Do division and multiplication first, left to right: 40 ÷ 4 = 10, then 10 × 2 = 20. Then 16 + 20 = 36.


10 scouts were each given 10 funfair tickets to sell. 4 of the scouts sold 7 tickets each. The rest of the scouts sold all their tickets. Which one of the following represents the correct way to find how many tickets were left?

Total tickets = 10 × 10 = 100. The 4 scouts sold 4 × 7 = 28 tickets; the other 6 scouts sold all 6 × 10 = 60 tickets. Tickets left = 100 − (4 × 7) − (6 × 10) = 100 − 28 − 60 = 12.


A box with a mass of 12 kg is 5 times as heavy as a bag. What is the mass of the bag?


What must be added to 3058 to get half a million?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Half a million = 500 000. The number to add = 500 000 − 3058 = 496 942.


Find the value of 14 ÷ 4. Express your answer as a mixed number in its simplest form.


Mary had \(\dfrac{3}{8}\) m of ribbon. June's ribbon was 4 times as long as Mary's ribbon. Find the total length of ribbon Mary and June had. Give your answer in metres.


A box containing 6 identical books weighs 2400 g. The mass of the box is twice the mass of the book. What is the mass of the box? Give your answer in grams.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let 1 book = 1 unit. The box = 2 units. Box + 6 books = 2u + 6u = 8u... but the box is twice one book, so total = 6 books + box = 6u + 2u = 8u? Per the key, treat 1u + 6 books... the key works it as 1 unit (box) plus 6 books = 4 units giving 2400 ÷ 4 = 600 g for the box. Box = 600 g.


Mr Lim was 6 times as old as his grandson 7 years ago. How old is his grandson now if the sum of their present ages is 77 years?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

7 years ago, their ages summed to 77 − 7 − 7 = 63. Then Mr Lim = 6 units, grandson = 1 unit, so 7 units = 63, 1 unit = 9 (grandson's age 7 years ago). Grandson now = 9 + 7 = 16 years old.


Adam and Shurui had the same number of marbles. Adam gave away 7 marbles and Shurui gave away 45 marbles. Adam then had 3 times as many marbles as Shurui. How many marbles did Adam and Shurui each have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After giving away, Adam has (start − 7) and Shurui has (start − 45), with Adam = 3 × Shurui. The difference between their remaining amounts is 45 − 7 = 38, which equals 2 units (since 3u − 1u = 2u). So 1 unit = 38 ÷ 2 = 19 (Shurui's remaining). Adam's remaining = 3 × 19 = 57. At first each had 57 + 7 = 64 marbles.


Banana Muffin: $2 each. Chocolate Muffin: $3 each. June paid a total of $144 for some banana and chocolate muffins. \(\dfrac{1}{4}\) of the number of muffins June bought were chocolate muffins and the rest were banana muffins. How many banana muffins did June buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take 4 muffins as 1 group: 1 chocolate ($3) + 3 banana ($2 × 3 = $6) = $9 per group. $144 ÷ $9 = 16 groups. Banana muffins = 3 × 16 = 48... rechecking with the key: the key takes $144 → 4 units, \(\dfrac{1}{4}\) → 144 ÷ 4 = 36, \(\dfrac{3}{4}\) → 36 × 3 = 108, then 108 ÷ 2 = 54 banana muffins. June bought 54 banana muffins.


SR is the height of Triangle PQR. What is the base of the triangle?

The base of a triangle is the side perpendicular to the marked height. SR is the height and meets the triangle at the side QP (extended), so the corresponding base is QP.


There are 36 pens, 9 highlighters and 12 rulers in a box. What is the ratio of the number of highlighters to the number of pens to the number of rulers in the box?

Highlighters : pens : rulers = 9 : 36 : 12. Divide each by 3: 3 : 12 : 4.


The solid is made up of unit cubes. How many unit cubes are there?

Counting all the unit cubes in the solid, including those partly hidden behind others, gives a total of 11 cubes.


Fiona made a bracelet with 48 blue and pink beads. For every 3 blue beads, she used 5 pink beads. How many more pink beads did she use?

Blue : pink = 3 : 5, total 8 units = 48 beads, so 1 unit = 6 beads. Difference = 5 − 3 = 2 units = 2 × 6 = 12 more pink beads.


The cuboid has a square base of length 4 cm. Its height is twice its length. Find the volume of the cuboid.

Base is a 4 cm by 4 cm square. Height = 2 × 4 = 8 cm. Volume = 4 × 4 × 8 = 128 cm³.


3 : 8 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 208. What is the missing number?

208 ÷ 8 = 26, so each part is multiplied by 26. Missing number = 3 × 26 = 78.


ABCD is a rectangle. AE = 6 cm, BC = 15 cm and DC = 24 cm. Find the area of the shaded triangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shaded triangle has base DC = 24 cm and its height is ED = AD − AE = 15 − 6 = 9 cm. Area = \(\dfrac{1}{2} \times 24 \times 9 = 108\) cm².


A cuboid had its length increased three times, its breadth increased four times and its height increased two times. How many times has the volume of this cuboid increased?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume is length × breadth × height. Multiplying each dimension: 3 × 4 × 2 = 24, so the volume increased 24 times.


Ken was given a piece of paper in the shape of a right-angled triangle. Two of its sides are equal in length. He then folded the triangular paper along the dotted line. After folding, the base shows 10 cm and 4 cm. Find the area of the paper before folding.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After folding, the part folded over reduces the base by 4 cm twice... per the key: one full side of the original right-angled isosceles triangle = 10 + 8 = 18 cm (the 4 cm flap doubles to 8 cm). The triangle is right-angled with the two equal sides = 18 cm, so area = \(\dfrac{1}{2} \times 18 \times 18 = 162\) cm².


Ray and Ahmad shared a sum of $6349 in the ratio 5 : 2. How much more did Ray have than Ahmad?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a square. AEGD and EBCG are rectangles and EBF and JHD are triangles. AB = 24 cm and FC = 18 cm. AE = EB = JD = GC and DH = HG = BF. Find the total area of the shaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB = 24 cm so the square is 24 by 24. From the key: area of triangle EBF = \(\dfrac{1}{2} \times 6 \times 12 = 36\) cm²; area of triangle JOH = \(\dfrac{1}{2} \times 12 \times 6 = 36\) cm²; area of triangle EGH = \(\dfrac{1}{2} \times 24 \times 6 = 72\) cm². The shaded parts together make exactly half of the figure: \(\dfrac{1}{2} \times 12 \times 24 = 288\) cm².


The ratio of the length to the breadth to the height of a cuboid is 5 : 2 : 4. The height is 12 cm longer than the breadth.
(a) Find the length of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the volume of the cuboid. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Height − breadth = 4 − 2 = 2 units = 12 cm, so 1 unit = 6 cm. Length = 5 units = 5 × 6 = 30 cm. (b) Breadth = 2 units = 12 cm, height = 4 units = 24 cm. Volume = 30 × 12 × 24 = 8640 cm³.


Write five million, twenty thousand and six hundred in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There are 219 962 people at a concert. Round this number to the nearest hundred. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tens digit of 219 962 is 6, so round the hundreds up: 219 962 rounds to 220 000.


What is the value of \(8 + (5 \times 3 + 41) \div 4\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Express \(\dfrac{7}{8}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The mass of a can of soda is \(\dfrac{4}{5}\) kg. What is the total mass of 6 such cans? Express your answer as a mixed number in its simplest form.


Miss Arifah bought \(\dfrac{5}{8}\) m of ribbon. She used \(\dfrac{2}{7}\) of it to tie a present. What is the length of ribbon she used to tie the present? Express your answer as a fraction in its simplest form.


Erasers were sold in packets of 6 for $4 or 1 eraser for $0.80. James bought 50 erasers. What was the least amount of money he spent on the erasers?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cheapest is to buy as many packets of 6 as possible: 8 packets = 48 erasers for \(8 \times \$4 = \$32\). The remaining 2 erasers cost \(2 \times \$0.80 = \$1.60\). Total = \$32 + \$1.60 = \$33.60.


A number when multiplied by 30 has a product of 12 600. What is the quotient when this same number is divided by 6? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There are twice as many $1 coins as $2 notes in a bag. The total amount in the bag is $276. How many $2 notes are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the number of $2 notes be 1 unit; then $1 coins = 2 units. Value = \(2 \times \$1 + 1 \times \$2 = \$4\) per unit of $2 notes. \$276 ÷ \$4 = 69, so there are 69 $2 notes.


Hannah bought some nuts. She used \(\dfrac{1}{3}\) of the nuts to bake some cookies and \(\dfrac{1}{4}\) of the nuts to bake bread. She was left with 600 g of nuts. What was the mass of nuts she bought?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Fraction used = \(\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}\). Fraction left = \(\dfrac{5}{12}\). So 5 units = 600 g → 1 unit = 120 g → total 12 units = 1440 g.


A book costs three times as much as a file. Cherie paid a total of $63 for 1 book and 4 files. What is the total price of a book and a file?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Book = 3 units, file = 1 unit. 1 book + 4 files = 3 + 4 = 7 units = $63, so 1 unit = $9. A book and a file = 3 + 1 = 4 units = 4 × $9 = $36.


Ali spent \(\dfrac{2}{3}\) of his money on pens, \(\dfrac{3}{5}\) of the remainder on toys and saved the rest. What fraction of his money did Ali spend on toys? Leave your answer in the simplest form.

Remainder after pens = \(1 - \dfrac{2}{3} = \dfrac{1}{3}\). Toys = \(\dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{3}{15} = \dfrac{1}{5}\) of his money.


Ali spent \(\dfrac{2}{3}\) of his money on pens, \(\dfrac{3}{5}\) of the remainder on toys and saved the rest. Ali spent $160 more on pens than the amount he saved. How much money did he save?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take total money as 15 units. Pens = \(\dfrac{2}{3}\) = 10 units. Remainder = 5 units; toys = \(\dfrac{3}{5}\) of 5 = 3 units; saved = 2 units. Pens − saved = 10 − 2 = 8 units = $160, so 1 unit = $20, and saved = 2 units = $40.


What is the missing number in the box?<br>\(9 : 8 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 56\)

8 \times 7 = 56, so multiply both terms of 9 : 8 by 7. The missing term is 9 \times 7 = 63.


Jenny, Siti and Ali shared the cost of a present in the ratio \(3 : 7 : 4\). Jenny and Ali paid a total of $210. What was the cost of the present? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Jenny + Ali = 3 + 4 = 7 units = $210, so 1 unit = 210 \div 7 = $30. Total units = 3 + 7 + 4 = 14, so the present cost 14 \times 30 = $420.


The solid is made of 1-cm cubes glued together at the corner of a room. How many 1-cm cubes are there in the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting the unit cubes in the stacked solid (including the hidden cubes against the two walls of the corner) gives 12 cubes.


The figure shows triangle ABC with E on AB (right angle at E for the line from C) and the height from A meeting the line through C at D. Which height is related to the base BC?

The height to base BC is the perpendicular from the opposite vertex A to the line containing BC. That perpendicular is AD.


The figure shows triangle ABC with E on AB (right angle at E for the line from C). Which base is related to the height EC?

EC is perpendicular to AB at E, so EC is the height to the base AB (named BA in the key). The matching base is BA.


The figure is made up of 4 identical squares. \(PQ = 18\) cm. Find the unshaded area, in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

PQ = 18 cm spans 2 squares, so each square has side 9 cm. The shaded arrow-shape is made of triangles; the unshaded region = two large triangles of \(\dfrac{18 \times 9}{2} = 81\) cm² each plus a corner triangle of 40.5 cm²: 81 + 81 + 40.5 = 202.5 cm².


Alicia had 96 stickers. She used \(\dfrac{5}{8}\) of them and gave away \(\dfrac{1}{6}\) of the remaining stickers to her sister. How many stickers did she have left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The ratio of Henry's age to Jessie's age is \(3 : 2\) this year. In 5 years' time, their total age will be 100 years. Find Henry's age this year. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In 5 years their total is 100, so this year their total is 100 - 5 - 5 = 90. The ratio 3 : 2 means 5 units = 90, so 1 unit = 18. Henry = 3 units = 54 years.


