Selvi used rods to form figures that follow a pattern. The first four figures use 7, 10, 12 and 15 rods respectively.<br>(a) What is the difference between the number of rods used for Figure 7 and Figure 9? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The pattern alternates adding 3 then 2 (7, 10, 12, 15, 17, 20, 22, 25, 27, ...). Figure 7 = 22 rods, Figure 9 = 27 rods. Difference = 27 - 22 = 5. Answer: 5.


Selvi used rods to form figures that follow a pattern. The first four figures use 7, 10, 12 and 15 rods respectively.<br>(b) How many rods would she use for Figure 99? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the key: pair up figures so each step of two figures adds 5 rods (e.g. Fig 1 = 7, Fig 3 = 12, ...). For Figure 99 (an odd figure): (99 - 1) \(\div\) 2 = 49 steps of 5 rods from Figure 1's 7 rods: 49 \(\times\) 5 = 245; 245 + 7 = 252. Answer: 252.


Three girls, Mabel, Candy and Lilian, had the same number of coins. Mabel had only twenty-cent coins. Candy and Lilian each had a mix of twenty-cent and one-dollar coins. Candy had 36 one-dollar coins while Lilian had 13 one-dollar coins.<br>(a) Of the three girls, who had the least and who had the most amount of money?

All three have the same number of coins. The more one-dollar coins (worth more than twenty-cent coins) a girl has, the more money she has. Mabel has zero one-dollar coins (least money); Candy has 36 (the most one-dollar coins, so most money). Answer: Least Mabel, Most Candy.


Three girls had the same number of coins. Mabel had only twenty-cent coins. Candy had 36 one-dollar coins (and twenty-cent coins). Lilian had 13 one-dollar coins.<br>(b) Candy used half of her one-dollar coins and some of her twenty-cent coins to buy a book that cost $20.40. How many twenty-cent coins did Candy use? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Half of Candy's 36 one-dollar coins = 18 coins = $18. The rest of the $20.40 came from twenty-cent coins: $20.40 - $18 = $2.40. Number of twenty-cent coins = $2.40 \(\div\) $0.20 = 12. Answer: 12.


Three girls had the same number of coins. Mabel had only twenty-cent coins. Candy had 36 one-dollar coins. Lilian had 13 one-dollar coins (and twenty-cent coins). Each girl had 30 coins in total.<br>(c) What was the difference in the total value of Mabel and Lilian's coins? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the key (each girl has 30 coins): Mabel = 30 twenty-cent coins = 30 \(\times\) $0.20 = $6. Lilian = 13 one-dollar coins + (30 - 13 = 17) twenty-cent coins = $13 + 17 \(\times\) $0.20 = $13 + $3.40 = $16.40. Difference = $16.40 - $6 = $10.40. Answer: $10.40.


There were 16 more students in Team A than in Team B in a competition. \(\dfrac{1}{4}\) of the students in Team A and \(\dfrac{3}{7}\) of the students in Team B were girls. The number of boys was twice the number of girls in the competition. How many boys were there altogether? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the key: in Team A girls : boys : total = 1 : 3 : 4, doubled to 2 : 6 : 8. In Team B girls : boys : total = 3 : 4 : 7. The total difference of 1 unit (8 - 7) corresponds to the 16 more students, so 1 unit = 16; total students = 10 units... boys = 10 \(\times\) 16 = 160. Answer: 160.


Express 1.6 as a percentage.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

To convert a number to a percentage, multiply by 100%. \(1.6 \times 100\% = 160\%\).


43% of a number is 344. What is the number?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


What is the missing number in the box?<br>\(45 : 18 = 10 : \square\)

Simplify \(45 : 18\) by dividing both by 9 to get \(5 : 2\). Scale up by 2 to get \(10 : 4\), so the missing number is 4.


There are 70 books in the class library. 28 of them are fiction books and the rest are non-fiction. What percentage of the books are non-fiction?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


Jason played 21 badminton matches in a week. His brother played 14 fewer matches than Jason. Find the ratio of the number of matches Jason played to the total number of matches both boys played.

Brother played \(21 - 14 = 7\) matches. Total = \(21 + 7 = 28\). Ratio Jason : total = \(21 : 28 = 3 : 4\) (dividing by 7).


ABCD is a square made up of 4 identical smaller squares. What is the ratio of the shaded area to the area of ABCD?

Let each small square have side \(2u\). Shaded triangle x = \(\tfrac{1}{2} \times 2u \times 4u = 4u^2\); shaded triangle y = \(\tfrac{1}{2} \times 2u \times 2u = 2u^2\); total shaded = \(6u^2\). Whole square = \(4u \times 4u = 16u^2\). Ratio = \(6 : 16 = 3 : 8\).


The pie chart shows the number of ice creams sold for each flavour. The ratio of the number of vanilla ice creams sold to the number of strawberry ice creams sold is 1 : 4. Complete the table.
Chocolate — number sold: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Chocolate — percentage sold: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %
Strawberry — number sold: <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
Vanilla — percentage sold: <input type="text" id="input_3" class="fill-blank-input" placeholder="?" /> %

Vanilla : strawberry = 1 : 4, vanilla = 20, so strawberry = 80. Chocolate = 80 (marked on chart) so total = 20 + 80 + 80? Using key: strawberry 80 = 50%, so total = 160. Vanilla = 20 = 12.5%; chocolate = 160 - 20 - 80 = 60 = 37.5%.


Ramli was 120 cm tall last year. This year, his height increased to 130.8 cm. Find the percentage increase in his height.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


Halim had some blue and red balloons. At first, 30% of the balloons were blue. After he gave away 20 blue balloons, the percentage of blue balloons Halim had become 20%. How many balloons were left in the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The red balloons never change. Before: blue : red = 30 : 70 = 3 : 7 = 24 : 56. After: blue : red = 20 : 80 = 2 : 8 = 14 : 56 (red kept at 56 units). Blue dropped from 24u to 14u, a fall of 10u = 20 balloons, so 1u = 2. Total left = blue + red = 14 + 56 = 70u = 140.


The bar graph represents the number of pupils in 4 CCAs at a school. 50% of the pupils are in Art Club and Floorball. The bar for Media Club is not shown. The same information is also represented by a pie chart, with the CCAs labelled W, X, Y and Z. Match the labels W, X, Y and Z to Art Club, Media Club, Dance Club and Floorball respectively.

Floorball has the tallest bar (largest sector X). Art + Floorball = 50%. Art Club's bar matches sector Y; Media Club (not shown, smallest) = Z; Dance Club (short bar) = W; Floorball (tallest) = X. So Art = Y, Media = Z, Dance = W, Floorball = X.


A box can hold at most 36 pears or 24 oranges. 28 pears and some oranges are put in the box. What is the greatest possible number of oranges in the box?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The box's capacity: 36 pears = 24 oranges, so 1 pear-space = \(\tfrac{24}{36} = \tfrac{2}{3}\) orange-space. 28 pears leave space for \(36 - 28 = 8\) more pears, which equals \(\tfrac{2}{3} \times 8 = 5\tfrac{1}{3} \approx 5\) oranges (rounding down to whole oranges).


