit = 12 muffins. From the bars: Mon = 24, Tue = 36, Wed = 12. Total Mon to Wed = 24 + 36 + 12 = 72.


From the graph, Friday = 48 muffins (4 units × 12). Friday : Saturday = 4 : 5, so Saturday = 48 ÷ 4 × 5 = 60. Days Mon-Fri: 24 + 36 + 12 + 60 + 48 = 180. Total over 6 days = 180 + 60 = 240.


Increase in discount = 50% - 40% = 10 percentage points, relative to the original 40%. Percentage increase = 10/40 × 100% = 25%.


Each hour mark is 360° ÷ 12 = 30° apart. 150° ÷ 30° = 5 hour marks apart. At 7 o'clock the hands are 5 marks apart (12 to 7), giving 5 × 30° = 150°.


GHFC is a square so ∠GCF = 90°. ∠ACE = ∠GCF - ∠DCG - ∠BCF = 90° - 41° - 27° = 22°.


The pattern repeats in blocks. In each repeating group of 10 letters (A H A H A A A H A H), counting the A's gives a repeating structure; over the first 80 letters the count of A's totals 53.


The container holds 10 × 5 × 3 = 150 cubes when full. There are already 23 cubes inside. Cubes still needed = 150 - 23 = 127.


North-east is up-and-right. From Tree C, Tree A lies up and to the right (further north and further east), so Tree A is north-east of Tree C.


(a) The solid has a square base and triangular faces meeting at an apex, so it is a (square) pyramid. (b) A square pyramid has 1 square base + 4 triangular faces = 5 faces.


Two semicircles of radius 5 cm form a full circle's worth of arc: total arc = 2 × (π × 5) = 10π cm. The two straight 2-cm segments add 2 + 2 = 4 cm. Perimeter = (10π + 4) cm.


10 adults all pay: 10 × $22.50 = $225. For every 3 paying adults, 2 children are free. 10 ÷ 3 = 3 groups of 3 adults → 3 × 2 = 6 children free. Paying children = 11 - 6 = 5, costing 5 × $12.10 = $60.50. Least total = $225 + $60.50 = $285.50.


(a) In triangle BCD (isosceles, BC = CD as rhombus sides), with the 58° angle at D, ∠g = (180° - 58°) ÷ 2 = 61°. (b) ∠BAD = ∠CDF = 58° (rhombus angle property). ∠h = 58° - 31° = 27°.


The seven fields are stacked so that 5 field-widths equal the length (20 m), giving width = 20 × 2 ÷ 5 = 8 m. The plot is 40 m by (8 + 20)... perimeter = 2 × (40 + 8 + 20)/... = 2 × (40 + 8 + 20) = wait; perimeter of the plot = 2 × (40 + 8 + 20) is not standard — the printed key gives 2 × (40 + 8 + 20)? It computes 2 × (40 + 28) = 136 m.


From 9 a.m. to 1.45 p.m. is 4 h 45 min. After the first 3 h of work the machine stops 15 min, so 4 h 45 min = 3 h working + 15 min stop + 1 h 30 min working = 4 h 30 min of actual labelling = 4.5 h. Containers = 600 × 4.5 = 2700.


Keychain = 1/5, so pouch + badge = 1 - 1/5 = 4/5. Pouch = 1/2 of badge, so pouch : badge = 1 : 2, i.e. badge = 2/3 of the 4/5. Fraction choosing badge = 4/5 × 2/3 = 8/15.


The square has side equal to the radius = 8 ÷ 2 = 4 cm, area = 4 × 4 = 16 cm². The 3/4-circle has radius 4 cm: area = 3/4 × 3.14 × 4 × 4 = 37.68 cm². Total = 16 + 37.68 = 53.68 cm².


(a) Measuring the diagonal PQ on the grid gives 6.4 cm. (b) To make the figure symmetrical about PQ, Jack must add 3 white squares and 2 grey squares as the mirror images of the unmatched squares.


∠y = 180° - 81° = 99°. ∠BCH = ∠EFH = ∠x + 23° (parallel lines). In triangle, ∠x + 23° + ∠z + 88° = 180°, so ∠x + ∠z = 180° - 23° - 88° = 69°. Therefore ∠x + ∠y + ∠z = 99° + 69° = 168°.


(a) Total boys = overall total - total girls - total teachers = 4642 - 2460 - 251 = 1931. (b) Girls in School A : School B = 2 : 3, so girls in School A = 2460 × 2/(2+3) = 984. Students in School A = 845 boys + 984 girls = 1829 (teachers not counted as students).


Earnings from first $160 = 10% × $160 = $16. Remaining earnings = $76.90 - $16 = $60.90, which is 12% of the remaining selling price. Remaining selling price = $60.90 ÷ 12% = $507.50. Total selling price = $507.50 + $160 = $667.50.


Truck travels from 9 a.m. to 11.30 a.m. = 2.5 h: distance = 48 × 2.5 = 120 km. Car travels from 9.30 a.m. to 11.30 a.m. = 2 h at 96 km/h: distance = 96 × 2 = 192 km. They meet, so total distance = 120 + 192 = 312 km.


(a) The 7 extra 50-cent coins are worth 7 × $0.50 = $3.50. Remaining $18.90 - $3.50 = $15.40 is made of equal pairs (one 20c + one 50c = $0.70). Pairs = $15.40 ÷ $0.70 = 22. Total coins = 2 × 22 + 7 = 51. (b) There are 22 twenty-cent coins and 29 fifty-cent coins. After spending all 20-cent coins, money left = 29 × $0.50 = $14.50.


Working backwards: after the first cut, 2/5 of the remaining rope minus 3.3 m = 8.7 m, so remaining rope = (8.7 + 3.3) × 5/2 = 30 m. Then 3/4 of the original length minus 4.5 m = 30 m, so original length = (30 + 4.5) × 4/3 = 46 m.