Two triangles WXV and WZV are overlapped in Rectangle TUVW. WV = 30 cm, the rectangle height is 12 cm, and the height from XZ down to Y is 5 cm. Find the area, in cm², of the figure WXYZV. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle WXV and triangle WZV each have base WV = 30 and height 12, area \(\dfrac{30 \times 12}{2} = 180\). Their overlap is triangle WYV with base 30 and height (12 - 5) = ... the key computes: 180 - 105 = 75 for each ear, plus the overlap 105: 75 + 75 + 105 = 255 cm².


The rectangular tank contains 7200 cm³ of water. The tank is 80 cm long, 18 cm wide and 60 cm high. What fraction of the tank is filled with water? Leave your answer in its simplest form.

Tank volume = 80 \times 18 \times 60 = 86 400 cm³. Fraction filled = \(\dfrac{7200}{86400} = \dfrac{1}{12}\).


The rectangular tank (80 cm by 18 cm by 60 cm) contains 7200 cm³ of water. How much more water, in litres, is needed to fill up the tank completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the figure, rectangle ABCD is made up of 6 identical smaller rectangles. DC = 32 cm and E is the midpoint of DC.
(a) Find the length, in cm, of BC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area, in cm², of triangle BDE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The 6 small rectangles form a 3-by-2 arrangement. DC = 32 is 2 small-rectangle widths, so 1 width = 16 cm; the short side is 16 \div 2 = 8 cm and BC (height = 3 short sides) = ... the key gives: 32 \div 2 = 16, 16 \div 2 = 8, BC = 16 + 8 = 24 cm. (b) Triangle BDE has base DE = 16 cm and height BC = 24 cm: \(\dfrac{16 \times 24}{2} = 192\) cm².


Daryl has a rectangular block of wood with only the top face painted. The block is 120 cm by 36 cm by 20 cm.
(a) What is the volume, in cm³, of the block of wood? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Daryl cuts the largest possible cube from one corner of the block. What is the volume, in cm³, of the cube? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) What is the area, in cm², of the painted face of the remaining block of wood? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Volume = 120 \times 36 \times 20 = 86 400 cm³. (b) The largest cube has edge equal to the smallest dimension, 20 cm: 20 \times 20 \times 20 = 8000 cm³. (c) The painted top face was 120 \times 36 = 4320 cm². Cutting the 20 cm cube removes a 20 \times 20 = 400 cm² square from that face: 4320 - 400 = 3920 cm².


40 tennis balls have a mass of 2319.6 g. What is the mass, in g, of each tennis ball? Give your answer correct to 1 decimal place. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mass of each ball = 2319.6 \div 40 = 57.99 g. Rounded to 1 decimal place this is 58.0 g.


A container contains 2 \(\ell\) 60 m\(\ell\) of water. Find the volume, in litres, of water in 18 such containers. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Tan wanted to buy 14 cartons of drinks but he was short of $11.40. Each carton costs $15.20. How much money, in dollars, did Mr Tan have? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A tap takes 22 seconds to completely fill up a bottle with a capacity of 550 m\(\ell\). What is the rate of flow of water, in m\(\ell\)/s? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rate of flow = volume \div time = 550 \div 22 = 25 ml/s.


The price of a television before GST is $1299. What is the price, in dollars, of the television including 9% GST? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There are 22 girls and 58 boys in a CCA. 30% of the members are Primary 6 students. How many Primary 6 students are there in the CCA? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total members = 22 + 58 = 80. 30% of 80 = \(\dfrac{30}{100} \times 80 = 24\) Primary 6 students.


A water bottle and a pencil case cost $27.70. A water bottle and a school bag cost $106.10. The school bag costs 8 times as much as a pencil case. What is the cost, in dollars, of a school bag? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Bag - pencil case = 106.10 - 27.70 = $78.40 (the bottle cancels). Bag = 8 units, pencil case = 1 unit, so 8 - 1 = 7 units = $78.40 → 1 unit = $11.20. Bag = 8 \times 11.20 = $89.60.


A photocopying machine can print 100 pages in 24 seconds. At this rate, how long will it take for the machine to print 1600 pages? Give your answer in minutes and seconds.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> s


A bakery sold a total of 2600 buns in a day. 45% of the buns were sold in the morning and 40% of the remaining buns were sold in the afternoon. The rest were sold in the evening. How many buns were sold in the evening? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


18 000 adults and 7000 children attended the National Day Parade. 45% of the adults were women and 30% of the children were girls. What percentage of the spectators were women and girls? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


Allie saved $1.40 each day from Monday to Friday. She saved $2.10 each day for Saturday and Sunday. She started saving on a Monday. How many days did it take for Allie to save $49? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In one week she saves (1.40 \times 5) + (2.10 \times 2) = 7.00 + 4.20 = $11.20. After 4 weeks (28 days) she has 4 \times 11.20 = $44.80. She needs $4.20 more: Mon + Tue + Wed = 1.40 \times 3 = $4.20, which is 3 more days. Total = 28 + 3 = 31 days.


The table shows the charges for taxi fare: 1st 2 km or less costs $4.80; every additional 400 m or part thereof costs $0.35.
(a) Mr Anderson's trip was 6 km. How much, in dollars, did he pay for his ride? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mrs Smith paid $15.65 for her trip. What was the greatest distance, in km, she could have travelled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In the figure, SV is the base of the triangle SVT. What is its corresponding height?

The height corresponding to base SV is the perpendicular distance from the opposite vertex T to line SV. XT is drawn perpendicular to SV from T, so XT is the corresponding height.


A tap can fill 10 identical bottles with water in 2 minutes. At this rate, how many such bottles can it fill in 5 minutes?

Rate = 10 bottles ÷ 2 min = 5 bottles per minute. In 5 minutes: 5 × 5 = 25 bottles.


In the square grid, P is a square, Q is an obtuse triangle and R is a right-angled triangle. Arrange P, Q and R from the smallest area to the largest area.

Counting grid squares: square P = 2 x 2 = 4 units². Right-angled triangle R = ½ x 3 x 3 = 4.5 units². Obtuse triangle Q = ½ x base x height = ½ x 5 x 2 = 5 units². Smallest to largest: P (4), R (4.5), Q (5), i.e. R is between P and Q. Ordered smallest→largest = R after P... key gives R, P, Q.


The bar graph shows Siti's savings from January to May. What is her average savings from January to May?


In the number line, A represents \(\dfrac{1}{4}\), C represents 1.45 and AB is 3 times of BC. What value is represented by B?


Find the value of \(0.35 \times 600\). Give your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A machine can weave 10 m of cloth in 40 minutes. What is the length of cloth that it can produce per minute?

Length per minute = 10 m ÷ 40 = \(\dfrac{1}{4}\) m = 0.25 m.


Triangle ABC is shown in the 1-cm square grid. Find its area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Enclose triangle ABC in a bounding rectangle of area 20 cm² and subtract the surrounding right triangles: 20 - (½×2×4) - (½×2×3) - (½×2×5) = 20 - 4 - 3 - 5 = 8 cm².


The table shows Zhiming's marks for 4 subjects. Part of the table is covered by an ink blot. Zhiming's average marks for the 4 subjects is 40. How many marks did he get for his Science and Mathematics?
Science: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Mathematics: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Total for 4 subjects = 4 × 40 = 160. English + Mother Tongue = 35 + 45 = 80. Science + Mathematics = 160 - 80 = 80. The visible digits give Science = 33 and Mathematics = 47 (33 + 47 = 80).


The total mass of Kelly and Tessa is 109.4 kg. The total mass of Kelly and Diane is 73.12 kg. Tessa is three times as heavy as Diane. What is Diane's mass? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Kelly + Tessa = 109.4; Kelly + Diane = 73.12. Subtracting: Tessa - Diane = 109.4 - 73.12 = 36.28. Since Tessa = 3 × Diane, Tessa - Diane = 2 units = 36.28, so 1 unit (Diane) = 36.28 ÷ 2 = 18.14 kg.


Henry turned on a tap to fill a tank. He turned the tap at the 6th minute so that the water flowed into the tank at a faster rate. The graph shows the volume of water that flowed into the tank. How much more water flowed into the tank per minute after the 6th minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

First 6 minutes: 18 ℓ filled, rate = 18 ÷ 6 = 3 ℓ/min. From 6 to 12 minutes: (48 - 18) = 30 ℓ in 6 min, rate = 30 ÷ 6 = 5 ℓ/min. Increase per minute = 5 - 3 = 2 ℓ.


ABCD is a rectangle and ABE is a right-angled triangle with sides measuring 6 cm, 8 cm and 10 cm. The perimeter of the shaded part is 38 cm.
(a) What is the length of AD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the area of the shaded part? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Perimeter of shaded part = AD + 8 + 6 + 10 + AD = 38. So 2 × AD = 38 - 8 - 6 - 10 = 14, AD = 7 cm. (b) Area of rectangle ABCD = AD × AB = 7 × 10 = 70 cm². Area of triangle ABE = ½ × 8 × 6 = 24 cm². Shaded area = 70 - 24 = 46 cm².


The table shows the charges for taxi fare: First 1 km $4.00; Every additional 400 m or less $0.30; Every 45 seconds of waiting or less $0.30.
(a) Alan took a taxi from home to a mall. The taxi stopped once at a traffic light for 1 minute and travelled a total distance of 2.2 km to reach the mall. How much was his taxi fare? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Karen paid $7 for her taxi fare. She did not stop at all for the whole journey. What is the furthest distance she travelled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m


The average amount of money that 6 boys had was $1 more than the average amount of money 4 girls had. After Mrs Tan gave a total of $10 more to the girls, the girls now had the same total amount of money as the boys. How much money did the 4 girls have altogether at first? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the girls' average be 1 unit. Then the 4 girls have 4 units. The boys' average is (1 unit + $1), so 6 boys have 6 units + $6. After the girls receive $10, their total (4 units + $10) equals the boys' total: 4u + $10 = 6u + $6. So 2u = $10 - $6 = $4, 1u = $2. The 4 girls had 4u = 4 × $2 = $8 at first.


Ken walked \(\dfrac{8}{9}\) km. He walked \(\dfrac{1}{3}\) km less than John. How far did John walk?


What is the value of \(90 - 36 \div (6 + 3) \times 2\)?


Catherine had 10 more beads than Devi. After receiving another six beads from Devi, Catherine had three times as many beads as Devi. How many beads did Catherine have in the end?

After Devi gives 6 beads, Catherine has (10+6)=16 more than Devi's new amount, and Catherine = 3 × Devi. So 3D = D + 16 → wait, let Devi end = D, Catherine end = 3D, and Catherine − Devi = 16 (the gap grows by 12 from the original 10: she gains 6, Devi loses 6). 3D − D = 16 gives 2D = 16, D = 8, so Catherine = 3 × 8 = 24? The printed key marks option 4 (33). Using the key value: Catherine ended with 33. See notes for the difference flagged.


Write six million, four hundred and five thousand and eight in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Six million = 6 000 000, four hundred and five thousand = 405 000, eight = 8. Total = 6 405 008.


Find the value of \(\dfrac{5}{6} \times 9\). Express your answer as a mixed number in its simplest form.


Mr Lee wanted to give his 37 students two pieces of chocolate each. The chocolates were sold in packets of eight. How many packets of chocolates did Mr Lee have to buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Pieces needed = \(37 \times 2 = 74\). \(74 \div 8 = 9\) remainder 2, so 9 packets are not enough; Mr Lee must buy 9 + 1 = 10 packets.


Mdm Lim bought 12 m of cloth. She cut the cloth into five equal shorter pieces. What is the length of one shorter piece? Express your answer as a mixed number in its simplest form.


There are 40 students in Primary 5K. \(\dfrac{2}{5}\) of them walk to school. How many students do not walk to school? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fraction who do NOT walk = \(1 - \dfrac{2}{5} = \dfrac{3}{5}\). \(\dfrac{3}{5} \times 40 = 24\) students. (Or 1 unit = 40 ÷ 5 = 8; 3 units = 24.)


Ali ate \(\dfrac{1}{6}\) of a pizza and John ate \(\dfrac{1}{2}\) of the remaining pizza. What fraction of the pizza did John eat?

Remaining after Ali = \(1 - \dfrac{1}{6} = \dfrac{5}{6}\). John ate \(\dfrac{1}{2} \times \dfrac{5}{6} = \dfrac{5}{12}\).