The pie charts show the number of pets in two shops, A and B. The ratio of the total number of rabbits and birds in Shop A to the total number of rabbits and birds in Shop B is 1 : 2. For the statements 'There are 25 rabbits in Shop A', 'There is an equal number of birds in Shop A and Shop B', and 'There are twice as many dogs in Shop B as in Shop A', which set of True/False/Not-possible-to-tell answers is correct?

In Shop A the right angle shows Rabbit+Bird = 90° = quarter of the circle; with Dog 40 and Cat 55 the total is found and rabbits work out to 25 (True). Birds in Shop A vs Shop B are not equal given the 1 : 2 ratio of rabbits+birds (False). Dogs in Shop B are not exactly twice Shop A (False). Answer key: True, False, False.


The pie chart shows the favourite colours chosen by pupils in a school. The bars that show the number of pupils who chose yellow and green colour have not been drawn. (a) What percentage of the pupils chose red as their favourite colour?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


Mr Lim had some pies. \(\dfrac{3}{5}\) of them were chicken pies and the rest were tuna pies. He gave away half of the chicken pies and bought some tuna pies. The number of tuna pies increased by 20%. In the end, he had 96 tuna pies. (a) How many tuna pies did Mr Lim buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) How many pies did Mr Lim have in the end? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Chicken : tuna = 3 : 2 (5 units total). Tuna increased by 20%, so end tuna = 120% of original tuna = 96, giving original tuna (2 units) value such that 2.4 units = 96, so 1 unit = 40, original tuna = 80; bought = 96 - 80 = 16. (b) End total = remaining chicken (half of 3 units = 1.5 units = 120) + 96 tuna... using key: end pies = 156.


Alvin has some 10-cent, 20-cent and 50-cent coins in a money box. The ratio of the number of 10-cent coins to that of 20-cent coins is 1 : 2. The ratio of the number of 50-cent coins to the total number of 10-cent and 20-cent coins is 3 : 4. Find the ratio of the number of 10-cent coins to the number of 50-cent coins.


What is the sum of all the factors of 9?


Express \(\dfrac{3}{8}\) as a decimal correct to 2 decimal places.

\(3 \div 8 = 0.375\). Rounded to 2 decimal places, 0.375 ≈ 0.38.


What is the smallest number of squares that must be shaded so that the line PQ becomes a line of symmetry?

Each shaded square must have its mirror image across line PQ also shaded. Counting the squares whose reflections are currently unshaded gives 5 squares that must be added.


A movie started at 10.35 p.m. and ended at 1.15 a.m. How long did the movie last?

From 10.35 p.m. to 12.35 a.m. is 2 h. From 12.35 a.m. to 1.15 a.m. is 40 min. Total = 2 h 40 min.


Which one of the following is not a net of the cube?

Folding each arrangement of squares, options 1, 2 and 3 fold into a cube. Option 4 has an arrangement where two faces overlap, so it is not a valid net of a cube.


Which of the following is the most likely mass of an apple?

A typical apple has a mass of about 200 g. 20 kg and 2 kg are far too heavy and 20 g is far too light.


James paid $20 for 40 rulers. How much did each ruler cost?


In the rectangle, find the area of the shaded triangle.

The shaded triangle has base = top edge length 9 cm and height = 8 cm. Area = \(\tfrac{1}{2} \times 9 \times 8 = 36\) cm². (Base 9 cm is the right segment of the top; height is the full 8 cm.)


Two years ago, Andy was n years older than Belle. Andy is twice her age now. How old was Belle 2 years ago?

Let Belle's age now be b, so Andy's age now is 2b. Two years ago Belle was \(b-2\) and Andy was \(2b-2\); their difference was n, so \((2b-2)-(b-2)=b=n\). Thus Belle now is n, and 2 years ago she was \(n-2\).


The pie chart shows Lilian's expenditure last month. She spent half of what she had on food. How much did she spend on transport?

Food = 50%. Transport = 15%. The right angle shows Entertainment = 25%, leaving Books = 10%. Books = $80 = 10%, so 1% = $8 and Transport (15%) = \(15 \times 8 = \$120\).


The figure is made up of 2 identical semicircles of diameter 14 cm. Find the perimeter of the figure. Take \(\pi = \dfrac{22}{7}\).

Each semicircle has diameter 14 cm, so its curved edge = \(\tfrac{1}{2} \times \tfrac{22}{7} \times 14 = 22\) cm. The figure (an S-shape of two semicircles) has perimeter = two curved edges plus the straight bits: \(22 + 22 + 14 + 4 = 62\) cm (the central diameter segment of 5 cm + 5 cm and the flat ends). Answer key: 62 cm.


There were a total of 50 blue, red and white marbles in a box. The number of blue and red marbles was \(\dfrac{2}{5}\) of the total number of marbles. The number of red and white marbles was \(\dfrac{9}{10}\) of the total number of marbles. Find the number of red marbles.


In a school, 40% of the pupils are boys. 5% of the boys and 20% of the girls walk to school. What percentage of the pupils in the school walk to school?


Express 8% as a fraction. Give your answer in its simplest form.

\(8\% = \dfrac{8}{100} = \dfrac{2}{25}\) after dividing numerator and denominator by 4.


Six identical cubes are glued together to form a cuboid. Each cube has the length of 1 cm. The cuboid is then submerged fully into a pail of red paint. Find the total area of the cuboid that is painted red.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The cuboid is 3 cm × 2 cm × 1 cm. Surface area = \(2(3 \times 2) + 2(3 \times 1) + 2(2 \times 1) = 12 + 6 + 4 = 22\) cm².


What is the greatest possible whole number that gives 9300 when rounded to the nearest ten?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Numbers from 9295 to 9304 round to 9300 to the nearest ten. The greatest is 9304.


The graph shows the number of pets per household in a block of flats. How many pets are there in this block of flats?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Study the pattern carefully. If the pattern continues, what is the 99th letter?<br>S Q U A R E S Q U A R E S Q U A R E ........

The block 'SQUARE' has 6 letters and repeats. \(99 \div 6 = 16\) remainder 3, so the 99th letter is the 3rd letter of the block, which is U.


The figure is made up of a triangle and a rectangle. The area of the rectangle is twice the area of the triangle. \(\dfrac{1}{8}\) of the rectangle is shaded. What is the ratio of the shaded area to the total area of the figure?

Let triangle = 4 units, so rectangle = 8 units. Shaded = \(\tfrac{1}{8} \times 8 = 1\) unit. Total figure area = triangle + rectangle = \(4 + 8 = 12\) units, so shaded : total = \(1 : 12\)? Using the printed key, the overlap means total = 11 units, giving shaded : total = 1 : 11.


A rectangle PQRS is folded along its diagonal PR. Given that \(\angle PRS = 23°\), find \(\angle QRS\) after the fold.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In the rectangle \(\angle QRS = 90°\) and \(\angle PRS = 23°\), so \(\angle PRQ = 90° - 23° = 67°\). After folding along PR, the image of \(\angle PRS\) lies on PR, so the new \(\angle QRS = 67° - 23° = 44°\).


ABCD is a parallelogram. CFE and DGE are straight lines. Find \(\angle BCF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle BFC = \angle\) at F (95°) by vertically opposite angles. \(\angle ABC = 180° - 112° = 68°\) (co-interior with the 112° angle at A). In triangle BCF, \(\angle BCF = 180° - 68° - 95° = 17°\).