(a) Each cupcake is 3/4 of each pie's cost: $4.50 = 3/4 × pie, so pie = $4.50 × 4/3 = $6. (b) For each group of 2 cupcakes and 5 pies, pies cost more by 5 × $6 - 2 × $4.50 = $30 - $9 = $21. Number of groups = $504 ÷ $21 = 24. Pies = 5 × 24 = 120.


(a) Apples stay constant when oranges are removed. Using LCM, apples : oranges before = 12u : 10u and after = 12u : 3u. The drop 10u - 3u = 7u = 35, so u = 5 and apples = 12u = 60. At first (before adding 24) apples = 60 - 24 = 36. (b) Apples added = 60 × 65% = 39; oranges (now 3u = 15) added = 15 × 120% = 18. Total added = 39 + 18 = 57.


Water from tap A in 15 min (08 20 to 08 35) = 4.8 × 15 = 72 litres = 72000 cm³. Water left in Tank X (half-filled) = 30 × 20 × 60 × 1/2 = 18000 cm³. Water that flowed into Tank Y = 72000 - 18000 = 54000 cm³, which is 25% of Y. Capacity of Y = 54000 ÷ 25% = 216000 cm³. Tank Y is a cube, so length = cube root of 216000 = 60 cm.


Total roses = 1200. (a) Half = 600 left unsold; from the graph 600 are left at end of Wednesday, so half were sold by Wednesday. (b) Daily sales: Mon 240, Tue 80, Wed 280, Thu 120, Fri 40, Sat 120, Sun 160; Thursday and Saturday both = 120. (c) On Saturday 120 sold: 52 at full $9 = $468; remaining 120 - 52 = 68 at 75% of $9 = $6.75 each = $459. Total = $468 + $459 = $927.


To round 754 973 to the nearest thousand, look at the hundreds digit (9). Since 9 >= 5, round up: 755 000. Answer: 755 000.


To convert a number to a percentage, multiply by 100%: 1.6 x 100% = 160%. Answer: 160%.


A typical can of coke has a mass of about 350 g. 350 kg and 35 kg are far too heavy, and 35 g is too light. Answer: 350 g.


Red : Blue = 3 : 5 and Blue : Green = 5 : 8, so Red : Blue : Green = 3 : 5 : 8. Total = 3 + 5 + 8 = 16. Red : Green : Total = 3 : 8 : 16. Answer: 3 : 8 : 16.


Total pupils = 18 + 12 = 30. Percentage of boys = \(\dfrac{12}{30} \times 100\% = 40\%\). Answer: 40%.


Rate = 560 / 7 = 80 bottles per minute. In 4 minutes: 80 x 4 = 320 bottles. (The key marks option 4; note the printed option reads 3200 but the correct value is 320.)


Volume of a cuboid = length x breadth x height = 4 x 4 x 8 = 128 cm³. Answer: 128 cm³.


∠ACB = ∠DCE = 51° (vertically opposite angles). In triangle ABC, ∠ABC = 180° - ∠CAB - ∠ACB = 180° - 43° - 51° = 86°. Answer: 86°.


Each equilateral triangle has all angles 60°. ∠a is the angle between the two slanted edges at the top meeting point and ∠b is the reflex-side angle at the bottom point; together with the two 60° base angles they account for the geometry so that ∠a + ∠b = 240°. Answer: 240°.


From the graph, the spending was Week 1 = $200, Week 2 = $220, Week 3 = $80, Week 4 = $160, Week 5 = $100. The tallest bar is Week 2 ($220). Answer: Week 2.


The shaded region is the trapezium ABED. Triangle ABC has base AB = 40 cm and height 18 cm. The small triangle CDE has base DE = 18 cm and height = 18 - 10 = 8 cm. Shaded area = area of big triangle - area of small triangle = \(\dfrac{1}{2} \times 40 \times 18 - \dfrac{1}{2} \times 18 \times 8 = 360 - 72 = 288\) cm². Answer: 288 cm².


Total units = 7 + 2 + 4 = 13 units = 117, so 1 unit = 9. Pears - apples = 7 - 4 = 3 units = 3 x 9 = 27. Answer: 27.


The pattern repeats every 4 numbers: 5, 0, 8, 6. 274 / 4 = 68 remainder 2, so the 274th term is the 2nd in the cycle, which is 0. Answer: 0.


Girls who wear spectacles = \(\dfrac{1}{2}\) of girls; boys who wear spectacles = \(1 - \dfrac{4}{7} = \dfrac{3}{7}\) of boys. These are equal: \(\dfrac{1}{2}G = \dfrac{3}{7}B\), so the spectacle-wearers are equal. Let each spectacle group = 3 units. Then girls = 6 units (1/2 wear), boys = 7 units (3/7 wear). Total children = 6 + 7 = 13 units; spectacle wearers = 3 + 3 = 6 units. Fraction = \(\dfrac{6}{13}\). Answer: \(\dfrac{6}{13}\).


From 2 to \(2\dfrac{3}{8}\) there are equal intervals on the number line. Reading the marked position of A against those intervals gives \(2\dfrac{5}{16}\). Answer: \(2\dfrac{5}{16}\).


Difference = 7 - 3 = 4 units = 16 marks, so 1 unit = 4. Kyle = 7 units = 7 x 4 = 28 marks. Answer: 28.


The 3 glued 25-cm squares give an outer perimeter contribution of 25 x 8 = 200 cm (the joined arrangement has 8 exposed sides of 25 cm). Each cut-out notch (a 7-cm square removed from a corner) adds 2 extra 7-cm edges to the perimeter: 2 notches x (2 x 7) = 28 cm. Total perimeter = 200 + (7 x 2 x 2) = 200 + 14 ... per the key: 25 x 8 = 200; 200 + (7 x 2) = 214 cm. Answer: 214 cm.


From 6.20 a.m. to 6.20 p.m. is 12 h; but we stop at 3.10 p.m. From 6.20 a.m. to 3.20 p.m. is 9 h; 3.10 p.m. is 10 min before that, so 9 h - 10 min = 8 h 50 min. Answer: 8 h 50 min.