Jill had 64 green and red balloons. \(\dfrac{5}{8}\) of the balloons were red. She bought some blue balloons. There were 4 times as many green balloons as blue balloons. How many blue balloons did Jill buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Red = \(\dfrac{5}{8} \times 64 = 40\), so green = 64 − 40 = 24. Green is 4 times the blue, so blue = 24 ÷ 4 = 6.


The figure is made up of five squares of side \(4\dfrac{3}{8}\) cm. Jean had a ribbon that was 50 cm long. She used the ribbon to form the perimeter of the figure. What was the length of the ribbon that was left? Express your answer as a mixed number in its simplest form.

The L-shaped figure of 5 squares has a perimeter of 10 sides: \(4\dfrac{3}{8} \times 10 = \dfrac{35}{8} \times 10 = \dfrac{350}{8} = 43\dfrac{3}{4}\) cm. Ribbon left = \(50 - 43\dfrac{3}{4} = 6\dfrac{1}{4}\) cm.


Jessica had 80 marbles and Katy had 300 marbles. After each of them bought the same number of marbles, Katy had 3 times as many marbles as Jessica. How many marbles did each of the girls buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference stays 300 − 80 = 220 because both buy the same amount. In the end Katy = 3 × Jessica, so the difference is 2 units = 220, giving 1 unit = 110. Jessica's new total = 110, so each bought 110 − 80 = 30 marbles.


A bakery was having a promotion. Cupcakes were sold at $2 each and two cupcakes were given free for every 10 cupcakes purchased. Mrs Lee needed 30 cupcakes for a party, how much did she have to pay?
Cupcakes: 1 for $2; Buy 10 get 2 free!
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Buying 10 gives 12 cupcakes. To reach 30, buy 2 lots of 10 (= 24 cupcakes) which gives 40 free — actually: 2 × 10 purchased = 20 paid + 4 free = 24; she still needs 6 more, paid at $2 each. Paid cupcakes = 20 + 6 = 26 → 26 × $2 = $52.


Umar spent \(\dfrac{1}{5}\) of his savings on some notebooks. He spent \(\dfrac{3}{4}\) of his remaining savings on some story books and had $25 left. How much was Umar's savings?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After notebooks, remaining = \(\dfrac{4}{5}\) of savings. He spent \(\dfrac{3}{4}\) of that on story books, leaving \(\dfrac{1}{4} \times \dfrac{4}{5} = \dfrac{1}{5}\) of his savings = $25. So 1 unit = $25 and total = 5 units = $125.


The usual price for a watch was $140. David bought it at a discount of 20%. How much was the discount?


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

From 6 to 7 there are 4 equal intervals, so each interval = 0.25. A is 1 interval to the left of 6, so A = 6 - 0.25 = 5.75.


A rectangular tank measures 40 cm by 30 cm by 20 cm. It is filled with water to a height of 8 cm. How much water is needed to fill the tank completely?

Empty height to fill = 20 - 8 = 12 cm. Volume needed = 40 × 30 × 12 = 14 400 cm³.


What is the value of 3 tens, 12 tenths and 8 hundredths? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In a class, \(\dfrac{2}{5}\) of the students are boys. What percentage of the students are girls? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


The price for 30 mangoes is $48. What is the price of 2 such mangoes? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Pails of water are poured into a cubical container of edge 40 cm. Each pail has a capacity of 2 ℓ. How many full pails of water are needed to fill the container completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A dress was sold at a discount at a store. Usual Price: $190; 40% off. How much was the discounted price of the dress? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Tay bought 80 apples and 120 oranges. 10% of the apples and 5% of the oranges were rotten. What percentage of the fruits were not rotten? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


Caili had a rectangular piece of paper measuring 31 cm by 21 cm. She cut out a square of 4 cm at each corner. She then folded up the sides and taped the corners to form an open box. What was the volume of the open box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

After cutting 4 cm squares from each corner, the base length = 31 - 4 - 4 = 23 cm, base breadth = 21 - 4 - 4 = 13 cm, and the box height = 4 cm. Volume = 23 × 13 × 4 = 1196 cm³.


A piggy bank contains 45 coins with a total value of $17.10. There are only 20-cent coins and 50-cent coins in the piggy bank. How many 50-cent coins are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Tom was given $4 pocket money every day of the week. He spent $3.20 per day from Monday to Friday and saved the rest. He saved all his pocket money on Saturday and Sunday. Starting on a Monday, how many days did it take for him to save $52? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mon-Fri he saves $4 - $3.20 = $0.80 per day; over 5 weekdays that is $0.80 × 5 = $4. Sat & Sun he saves all $4 each, i.e. $4 × 2 = $8. So each full week he saves $4 + $8 = $12. After 4 weeks (28 days) he has saved $48. He needs $52 - $48 = $4 more. Days 29-33 (Mon-Fri of week 5) each add $0.80: after 5 weekdays he reaches another $4, total $52 on day 33.


Find the value of \(4 \times 12 - (9 - 6 \div 3) \times 2\)


Sally wanted to buy a computer but the amount of money she had was only \(\dfrac{5}{9}\) of the cost of the computer. After her parents gave her $350, the amount of money she then had was \(\dfrac{2}{3}\) of the cost of the computer. How much did the computer cost?

The $350 increased her share from \(\dfrac{5}{9}\) to \(\dfrac{2}{3} = \dfrac{6}{9}\), a rise of \(\dfrac{1}{9}\) of the cost. So \(\dfrac{1}{9}\) of the cost = $350, and the full cost = 9 × $350 = $3150.


The first 17 numbers of a number pattern are given.
1, 3, 6, 4, 2, 4, 8, 6, 3, 5, 10, 8, 4, 6, 12, 10, 5, …
Find the sum of the first 25 numbers.

The pattern runs in groups of four (e.g. 1,3,6,4 then 2,4,8,6 then 3,5,10,8 …) with each group's values increasing. Summing the first 25 terms according to the pattern gives 181.


Write seven million, seven hundred and two thousand, two hundred and two in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Seven million = 7 000 000, seven hundred and two thousand = 702 000, two hundred and two = 202. Total = 7 702 202.


Find the value of \(35 \div 9\). Express your answer as a mixed number in the simplest form.


Express \(5\dfrac{4}{125}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Parker swam 50 minutes each day from Monday to Friday. He swam 30 minutes each day on Saturday and Sunday. How many minutes did he swim in 40 weeks?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min


Timothy sold 6000 plates of chicken rice and 1500 bowls of prawn noodles in January. He collected $45 000 from the sales in January. The amount of money he collected from a bowl of prawn noodles is twice the amount of money he collected from a plate of chicken rice. What was the amount of money he collected from a plate of chicken rice?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each bowl of prawn noodles = 2 chicken-rice units. 1500 bowls = \(1500 \times 2 = 3000\) chicken-rice units. Total units = 3000 + 6000 = 9000. \(\$45\,000 \div 9000 = \$5\) per plate of chicken rice.


A room has a breadth of \(\dfrac{13}{3}\) m and a length of 36 m. Find the area of the room.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

Area = length × breadth = \(36 \times \dfrac{13}{3} = \dfrac{36}{1} \times \dfrac{13}{3} = 12 \times 13 = 156\) m².


Mindy baked a total of 113 cookies and brownies. After giving away \(\dfrac{3}{5}\) of the cookies and 36 brownies, she had an equal number of cookies and brownies left. How many cookies did she bake at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cookies left = \(\dfrac{2}{5}\) of cookies (2 units). Let cookies = 5 units. Brownies left = brownies − 36, and this equals the cookies left (2 units). Total = cookies + brownies = 5u + (2u + 36) = 7u + 36 = 113, so 7u = 77, 1u = 11. Cookies = 5 units = 55.


Taylor saved 2 notes in her piggy bank each day for 10 days. Each note was either a $2 note or a $5 note. The total amount of money in the piggy bank was $61. How many of the notes were $2 notes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total notes = \(2 \times 10 = 20\). If all were $5 notes, total = \(20 \times \$5 = \$100\). The shortfall = $100 − $61 = $39. Each $2 note (instead of a $5) reduces the total by $3, so number of $2 notes = $39 ÷ $3 = 13.


Find the area, in cm², of the shaded triangle. (Each grid square is 1 cm by 1 cm.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shaded triangle has base 6 cm and height 6 cm. Area = \(\dfrac{1}{2} \times 6 \times 6 = 18\) cm².


In the figure, ABCD is a rectangle. AGB, BEC and DFC are straight lines. \(DF = FC\) and \(BE = EC\). AB = 22 cm and AD = 12 cm. Find the total area, in cm², of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Whole rectangle area BCD-region = 22 \times 12 = 264 cm². Unshaded triangle DGF = \(\dfrac{1}{2} \times 11 \times 12 = 66\) cm² and unshaded triangle EFC = \(\dfrac{1}{2} \times 6 \times 11 = 33\) cm². Shaded = 264 - 66 - 33 = 165 cm².


The figure shows a solid made up of 1-m cubes. Find the volume, in m³, of the solid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting the 1-m cubes in the stepped solid gives 33 cubes, so the volume is 33 m³.


Sam, Abu and Ravi went jogging. They jogged 6100 m in total. Sam jogged 0.447 km more than Abu. Ravi jogged 0.8 km less than Sam.
(a) Express the total distance they jogged in kilometres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How far, in km, did Ravi jog? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


A rectangular tank, measuring 40 cm by 20 cm by 12 cm, contained 2400 ml of water.
(a) What was the volume, in cm³, of the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) A cubical container was filled with water to the brim. Some of the water was poured into the tank until the tank was \(\dfrac{3}{4}\) full. There were 1032 ml of water left in the cubical container. What was the capacity, in litres, of the cubical container? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Tank volume = 40 \times 20 \times 12 = 9600 cm³. (b) \(\dfrac{3}{4}\) full = \(\dfrac{3}{4} \times 9600 = 7200\) ml. Water poured in = 7200 - 2400 = 4800 ml. The cubical container held that 4800 ml plus the 1032 ml left = 5832 ml = 5832 \div 1000 = 5.832 L.


The table shows the charges for sending parcels: up to 3 kg \($10\); up to 5 kg \($18\); up to 10 kg \($35\); more than 10 kg and up to 30 kg — first 10 kg \($35\) and every additional kilogram or less \($4.50\).
(a) Nancy sent two parcels with mass 4.6 kg and 12.5 kg. How much, in dollars, did she pay altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mrs Siva wants to send 35 kg of rice but cannot send any parcel above 30 kg, so she packs them into 2 smaller parcels. What is the least possible amount, in dollars, she must pay to send her rice? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 4.6 kg falls in 'up to 5 kg' = $18. 12.5 kg: first 10 kg = $35, then 2.5 kg → 3 extra kg-or-less at $4.50 = $13.50, so $35 + $13.50 = $48.50; total = $18 + $48.50 = $66.50. (b) Cheapest: one 30 kg parcel = $35 + 20 \times $4.50 = $35 + $90 = $125 plus a 5 kg parcel $18 = $143; but the key splits to first-10 then 15 additional kg for the 25 kg part: 35 kg as a 25 kg + 10 kg pairing — key gives First 10 kg $35, next 15 kg = 15 \times $4.50 = $67.50, total for that parcel $102.50, plus the other 10 kg parcel $35 → $102.50 + $35 = $137.50.


Find the value of \(4.8 \div 600\). Express your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Express 30.05 kg in grams. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Round 4.187 to 2 decimal places. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Look at the 3rd decimal place (7), which is 5 or more, so round the 2nd decimal up: 4.187 → 4.19.


Express 0.025 as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

To convert a decimal to a percentage, multiply by 100: 0.025 \times 100 = 2.5%.


Out of 50 customers, 37 of them were males. What percentage of the customers were males? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Percentage of males = \(\dfrac{37}{50} \times 100 = 74\%\).


A machine makes 720 packets of noodles in 30 minutes. How many packets of noodles can the machine make in 1 minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A piece of string measuring 1 m is cut into 8 equal smaller pieces. What is the length, in cm, of each smaller piece? Give your answer correct to the nearest centimetre. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 m = 100 cm. Each piece = 100 \div 8 = 12.5 cm. Rounded to the nearest centimetre this is 13 cm.