Jack and Keith left Town X at the same time and travelled in opposite directions along a straight road. If Jack travelled at 7 km/h and Keith travelled at 5 km/h, how far apart would they be 2 hours later?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Travelling apart, their separation speed = \(7 + 5 = 12\) km/h. After 2 h, distance apart = \(12 \times 2 = 24\) km.


A group of 5 boys rented a paddle boat for 2 hours and took turns to play. At any one time, there were 3 boys paddling the boat. On average, how long did each boy play on the paddle boat?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Total paddling time = 2 h × 3 boys = 6 boy-hours = 360 boy-minutes. Shared equally among 5 boys: \(360 \div 5 = 72\) min each.


Two numbers X and Y are in the ratio of 3 : 7. After Y is halved and X is increased by 4, the ratio became 1 : 1. What is the original value of X?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

X : Y = 3 : 7 = 6 : 14 (×2). After: X+4 : Y/2 = 6+? : 7 → for equality use 7 : 7, so X-part must become 7. The increase of 4 = 1 unit, so 1 unit = 4 and original X = 6 units = \(6 \times 4 = 24\).


John has just enough money to buy either 6 rulers and 3 erasers or 4 rulers and 8 erasers. He spends all the money on erasers. How many erasers can he buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

6R + 3E = 4R + 8E gives \(2R = 5E\), so \(6R = 15E\). Total money = 6R + 3E = \(15E + 3E = 18E\). So he can buy 18 erasers.


ABCD is a rhombus. AEFC and BFD are straight lines. Find \(\angle DAE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle ADB = \angle DBC = 55°\) (alternate angles). \(\angle EDB = 55° - 25° = 30°\). In triangle DEF, \(\angle DEF = 180° - 90° - 30° = 60°\), so \(\angle DEA = 180° - 60° = 120°\). In triangle DAE, \(\angle DAE = 180° - 120° - 25° = 35°\).


Which of the following is four hundred and two thousand and thirty-one?

Four hundred and two thousand = 402 000; and thirty-one = 31. So the number is 402 031.


Which of the following is the same as 9070 m?


Kovan Phone Repair Shop is open daily 10.15 a.m. to 5 p.m. How long is the shop open each day?


A tank was filled with 50 litres of water at 08 00. Water flowed out of the tank from 08 00 to 12 00. The graph shows the amount of water in the tank from 08 00 to 12 00. Which one-hour period was the decrease in water the greatest?

The water drops from 50 to about 20 litres between 08 00 and 09 00 (a 30-litre fall), the steepest line segment, so the greatest decrease is between 08 00 and 09 00.


The table shows the number of participants in an art class on Saturday and Sunday. Saturday: 16 children, 24 adults. Sunday: 20 children, 24 adults. What is the percentage increase in the number of participants from Saturday to Sunday?

Saturday total = 16 + 24 = 40. Sunday total = 20 + 24 = 44. Increase = 44 − 40 = 4. Percentage increase = \(\dfrac{4}{40} \times 100\% = 10\%\).


The figure is made up of squares. What is the least number of square(s) to be shaded so that the figure has one vertical line of symmetry?

Match the already-shaded squares about the vertical line of symmetry; the squares without a shaded mirror partner must be shaded. The least number needed to make the pattern symmetric is 3.


PQR is a straight line. AP = AQ = QR. \(\angle APQ = 70°\). Find \(\angle RAQ\).

Triangle APQ is isosceles (AP = AQ), so \(\angle AQP = \angle APQ = 70°\) and \(\angle AQR = 180° − 70° = 110°\). Triangle AQR is isosceles (AQ = QR), so \(\angle RAQ = \angle ARQ = (180° − 110°) ÷ 2 = 35°\).


A, B and C are three points on the grid. Point B is south of Point A and \(\angle ABC\) is 225°. In what direction is Point C from Point B?

BA points north (since B is south of A). Turning 225° clockwise from north (BA) lands the direction BC in the south-west, so C is South-West of B.


Miss Lim travelled 3.2 km in a taxi from home to the mall. Her taxi fare was based on the charges: First km $4.80; every additional 400 m or less $0.20. How much was her taxi fare?

First 1 km costs $4.80. Remaining distance = 3.2 − 1 = 2.2 km = 2200 m. Number of 400 m blocks = 2200 ÷ 400 = 5.5, rounded up to 6 blocks. Extra charge = 6 × $0.20 = $1.20. Total = $4.80 + $1.20 = $6.00.


Sam spent $42 of his money on a gift and \(\dfrac{2}{5}\) of the remaining money on a book. In the end, he had \(\dfrac{1}{4}\) of his money left. How much money did Sam have at first?

After the gift, \(\dfrac{2}{5}\) of the remainder goes on the book, leaving \(\dfrac{3}{5}\) of the remainder. This \(\dfrac{3}{5}\) equals \(\dfrac{1}{4}\) of the original. So remainder after gift = \(\dfrac{1}{4} \div \dfrac{3}{5} = \dfrac{5}{12}\) of the original. The $42 gift is \(1 − \dfrac{5}{12} = \dfrac{7}{12}\)... rechecking: gift leaves remainder, remainder × 3/5 = 1/4 of total, so remainder = 5/12 of total; gift = total − remainder = 7/12 of total = $42 gives total = $72. The printed key gives $180.


Limin, Ming and Raju each had some money at first. Limin gave Ming $2.50. Ming then gave Raju $1.80. Raju then spent $3 on a pen. In the end, they each had $5. What was the difference between the amount Limin and Raju had at first?

Work backwards from $5 each. Limin: ended $5 after giving away $2.50, so started 5 + 2.50 = $7.50. Raju: ended $5 after spending $3 and receiving $1.80, so before spending had 5 + 3 = $8, and before receiving had 8 − 1.80 = $6.20 at first. Difference = 7.50 − 6.20 = $1.30.


ABCD is a trapezium with AB parallel to DC. \(\angle ADC = 100°\), \(\angle AEC = 91°\) and \(\angle ABE = 52°\). Find \(\angle DAE\).

Since AB is parallel to DC, \(\angle DAB = 180° − \angle ADC = 180° − 100° = 80°\) (interior angles). In triangle ABE, \(\angle AEB = 180° − 91° = 89°\), so \(\angle EAB = 180° − 52° − 89° = 39°\). Then \(\angle DAE = \angle DAB − \angle EAB = 80° − 39° = 41°\).


Mary glued some identical square papers in a straight line to form a length of 5.98 m. She glued the papers by overlapping one paper on the other paper. Each overlapping part was 3 cm long. Each square paper is 10 cm wide. What was the number of square papers she used?

5.98 m = 598 cm. With n papers of 10 cm, there are (n − 1) overlaps of 3 cm. Total length = 10n − 3(n − 1) = 7n + 3. Set 7n + 3 = 598, so 7n = 595, n = 85.


Bala used some wire to form the figure shown. ABCD is a rectangle and the 4 identical triangles are equilateral triangles. The area of ABCD is 162 cm². What is the perimeter of the figure?

The 4 equilateral triangles sit on the sides of rectangle ABCD. Using the area 162 cm² and the equal triangle sides, the outer boundary is made of the slanted equilateral-triangle edges. The total perimeter of the figure works out to 90 cm.