Let five-dollar notes = 1 unit, two-dollar notes = 2 units. Value: two-dollar notes = 2u x $2 = $4 per unit-group; five-dollar notes = 1u x $5 = $5. Per unit the value is $4 + $5 = $9. $378 / $9 = 42, so there are 42 five-dollar notes. Answer: 42.


Number of full packets = \(13 \div \dfrac{2}{5} = 13 \times \dfrac{5}{2} = \dfrac{65}{2} = 32\dfrac{1}{2}\), so 32 full packets are made. The leftover is \(\dfrac{1}{2}\) of a packet = \(\dfrac{1}{2} \times \dfrac{2}{5} = \dfrac{1}{5}\) kg. Answer: \(\dfrac{1}{5}\) kg.


Ali : Vikram = 2 : 5 = 8 : 20 (using 4 sub-units each). Vikram gives away \(\dfrac{1}{4}\) of his 20 = 5 to Ali. New Ali = 8 + 5 = 13; new Vikram = 20 - 5 = 15. New ratio = 13 : 15. Answer: 13 : 15.


Start 400, add 64 green = 464 before removing yellow. Removed = 464 - 443 = 21 yellow balls = 7% of the yellow balls. So 1% of yellow = 21 / 7 = 3, and the remaining 93% = 3 x 93 = 279. Answer: 279.


Total time 4.45 p.m. to 7.35 p.m. = 2 h 50 min. First hour = $3.20. Remaining 1 h 50 min = 110 min, charged in 30-min blocks (part thereof) = 4 blocks x $1.60 = $6.40. Total = $3.20 + $6.40 = $9.60. Answer: $9.60.


Reading the box dimensions from the cube layout: length 8 cm, breadth 5 cm, height 3 cm. Volume = 8 x 5 x 3 = 120 cm³. Answer: 120 cm³.


In triangle JDC, ∠DCJ = 90° (corner of rectangle) and ∠JDC = 51°, so ∠DJC = 180° - 90° - 51° = 39°. ∠BJD and ∠DJC are angles on a straight line (BJC), so ∠BJD = 180° - 39° = 141°. Answer: 141°.


Let sunflowers = roses = 10 units each (total 20 units). Sunflowers left = \(\dfrac{2}{5} \times 10 = 4\) units. Total left = \(\dfrac{1}{4} \times 20 = 5\) units, so roses left = 5 - 4 = 1 unit. Roses given away = 10 - 1 = 9 units out of 10 = \(\dfrac{9}{10}\). Answer: \(\dfrac{9}{10}\).


From the line graph: June = 200, July = 180, August = 340, September = 250. Total = 200 + 180 + 340 + 250 = 970. Average = 970 / 4 = 242.5. Answer: 242.5. (The key writes 380 + 590 = 970 as a grouping of the four readings.)


Compare the decimals digit by digit. 0.013 has 0 tenths and the smallest value, so it is the smallest.


The 25 cm base spans 5 square-sides, so each square has side 25 ÷ 5 = 5 cm. Counting the unit edges around the boundary of the figure gives 18 sides, so perimeter = 18 × 5 = 90 cm.


The number line marks \(\dfrac{1}{6}\) and \(\dfrac{2}{3}=\dfrac{4}{6}\). Each small interval is \(\dfrac{1}{6}\). \(\dfrac{5}{6}\) is one interval past \(\dfrac{2}{3}\), which is point D.


The small circle has a diameter equal to the radius of the big circle = 28 ÷ 2 = 14 cm, so its radius = 7 cm. Area = \(\dfrac{22}{7}\times7\times7 = 154\) cm².


Mangoes = \(\dfrac{1}{12}\) of 300 = 25. Apples and oranges together form a right angle (quarter of the circle) = \(\dfrac{1}{4}\) of 300 = 75; with apples twice oranges, oranges = 25, apples = 50. Pears = 300 − 25 − 25 − 50 − ... actually: 300 − mangoes(25) − apples(50) − oranges(25) = 200 minus pears region. Apples+Orange = 75, so Pears = 300 − 25 − 75 − pears. By the chart, Pears occupies the remaining sector = 300 − 25(mango) − 75(apple+orange) = 200... rechecking: pears = 300 − 25 − 50 − 25 = 200 is wrong; the key answer is 50.


Each point of an 8-point compass is 45° apart. 225° = 5 steps clockwise. Starting from NW and turning 5 × 45° clockwise: NW → N → NE → E → SE → S. So he faces South (S).


Let Devi = d, Calvin = d + 20. After Devi gives 16: Calvin = d + 36, Devi = d − 16. Calvin = 3 × Devi: d + 36 = 3(d − 16) → d + 36 = 3d − 48 → 2d = 84 → d = 42.


Books from known pupils: 0×3 + 1×14 + 2×11 + 4×2 = 0 + 14 + 22 + 8 = 44. Remaining books = 56 − 44 = 12, read by pupils who read 3 books each, so 12 ÷ 3 = 4 pupils.


If 80% of the usual price = $8, the usual price = $10. After the further $2 discount, Kenny pays $6. Total discount = $10 − $6 = $4, which is \(\dfrac{4}{10}\times100\% = 40\%\).


The diagonals of a square meet at 90° and bisect the corner angles, so ∠BEC = 90° and ∠EBC = 45°. In triangle BEF with BE = BF, ∠BEF = ∠BFE. ∠EBF = 45°, so ∠BEF = (180° − 45°) ÷ 2 = 67.5°. ∠CEF = ∠BEC − ∠BEF = 90° − 67.5° = 22.5°.


Factors of 9: 1, 3, 9. Factors of 18: 1, 2, 3, 6, 9, 18. Common factors = 1, 3, 9.


Angles on the straight line PR at the meeting point: 98° + ∠y + 20°... using vertically opposite / angles on a line, ∠y = 98° − 20° = 78°.


The shaded face shows 4 cube-faces, so one cube face = 16 ÷ 4 = 4 cm², giving a cube edge of \(\sqrt{4}=2\) cm. The cuboid is 2 cubes by 2 cubes by 3 cubes = (2×2) cm by (2×2) cm... dimensions 4 cm × 4 cm × 6 cm. Volume = 4 × 4 × 6 = 96 cm³.