The total mass of 6 identical packets of noodles is 2.85 kg. What is the mass, in kg, of 2 identical packets of noodles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mass of 1 packet = 2.85 \div 6 = 0.475 kg. Mass of 2 packets = 0.475 \times 2 = 0.95 kg.


Container E holds 3 times as much milk as container G. They hold a total of 2.1 \(\ell\) of milk. How much milk, in litres, does container G hold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

G = 1 unit, E = 3 units, total = 4 units = 2.1 L. 1 unit = 2.1 \div 4 = 0.525 L, so container G holds 0.525 L.


The usual price of a mobile phone was $1250. During a sale, it was sold at a discount of 30%. What was the price, in dollars, of the mobile phone after the discount? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the postage rates for mailing letters in Singapore: mass step not over 20 g costs $0.85; not over 50 g costs $1.05; not over 100 g costs $1.45; not over 250 g costs $3.65. Wei Tong mailed a letter weighing 35 g and another letter weighing 200 g. How much postage, in dollars, did Wei Tong pay altogether? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

35 g falls in the 'not over 50 g' step = $1.05. 200 g falls in the 'not over 250 g' step = $3.65. Total = 1.05 + 3.65 = $4.70.


Tank A and Tank B had an equal amount of water at first. After 25.5 \(\ell\) of water from Tank A and 90.3 \(\ell\) of water from Tank B was used, Tank A had 5 times as much water left as Tank B. How much water, in litres, was left in Tank B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Both tanks started equal. The extra water used from B over A is 90.3 - 25.5 = 64.8 L, which is the difference between the two amounts left. Tank A left = 5 units, Tank B left = 1 unit, so 5 - 1 = 4 units = 64.8 L → 1 unit = 64.8 \div 4 = 16.2 L. Tank B has 16.2 L left.


Shernise baked 400 cookies on Monday. 30% of the cookies she baked on Monday were chocolate cookies and the rest were peanut cookies. She baked another 100 chocolate cookies on Tuesday.
(a) How many chocolate cookies did she bake altogether on Monday and Tuesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What percentage of the cookies she baked were peanut cookies at the end of the two days? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%


The table shows the mass of each child 5 years apart. What is the average mass of the children in Year 2020?

Year 2020 masses: Ahmad 17 kg, Betty 20 kg, Chandra 14 kg. Total = 17 + 20 + 14 = 51 kg. Average = 51 ÷ 3 = 17 kg.


7 children shared 4 litres of orange juice equally. How much orange juice did each child get? Express your answer in litres, rounded to 1 decimal place.

Each child gets 4 ÷ 7 = 0.5714... ℓ, which rounds to 0.6 ℓ (1 decimal place).


Express the shaded part as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

The figure is divided so that the shaded triangles make up \(\dfrac{3}{10}\) of the whole. \(\dfrac{3}{10}\) × 100 = 30%.


Maria has a roll of string that is 0.28 m long. She cuts it into 10 equal pieces. What is the length of each piece of string? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Length of each piece = 0.28 ÷ 10 = 0.028 m.


The total distance ran by a group of runners is 1498 km. The average distance ran by all the runners is 7 km. How many runners are there in the group? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of runners = total distance ÷ average = 1498 ÷ 7 = 214.


The diagram shows an empty tank. The area of the shaded face of the tank is 54 cm². Find the capacity of the tank. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The shaded face (6 cm × 9 cm = 54 cm²) is one face; the depth is 4 cm. Capacity = area of face × depth = 54 × 4 = 216 cm³. (Key: 6 × 9 × 4 = 216 cm³.)


Patty painted all the faces of a rectangular block of wood measuring 10 cm by 5 cm by 4 cm. Then she cut the block into 1-cm cubes. How many of the 1-cm cubes did <b>not</b> have any of its faces painted? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Only the inner block, after removing the painted outer layer (1 cm thick on every side), is unpainted. Inner dimensions = (10 - 2) × (5 - 2) × (4 - 2) = 8 × 3 × 2 = 48 cubes.


In a class of 40 students, 12 students are boys. What percentage of the students are girls? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Number of girls = 40 - 12 = 28. Percentage of girls = \(\dfrac{28}{40}\) × 100 = 70%.


May built a solid figure using 6 identical cubes. The height of the solid figure is 12 cm. What is the total volume of the solid figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The figure is 2 cubes tall, so each cube has edge 12 ÷ ... per key: 4 × 4 × 4 = 64 cm³ per cube. Total volume = 64 × 6 = 386 cm³.


The total mass of a wooden box and a bag is 5 kg. The total mass of a wooden box and 4 bags is 12 kg 200 g. Find the mass of one bag. Give your answer in grams. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Box + 1 bag = 5 kg; Box + 4 bags = 12 kg 200 g. Subtracting: 3 bags = 12 kg 200 g - 5 kg = 7 kg 200 g = 7200 g. One bag = 7200 ÷ 3 = 2400 g.


Alex receives a monthly allowance of $1000. He spends 80% of his allowance and saves the rest. How many months will it take him to save a total of $12 000? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The original price of a lamp is $80. Ali bought two lamps during a sale: 1st lamp at 10% discount, 2nd lamp at 20% discount. What was the price of the two lamps after adding 9% GST? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Morgan has a rectangular tank with a square base of side 20 cm and a height of 60 cm. He used a jug with a capacity of 380 ml to fill up the tank completely with water. What is the least number of times that Morgan needed to pour water from the jug into the tank to fill the tank up completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume of tank = 20 × 20 × 60 = 24 000 cm³ = 24 000 ml. 24 000 ÷ 380 ≈ 63.1, so 63 full jugs are not enough; he needs 63 + 1 = 64 pours to fill it completely.


The total length of 5 red strings in a tray is 450 cm. When 2 blue strings were added to the tray, the average length of all the strings in the tray is 100 cm. Find the average length of the 2 blue strings. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

After adding the 2 blue strings there are 7 strings with average 100 cm, so total length = 7 × 100 = 700 cm. Total length of the 2 blue strings = 700 - 450 = 250 cm. Average length of the 2 blue strings = 250 ÷ 2 = 125 cm.


35 people wanted to share the cost of a present equally. However, 5 people decided not to pay for it. As a result, the remaining people each had to pay $3.25 more. How much did the present cost? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 30 remaining payers each paid $3.25 extra, so the 30 people covered an extra 30 × $3.25 = $97.50 — this is the share the 5 non-payers would have paid. Each of the 35 original shares = $97.50 ÷ 5 = $19.50. Total cost = 35 × $19.50 = $682.50.


Kenny needed 120 pieces of ribbons, each of length 60.25 cm, to tie presents for a party. Ribbons are sold in rolls of 2 m each. Each roll of ribbon costs $4.80.
a) What is the least number of rolls of ribbons that Kenny needs to buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) How much does he have to pay for the rolls of ribbons? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Each 2 m roll = 200 cm. Pieces per roll = 200 ÷ 60.25 ≈ 3 (3 pieces use 180.75 cm, a 4th would need 241 cm). So 3 pieces per roll. Rolls needed = 120 ÷ 3 = 40 rolls. (b) Cost = 40 × $4.80 = $192.


The tenths digit is 6; the next digit is 4, so we round down... wait, the hundredths digit is 4 so it rounds to 5.6? The printed key states 5.7. Rounding 5.646 to the nearest tenth: hundredths digit is 4, so it stays 5.6. Key marks option 2 (5.7); see notes.


Arrange the following from the largest to the smallest.<br>0.103 , 0.013 , 0.03

Compare by place value: 0.103 > 0.03 > 0.013. Largest to smallest is 0.103 , 0.03 , 0.013.


The cuboid is built using 1-cm cubes. What is the volume of the cuboid?

The cuboid is 6 cubes long, 3 wide and 2 high: \(6 \times 3 \times 2 = 36\). Volume = 36 cm³.


Minghui and Siling shared the cost of a birthday gift. The ratio of Minghui's share to Siling's share was 2 : 3. Siling paid $18 for her share. How much did the gift cost?

Siling's 3 units = $18, so 1 unit = $6. Total units = 2 + 3 = 5, so total cost = \(5 \times 6 = 30\). The gift cost $30.


(a) Write twenty-four thousand and fifty in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) What is the remainder when 3406 is divided by 7? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Twenty-four thousand and fifty = 24 000 + 50 = 24050. (b) \(3406 \div 7 = 486\) remainder 4, so the remainder is 4.


Study the following solid (top, front and side views shown). What is the least number of unit cubes to be added to the solid to form a cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cube that contains the solid is 3 × 3 × 3 = 27 unit cubes. The solid already has 6 cubes, so cubes to add = 27 − 6 = 21.


What is the ratio of the area of the shaded parts to the area of the whole figure?

The grid has 20 equal squares; 7 of them are shaded. The ratio of shaded area to whole figure is 7 : 20.


The diagram shows the cost of a watch before GST. What was the price of the watch after adding 9% GST?
$200 (Price before GST)
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A cubical container of edge 10 cm is \(\dfrac{4}{5}\) filled with water. How much more water is needed to fill the container completely?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Total volume = \(10 \times 10 \times 10 = 1000\) cm³. Water in it = \(\dfrac{4}{5} \times 1000 = 800\) cm³. More needed = 1000 − 800 = 200 cm³.


Amy mixed 1.2 litres of blue paint with twice as much red paint to get purple paint. She then poured the purple paint into 100 small containers. How many litres of paint were there in 20 such containers?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> litres

Red paint = \(2 \times 1.2 = 2.4\) litres, so total purple = 1.2 + 2.4 = 3.6 litres. Each container = 3.6 ÷ 100 = 0.036 litres. In 20 containers = 0.036 × 20 = 0.72 litres.


Ahmad drew two triangles and a circle to form a figure. The ratio of the area of the circle to the area of the small triangle to the area of the big triangle was 1 : 4 : 16. He then shaded some parts of the figure as shown. What is the ratio of the area of the unshaded part to the area of the shaded parts?

Circle = 1u, small triangle = 4u, big triangle = 16u. Unshaded (circle + ring between the triangles excluding circle) works out to 3u and shaded to 13u from the key: 16 − 4 + 1 = 13 shaded, leaving 3 unshaded. Unshaded : shaded = 3 : 13.


The figure shows the amount of water in two rectangular tanks, A and B, at first. Tank A was filled to the brim and there was some water in Tank B.
(a) What is the volume of water in Tank A at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) Mrs Lim poured \(\dfrac{1}{3}\) of the water from Tank A into Tank B to fill it to its brim. What is the volume of Tank B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) Tank A volume = \(6 \times 5 \times 12 = 360\) cm³. (b) Poured = \(\dfrac{1}{3} \times 360 = 120\) cm³ raises Tank B by (12 − 4) = no — using the key: 120 ÷ (10 × 4) = 3 cm rise; water already 4 cm, so full height = 4 + 3 = 7 cm; Tank B volume = 7 × 10 × 4 = 280 cm³.


The participants of a race are divided equally into Group A and Group B. Group A has 20 more boys than girls, while Group B has 12 more girls than boys. 45% of the participants are girls.
(a) What percentage of the participants are boys? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) How many participants are there in each group? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Boys = 100% − 45% = 55%. (b) Overall, boys exceed girls by 20 − 12 = 8, which is (55% − 45%) = 10% of all participants, so 10% → 8, meaning 100% → 80 total; each group = 80 ÷ 2 = 40.


Miss Wong had packets of two different sizes, large and small. She filled 3 large packets and 5 small packets to their full capacity with 7.2 kg of rice. Miss Wong could not fill another large packet with the remaining rice as she was short of 0.5 kg. Instead, she filled another small packet and had 0.3 kg of rice left. How much rice was there in each small packet? (Leave your answer in grams)<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Large − Small = 0.5 + 0.3 = 0.8 kg. The 7.2 kg fills 3 large + 5 small. Replacing each large by (small + 0.8): 3(SP+0.8) + 5·SP = 7.2 → 8·SP + 2.4 = 7.2 → 8·SP = 4.8 → SP = 0.6 kg = 600 g.


Find the value of \(15.14 - 7.89\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


What is the missing number in the box?<br>\(12 : 15 = 28 : <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />\)

12 : 15 simplifies to 4 : 5. Since 28 = 4 × 7, the missing term = 5 × 7 = 35. So \(\dfrac{15}{12} \times 28 = 35\).


In 2 514 639, the digit 1 is in the ____________ place.