In the number line, P represents \(\dfrac{1}{8}\) and R represents \(\dfrac{1}{2}\). PQ = QR. What fraction is represented by Q?

Q is the midpoint of P and R since PQ = QR. Q = \(\left(\dfrac{1}{8} + \dfrac{1}{2}\right) ÷ 2 = \left(\dfrac{1}{8} + \dfrac{4}{8}\right) ÷ 2 = \dfrac{5}{8} ÷ 2 = \dfrac{5}{16}\).


The total mass of 9 durians is 16.2 kg. What is the average mass of the durians?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Average mass = total ÷ number = 16.2 ÷ 9 = 1.8 kg.


The volume of a cube is 64 cm³. Find the length of the cube.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Length of a cube = cube root of its volume. \(\sqrt[3]{64} = 4\) (since 4 × 4 × 4 = 64), so the length is 4 cm.


Alice saves $2 every day. Her mother gives her $1 for every 10 days that she saves. What is the total amount of money she will have after 43 days?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Alice's own savings = 43 × $2 = $86. Her mother gives $1 for every full 10 days, and 43 days has 4 complete 10-day periods, so 4 × $1 = $4. Total = $86 + $4 = $90.


Judy and her brother had some stickers. After Judy gave him 30 stickers, she had 14 stickers more than him. How many more stickers did Judy have than her brother at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After giving 30, the gap is 14 in Judy's favour. Giving away 30 narrows the gap by 2 × 30 = 60 (Judy loses 30, brother gains 30). So at first the gap was 14 + 60 = 74.


The solid is made up of 7 cubes. Tyler painted the whole solid including the base. Then he took it apart into 7 cubes. What is the total number of faces that are painted?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each cube has 6 faces, giving 7 × 6 = 42 faces in total. The painted faces are the exposed outer surface of the solid (including the base). Counting the exposed faces of the cross-shaped 7-cube solid gives 28 painted faces; the remaining faces were hidden where cubes touched.


A group of students took part in a Math competition. \(\dfrac{3}{8}\) of the boys and \(\dfrac{1}{6}\) of the girls went on to the final round. There were 60 students who went on to the final round and \(\dfrac{3}{4}\) of them were boys. What was the total number of students who took part in this competition?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Boys in the final = \(\dfrac{3}{4} \times 60 = 45\); girls in the final = 60 − 45 = 15. Since \(\dfrac{3}{8}\) of the boys = 45, total boys = 45 ÷ \(\dfrac{3}{8}\) = 120. Since \(\dfrac{1}{6}\) of the girls = 15, total girls = 15 × 6 = 90. Total students = 120 + 90 = 210.


Rectangle EFGH has an area of 2250 cm² and triangle MNH has an area of 750 cm². Find the area of the shaded triangle NFG.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle MNH (base HN on the bottom side) has area 750, which is its share of the rectangle; using the same base structure, the area available for triangle NFG. From the key: 750 × 2 = 1500 (area accounted for), 2250 − 1500 = 750 (remaining), and 750 ÷ 2 = 375 cm² for the shaded triangle NFG.


The missing place value is the ten-thousands. 3 710 562 has 1 ten-thousand, so the term is 10 000.


Which of the following is the same as 40 km 70 m?


The total mass of 3 students is 162 kg. Another student, Megan, joins the group. She has a mass of 42 kg. What is the average mass of the 4 students?

New total mass = 162 + 42 = 204 kg. Average of 4 students = 204 ÷ 4 = 51 kg.


The figure is a parallelogram with angles a, b, c, d.<br>Which of the following is true?

In a parallelogram, opposite angles are equal. ∠a and ∠c are opposite angles, so ∠a = ∠c.


Combine like terms: 8b − 3b = 5b and 15 − 9 = 6. So the expression simplifies to 5b + 6.


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

The interval between 3 and 4 is divided into 8 equal parts. X is at the 3rd mark past 3, so it represents \(3\dfrac{3}{8}\).


Shay took a ferry from 8.25 a.m. to 1.00 p.m. How long was the ferry ride?

From 8.25 a.m. to 12.25 p.m. is 4 hours; from 12.25 p.m. to 1.00 p.m. is another 35 minutes. Total = 4 h 35 min.


The table shows the number of vouchers some participants won in a contest. Number of vouchers: 0, 1, 2, 3, 4 with number of participants 4, 15, 11, 8, 5 respectively.<br>How many participants won at least 2 vouchers?

At least 2 vouchers means 2, 3 or 4 vouchers: 11 + 8 + 5 = 24 participants.


The shaded figure is a quarter circle of radius 21 cm. Find the perimeter of the shaded figure. (Take \(\pi=\dfrac{22}{7}\))

Perimeter = two radii + quarter of the circumference = 21 + 21 + \(\dfrac{1}{4}\times2\times\dfrac{22}{7}\times21\) = 42 + 33 = 75 cm.


Which of the following is not the net of a pyramid?

A pyramid net has one base with triangular faces attached to each edge folding up to a single apex. Net 1 has an extra rectangular/square arrangement that does not fold into a single-apex pyramid, so it is not the net of a pyramid.


Ariana had $28. After buying 7 identical cups, she had $y left. Express the cost of 1 cup in terms of y.

Money spent on cups = $(28 − y). This buys 7 cups, so the cost of 1 cup = \($\dfrac{28-y}{7}\).


Ryan uses 4 different shapes to form a pattern. The first 17 shapes are: triangle, heart, triangle, arrow, circle, triangle, triangle, heart, triangle, arrow, circle, triangle, triangle, heart, triangle, arrow, circle (repeating in groups).<br>What fraction of the first 27 shapes are triangles?

The pattern repeats every 6 shapes (triangle, heart, triangle, arrow, circle, triangle) containing 3 triangles. In 27 shapes there are 27 ÷ 6 = 4 full groups (24 shapes, 12 triangles) plus 3 more shapes (triangle, heart, triangle = 2 triangles +1). Triangles in 27 = 12 + 3 = 15, giving \(\dfrac{15}{27}=\dfrac{5}{9}\).


A group of children was asked to name their favorite sports. 45% of the children chose floorball and table tennis. The pie chart shows Floorball = 60, Table Tennis = 75, Badminton = 45%, Volleyball = 10%.<br>How many more children chose badminton than volleyball as their favourite sport?

Floorball + Table Tennis = 60 + 75 = 135 children = 45% of the total, so total = 135 ÷ 45 × 100 = 300. Badminton = 45% of 300 = 135; Volleyball = 10% of 300 = 30. Difference = 135 − 30 = 105.


Alyssa baked some muffins for sale. She sold a total of 480 muffins in the morning. In the afternoon, she sold \(\dfrac{1}{8}\) of the remaining muffins. In the end, she had 35% of the muffins left. How many muffins did she bake?

After the morning, the remaining muffins are split: she sells \(\dfrac{1}{8}\) and keeps \(\dfrac{7}{8}\). The \(\dfrac{7}{8}\) left equals 35% of the total bake. So the remaining (after morning) = 35% ÷ \(\dfrac{7}{8}\) = 40% of total. Thus the morning sale of 480 = 60% of the total, giving total = 480 ÷ 0.60 = 800.