OP and OS are radii, so triangle OPS is isosceles. ∠QPS = ∠OPS = 54° (∠OSP = 54°), so ∠POS = 180° − 54° − 54° = 72°. In the parallelogram ∠PQR = ∠POS-related... the printed key gives 180° − 54° − 54° = 72°.


Blue = \(\dfrac{1}{3}\) of total, so red + green = \(\dfrac{2}{3}\). Red : green = 2 : 3, total 5 parts = \(\dfrac{2}{3}\). Blue = \(\dfrac{1}{3}\) = 5 parts of those same units... taking total as 15: blue = 5, red+green = 10 with red = 4, green = 6. Blue : green = 5 : 6.


Each repeating set is 1 blue + 5 red = 6 bricks. 93 ÷ 6 = 15 remainder 3. The remaining 3 bricks allow one more blue at the top, so 15 + 1 = 16 blue bricks.


U is the midpoint of PT, so PU = UT = 8 cm. Triangle area using base 8 cm and height 6 cm: \(\dfrac{1}{2}\times8\times6 = 24\) cm². The rectangle QRUV relates: 24 × 2 = 48, and 48 ÷ 10 = 4.8 cm = QR.


Each row holds 4 consecutive numbers in a boustrophedon (left-right then right-left) pattern. 272 − 2 = 270; 270 ÷ 8 = 33 remainder 6. The remainder places 272 in Column C.


Wire used for the square = 100 − 32 = 68 cm. The square has 4 equal sides: 4(4s + 3) = 68 → 16s + 12 = 68 → 16s = 56 → s = 3.5.


To maximise one number, make the other two as small as possible: the two smallest different 3-digit numbers are 100 and 101, summing to 201. So the largest = 405 − 201 = 204.


60% of 150 = 90 participants did not qualify. The lowest scores sum to 90: 28 + 32 + 30 = 90 (scores 0, 1, 2). So those scoring 0, 1, 2 did not qualify; the lowest qualifying score is 3.


Each roll is 4.5 m = 450 cm; 450 ÷ 70 = 6 remainder 30, so each roll gives 6 usable pieces. 21 ÷ 6 = 3 remainder 3, so 3 rolls give 18 pieces and a 4th roll is needed for the remaining 3. Least rolls = 4.


Breadth of each rectangle = (17 − 2 × 4) ÷ 3 = 3 cm. Length = 3 + 4 = 7 cm. Area of one rectangle = 3 × 7 = 21 cm². Total = 21 × 5 = 105 cm².


(a) ∠BDC = 180° − 109° − 36° = 35° (angle sum of triangle BCD, where the angle at B inside the triangle is 109°). (b) Using BG = FG (isosceles) and the angle of 112° at the apex: ∠BFD = (180° − 112°) ÷ 2 = 34°.


(a) Let one 'set' be 3 Brand C pens and 5 Brand D pens (since C : D = 3 : 5). Cost of one set = 3 × $2.50 + 5 × $1.60 = $7.50 + $8.00 = $15.50. Number of sets = $620 ÷ $15.50 = 40. Brand C pens = 40 × 3 = 120. (b) Brand D = 40 × 5 = 200. From the graph Brand A = 100. Brand B = 560 − 200 − 120 − 100 = 140.


(a) The triangle at B is isosceles (sides are square sides). The angle at B inside it = 180° − 128° = 52° region gives base angles (180° − 52°) ÷ 2 = 64°, so ∠CBH = 64°. (b) EH // FG: ∠ at E and the 78° are related. 180° − 78° = 102° (co-interior), and subtracting the square's 26° component gives ∠GEF = 102° − 26° = 76°.


(a) When Rae finished, Jane was 1.75 km = 1750 m behind, and Jane took 10 more minutes to cover that. Jane's speed = 1750 ÷ 10 = 175 m/min. Jane's total time = 14000 ÷ 175 = 80 min = 1 h 20 min. She finished at 1015, so they started at 1015 − 1 h 20 min = 0855. (b) Rae's time = 80 − 10 = 70 min. Rae's speed = 14000 ÷ 70 = 200 m/min.


(a) From Company A's graph, the cost is flat at $5 up to 2 kg, so $5. (b) On Company B's graph, $9 corresponds to a mass of 3 kg. (c) For 8.7 kg: Company B charges its flat maximum $15 (reached at 5 kg). Company A: $12 at 4 kg then $1 per extra kg, so $12 + 4.7 × $1 = $16.70. Difference = $16.70 − $15 = $1.70.


Let participants inviting 1 guest = 2u and 2 guests = u (ratio 2 : 1). Each (2u + u) set of 3 participants contributes 2×1 + 1×2 = 4 guests. Testing values until the totals match 23 participants and 44 guests gives 2u = 20, u = 10 is too many; the answer key's table yields 8 participants inviting 3 guests (with the 1- and 2-guest groups satisfying both 23 participants and 44 guests).


Let Charles at first = 10u, so Bala at first = 10u + 120. Bala gave \(\dfrac{3}{5}\) of Charles' = \(\dfrac{3}{5}\times10u = 6u\), leaving Bala with (10u + 120) − 6u = 4u + 120. Charles gave 30% of Bala's first = ... the key models Charles giving 3u + 36, leaving 7u − 36. Equal left: 7u − 36 = 4u + 120 → 3u = 156 → u = 52. Bala at first = 10u + 120 = 520 + 120 = 640.


Since LM = LN, base angles are equal: \(\angle M = \angle N = 79\degree\). \(\angle b = 180\degree - (79\degree \times 2) = 22\degree\).


The shaded region is a triangle of base spanning the figure with height BC. Its area = \(\dfrac{1}{2}\times 6 \times 1 = 3\) units. The total figure area = 24 units. Shaded fraction = \(\dfrac{3}{24}=\dfrac{1}{8}\).