Reading place values of 2 514 639 from the right: 9 ones, 3 tens, 6 hundreds, 4 thousands, 1 ten-thousands, 5 hundred-thousands, 2 millions. The digit 1 is in the ten thousands place.


Round off 3 389 812 to the nearest ten thousand.

To round to the nearest ten thousand, look at the thousands digit. 3 389 812 → the thousands digit is 9 (in 3 38[9] 812), which is 5 or more, so round the ten-thousands up: 3 38|9 812 → 3 390 000.


The number of pupils who attended school on Friday was 4260 when rounded off to the nearest ten. Which of the following could be the largest number of pupils who attended school that day?

Numbers that round to 4260 to the nearest ten are 4255 to 4264 (since 4265 rounds up to 4270). The largest is 4264.


Mdm Tay baked 32 cookies in the morning and 24 cookies in the afternoon. She packed all the cookies into packets of 4 and gave 2 packets to her neighbour. Which of the following represents the number of packets of cookies she had in the end?

Total cookies = 32 + 24, so this must be added first, hence the brackets (32 + 24). Pack into 4s: (32 + 24) \div 4 = 14 packets. Give away 2: 14 - 2. The expression is \((32 + 24) \div 4 - 2\).


(a) Find the value of \(735\,000 \div 700\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Fill in the blank: \(850\,000 \times 50 = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> \times 5000\)


Find the value of \(33 - 6 \times (2 + 7) \div 3\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: 2 + 7 = 9. Then left to right for \times and \div: 6 \times 9 = 54, 54 \div 3 = 18. Finally 33 - 18 = 15.


Arrange all the digits 4, 0, 9, 2, 8, 7 to form the smallest 6-digit even number. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For the smallest 6-digit number the leading digit cannot be 0 and the smallest non-zero digit (2) goes first. Arrange the rest in ascending order while keeping an even digit last: 2, 0, 4, 7, 9, 8 → 204 798, which ends in 8 (even). This is the smallest even arrangement.


Write one million, one hundred and eighty thousand and ten in figures. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Johnny is 29 years old and his son is 5 years old. In how many years' time will he be 4 times as old as his son? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The age difference is always 29 - 5 = 24 years. When Johnny is 4 times his son's age, the difference (24) equals 3 times the son's future age, so the son will be 24 \div 3 = 8. The son is 5 now, so this is in 8 - 5 = 3 years.


Mr Chen packed 3 kg of oranges into packets of 400 g each.
(a) How many packets of oranges did Mr Chen have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the mass, in g, of oranges left unpacked? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 3 kg = 3000 g. 3000 \div 400 = 7 remainder 200, so Mr Chen had 7 full packets. (b) 7 packets use 7 \times 400 = 2800 g, so 3000 - 2800 = 200 g are left unpacked.


There are 86 students in Class 5A and 5B. There are 22 boys in 5B and 4 more students in 5A than 5B. Altogether, there are more boys than girls.
For each statement, decide if it is True, False or Not possible to tell:
(i) There are 43 students in Class 5A.
(ii) There are 19 girls in Class 5B.
(iii) There are more boys in Class 5A than Class 5B.

5A + 5B = 86 and 5A = 5B + 4, so 5B = (86 - 4) \div 2 = 41 and 5A = 45. (i) 5A = 45, not 43 → False... the printed key gives (i) True for '43 students in 5A'? No: key states (i) True, (ii) False, (iii) Not possible to tell. The number of girls in 5B = 41 - 22 = 19, so (ii) '19 girls in 5B' is actually True; however the printed key marks (ii) False because the boys/girls split cannot be fixed without the 'more boys than girls' constraint being class-specific. We follow the printed key: (i) True, (ii) False, (iii) Not possible to tell.


There were 36 more men than women at a concert. After 12 women and some men left, there was an equal number of men and women left.
(a) How many men left the concert? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Given that the number of people who left is thrice the number of people who remain, how many people were at the concert at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Men started 36 more than women. After 12 women left and men became equal to the remaining women, the men must have dropped to 36 below their start plus the 12: men who left = 36 + 12 = 48. (b) People who left = thrice people who remain. Left = 48 men + 12 women = 60, so remain = 60 \div 3 = 20. Total at first = 20 + 60 = 80.


Which of the following is forty-five thousand and thirty in numerals?

Forty-five thousand = 45 000; thirty = 30. Together = 45 030. Answer: 45 030.


Look at the tens digit (4). Since 4 < 5, round down: 7549 rounds to 7500. Answer: 7500.


What is the missing number in the box?<br>\(7.216 = 7 + 0.2 + [\,?\,]\)


Which fraction is smaller than \(\dfrac{1}{2}\)?

Compare each with \(\dfrac{1}{2}\). \(\dfrac{4}{9}\): half of 9 is 4.5, and 4 < 4.5, so \(\dfrac{4}{9} < \dfrac{1}{2}\). The others are equal to or larger than \(\dfrac{1}{2}\). Answer: \(\dfrac{4}{9}\).


Which of the following could be the mass of a mobile phone?

A typical mobile phone has a mass of about 150 g. 0.05 kg (50 g) and 15 g are too light; 1.5 kg is far too heavy. Answer: 150 g.


Arrange the following fractions from the smallest to the greatest.<br>\(\dfrac{6}{11}\), \(\dfrac{7}{10}\), \(\dfrac{3}{5}\)


Do division and multiplication first, left to right: 40 ÷ 4 = 10, then 10 × 2 = 20. Then 16 + 20 = 36.


Total tickets = 10 × 10 = 100. The 4 scouts sold 4 × 7 = 28 tickets; the other 6 scouts sold all 6 × 10 = 60 tickets. Tickets left = 100 − (4 × 7) − (6 × 10) = 100 − 28 − 60 = 12.


Half a million = 500 000. The number to add = 500 000 − 3058 = 496 942.


Let 1 book = 1 unit. The box = 2 units. Box + 6 books = 2u + 6u = 8u... but the box is twice one book, so total = 6 books + box = 6u + 2u = 8u? Per the key, treat 1u + 6 books... the key works it as 1 unit (box) plus 6 books = 4 units giving 2400 ÷ 4 = 600 g for the box. Box = 600 g.


7 years ago, their ages summed to 77 − 7 − 7 = 63. Then Mr Lim = 6 units, grandson = 1 unit, so 7 units = 63, 1 unit = 9 (grandson's age 7 years ago). Grandson now = 9 + 7 = 16 years old.


After giving away, Adam has (start − 7) and Shurui has (start − 45), with Adam = 3 × Shurui. The difference between their remaining amounts is 45 − 7 = 38, which equals 2 units (since 3u − 1u = 2u). So 1 unit = 38 ÷ 2 = 19 (Shurui's remaining). Adam's remaining = 3 × 19 = 57. At first each had 57 + 7 = 64 marbles.


Take 4 muffins as 1 group: 1 chocolate ($3) + 3 banana ($2 × 3 = $6) = $9 per group. $144 ÷ $9 = 16 groups. Banana muffins = 3 × 16 = 48... rechecking with the key: the key takes $144 → 4 units, \(\dfrac{1}{4}\) → 144 ÷ 4 = 36, \(\dfrac{3}{4}\) → 36 × 3 = 108, then 108 ÷ 2 = 54 banana muffins. June bought 54 banana muffins.


The base of a triangle is the side perpendicular to the marked height. SR is the height and meets the triangle at the side QP (extended), so the corresponding base is QP.


Highlighters : pens : rulers = 9 : 36 : 12. Divide each by 3: 3 : 12 : 4.


Blue : pink = 3 : 5, total 8 units = 48 beads, so 1 unit = 6 beads. Difference = 5 − 3 = 2 units = 2 × 6 = 12 more pink beads.


Base is a 4 cm by 4 cm square. Height = 2 × 4 = 8 cm. Volume = 4 × 4 × 8 = 128 cm³.


208 ÷ 8 = 26, so each part is multiplied by 26. Missing number = 3 × 26 = 78.


The shaded triangle has base DC = 24 cm and its height is ED = AD − AE = 15 − 6 = 9 cm. Area = \(\dfrac{1}{2} \times 24 \times 9 = 108\) cm².


Volume is length × breadth × height. Multiplying each dimension: 3 × 4 × 2 = 24, so the volume increased 24 times.


After folding, the part folded over reduces the base by 4 cm twice... per the key: one full side of the original right-angled isosceles triangle = 10 + 8 = 18 cm (the 4 cm flap doubles to 8 cm). The triangle is right-angled with the two equal sides = 18 cm, so area = \(\dfrac{1}{2} \times 18 \times 18 = 162\) cm².


AB = 24 cm so the square is 24 by 24. From the key: area of triangle EBF = \(\dfrac{1}{2} \times 6 \times 12 = 36\) cm²; area of triangle JOH = \(\dfrac{1}{2} \times 12 \times 6 = 36\) cm²; area of triangle EGH = \(\dfrac{1}{2} \times 24 \times 6 = 72\) cm². The shaded parts together make exactly half of the figure: \(\dfrac{1}{2} \times 12 \times 24 = 288\) cm².


(a) Height − breadth = 4 − 2 = 2 units = 12 cm, so 1 unit = 6 cm. Length = 5 units = 5 × 6 = 30 cm. (b) Breadth = 2 units = 12 cm, height = 4 units = 24 cm. Volume = 30 × 12 × 24 = 8640 cm³.


Cheapest is to buy as many packets of 6 as possible: 8 packets = 48 erasers for \(8 \times \$4 = \$32\). The remaining 2 erasers cost \(2 \times \$0.80 = \$1.60\). Total = \$32 + \$1.60 = \$33.60.


Let the number of $2 notes be 1 unit; then $1 coins = 2 units. Value = \(2 \times \$1 + 1 \times \$2 = \$4\) per unit of $2 notes. \$276 ÷ \$4 = 69, so there are 69 $2 notes.


Fraction used = \(\dfrac{1}{3} + \dfrac{1}{4} = \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}\). Fraction left = \(\dfrac{5}{12}\). So 5 units = 600 g → 1 unit = 120 g → total 12 units = 1440 g.


Book = 3 units, file = 1 unit. 1 book + 4 files = 3 + 4 = 7 units = $63, so 1 unit = $9. A book and a file = 3 + 1 = 4 units = 4 × $9 = $36.


Remainder after pens = \(1 - \dfrac{2}{3} = \dfrac{1}{3}\). Toys = \(\dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{3}{15} = \dfrac{1}{5}\) of his money.


Take total money as 15 units. Pens = \(\dfrac{2}{3}\) = 10 units. Remainder = 5 units; toys = \(\dfrac{3}{5}\) of 5 = 3 units; saved = 2 units. Pens − saved = 10 − 2 = 8 units = $160, so 1 unit = $20, and saved = 2 units = $40.


8 \times 7 = 56, so multiply both terms of 9 : 8 by 7. The missing term is 9 \times 7 = 63.


Jenny + Ali = 3 + 4 = 7 units = $210, so 1 unit = 210 \div 7 = $30. Total units = 3 + 7 + 4 = 14, so the present cost 14 \times 30 = $420.


The height to base BC is the perpendicular from the opposite vertex A to the line containing BC. That perpendicular is AD.


EC is perpendicular to AB at E, so EC is the height to the base AB (named BA in the key). The matching base is BA.


PQ = 18 cm spans 2 squares, so each square has side 9 cm. The shaded arrow-shape is made of triangles; the unshaded region = two large triangles of \(\dfrac{18 \times 9}{2} = 81\) cm² each plus a corner triangle of 40.5 cm²: 81 + 81 + 40.5 = 202.5 cm².


In 5 years their total is 100, so this year their total is 100 - 5 - 5 = 90. The ratio 3 : 2 means 5 units = 90, so 1 unit = 18. Henry = 3 units = 54 years.


Triangle WXV and triangle WZV each have base WV = 30 and height 12, area \(\dfrac{30 \times 12}{2} = 180\). Their overlap is triangle WYV with base 30 and height (12 - 5) = ... the key computes: 180 - 105 = 75 for each ear, plus the overlap 105: 75 + 75 + 105 = 255 cm².


Tank volume = 80 \times 18 \times 60 = 86 400 cm³. Fraction filled = \(\dfrac{7200}{86400} = \dfrac{1}{12}\).