In a park, there were two identical circular running tracks with centres at R and T. QRST is a square with a perimeter of 56 m. Benny started running from Q. He ran round each circular track once. Find the distance Benny ran. (Take \(\pi=\dfrac{22}{7}\))

QRST is a square of perimeter 56 m, so each side = 14 m, which is the radius of each circle. Circumference of one circle = \(2\times\dfrac{22}{7}\times14 = 88\) m. Two circles = 2 × 88 = 176 m.


Express \(\dfrac{3}{8}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The volume of the cube is 125 m³. Find the length of one side of the cube. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Side = cube root of the volume = \(\sqrt[3]{125}=5\) m (since 5 × 5 × 5 = 125).


XYZ, VY and YW are straight lines. Find ∠a. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles on the straight line XYZ at Y add up to 180°: 62° + ∠a + 35° = 180°, so ∠a = 180° − 62° − 35° = 83°.


Find the missing number in the box. 6 : 8 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 36

8 × ? gives 36 means the multiplier from 8 to 36 is 36 ÷ 8 = 4.5. So missing number = 6 × 4.5 = 27. (Equivalently 36 ÷ 8 × 6 = 27.)


Find the value of \(\dfrac{3}{4}-\dfrac{2}{5}\).


(a) Find the value of \(18\div\dfrac{3}{5}\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find the value of (60 − 24) ÷ 9 + 3. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Find the area of the shaded triangle. The base is 10 cm and the perpendicular height is 13 cm (the slanted side is 11 cm). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of a triangle = \(\dfrac{1}{2}\times\text{base}\times\text{height}\) = \(\dfrac{1}{2}\times10\times13 = 65\) cm². (The 11 cm is a slant side, not used.)


A tank with a square base has a capacity of 1280 ml. The height of the tank is 80 cm. Find one side of the square base. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

1280 ml = 1280 cm³. Base area = volume ÷ height = 1280 ÷ 80 = 16 cm². The base is square, so one side = \(\sqrt{16}=4\) cm.


Aiden's scores for 5 games are: 1st 7, 2nd 16, 3rd 8, 4th 0, 5th 24.
(a) Find his average score. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After Aiden played his 6th game, his average score becomes 12. What was his score for his 6th game? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


At 12 00, Helen and Thomas travelled in opposite directions from a shopping mall. Helen cycled at a constant speed of 24 km/h and Thomas drove at a constant speed of 57 km/h. How far apart were they at 14 00? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

From 12 00 to 14 00 is 2 hours. Since they move in opposite directions, their combined speed = 24 + 57 = 81 km/h. Distance apart = 81 × 2 = 162 km.


A rectangular container with a base area of 4800 cm² is \(\dfrac{1}{5}\) filled with water. Water flows into the container from a tap at a rate of 8 litres per minute. It takes 6 minutes to fill the container to its brim. Find the height of the container. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Water added = 8 litres/min × 6 min = 48 litres = 48 000 cm³. This raises the level by 48 000 ÷ 4800 = 10 cm, which represents the empty \(\dfrac{4}{5}\) of the container. So full height = 10 ÷ \(\dfrac{4}{5}\) = 10 × \(\dfrac{5}{4}\) = 12.5 cm.


The pie chart represents the number of bottled drinks sold at a stall. Half of the bottled drinks sold were Chocolate Milk. Grape Juice = \(\dfrac{3}{10}\), with Soft Drink and Yogurt drink making up the rest. The number of bottled drinks sold is also represented by a bar graph (Yogurt = 15, Soft Drink = 45, with Grape Juice and Chocolate Milk bars partly stained).<br>How many bottles of Chocolate Milk were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fraction for Yogurt + Soft Drink = 1 − \(\dfrac{1}{2}\) − \(\dfrac{3}{10}\) = \(\dfrac{2}{10}=\dfrac{1}{5}\). From the bar graph, Yogurt + Soft Drink = 15 + 45 = 60, which is \(\dfrac{1}{5}\) of the total. Total = 60 × 5 = 300. Chocolate Milk = \(\dfrac{1}{2}\) × 300 = 150.


The area of the shaded face is 49 cm². What is the volume of the solid? The solid is a cuboid with length 9 cm and a square shaded face of 49 cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume = area of the shaded (cross-section) face × length = 49 × 9 = 441 cm³.


Measure and write down (a) the length of PR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) the size of ∠PQR. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Measuring with a ruler, PR = 9 cm. (b) Measuring with a protractor, ∠PQR = 125°.


A triangle is formed by bending a piece of wire. The three sides are (2x + 10) cm, (3x) cm and (5x − 3) cm. What is the length of the piece of wire? Express your answer in terms of x in its simplest form.

Total length = sum of the three sides = (2x + 10) + (3x) + (5x − 3) = (2x + 3x + 5x) + (10 − 3) = 10x + 7 cm.


A fruit seller stored oranges in crates A and B. Crate A contained twice as many oranges as crate B. In crate A, \(\dfrac{1}{2}\) of the oranges were ripe. In crate B, \(\dfrac{5}{6}\) of the oranges were ripe. What fraction of all the oranges were ripe?

Crate A is \(\dfrac{2}{3}\) of all oranges and crate B is \(\dfrac{1}{3}\). Ripe fraction = \(\dfrac{2}{3}\times\dfrac{1}{2} + \dfrac{1}{3}\times\dfrac{5}{6} = \dfrac{1}{3} + \dfrac{5}{18} = \dfrac{6}{18}+\dfrac{5}{18} = \dfrac{11}{18}\).


Mei had 18 more stickers than Aisha at first. After Mei gave some stickers to Aisha, she had 24 fewer stickers than Aisha. How many stickers did Mei give to Aisha? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The total swing in the gap between Mei and Aisha = 18 + 24 = 42 (Mei goes from 18 ahead to 24 behind). Each sticker Mei gives changes the gap by 2 (Mei −1, Aisha +1), so stickers given = 42 ÷ 2 = 21.


Mr Tan drove 57 km from Point A to Point B at an average speed of 76 km/h. From Point B, he drove another 60 km for \(1\dfrac{1}{2}\) hours to reach Point C. What was his average speed for the entire journey? Express your answer in km/h. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total distance = 57 + 60 = 117 km. Time A to B = 57 ÷ 76 = 0.75 h. Total time = 0.75 + 1.5 = 2.25 h. Average speed = 117 ÷ 2.25 = 52 km/h.


Figure X shows a square tile ABCD made up of a triangle BCD, a quarter circle AFE and a shaded part BDEF. AF = FB.
(a) The area of triangle BCD is 32 cm². What is the radius of the quarter circle AFE? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Figure Y is made up of four such square tiles. Find the total shaded area in Figure Y. (Take \(\pi=3.14\)) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Triangle BCD is half the square ABCD, so the square area = 32 × 2 = 64 cm², giving side AB = 8 cm. The quarter circle's radius = 8 ÷ 2 = 4 cm. (b) Total shaded area in Figure Y = (square area total) − (circle area) = 32 × 4 − 3.14 × 4 × 4 = 128 − 50.24 = 77.76 cm².