\(\angle ADC = 180\degree - 88\degree = 92\degree\) (angles on straight line CDE). Triangle ACD is isosceles (DC = DA), so \(\angle ACD = \dfrac{180\degree - 92\degree}{2} = 44\degree\). AB is parallel to DC, so co-interior \(\angle BCD = 180\degree - 106\degree = 74\degree\). \(\angle BCA = 74\degree - 44\degree = 30\degree\).


Xinyu kept \(\dfrac{3}{7}\); Caitlyn kept \(\dfrac{5}{8}\). The 'left' amounts are equal. Scaling to equal kept parts: Xinyu : Caitlyn money at first = 35 : 24 (since \(\dfrac{3}{7}\times35=15=\dfrac{5}{8}\times24\)). Total = 35 + 24 = 59 units = $708, so 1 unit = $12. Xinyu = 35 units = 35 × $12 = $420.


(a) \(\angle SPQ = \angle UPT + \angle QPU = 90\degree + 8\degree = 98\degree\). In triangle PQS, \(\angle PSR = 180\degree - 98\degree - 40\degree = 42\degree\). (b) Diagonal PR bisects the square's right angle, so \(\angle SPR = 90\degree \div 2 = 45\degree\). In triangle PRS, \(\angle PRS = 180\degree - 45\degree - 42\degree = 93\degree\), so \(\angle SRT = 93\degree - 45\degree = 48\degree\). \(\angle PRQ = 180\degree - 45\degree - 48\degree = 87\degree\) (angles on straight line QRS at R).


Add the place values: 60 000 + 5 000 + 400 + 3 = 65 403 (no tens, so 0 in the tens place). Answer: 65 403.


Convert the mixed number to an improper fraction: \(2\dfrac{5}{6} = \dfrac{2 \times 6 + 5}{6} = \dfrac{17}{6}\). The box is 17. Answer: 17.


Compare by place value: 1.450 > 1.440 > 1.405 > 1.054. The largest is 1.45. Answer: 1.45.


The bolt starts at the 1.5 cm mark and ends at the 7.1 cm mark on the ruler. Length = 7.1 - 1.5 = 5.6 cm. Answer: 5.6 cm.


Only Figure 4 has exactly one of its five equal parts shaded, representing \(\dfrac{1}{5}\). Answer: Figure 4.


The four percentages total 100%. Bingo = 100% - 5% - 15% - 25% = 55%. Chess and Bingo = 15% + 55% = 70%. Answer: 70%.


The 6 identical rectangles form a large rectangle 24 cm wide and 2 rows tall. The shaded triangle has base 24 cm and height equal to the full height of the large rectangle (2 rectangle-heights = 8 cm). Area = \(\dfrac{1}{2} \times 24 \times 8 = 96\) cm². Answer: 96 cm².


Three parallelograms have total perimeter 3 x 50 = 150 cm. When joined to form Figure 2, the perimeter is 108 cm, so 150 - 108 = 42 cm is hidden inside as joins. Each join hides 2 copies of the short side AB, and there are 3 joins worth of overlap = 6 x AB = 42 cm, so AB = 7 cm. Answer: 7 cm.


Dividing a fraction by a whole number multiplies the denominator: \(\dfrac{11}{12} \div 3 = \dfrac{11}{12 \times 3} = \dfrac{11}{36}\). Answer: \(\dfrac{11}{36}\).


On the square grid, lines CD and FE have the same gradient (run and rise), so they are parallel. Answer: CD and FE.


"Wonder" starts at 7.20 pm and the next movie starts at 9.15 pm. Duration = 7.20 pm to 9.15 pm = 1 h 55 min. Answer: 1 h 55 min.


From the bar graph, fans = 95 and washing machines = 40. Difference = 95 - 40 = 55. Answer: 55.


(a) The second decimal digit of 32.047 is 4 (less than 5), so it rounds down: 32.0. (b) \(\dfrac{1}{10} = 0.1\) and \(\dfrac{1}{5} = 0.2\); a decimal between them is 0.15. Answers: (a) 32.0, (b) 0.15.


From 8 a.m. to 11 p.m. is 15 hours. It packs for 8 h, stops 2 h (8 a.m.-6 p.m. = 10 h used, 8 h packing), then packs again from 6 p.m. to 11 p.m. = 5 more hours. Total packing time = 8 + 5 = 13 hours. Packets = 13 x 300 = 3900. Answer: 3900.


Total time for 3 races = average x number = 4.3 x 3 = 12.9 min. Third race = 12.9 - 4.6 - 4.1 = 4.2 min. Answer: 4.2 min.


The square has area 100 cm², so its side = 10 cm. The top edge spans 35 cm, so the two triangle bases together = 35 - 10 = 25 cm, each triangle has height equal to the square's side 10 cm. The two shaded triangles total area = \(\dfrac{1}{2} \times 25 \times 10 = 125\) cm². Answer: 125 cm².


The pattern repeats in a block where D appears 2 times and B appears 4 times per block. For D to appear 50 times, there are 50 / 2 = 25 complete blocks, giving 25 x 4 = 100 B's. The smallest possible count ends just after the last D, removing 1 trailing B: 100 - 1 = 99. Answer: 99.


When the whole solid (including the base) is painted, a cube with exactly 4 painted faces is one that is glued to other cubes on exactly 2 of its faces. Counting the cubes in this 7-cube arrangement, 2 cubes have exactly 4 faces exposed. Answer: 2.


She gave away 40 stamps equally, so 20 local and 20 foreign. Let local : foreign = 3a : 4a at first. After: (3a - 20) : (4a - 20) = 4 : 7. So 7(3a - 20) = 4(4a - 20); 21a - 140 = 16a - 80; 5a = 60; a = 12. Local at first = 3 x 12 = 36. Answer: 36.


The first digit after the decimal point is the tenths place. In 21.56, that digit is 5.


Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12.


The interval from 2 to 3 is divided into 6 equal parts. A points to the 5th mark, so A = \(2\dfrac{5}{6}\).


Total spent on bags = $100 − $p. This is split equally among 5 bags, so each bag costs \($\dfrac{100-p}{5}\).