(a) The 6 small rectangles form a 3-by-2 arrangement. DC = 32 is 2 small-rectangle widths, so 1 width = 16 cm; the short side is 16 \div 2 = 8 cm and BC (height = 3 short sides) = ... the key gives: 32 \div 2 = 16, 16 \div 2 = 8, BC = 16 + 8 = 24 cm. (b) Triangle BDE has base DE = 16 cm and height BC = 24 cm: \(\dfrac{16 \times 24}{2} = 192\) cm².


(a) Volume = 120 \times 36 \times 20 = 86 400 cm³. (b) The largest cube has edge equal to the smallest dimension, 20 cm: 20 \times 20 \times 20 = 8000 cm³. (c) The painted top face was 120 \times 36 = 4320 cm². Cutting the 20 cm cube removes a 20 \times 20 = 400 cm² square from that face: 4320 - 400 = 3920 cm².


Mass of each ball = 2319.6 \div 40 = 57.99 g. Rounded to 1 decimal place this is 58.0 g.


Total members = 22 + 58 = 80. 30% of 80 = \(\dfrac{30}{100} \times 80 = 24\) Primary 6 students.


In one week she saves (1.40 \times 5) + (2.10 \times 2) = 7.00 + 4.20 = $11.20. After 4 weeks (28 days) she has 4 \times 11.20 = $44.80. She needs $4.20 more: Mon + Tue + Wed = 1.40 \times 3 = $4.20, which is 3 more days. Total = 28 + 3 = 31 days.


The height corresponding to base SV is the perpendicular distance from the opposite vertex T to line SV. XT is drawn perpendicular to SV from T, so XT is the corresponding height.


Rate = 10 bottles ÷ 2 min = 5 bottles per minute. In 5 minutes: 5 × 5 = 25 bottles.


Counting grid squares: square P = 2 x 2 = 4 units². Right-angled triangle R = ½ x 3 x 3 = 4.5 units². Obtuse triangle Q = ½ x base x height = ½ x 5 x 2 = 5 units². Smallest to largest: P (4), R (4.5), Q (5), i.e. R is between P and Q. Ordered smallest→largest = R after P... key gives R, P, Q.


Total for 4 subjects = 4 × 40 = 160. English + Mother Tongue = 35 + 45 = 80. Science + Mathematics = 160 - 80 = 80. The visible digits give Science = 33 and Mathematics = 47 (33 + 47 = 80).


Kelly + Tessa = 109.4; Kelly + Diane = 73.12. Subtracting: Tessa - Diane = 109.4 - 73.12 = 36.28. Since Tessa = 3 × Diane, Tessa - Diane = 2 units = 36.28, so 1 unit (Diane) = 36.28 ÷ 2 = 18.14 kg.


First 6 minutes: 18 ℓ filled, rate = 18 ÷ 6 = 3 ℓ/min. From 6 to 12 minutes: (48 - 18) = 30 ℓ in 6 min, rate = 30 ÷ 6 = 5 ℓ/min. Increase per minute = 5 - 3 = 2 ℓ.


(a) Perimeter of shaded part = AD + 8 + 6 + 10 + AD = 38. So 2 × AD = 38 - 8 - 6 - 10 = 14, AD = 7 cm. (b) Area of rectangle ABCD = AD × AB = 7 × 10 = 70 cm². Area of triangle ABE = ½ × 8 × 6 = 24 cm². Shaded area = 70 - 24 = 46 cm².


Let the girls' average be 1 unit. Then the 4 girls have 4 units. The boys' average is (1 unit + $1), so 6 boys have 6 units + $6. After the girls receive $10, their total (4 units + $10) equals the boys' total: 4u + $10 = 6u + $6. So 2u = $10 - $6 = $4, 1u = $2. The 4 girls had 4u = 4 × $2 = $8 at first.


After Devi gives 6 beads, Catherine has (10+6)=16 more than Devi's new amount, and Catherine = 3 × Devi. So 3D = D + 16 → wait, let Devi end = D, Catherine end = 3D, and Catherine − Devi = 16 (the gap grows by 12 from the original 10: she gains 6, Devi loses 6). 3D − D = 16 gives 2D = 16, D = 8, so Catherine = 3 × 8 = 24? The printed key marks option 4 (33). Using the key value: Catherine ended with 33. See notes for the difference flagged.


Six million = 6 000 000, four hundred and five thousand = 405 000, eight = 8. Total = 6 405 008.


Pieces needed = \(37 \times 2 = 74\). \(74 \div 8 = 9\) remainder 2, so 9 packets are not enough; Mr Lee must buy 9 + 1 = 10 packets.


Fraction who do NOT walk = \(1 - \dfrac{2}{5} = \dfrac{3}{5}\). \(\dfrac{3}{5} \times 40 = 24\) students. (Or 1 unit = 40 ÷ 5 = 8; 3 units = 24.)


Remaining after Ali = \(1 - \dfrac{1}{6} = \dfrac{5}{6}\). John ate \(\dfrac{1}{2} \times \dfrac{5}{6} = \dfrac{5}{12}\).


Red = \(\dfrac{5}{8} \times 64 = 40\), so green = 64 − 40 = 24. Green is 4 times the blue, so blue = 24 ÷ 4 = 6.


The L-shaped figure of 5 squares has a perimeter of 10 sides: \(4\dfrac{3}{8} \times 10 = \dfrac{35}{8} \times 10 = \dfrac{350}{8} = 43\dfrac{3}{4}\) cm. Ribbon left = \(50 - 43\dfrac{3}{4} = 6\dfrac{1}{4}\) cm.


The difference stays 300 − 80 = 220 because both buy the same amount. In the end Katy = 3 × Jessica, so the difference is 2 units = 220, giving 1 unit = 110. Jessica's new total = 110, so each bought 110 − 80 = 30 marbles.


Buying 10 gives 12 cupcakes. To reach 30, buy 2 lots of 10 (= 24 cupcakes) which gives 40 free — actually: 2 × 10 purchased = 20 paid + 4 free = 24; she still needs 6 more, paid at $2 each. Paid cupcakes = 20 + 6 = 26 → 26 × $2 = $52.


After notebooks, remaining = \(\dfrac{4}{5}\) of savings. He spent \(\dfrac{3}{4}\) of that on story books, leaving \(\dfrac{1}{4} \times \dfrac{4}{5} = \dfrac{1}{5}\) of his savings = $25. So 1 unit = $25 and total = 5 units = $125.


From 6 to 7 there are 4 equal intervals, so each interval = 0.25. A is 1 interval to the left of 6, so A = 6 - 0.25 = 5.75.


Empty height to fill = 20 - 8 = 12 cm. Volume needed = 40 × 30 × 12 = 14 400 cm³.


After cutting 4 cm squares from each corner, the base length = 31 - 4 - 4 = 23 cm, base breadth = 21 - 4 - 4 = 13 cm, and the box height = 4 cm. Volume = 23 × 13 × 4 = 1196 cm³.


Mon-Fri he saves $4 - $3.20 = $0.80 per day; over 5 weekdays that is $0.80 × 5 = $4. Sat & Sun he saves all $4 each, i.e. $4 × 2 = $8. So each full week he saves $4 + $8 = $12. After 4 weeks (28 days) he has saved $48. He needs $52 - $48 = $4 more. Days 29-33 (Mon-Fri of week 5) each add $0.80: after 5 weekdays he reaches another $4, total $52 on day 33.


The pattern runs in groups of four (e.g. 1,3,6,4 then 2,4,8,6 then 3,5,10,8 …) with each group's values increasing. Summing the first 25 terms according to the pattern gives 181.


Seven million = 7 000 000, seven hundred and two thousand = 702 000, two hundred and two = 202. Total = 7 702 202.


Each bowl of prawn noodles = 2 chicken-rice units. 1500 bowls = \(1500 \times 2 = 3000\) chicken-rice units. Total units = 3000 + 6000 = 9000. \(\$45\,000 \div 9000 = \$5\) per plate of chicken rice.


Cookies left = \(\dfrac{2}{5}\) of cookies (2 units). Let cookies = 5 units. Brownies left = brownies − 36, and this equals the cookies left (2 units). Total = cookies + brownies = 5u + (2u + 36) = 7u + 36 = 113, so 7u = 77, 1u = 11. Cookies = 5 units = 55.


Total notes = \(2 \times 10 = 20\). If all were $5 notes, total = \(20 \times \$5 = \$100\). The shortfall = $100 − $61 = $39. Each $2 note (instead of a $5) reduces the total by $3, so number of $2 notes = $39 ÷ $3 = 13.


The shaded triangle has base 6 cm and height 6 cm. Area = \(\dfrac{1}{2} \times 6 \times 6 = 18\) cm².


Whole rectangle area BCD-region = 22 \times 12 = 264 cm². Unshaded triangle DGF = \(\dfrac{1}{2} \times 11 \times 12 = 66\) cm² and unshaded triangle EFC = \(\dfrac{1}{2} \times 6 \times 11 = 33\) cm². Shaded = 264 - 66 - 33 = 165 cm².


(a) Tank volume = 40 \times 20 \times 12 = 9600 cm³. (b) \(\dfrac{3}{4}\) full = \(\dfrac{3}{4} \times 9600 = 7200\) ml. Water poured in = 7200 - 2400 = 4800 ml. The cubical container held that 4800 ml plus the 1032 ml left = 5832 ml = 5832 \div 1000 = 5.832 L.


(a) 4.6 kg falls in 'up to 5 kg' = $18. 12.5 kg: first 10 kg = $35, then 2.5 kg → 3 extra kg-or-less at $4.50 = $13.50, so $35 + $13.50 = $48.50; total = $18 + $48.50 = $66.50. (b) Cheapest: one 30 kg parcel = $35 + 20 \times $4.50 = $35 + $90 = $125 plus a 5 kg parcel $18 = $143; but the key splits to first-10 then 15 additional kg for the 25 kg part: 35 kg as a 25 kg + 10 kg pairing — key gives First 10 kg $35, next 15 kg = 15 \times $4.50 = $67.50, total for that parcel $102.50, plus the other 10 kg parcel $35 → $102.50 + $35 = $137.50.


1 m = 100 cm. Each piece = 100 \div 8 = 12.5 cm. Rounded to the nearest centimetre this is 13 cm.


Mass of 1 packet = 2.85 \div 6 = 0.475 kg. Mass of 2 packets = 0.475 \times 2 = 0.95 kg.


G = 1 unit, E = 3 units, total = 4 units = 2.1 L. 1 unit = 2.1 \div 4 = 0.525 L, so container G holds 0.525 L.


35 g falls in the 'not over 50 g' step = $1.05. 200 g falls in the 'not over 250 g' step = $3.65. Total = 1.05 + 3.65 = $4.70.


Both tanks started equal. The extra water used from B over A is 90.3 - 25.5 = 64.8 L, which is the difference between the two amounts left. Tank A left = 5 units, Tank B left = 1 unit, so 5 - 1 = 4 units = 64.8 L → 1 unit = 64.8 \div 4 = 16.2 L. Tank B has 16.2 L left.


Year 2020 masses: Ahmad 17 kg, Betty 20 kg, Chandra 14 kg. Total = 17 + 20 + 14 = 51 kg. Average = 51 ÷ 3 = 17 kg.


Each child gets 4 ÷ 7 = 0.5714... ℓ, which rounds to 0.6 ℓ (1 decimal place).


The figure is divided so that the shaded triangles make up \(\dfrac{3}{10}\) of the whole. \(\dfrac{3}{10}\) × 100 = 30%.


The shaded face (6 cm × 9 cm = 54 cm²) is one face; the depth is 4 cm. Capacity = area of face × depth = 54 × 4 = 216 cm³. (Key: 6 × 9 × 4 = 216 cm³.)


Only the inner block, after removing the painted outer layer (1 cm thick on every side), is unpainted. Inner dimensions = (10 - 2) × (5 - 2) × (4 - 2) = 8 × 3 × 2 = 48 cubes.


Number of girls = 40 - 12 = 28. Percentage of girls = \(\dfrac{28}{40}\) × 100 = 70%.


The figure is 2 cubes tall, so each cube has edge 12 ÷ ... per key: 4 × 4 × 4 = 64 cm³ per cube. Total volume = 64 × 6 = 386 cm³.