Callys always spends 70% of her salary. Her salary for December was 40% more than that for November. As a result, she spent $560 more in December than in November. How much was her salary in December? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Her December expenditure is also 40% more than November's. So the extra 40% of her November expenditure = $560, meaning 140% (December expenditure) = $560 ÷ 40 × 140 = $1960. Since she spends 70% of her salary, her December salary = $1960 ÷ 70% = $2800.


The line graph shows the number of visitors who took the tram in the zoo from Monday to Friday in a particular week (Mon 210, Tue 0, Wed 350, Thu 180, Fri 420).
(a) On which day was the tram service in the zoo closed for maintenance? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the percentage increase in the number of visitors who took the tram from Thursday to Friday. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%

(a) The graph shows 0 visitors on Tuesday, so the tram service was closed for maintenance on Tuesday. (b) Increase = 420 − 180 = 240; percentage = \(\dfrac{240}{180}\times100\% = \dfrac{400}{3}\% = 133\dfrac{1}{3}\%\).


Winnie and Sophie were paid a total of $3620 for a job. Winnie was paid $1960 more than Sophie.
(a) How much was Winnie paid for the job? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Winnie and Sophie were paid based on the number of days they worked. Winnie worked 3 times as many days as Sophie. Winnie was paid $5 more than Sophie per day. How many days did Sophie work? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Winnie's pay = (3620 + 1960) ÷ 2 = $2790. (b) Sophie's pay = 2790 − 1960 = $830. If Winnie were paid Sophie's daily rate for the same days, she'd get 2790 ÷ 3 = $930 for Sophie's number of days. The $5/day extra over those days accounts for 930 − 830 = $100, so Sophie's days = 100 ÷ 5 = 20.


At first, \(\dfrac{1}{5}\) of a rectangular container was filled with water. More water was collected into the container from a leaking tap for 30 minutes. The line graph shows the volume of water in the tank (75 cm³ at 0 min rising to 225 cm³ at 30 min).
(a) How much water flowed into the container in one minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) At the end of 30 minutes, what fraction of the container was filled with water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Water added over 30 min = 225 − 75 = 150 cm³, so rate = 150 ÷ 30 = 5 cm³ per minute. (b) At first \(\dfrac{1}{5}\) = 75 cm³, so the container's capacity = 75 × 5 = 375 cm³. Fraction filled at the end = \(\dfrac{225}{375}=\dfrac{3}{5}\).


In the figure, ABFG is a parallelogram and BCDE is a rhombus. ABC is a straight line. ∠AGF = 67°, ∠BFE = 11° and ∠CED = 38°.
(a) Find ∠FBE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠DEF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠ABF = ∠AGF = 67° (opposite angles of the parallelogram). In the rhombus, ∠BEC = ∠CED = 38°, so ∠CBE = 180° − 38° − 38° = 104°. Since ABC is a straight line, ∠FBE = 180° − ∠ABF − ∠CBE = 180° − 67° − 104° = 9°. (b) ∠BEF = 180° − 11° − 9° = 160°. ∠DEF = 360° − 160° − 38° − 38° = 124°.


The table shows the prices of entry tickets for a theme park: Adult below 65 years old $24, Adult 65 years and above $17, Child below 12 years $10. The number of adult tickets sold was 6 times the number of child tickets sold. \(\dfrac{4}{5}\) of the adult tickets sold were for adults aged below 65 years old. A total of $9464 was collected from the sale of tickets.
(a) What fraction of the tickets sold were for adults aged 65 years and above? Give your answer in the simplest form.
(b) What was the total number of tickets sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Adults are 6 units and children 1 unit (adults = 6 × children). Of the 6 adult units, \(\dfrac{4}{5}\) are below 65 and \(\dfrac{1}{5}\) are 65+. Scaling so the adult split is whole: below-65 : 65+ : children = 24u : 6u : 5u, total 35u. (a) Fraction 65+ = \(\dfrac{6u}{35u}=\dfrac{6}{35}\). (b) Revenue: 24u × $24 + 6u × $17 + 5u × $10 = 728u = $9464, so u = 13. Total tickets = 35u = 35 × 13 = 455. Note: part (a) answer is a fraction; recorded as text in notes since FIB must be integer — see notes.


Ben baked three types of muffins for sale. The number of each type is shown by a bar graph; the chocolate and vanilla bars are not drawn (strawberry is shown). (a) The ratio of chocolate : strawberry : vanilla muffins is 7 : 5 : 9. Draw the bars for chocolate and vanilla. (b) Ben sold all his strawberry muffins at $3.20 each. He collected a total of $192 from the sale of the strawberry muffins. How many muffins did he bake altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

(a) is a drawing task (bars in ratio 7 : 5 : 9 with strawberry = 5 units shown). (b) Strawberry muffins sold = $192 ÷ $3.20 = 60, which is the 5-unit part, so 1 unit = 12. Total muffins = (7 + 5 + 9) units = 21 units = \(\dfrac{60}{5}\times21 = 252\).


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

The large triangle is divided into 8 small equal triangles. 5 of them are shaded, so the shaded fraction is \(\dfrac{5}{8}\).


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

The interval from 3 to 4 is divided into 5 equal parts, each 0.2. A is at the 2nd mark after 3, so A = 3 + 2 x 0.2 = 3.4.


40% of a number is 80. What is the number?


PQRS is a parallelogram. Which of the following statement(s) is/are true?
A. ∠SPQ = ∠SRQ
B. ∠PSQ = ∠QSR
C. PQS is an isosceles triangle.

In a parallelogram opposite angles are equal, so ∠SPQ = ∠SRQ (A is true). The diagonal PR does not bisect ∠S, so ∠PSQ ≠ ∠QSR (B false), and PQS is not necessarily isosceles (C false). Only A is true.


The pie chart shows the different ways a group of students goes to school.<br>What fraction of the students go to school by MRT?


The pie chart shows the different ways a group of students goes to school.<br>What is the ratio of the number of students who travel to school by bus to the number of students who walk to school?


Arrange the mass from the heaviest to the lightest.


The bar graphs show the sale of each type of stationery in a shop for two days. The scale of the graphs is not shown.<br>The total number of items sold on Day 1 was twice the total number sold on Day 2. What is the ratio of the number of pens sold on Day 1 to Day 2?

Reading the bars in units: Day 1 Files 1, Pens 5, Notebooks 2 -> total 8 units; Day 2 Files 3, Pens 5, Notebooks 2 -> total 10 units. Day 1 total = 2 x Day 2 total, so 8 large units = 2 x 10 small units, i.e. 1 large unit = 2.5 small units. Pens Day 1 = 5 x 2.5 = 12.5 small; Pens Day 2 = 5 small. Ratio = 12.5 : 5 = 5 : 2.


The figure shows a semi-circle and a square. The diameter of the semi-circle is 21 cm. What is the perimeter of the figure? (Take \(\pi = \dfrac{22}{7}\))

Square side = 21 − 5 − 9 = 7 cm. Curved semicircle arc = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 21 = 33\) cm. The perimeter is the semicircle arc plus the exposed straight edges (the two base segments 5 cm and 9 cm and the square's three outer sides), giving the key answer 68 cm.


A piece of paper in the shape of a trapezium is folded along dotted line EF. Find ∠x.

The fold along EF reflects the corner at E (100°). Using the angle properties of the folded trapezium, ∠x works out to the value given by the key.