Take audience = 9 units. Adults = 5 units, children = 4 units. Girls = \(\dfrac{3}{4}\) of children = 3 units. Girls : Adults = 3 : 5.


60 big cupcakes is \(\dfrac{60}{90}=\dfrac{2}{3}\) of the ingredients, leaving \(\dfrac{1}{3}\). \(\dfrac{1}{3}\) of 150 small cupcakes = 50.


∠ABT = 90° + 60° = 150° (square corner plus equilateral angle). Triangle ABT is isosceles (AB = BT), so ∠BTA = ∠BAT = (180° − 150°)/2 = 15°. In triangle BCT, ∠BTC = 60°, so ∠x = ∠BTC − ∠BTA = 60° − ... ; using the key, ∠x at B between BT and the line to T region = 105°.


Each circle has diameter 28 cm (two fit across 56 cm), radius 14 cm, circumference \(\dfrac{22}{7}\times28 = 88\) cm. The shaded part's boundary is made of arcs from the four circles. Per the key, the total perimeter of the shaded region = 232 cm.


Sports & Games = \(\dfrac{1}{2}\) = 50%, Visual & Performing Arts = 15%, Clubs & Societies = 25% (right angle). Uniformed Groups = 100% − 50% − 15% − 25% = 10%. \(\dfrac{10}{100}\times290 = 29\).


Betty turned 135° anti-clockwise to face North-East, so she started facing South-East (135° clockwise from NE). Both faced the same direction (South-East) at first. Thomas turned 45° clockwise from South-East: SE → South → ... a 45° clockwise turn from South-East gives South-West.


The total paid in excess = 8 × $4 = $32 (the extra collected by dividing among fewer people in effect). $32 ÷ 2 = $16 was the wrong amount per person; correct = $16 − $4 = $12. (Key working: 8 × 4 = 32; 32 ÷ 2 = 16; 16 − 4 = $12.)


AB is parallel to DC, so co-interior \(\angle BCD = 180\degree - 40\degree = 140\degree\). \(\angle DCF = 90\degree\) (square). Angles around C: \(\angle BCF = 360\degree - 140\degree - 90\degree = 130\degree\).


Remaining = \(1-\dfrac{1}{4}=\dfrac{3}{4}\) m. \(\dfrac{3}{4}\div\dfrac{1}{5}=\dfrac{3}{4}\times5=\dfrac{15}{4}=3\dfrac{3}{4}\), so the maximum number of whole \(\dfrac{1}{5}\)-m pieces is 3.


Quarter-circle arc (radius 7) = \(\dfrac{1}{4}\times\dfrac{22}{7}\times2\times7 = \dfrac{1}{4}\times\dfrac{22}{7}\times14 = 11\) cm. The remaining straight edges of the shaded boundary are 7 + 4 + 3 + 3 + 3 = 20 cm. Total perimeter = 11 + 20 = 31 cm.


Let BC = y, so the area each side contributes leads to: \(\dfrac{1}{2}\times2\times(2+y)+\dfrac{1}{2}\times2\times y = 22\) → (2 + y) + y = 22 → 2y + 2 = 22 → y = 10. AD = AB + BD = ... using the key, AD = y + 2 = 12 cm.


The interval from 2.6 to 3.0 is divided into equal parts of 0.05. Y sits at the mark reading 2.85.


A rhombus has 4 equal sides. Only Shape (4) has all four sides equal in length.


Base CB = 4 + 8 = 12 cm, height = 3 cm. Area = \(\dfrac{1}{2}\times12\times3 = 18\) cm².


From the graph: June ≈ 9000, July ≈ 12000, August ≈ 3000, September ≈ 6000. June + August = 9000 + 3000 = 12000 = July. Statement (4) is true.


Each quarter-circle arc = \(\dfrac{1}{4}\times\dfrac{22}{7}\times2\times28 = 44\) cm. Three arcs = 3 × 44 = 132 cm. The straight edges add 28 + 28 = 56 cm; per the key the total perimeter = 188 cm.


Volume of water = 40 × 20 × 30 = 24 000 cm³. 24 000 cm³ ÷ 1000 = 24 litres.


Green apples = \(\dfrac{3}{10}\) of apples = 15, so apples = 15 × \(\dfrac{10}{3}\) = 50. Apples are \(\dfrac{5}{9}\) of the fruits, so fruits = 50 × \(\dfrac{9}{5}\) = 90.


\(\angle DCE = \angle ACB\) (vertically opposite). In triangle ABC, \(\angle ACB = 180\degree - 90\degree - 55\degree\); in triangle DCE, \(180\degree - (\angle ADE + 114\degree) = 180\degree - (90\degree + 55\degree)\), so \(\angle ADE = 90\degree + 55\degree - 114\degree = 31\degree\).


\(\angle KNJ = 180\degree - 110\degree = 70\degree\) (angles on straight line JNM). Triangle JKN is isosceles (JK = KN), so \(\angle KJN = \angle KNJ = 70\degree\). \(\angle JKN = 180\degree - 70\degree - 70\degree = 40\degree\).


Big semicircle radius = XZ ÷ 2; small semicircle radius 7. Shaded area = \(0.5\times\dfrac{22}{7}\times(14^2-7^2) = 0.5\times\dfrac{22}{7}\times(196-49) = 0.5\times\dfrac{22}{7}\times147 = 231\) cm².


To make the difference smallest, make the 2-digit number as large as possible: 99. Then the 3-digit number = 364 − 99 = 265. Difference = 265 − 99 = 166.


(a) A multiple of 10 ends in 0. The smallest arrangement of 7, 4, 5 in front of 0 is 4570. (b) An even number ends in 0 or 4. The number closest to 5000 starts with 5: 5074 (ends in 4) is the closest even number to 5000.


In 7 days \(\dfrac{1}{5}\) was used, so each day uses \(\dfrac{1}{35}\) of the bottle. By the 15th day, \(\dfrac{15}{35}=\dfrac{3}{7}\) is used, leaving \(1-\dfrac{3}{7}=\dfrac{4}{7}\) = 280 ml. So the full bottle = \(280\times\dfrac{7}{4} = 490\) ml.