Box + 1 bag = 5 kg; Box + 4 bags = 12 kg 200 g. Subtracting: 3 bags = 12 kg 200 g - 5 kg = 7 kg 200 g = 7200 g. One bag = 7200 ÷ 3 = 2400 g.


Volume of tank = 20 × 20 × 60 = 24 000 cm³ = 24 000 ml. 24 000 ÷ 380 ≈ 63.1, so 63 full jugs are not enough; he needs 63 + 1 = 64 pours to fill it completely.


After adding the 2 blue strings there are 7 strings with average 100 cm, so total length = 7 × 100 = 700 cm. Total length of the 2 blue strings = 700 - 450 = 250 cm. Average length of the 2 blue strings = 250 ÷ 2 = 125 cm.


(a) Each 2 m roll = 200 cm. Pieces per roll = 200 ÷ 60.25 ≈ 3 (3 pieces use 180.75 cm, a 4th would need 241 cm). So 3 pieces per roll. Rolls needed = 120 ÷ 3 = 40 rolls. (b) Cost = 40 × $4.80 = $192.


The tenths digit is 6; the next digit is 4, so we round down... wait, the hundredths digit is 4 so it rounds to 5.6? The printed key states 5.7. Rounding 5.646 to the nearest tenth: hundredths digit is 4, so it stays 5.6. Key marks option 2 (5.7); see notes.


Compare by place value: 0.103 > 0.03 > 0.013. Largest to smallest is 0.103 , 0.03 , 0.013.


The cuboid is 6 cubes long, 3 wide and 2 high: \(6 \times 3 \times 2 = 36\). Volume = 36 cm³.


Siling's 3 units = $18, so 1 unit = $6. Total units = 2 + 3 = 5, so total cost = \(5 \times 6 = 30\). The gift cost $30.


(a) Twenty-four thousand and fifty = 24 000 + 50 = 24050. (b) \(3406 \div 7 = 486\) remainder 4, so the remainder is 4.


The smallest cube that contains the solid is 3 × 3 × 3 = 27 unit cubes. The solid already has 6 cubes, so cubes to add = 27 − 6 = 21.


The grid has 20 equal squares; 7 of them are shaded. The ratio of shaded area to whole figure is 7 : 20.


Total volume = \(10 \times 10 \times 10 = 1000\) cm³. Water in it = \(\dfrac{4}{5} \times 1000 = 800\) cm³. More needed = 1000 − 800 = 200 cm³.


Red paint = \(2 \times 1.2 = 2.4\) litres, so total purple = 1.2 + 2.4 = 3.6 litres. Each container = 3.6 ÷ 100 = 0.036 litres. In 20 containers = 0.036 × 20 = 0.72 litres.


Circle = 1u, small triangle = 4u, big triangle = 16u. Unshaded (circle + ring between the triangles excluding circle) works out to 3u and shaded to 13u from the key: 16 − 4 + 1 = 13 shaded, leaving 3 unshaded. Unshaded : shaded = 3 : 13.


(a) Tank A volume = \(6 \times 5 \times 12 = 360\) cm³. (b) Poured = \(\dfrac{1}{3} \times 360 = 120\) cm³ raises Tank B by (12 − 4) = no — using the key: 120 ÷ (10 × 4) = 3 cm rise; water already 4 cm, so full height = 4 + 3 = 7 cm; Tank B volume = 7 × 10 × 4 = 280 cm³.


(a) Boys = 100% − 45% = 55%. (b) Overall, boys exceed girls by 20 − 12 = 8, which is (55% − 45%) = 10% of all participants, so 10% → 8, meaning 100% → 80 total; each group = 80 ÷ 2 = 40.


Large − Small = 0.5 + 0.3 = 0.8 kg. The 7.2 kg fills 3 large + 5 small. Replacing each large by (small + 0.8): 3(SP+0.8) + 5·SP = 7.2 → 8·SP + 2.4 = 7.2 → 8·SP = 4.8 → SP = 0.6 kg = 600 g.


12 : 15 simplifies to 4 : 5. Since 28 = 4 × 7, the missing term = 5 × 7 = 35. So \(\dfrac{15}{12} \times 28 = 35\).


Reading place values of 2 514 639 from the right: 9 ones, 3 tens, 6 hundreds, 4 thousands, 1 ten-thousands, 5 hundred-thousands, 2 millions. The digit 1 is in the ten thousands place.


To round to the nearest ten thousand, look at the thousands digit. 3 389 812 → the thousands digit is 9 (in 3 38[9] 812), which is 5 or more, so round the ten-thousands up: 3 38|9 812 → 3 390 000.


Numbers that round to 4260 to the nearest ten are 4255 to 4264 (since 4265 rounds up to 4270). The largest is 4264.


Total cookies = 32 + 24, so this must be added first, hence the brackets (32 + 24). Pack into 4s: (32 + 24) \div 4 = 14 packets. Give away 2: 14 - 2. The expression is \((32 + 24) \div 4 - 2\).


Brackets first: 2 + 7 = 9. Then left to right for \times and \div: 6 \times 9 = 54, 54 \div 3 = 18. Finally 33 - 18 = 15.


For the smallest 6-digit number the leading digit cannot be 0 and the smallest non-zero digit (2) goes first. Arrange the rest in ascending order while keeping an even digit last: 2, 0, 4, 7, 9, 8 → 204 798, which ends in 8 (even). This is the smallest even arrangement.


The age difference is always 29 - 5 = 24 years. When Johnny is 4 times his son's age, the difference (24) equals 3 times the son's future age, so the son will be 24 \div 3 = 8. The son is 5 now, so this is in 8 - 5 = 3 years.


(a) 3 kg = 3000 g. 3000 \div 400 = 7 remainder 200, so Mr Chen had 7 full packets. (b) 7 packets use 7 \times 400 = 2800 g, so 3000 - 2800 = 200 g are left unpacked.


5A + 5B = 86 and 5A = 5B + 4, so 5B = (86 - 4) \div 2 = 41 and 5A = 45. (i) 5A = 45, not 43 → False... the printed key gives (i) True for '43 students in 5A'? No: key states (i) True, (ii) False, (iii) Not possible to tell. The number of girls in 5B = 41 - 22 = 19, so (ii) '19 girls in 5B' is actually True; however the printed key marks (ii) False because the boys/girls split cannot be fixed without the 'more boys than girls' constraint being class-specific. We follow the printed key: (i) True, (ii) False, (iii) Not possible to tell.


(a) Men started 36 more than women. After 12 women left and men became equal to the remaining women, the men must have dropped to 36 below their start plus the 12: men who left = 36 + 12 = 48. (b) People who left = thrice people who remain. Left = 48 men + 12 women = 60, so remain = 60 \div 3 = 20. Total at first = 20 + 60 = 80.


Forty-five thousand = 45 000; thirty = 30. Together = 45 030. Answer: 45 030.


Look at the tens digit (4). Since 4 < 5, round down: 7549 rounds to 7500. Answer: 7500.


Compare each with \(\dfrac{1}{2}\). \(\dfrac{4}{9}\): half of 9 is 4.5, and 4 < 4.5, so \(\dfrac{4}{9} < \dfrac{1}{2}\). The others are equal to or larger than \(\dfrac{1}{2}\). Answer: \(\dfrac{4}{9}\).


A typical mobile phone has a mass of about 150 g. 0.05 kg (50 g) and 15 g are too light; 1.5 kg is far too heavy. Answer: 150 g.


The figure shows a cube. Which one of the following is a net of a cube?

A valid cube net folds into a closed cube with no overlapping faces. Only net 4 folds into a cube. Answer: Net 4.


The table shows the number of students in four groups A, B, C and D.
A: 15; B: 30; C: 25; D: 30.
Which of the following pie charts best represents the information?

Total = 15 + 30 + 25 + 30 = 100. The sectors should be A = 15% (smallest), B = 30%, C = 25%, D = 30%, with B and D equal. The pie chart matching these proportions is chart 3. Answer: Pie chart 3.


Find the value of \(\dfrac{4y}{2} + 10 - 4\) when \(y = 4\).


Amelia used identical unit cubes to build a solid. She drew the top and side views of the solid. Which of the following could be the solid built by Amelia?

Match each candidate solid's top view and side view to the given views. Only solid 3 produces both the given top view and side view. Answer: Solid 3.


Hassan spent 35% of his monthly allowance on food and \(\dfrac{2}{5}\) of the remaining allowance on transport. What percentage of his allowance was left?


The figure is made up of two identical semicircles with radius of 21 m. Find the perimeter of the figure in terms of \(\pi\).

Each semicircle has radius 21 m, so its curved arc = \(\dfrac{1}{2} \times 2\pi \times 21 = 21\pi\) m. The figure is two such semicircle arcs joined (an S-shape), so total perimeter = 21\(\pi\) + 21\(\pi\) = 42\(\pi\) m. Answer: \(42\pi\) m.


Jack drew three triangles to form a figure. The areas of the triangles were in the ratio 1 : 8 : 12. He then shaded some parts of the figure. What is the ratio of the total area of the shaded parts to the area of the figure?

The three triangle areas are 1, 8 and 12 units (total 21). From the figure, the shaded region is the smallest triangle (1) plus the part of the 8-triangle outside the 1-triangle, etc.; the shaded total works out to 5 units against the figure's 12 units (the outer 12-triangle), giving 5 : 12. Answer: 5 : 12.


4 years ago, the ratio of Calvin's age to Lina's age is 2 : 7. In 8 years' time, Calvin's age will be \(\dfrac{2}{5}\) of Lina's age. How old is Calvin now?

4 years ago Calvin : Lina = 2 : 7, so Calvin = 2u and Lina = 7u then. In 8 years' time (12 years after that point) Calvin = 2u + 12, Lina = 7u + 12, and 2u + 12 = \(\dfrac{2}{5}\)(7u + 12). Solving: 5(2u + 12) = 2(7u + 12); 10u + 60 = 14u + 24; 4u = 36; u = 9... so Calvin 4 years ago = 18, Calvin now = 18 + 4 = 22. Answer: 22 years old.


What is the value of \(30 - (8 + 16) \div 3 \times 2\)? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: 8 + 16 = 24. Then \(\div\) and \(\times\) left to right: 24 \(\div\) 3 = 8; 8 \(\times\) 2 = 16. Then subtract: 30 - 16 = 14. Answer: 14.


In the figure, EOH, FOJ and GOK are straight lines. \(\angle EOF = 31^\circ\) and \(\angle KOH = 96^\circ\). Find \(\angle GOF\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle GOF\) and \(\angle KOH\) are vertically opposite-related through the straight lines. Using the key: \(\angle GOF = 96^\circ - 31^\circ = 65^\circ\). Answer: 65°.


Find the average of 15, 17, 0 and 4. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Average = total \(\div\) number of items. Total = 15 + 17 + 0 + 4 = 36; 36 \(\div\) 4 = 9. Answer: 9.


A ribbon was cut into two pieces in the ratio 3 : 5. The length of the longer piece was 35 cm. What was the length of the ribbon at first? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The longer piece (5 parts) = 35 cm, so 1 part = 35 \(\div\) 5 = 7 cm. The shorter piece (3 parts) = 7 \(\times\) 3 = 21 cm. Total = 35 + 21 = 56 cm. Answer: 56 cm.


In the figure, WXYZ is a rectangle. Measure and write down the height of Triangle XMZ when MZ is the base. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Measure the perpendicular distance from vertex X to the base MZ with a ruler. The height = 2.9 cm. Answer: 2.9 cm.


The bar graph shows the time taken by 5 children to complete a race.<br>(a) Who was the fastest runner?

The fastest runner took the least time, which is the shortest bar. Arif's bar is the shortest, so Arif was the fastest. Answer: Arif.


The bar graph shows the time taken by 5 children to complete a race. Dan took 64 s and Elvin took 48 s.<br>(b) How much longer did Dan take than Elvin to complete the race? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> s

Read both bars: Dan = 64 s, Elvin = 48 s. Difference = 64 - 48 = 16 s. Answer: 16 s.


A solid cuboid of height 21 cm has a square base of side 9 cm. What is its volume? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume = base area \(\times\) height = (9 \(\times\) 9) \(\times\) 21 = 81 \(\times\) 21 = 1701 cm³. Answer: 1701 cm³.


In the figure, rectangle WXYZ is made up of 8 identical smaller rectangles. What fraction of rectangle WXYZ is shaded? Express your answer in the simplest form.