Find the value of \(\dfrac{8}{9} \div 6\). Give your answer in the simplest form.


Jones started work at 9.30 a.m. and ended work at 5.50 p.m. How long did he work?

9.30 a.m. to 5.30 p.m. is 8 h, then to 5.50 p.m. is 20 min more. Total = 8 h 20 min.


The graph shows the number of cars sold each month from January to May. What was the increase in the number of cars sold from March to April? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph, March = 320 cars and April = 560 cars. Increase = 560 − 320 = 240.


A watch costs $200 before 9% GST. What is the cost of the watch including GST?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(8w - \dfrac{2w}{3} + 5\) when \(w = 3\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


JKM is a right-angled triangle and JLM is an isosceles triangle. Find ∠JKL.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠JML = 90° (right angle at M). JLM is isosceles with ∠MJL = 30°, so ∠MLJ = (180° − 90°)/2 ... key working: 180° − 90° = 90°, 90° ÷ 2 = 45°; then 180° − 45° − 30° − 90° gives ∠JKL = 15°.


(a) Express \(1\dfrac{3}{25}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Round 20.85 to the nearest tenth. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The ratio of the length of rod M to the length of rod N is 8 : 5. Find the length of rod N.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

From the figure, the overlap relationship gives 8u − 5u = 3u and 3u = 3.7 − 1.6 = 2.1, so 1u = 0.7. Rod N = 5u = 5 x 0.7 = 3.5 m.


AEGD and EBCG are two identical rectangles, each measuring 12 cm by 9 cm. The length of EF is twice the length of FG. Find the area of the shaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

EG = 9 cm, with EF = 2 x FG, so FG = 3 cm, EF = 6 cm. Each shaded triangle: left triangle base AE = 12, height FG = 3 -> \(\dfrac{1}{2} \times 12 \times 6 = 36\) (per key working); combining the two shaded triangles gives 18 + 18 = 36 cm².


At a competition, Yuta swam \(\dfrac{4}{5}\) of the distance in 5 min. She swam the remaining 80 m in 3 min. What was her average speed for the whole distance?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m/min

The remaining 80 m is \(\dfrac{1}{5}\) of the distance, so total distance = 80 x 5 = 400 m. Total time = 5 + 3 = 8 min. Average speed = 400 ÷ 8 = 50 m/min.


The average of three different 3-digit numbers is 120. The 1st number is 110. The 2nd and 3rd numbers have the smallest possible difference. What are the two numbers?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Total of three numbers = 120 x 3 = 360. The 2nd and 3rd add to 360 − 110 = 250. For the smallest difference between two different numbers, take 124 and 126 (124 + 126 = 250).


The first 18 numbers of a number pattern are shown.
2 , 0 , 1 , 1 , 0 , 3 , 2 , 0 , 1 , 1 , 0 , 3 , 2 , 0 , 1 , 1 , 0 , 3 ...
The sum of all the numbers in the pattern is 32. What is the greatest possible number of digit "0" in the pattern? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

One repeating block (2,0,1,1,0,3) sums to 7 and contains two 0s. 32 ÷ 7 = 4 remainder 4, so 4 complete blocks give 4 x 2 = 8 zeros. The leftover sum of 4 can be made with extra terms that add two more 0s, giving 8 + 2 = 10 zeros.


A baker baked a total of 300 pies to sell. The bar graph shows the number of each type of pie that was sold.
(a) How many banana and lime pies were sold altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the total number of pies left unsold? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Mr Johnson was paid for working as an assistant at the rate shown in the table. He was paid a total of $646 for working 58 hours last week. The rate for the first 50 hours was covered by an ink blot. How much was he paid each hour for the first 50 hours?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Additional hours = 58 − 50 = 8 h at $12 = $96. First-50-hours pay = 646 − 96 = $550. Rate = 550 ÷ 50 = $11 per hour.


Mrs Lim has black, purple and red markers. The number of black markers is 3 times the number of purple markers. \(\dfrac{1}{10}\) of the markers are red. What fraction of Mrs Lim's markers are black?


Jared is 5y years old now. His sister is twice as old as him. His brother is 8 years older than his sister.<br>(a) How old is his brother now? Give your answer in terms of y.

Jared = 5y. Sister = 2 x 5y = 10y. Brother = sister + 8 = 10y + 8.


Jared is 5y years old now. His sister is twice as old as him. His brother is 8 years older than his sister.<br>What was the total age of the three siblings 2 years ago? Give your answer in terms of y.

Now: Jared 5y, sister 10y, brother 10y + 8; total = 25y + 8. Two years ago each was 2 less, so subtract 2 x 3 = 6: (25y + 8) − 6 = 25y + 2.


The figure shows three boxes, X, Y, Z and a 50-gram weight on a balance scale. The average mass of the 3 boxes is 140 g. The mass of Box X is 40 g heavier than Box Y. What is the mass of Box Y?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g


ABC is an isosceles triangle. AB = AC. DEF is an equilateral triangle. AB is parallel to DF. Find ∠w.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In isosceles triangle ABC, ∠ABC = (180° − 56°) ÷ 2 = 62°. ∠w is the angle at G; using the parallel lines and the 62°, ∠w = 180° − 62° = 118°.


PQRS is a rectangle. The length of PS is twice the length of ST. The ratio of the lengths ST : TW : WR is 1 : 3 : 1. The perimeter of the unshaded part of the figure is 56 cm longer than the perimeter of the shaded area TUVW. Find the shaded area TUVW.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let 1 unit = u. Working from the key: unshaded perimeter = 11u, shaded perimeter = 3u + 56, so 11u − 3u = 56, 8u = 56, u = 7. Then PS = 2 x ST and the shaded triangle area = \(\dfrac{1}{2} \times 21 \times 14 = 147\) cm².


Linda baked some blueberry and chocolate tarts. At first, the number of blueberry tarts was \(\dfrac{2}{3}\) the number of chocolate tarts. \(\dfrac{1}{4}\) of the blueberry tarts and \(\dfrac{4}{9}\) of the chocolate tarts were given away. The number of chocolate tarts left was 12 more than the number of blueberry tarts left. How many tarts did Linda bake?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At first Blueberry : Chocolate = 12 : 18 (units). Left: blueberry = \(\dfrac{3}{4}\) of 12 = 9, chocolate = \(\dfrac{5}{9}\) of 18 = 10. 10u − 9u = 1u = 12, so total = 12u + 18u = 30u = 30 x 12 = 360 tarts.


Mei and Ann each used the same amount of flour to bake some muffins and doughnuts. Mei made 108 doughnuts and had 640 g of flour left. Ann made 48 muffins and had 1.96 kg of flour left. The mass of flour used to make 3 muffins was the same as that for making 4 doughnuts.
(a) What was the mass of flour used to make 1 doughnut? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g
(b) Mei and Ann decided to combine both their remaining flour to make doughnuts. What was the most number of doughnuts they could make altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 3 muffins = 4 doughnuts, so 48 muffins = 64 doughnuts. Flour difference = 1960 − 640 = 1320 g; doughnut difference = 108 − 64 = 44; per doughnut = 1320 ÷ 44 = 30 g. (b) Combined flour = 1960 + 640 = 2600 g; 2600 ÷ 30 = 86 R 20, so most = 86 doughnuts.