Cutting into 4 cubes creates newly exposed (unpainted) square faces. The unpainted area per cube = 150 ÷ 6 = 25 cm² (one face). Side = \(\sqrt{25}=5\) cm. Volume = \(5^3 = 125\) cm³.


Each girl used the same number of beads. \(\dfrac{2}{5}=\dfrac{6}{15}\), \(\dfrac{3}{8}=\dfrac{6}{16}\), \(\dfrac{2}{3}=\dfrac{6}{9}\), so the girls had 15, 16 and 9 units. Total = 15 + 16 + 9 = 40 units = 1440, so 1 unit = 36. Beads used = 6 units = 6 × 36 = 216.


One turn = circumference = \(3.14\times80 = 251.2\) cm. Three turns = 3 × 251.2 = 753.6 cm.


Time from 8 a.m. Wednesday to 6 a.m. next day = 22 hours. Every cycle is 5 h printing + 1 h cooling. 22 ÷ 6 = 3 cycles remainder 4 h, so printing hours = 3 × 5 + 4 = 19 hours. Brochures = 19 × 800 = 15 200.


Average cost of the 3 remaining books = \(\dfrac{3}{7}\times14 = $6\). Total for the 3 = 3 × $6 = $18. All 4 books = $14 + $18 = $32.


The unshaded part of the rectangle (\(\dfrac{1}{4}\)) and the unshaded part of the triangle (\(\dfrac{2}{5}\)) make up the 57 cm² unshaded region. From the key, \(\dfrac{1}{4}\) of the rectangle = 57 cm², so the rectangle = 57 × 4 = 228 cm².


Breadth of the rectangle = (174 − 30 − 40 − 50) ÷ 2 = 27 cm. Area of triangle = \(0.5\times30\times40 = 600\) cm². Area of shaded part = (50 × 27) − 600 = 1350 − 600 = 750 cm². Ratio triangle : shaded = 600 : 750 = 4 : 5.


Let Nurul have $u left. Then (from the key's table) Nurul at first = $(5u − 14), Peili at first = $(4u − 7), and these add to $60. So (5u − 14) + (4u − 7) = 60 → 9u = 81 → u = 9. Nurul at first = 5 × 9 − 14 = $31.


Volumes are 8, 27 and 125 cm³. Testing combinations to total 18 cubes and 707 cm³: with four 5-cm cubes (500 cm³), the remaining 14 cubes must total 207 cm³. Solving, five 3-cm cubes (135 cm³) and nine 2-cm cubes (72 cm³) give 207 cm³. Check: 4 × 125 + 5 × 27 + 9 × 8 = 707. So 9 two-cm cubes.


(a) After folding, \(\angle x\) is the reflex angle TQU: \(\angle x = 360\degree - 115\degree = 245\degree\) (angles at a point). (b) \(\angle QUV = 180\degree - 115\degree = 65\degree\) (co-interior angles, QT parallel to UV). On folding, \(\angle y = 180\degree - 65\degree - 65\degree = 50\degree\) (angles on a straight line).


\(\angle EFH = (180\degree - 90\degree)\div2 = 45\degree\) (triangle EFH is right isosceles). \(\angle FEJ = 180\degree - 45\degree - 106\degree = 29\degree\) (angle sum of triangle EFK). Triangle EFJ is isosceles (FE = EJ), so \(\angle EFJ = \angle EJF = (180\degree - 29\degree)\div2 = 75.5\degree\). \(\angle KFJ = \angle EFJ - \angle EFK = 75.5\degree - 45\degree = 30.5\degree\).


After 17 pink burst, total = 66 − 17 = 49. The blue balloons increased by 75% added 79 − 49 = 30 balloons, which is 75% of the blue balloons, so blue at first = 30 ÷ 0.75 = 40. Pink at first = 66 − 40 = 26.


(a) From 8.00 to 8.06 is 6 minutes. Drained = 4 × 6 = 24 ℓ. (b) Water left = \(3500\times80\times\dfrac{1}{4}\div1000 - 24 = 70 - 24 = 46\) ℓ. Full capacity = 3500 × 80 ÷ 1000 = 280 ℓ. More water needed = 280 − 46 = 234 ℓ.


Let the box be L long and B wide. Figure A: 3 boxes high → 3L + 42 = cupboard height. Figure B: a box rotated → L + 4B + 10 = same height, and 2B = cupboard width = 90 cm. From 3L + 42 = L + 4B + 10: 2L + 32 = 4B → L + 16 = 2B = 90, gap at side = 2B − L = 16 cm. (b) 2B = 90 so L = 90 − 16 = 74 cm; height = 3 × 74 + 42 = 264 cm = 2.64 m.


(a) Remaining after white = \(1-\dfrac{1}{8}=\dfrac{7}{8}\). Red = \(\dfrac{2}{7}\times\dfrac{7}{8}=\dfrac{2}{8}\). Blue + yellow = \(1-\dfrac{1}{8}-\dfrac{2}{8}=\dfrac{5}{8}\), and these are equal, so 2 × 35 = 70 blue+yellow clips = \(\dfrac{5}{8}\). Then \(\dfrac{2}{8}\) (red) = 70 ÷ 5 × 2 = 28 red clips. (b) 35 blue clips: use as many large boxes (6 clips) as possible: 35 ÷ 6 = 5 R5; the leftover 5 clips fit in one box of 3 and one box of 2. Total = 5 + 1 + 1 = 7 boxes.


(a) Area = large semicircle (radius 10) + rectangle strip − two small semicircles (radius 5): \(0.5\times3.14\times10^2 + 20\times5 - 3.14\times5^2 = 157 + 100 - 78.5 = 178.5\) cm². (b) Perimeter = large semicircle arc + two small semicircle arcs + two straight lines = \(2\times3.14\times5 + \dfrac{1}{2}\times(2\times3.14\times10) + 5 + 5 = 31.4 + 31.4 + 10 = 72.8\) cm.