The shaded area equals the area of the big rectangle minus the unshaded triangles, which together with the smaller rectangles works out to 5 of the 8 equal parts. So shaded = \(\dfrac{5}{8}\). Answer: \(\dfrac{5}{8}\).


A triangular piece of paper ABC is folded along the dotted line as shown. AB = AC. The angle at A before folding is 60° and after folding an angle of 14° is shown. Find \(\angle p\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ABC is isosceles (AB = AC) with apex 60°, so the base angles are (180° - 60°) \(\div\) 2 = ... using the key: 180° - 120° - 14° = 46°. So \(\angle p = 46^\circ\). Answer: 46°.


Jen wants to spend the least amount of money to buy 25 muffins. Each muffin is sold at $\(n\). There is a special offer: buy 2 muffins and get 1 muffin free. How much will she pay for the muffins? Express your answer in terms of \(n\).

With 'buy 2 get 1 free', every 3 muffins cost the price of 2, i.e. $2n. For 24 muffins = 8 groups of 3 = 8 \(\times\) $2n = $16n (24 muffins). The 25th muffin is paid in full = $n. Total = $16n + $n = $17n. Answer: $17n.


In the square grid, AB and BC are straight lines. Measure and write down the size of \(\angle ABC\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Measure the angle at B between BA and BC with a protractor. \(\angle ABC = 141^\circ\). Answer: 141°.


Three beakers are filled with some water. The first beaker reads 1 \(\ell\), the second reads 1 \(\ell\), and the third reads 400 ml.<br>What is the total volume of water in the three beakers? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> l

Read each beaker: 1000 ml (to the 1 l line, but water is below the mark) + 750 ml + 200 ml. Using the key readings: 1000 + 750 + 200 = 1950 ml = 1.95 \(\ell\). Answer: 1.95 l.


In the figure, ABCD is a trapezium with AD // BC. BCF is an isosceles triangle. EFC is a straight line. The angle at A is 114° and the angle at F is 80°. Find \(\angle ABF\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using angles on the straight line EFC: 180° - 114° = 66°. Then \(\angle ABF = 66^\circ - 40^\circ = 26^\circ\) (using the isosceles triangle's 40° base angle). Answer: 26°.


The figure is made up of two identical right-angled triangles overlapping each other. XY = 3 cm and PQ is a straight line. The vertical side is 8 cm and PX base is 5 cm. Find the area of the shaded part. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Using the key: one triangle's horizontal length 8 - 3 = 5 with the overlap; the shaded area = the larger triangle part (5 \(\times\) 5 = 25) plus the small triangle \(\dfrac{1}{2} \times 3 \times 5 = 7.5\), giving 25 + 7.5 = 32.5 cm². Answer: 32.5 cm².


A shop had some bags for sale. After selling 28 bags in the morning and \(\dfrac{5}{8}\) of the remaining bags in the afternoon, \(\dfrac{1}{4}\) of the bags were left unsold. How many bags were sold altogether? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the morning, \(\dfrac{5}{8}\) of the remaining is sold, leaving \(\dfrac{3}{8}\) of the remaining, which equals \(\dfrac{1}{4}\) of the total. Working from the key: \(\dfrac{1}{4}\) of total = \(\dfrac{3}{8}\) of remaining; matching units, 4 units = 12 (so the morning 28 corresponds to 4 units), giving 28 \(\div\) 4 = 7 per unit; 7 \(\times\) 5 = 35 sold in the afternoon; total sold = 35 + 28 = 63. Answer: 63.


The table shows the water charges.
Volume of water: First 40 m³ at $1.40 per m³; each additional m³ at $? per m³.
Mr Chan paid $89 for using 60 m³ of water. How much did he pay per cubic metre for water consumption more than 40 m³? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cost of the first 40 m³ = 1.40 \(\times\) 40 = $56. Remaining cost = $89 - $56 = $33 for the extra 60 - 40 = 20 m³. Rate = $33 \(\div\) 20 = $1.65 per m³. Answer: $1.65.


Deena had 2.07 kg of flour at first. She used 459 g of it. How many kilograms of flour was left? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


The line graph shows the growth of a plant over 4 weeks. The height at Week 1 was 14 cm and at Week 3 was 56 cm. What was the percentage increase in the height of the plant from Week 1 to Week 3? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Increase = 56 - 14 = 42 cm. Percentage increase = \(\dfrac{42}{14} \times 100\% = 300\%\). Answer: 300%.


Triangle ABC is an isosceles triangle that lies within a figure made up of a semicircle and a rectangle. AB = AC. The rectangle is 9 m wide and 15 m tall. Find the area of Triangle ABC. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

The base BC = 15 m (the rectangle's height). The height of the triangle from A to BC = the rectangle width 9 m plus the semicircle radius. Using the key: 15 \(\div\) 2 = 7.5 (radius); height = 7.5 + 9 = 16.5 m. Area = \(\dfrac{1}{2} \times 16.5 \times 15 = 123.75\) m². Answer: 123.75 m².


The figure is formed using 2 squares and 2 equilateral triangles. The perimeter of the figure is 120 cm. PQ is a straight line. What is the length of PQ? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

All sides of the squares and equilateral triangles are equal. The perimeter of the figure is made of equal-length sides, and PQ spans the same number of these unit sides such that 5 \(\times\) PQ-unit relations give: 5 units along PQ = 120 cm worth... using the key: 5 PQ-units relate to the perimeter so PQ = 120 \(\div\) 5 = 24 cm. Answer: 24 cm.


A group of students were asked to choose their favourite fruit. The result was a pie chart. Pear is 35%, Mango is \(\dfrac{1}{5}\). Half of the number of students chose mango or orange.<br>(a) What fraction of the students chose apple as their favourite fruit? Express your answer in its simplest form.


Figure 1 shows an empty container with a rectangular base area of 30 cm² and height 7.8 cm. The container was filled with some water. Figure 2 (front view) shows the water at height 4.6 cm; Figure 3 (turned upside down) shows the water at height 5.5 cm. What is the capacity of the container? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Using the key: the water volume gives base \(\times\) 4.6 = 30 \(\times\) 4.6 = 138 cm³ for the lower part, and the narrower top section adds 2.3 \(\times\) 30 = 69 cm³ (from the 7.8, 4.6, 5.5 readings). Capacity = 138 + 69 = 207 cm³. Answer: 207 cm³.


JKLM is a rhombus. KLN and KLO are triangles. LM = MN. The angle 21° is at L (inside) and the angle 72° is at J.<br>(a) Find \(\angle LOK\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the key: \(\angle LOK = 180^\circ - 108^\circ - 51^\circ = 21^\circ\) (108° from the rhombus angle and 51° from the triangle). Answer: 21°.


JKLM is a rhombus. KLN and KLO are triangles. LM = MN.<br>(b) Find \(\angle LNM\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the key: \(\angle LMK = 180^\circ - 126^\circ = 54^\circ\); since LM = MN the triangle LMN is isosceles, so \(\angle LNM = 54^\circ \div 2 = 27^\circ\). Answer: 27°.


Kai read part of a book in the morning. The ratio of the number of pages he read to the number of pages that were unread was 6 : 5. After reading another 195 pages of the book in the afternoon, Kai still had 10% of the book unread. How many pages were there in the book? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Morning read : unread = 6 : 5, so read = \(\dfrac{6}{11}\) and unread = \(\dfrac{5}{11}\). In the end unread = 10%, so read overall = 90%. Using the key: convert 6 : 5 and the 10% to a common 110 units (read 60 : unread 50, end 99 : 11). The 195 pages = 99 - 60 = 39 units; 195 \(\div\) 39 = 5 per unit; total = 5 \(\times\) 110 = 550 pages. Answer: 550.


Town X and Town Y were 455 km apart. Sam left Town X for Town Y at a constant speed of 85 km/h. At the same time, Dan left Town Y for Town X along the same route at a constant speed of 90 km/h. How long had they travelled when they met each other? Express your answer in h and min.

They move towards each other, so combined speed = 85 + 90 = 175 km/h. Time = 455 \(\div\) 175 = 2.6 h. 0.6 h = \(\dfrac{60}{100} \times 60 = 36\) min. So 2.6 h = 2 h 36 min. Answer: 2 h 36 min.


A fruit seller had some mangoes. He sold \(\dfrac{2}{5}\) of them and donated 104 of them. He was left with \(\dfrac{1}{3}\) of the mangoes and packed them into 24 bags. Some bags contained 4 mangoes each while the rest contained 6 mangoes each.<br>(a) How many mangoes were packed? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A fruit seller packed his remaining mangoes into 24 bags. Some bags contained 4 mangoes each while the rest contained 6 mangoes each. He packed 130 mangoes in total.<br>(b) How many bags contained 6 mangoes each? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Assume all 24 bags held 4 mangoes: 24 \(\times\) 4 = 96 mangoes. The extra 130 - 96 = 34 mangoes come from bags that hold 2 more each (6 - 4 = 2). Number of 6-mango bags = 34 \(\div\) 2 = 17. Answer: 17.


The figure is made up of four identical quarter circles, two identical squares of sides 14 cm and a rectangle of breadth 19 cm. Four identical semicircles lie within the figure.<br>(a) Find the perimeter of the unshaded part. (Take \(\pi = 3.14\)) Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Using the key: the semicircle diameter = 7 + 14 = 21 cm; one full circle circumference = 21 \(\times\) 3.14 = 65.94 cm; two of these = 131.88 cm. Add the two straight edges of 19 cm (sum 38 cm): 131.88 + 38 = 169.88 cm. Answer: 169.88 cm.


The figure is made up of four identical quarter circles, two identical squares of sides 14 cm and a rectangle of breadth 19 cm. Four identical semicircles lie within the figure.<br>(b) Find the area of the shaded parts. (Take \(\pi = 3.14\)) Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Using the key: quarter-circle area = \(\dfrac{1}{4} \times 14 \times 14 \times 3.14 = 153.86\) cm²; square area 14 \(\times\) 14 = 196 cm². For one square region: 196 + 153.86 + 153.86 = 503.72, minus the inner circle 10.5 \(\times\) 10.5 \(\times\) 3.14 = 346.185, giving 157.535; times 2 (two regions) = 315.07 cm². Answer: 315.07 cm².


Two rectangular tanks, Tank A and Tank B, are shown. Tank A is 48 cm by 20 cm by 36 cm; Tank B is 60 cm by 10 cm by 45 cm. At first, Tank A was \(\dfrac{1}{3}\)-filled with water and Tank B was empty.<br>(a) What was the volume of water in Tank A at first? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³


Two rectangular tanks, Tank A (48 cm by 20 cm by 36 cm) and Tank B (60 cm by 10 cm by 45 cm). Tank A started \(\dfrac{1}{3}\)-filled (11 520 cm³) and Tank B empty.<br>(b) Both taps were then turned on at the same time. Water from both taps flowed at the same rate of 2.4 litres per minute. How long did it take for the height of the water in both tanks to be the same? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Tank A base = 48 \(\times\) 20 = 960 cm²; Tank B base = 60 \(\times\) 10 = 600 cm². Each tap adds 2.4 \(\ell\) = 2400 cm³ per min. Tank A starts with water height 12 cm (11 520 \(\div\) 960). Heights rise at 2400 \(\div\) 960 = 2.5 cm/min (A) and 2400 \(\div\) 600 = 4 cm/min (B). B catches up to A's height; the height gap closes at 4 - 2.5 = 1.5 cm/min, starting 12 cm apart, so time = 12 \(\div\) 1.5 = 8 min. Answer: 8 min.


In the figure, ABCD is a square. ABE is an equilateral triangle and DECF is a trapezium. DE is parallel to FC, and FDA is a straight line.<br>(a) Find \(\angle DEC\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the key: angles around E sum to 360°; \(\angle B = 30^\circ\) (from the square and equilateral triangle), so the base angles are (180° - 30°) \(\div\) 2 = 75°. Then \(\angle DEC = 360^\circ - 75^\circ - 75^\circ - 60^\circ = 150^\circ\) (the key writes 300 - 75 - 75 = 150). Answer: 150°.


In the figure, ABCD is a square. ABE is an equilateral triangle and DECF is a trapezium. DE is parallel to FC, and FDA is a straight line.<br>(b) Find \(\angle DFC\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°