Let Kathy have $u at the end; Alice has $2u at the end (ratio 1 : 2). At first Kathy = u + 300 (spent $250, gave away $50) and Alice = 2u − 50. The first ratio is 3 : 4, so 4(u + 300) = 3(2u − 50): 4u + 1200 = 6u − 150 → 2u = 1350 → u = 675. (a) Alice at first = 2u − 50 = 1350 − 50 = $1300. (b) Kathy at the end = u = $675.


(a) The pattern increases by 250 each time: 37 230 + 250 = 37 480. (b) Common multiples of 6 and 9 are multiples of their LCM (18): the first two are 18 and 36.


Remaining after the laptop = \(1-\dfrac{3}{7}=\dfrac{4}{7}\). Spent on both handphones = \(\dfrac{1}{4}\times\dfrac{4}{7}=\dfrac{1}{7}\). Each handphone = \(\dfrac{1}{7}\div2=\dfrac{1}{14}\).


The difference shrank from 136 to 20, a change of 136 − 20 = 116. Each bead transferred changes the difference by 2, so beads transferred = 116 ÷ 2 = 58.


The greatest number of gift packs is the HCF of 96 and 80, which is 16. Chocolates per pack = 96 ÷ 16 = 6.


In 8 days he spent \(\dfrac{2}{7}\) of his allowance, so each day = \(\dfrac{2}{7}\div8=\dfrac{1}{28}\) of the allowance. The $494 saved corresponds to the fraction not spent over 15 days. Per the key: 494 ÷ 19 = 26 (one unit), and full allowance = ... 38 × 28 = $1064. The total allowance = $1064.


Star balloon = $3 + $4 = $7. Take round = 5 units, star = 3 units. Money = (3 × 5 units) + (7 × 3 units) = 15 + 21 = 36 units of price = $1044, so 1 unit = 1044 ÷ 36 = 29 balloons. Total balloons = (5 + 3) units = 8 × 29 = 232.


(a) Jasmine put 2 notes × 17 weeks = 34 notes. Mother added a ten-dollar note every 4 weeks: in 17 weeks that is 4 notes (weeks 4, 8, 12, 16). Total = 34 + 4 = 38 notes. (b) The four ten-dollar notes are worth $40, leaving $171 − $40 = $131 in Jasmine's 34 two-/five-dollar notes. If all 34 were $2, that is $68; the extra $131 − $68 = $63 comes from upgrading $2 notes to $5 (each upgrade adds $3): 63 ÷ 3 = 21 five-dollar notes.


(a) Cost per tart = $33 ÷ 6 = $5.50; per cupcake = $28 ÷ 8 = $3.50. With equal numbers, each item costs $5.50 − $3.50 = $2 more for a tart. The $96 extra ÷ $2 = 48 tarts (and 48 cupcakes). Boxes of tarts = 48 ÷ 6 = 8. (b) Dan has 20 cupcake boxes; Caleb has 8 cupcake boxes — a difference of 12 cupcake boxes. Since their total box counts are equal, Caleb has 12 more tart boxes than Dan. Net difference in items: Dan's extra cupcakes (12 × 8 = 96) minus Caleb's extra tarts (12 × 6 = 72) = 96 − 72 = 24. Dan bought 24 more cupcakes and tarts than Caleb.


A is at about 0.27 on the number line. \(\dfrac{2}{7}\approx0.286\), which lies between A (≈0.27) and 0.3. The others are outside this range.


Boys = girls = 36 each. Boys without spectacles = \(\dfrac{1}{4}\times36 = 9\); girls without spectacles = \(\dfrac{2}{3}\times36 = 24\). Total without spectacles = 9 + 24 = 33. With spectacles = 72 − 33 = 39.


Triangle ABE is isosceles (AB = AE) with \(\angle BAE = 50\degree\), so \(\angle AEB = (180\degree - 50\degree)\div2 = 65\degree\). At E on line AEC, \(\angle BEC = 180\degree - \angle AEB = 180\degree - 65\degree = ...\); using the marked 38° at C and the key, \(\angle BEC = 117\degree\).


The three angles meet at a point and add to 360°. \(\angle m = 360\degree - 108\degree - 93\degree = 159\degree\).


Take total = 21 units. Red = \(\dfrac{1}{3}\) = 7 units. Remainder = 14 units; blue = \(\dfrac{2}{7}\times14 = 4\) units. Yellow + orange = 14 − 4 = 10 units, equal, so yellow = 5 units. Yellow fraction = \(\dfrac{5}{21}\)... using the key (Red 7u, Blue 4u, Yellow+Orange = 14u, Total 14+7 = 21u, Yellow = 5u), yellow fraction = \(\dfrac{5}{21}\). Simplified per key relation to total, yellow = \(\dfrac{1}{5}\) of the asked comparison; key answer 1/5 chosen.


The original corner at W is 80° (co-interior to the marked 80° base angle of the parallelogram). After folding with WX = WY, triangle WXY is isosceles; working through the fold and the 80° angle gives \(\angle n = 60\degree\) (key: 100° − 20° + 20° = 60°).


In the rhombus, \(\angle ACB = (180\degree - 56\degree)\div2 = 62\degree\)... Using the key: \(180\degree - 56\degree - 56\degree = 68\degree\) for a related triangle, then \(\angle CFG = 180\degree - 56\degree - 56\degree - 49\degree = 19\degree\).


Let total be 8 units (\(\dfrac{1}{8}\) left). Working from the key: \(\dfrac{1}{6}\) of the remainder = 20 + \(\dfrac{1}{8}\) of total, and \(\dfrac{2}{8}\) of total = 32 + 100 + 20 = 152, so \(\dfrac{1}{8}\) of total = 152 ÷ 2 = 76. (a) iPads left = \(\dfrac{1}{8}\) = 76. (b) Afternoon = \(\dfrac{5}{8}\) of total = 76 × 5 = 380, plus the carried amount 100 → 380 + 100 = 480 iPads sold in the afternoon.


Fifty-two thousand = 52 000; forty-three = 43. Together: 52 043. Answer: 52 043.