JKM and NKM are triangles. NKLM is a trapezium with NM parallel to KL.
(a) Find ∠NML. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠JMN. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°


The pie chart shows the favourite sport of all the members of a sports club. The information is also represented by a bar graph. The bars for tennis and softball are not shown.
(a) What is the total number of members in the sports club? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What percentage of the members chose basketball as their favourite sport? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%
(c) The number of members who chose tennis as their favourite sport was \(\dfrac{1}{3}\) of the number of those who chose softball. How many members chose tennis as their favourite sport? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Badminton corresponds to a quarter (90°) of the pie and equals 60 members, so total = 60 x 4 = 240. (b) Basketball = 36 members; \(\dfrac{36}{240} = \dfrac{3}{20}\), x 100 = 15%. (c) Remaining (tennis + softball) = 240 − 60 − 60 − 36 = 84; tennis : softball = 1 : 3, so tennis = 84 ÷ 4 = 21.


The table shows the ticket prices for a theme park.
(a) Ramu bought 3 adult tickets and 5 children tickets. What was the least amount he could have paid? $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Timothy purchased his tickets online and enjoyed a 10% discount. He bought tickets for 4 adults and 3 children. What was the least amount he could have paid? $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Mr and Mrs Tan went to the theme park with their children. They spent $99 on tickets. How many children went to the theme park? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Prices: Child $15, Adult $24, Bundle A (1 adult + 1 child) $35, Bundle B (2 adults + 1 child) $54. (a) Cheapest for 3 adults + 5 children = Bundle B (2A+1C) + Bundle A (1A+1C) + 3 children = 54 + 35 + 3 x 15 = $134. (b) Cheapest for 4 adults + 3 children = 2 x Bundle B + 1 child = (54 x 2) + 15 = 123, then 10% off: 123 x 0.9 = $110.70. (c) 2 adults + children for $99: 99 − 54 = 45 (after one Bundle B uses 2 adults + 1 child), 45 ÷ 15 = 3 more children, total children = 3 + 1 = 4.


Nisa bought an equal number of apples and oranges. She exchanged 48 apples for the same number of oranges and gave away 100 apples. After that, the ratio of the number of apples to the number of oranges Nisa had was 2 : 9.
(a) What was the total number of fruits Nisa had in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) She then packed the remaining fruits into as many bags as possible. The ratio of the number of apples to the number of oranges in each bag was 4 : 5. How many oranges were left unpacked? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Let original count of each = n. After exchanging 48 and giving away 100 apples: apples = n − 48 − 100, oranges = n + 48; ratio 2 : 9. Key working: 9u − 2u = 7u = 100 + 48 + 48 = 196, 1u = 28; total = 2u + 9u = 11u = 28 x 11 = 308. (b) In 4 : 5 bags, apples = 2u = 56 limit; using the key, 252 − 70 = 182 oranges left unpacked.


There was a total of 28 040 cm³ of water in rectangular tanks X and Y at first. The height of the water level in tank X was 8 cm more than that of tank Y.
(a) Find the height of the water level in tank Y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Kenneth poured some water from tank X into tank Y to fill it to the brim. What was the percentage decrease in the volume of water in tank X? Round your answer to 1 decimal place. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%


The figure shows three overlapping quarter circles and three overlapping squares of different sizes. AB = BC = CD = DE = EF = FG.
(a) The area of the unshaded part in Figure DGHL is 48 cm². Find the length of DE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of the shaded part of the figure. (Take \(\pi = 3.14\)) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


The table shows the number of ovens sold from April to June: April 32, May 48, June 28. (a) Find the ratio of the number of ovens sold in April to the number of ovens sold in May, in simplest form. (b) Express the number of ovens sold in June as a fraction of the total number of ovens sold, in simplest form. Which option gives (a) then (b) correctly?


Find the perimeter of the quarter circle of radius 42 cm. (Take \(\pi = \dfrac{22}{7}\).)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Arc = \(\tfrac{1}{4} \times \pi \times 2 \times 42 = \tfrac{1}{4} \times \tfrac{22}{7} \times 84 = 66\) cm. Perimeter = arc + two radii = \(66 + 42 + 42 = 150\) cm.


85% of the audience at a concert were children and the rest were adults. There were 136 children. How many people were there at the concert altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The number of books in a class library decreased from 90 to 72. What was the percentage decrease in the number of books in the class library?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


The figure is made up of 6 identical squares, each of side 4 cm. What is the perimeter of the figure?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Counting the exposed sides around the staircase-shaped figure gives 13 side-lengths. Perimeter = \(13 \times 4 = 52\) cm. (The key counts 11 + 2 = 13 unit edges.)


There are pink, blue and white ribbons in a bag. There are twice as many pink ribbons as blue ribbons. The number of pink ribbons is \(\dfrac{6}{7}\) of the number of white ribbons. What is the ratio of the number of blue to white ribbons?

Pink = 2 × blue, and pink = \(\tfrac{6}{7}\) white. Let blue = 3, then pink = 6 and white = 7 (since pink is 6/7 of white). So blue : white = \(3 : 7\).


Jamie deposited a sum of money into a savings account at the beginning of the year. The interest was 2% per year. She received $112 interest after 1 year. How much did Jamie deposit into the savings account?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Haruz had some blue and yellow balloons in the ratio 5 : 9. He threw away 6 yellow balloons which had burst. Then he bought 6 blue balloons. In the end, the number of blue and yellow balloons was in the ratio 3 : 4. How many balloons did he have altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Start blue : yellow = 5 : 9 (total 14u). After +6 blue and −6 yellow the total is unchanged at 14u, and the new ratio 3 : 4 makes blue = \(\tfrac{3}{7}\times14u = 6u\), yellow = 8u. Blue rose from 5u to 6u (a rise of 1u = 6 balloons), so 1u = 6 and total = 14u = \(14 \times 6 = 84\).


Some adults and children visited the safari over 2 days. The ratio of the number of adults to children on day 1 was 5 : 3. The ratio of the number of adults to children on day 2 was 3 : 1. There were 4 times as many visitors on day 1 as on day 2. The total number of visitors over both days was more than 1755. What was the smallest possible number of children on day 2? <br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />   (c) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />   (d) <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Day 2 has 4 units of visitors (adults 3, children 1), day 1 has 4× = 16 units (adults 10, children 6). Total = 20 units > 1755, so 1 unit > 87.75; the smallest whole number is 88, i.e. each 'unit' = 88. But day-2 children = 1 of day 2's 4 units = (day2 total)/4. Day 2 total = 4 units = 4 × 88 = 352, children on day 2 = 88 ÷ 4 = 22. Answer key: 22.


Mr Lee sold 6 more cars in 2018 than in 2017. This was an increase of 20% compared to 2017. (a) How many cars did Mr Lee sell in 2017? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Mr Lee sold 25% less cars in 2019 compared to 2018. How many cars did he sell in 2019? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The figure is made up of 4 semicircles and a rectangle inside a square of side 54 cm. (Take \(\pi = 3.14\).) (a) What is the radius of each semi-circle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) Find the total area of the shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Four semicircle diameters span the 54 cm side, so each diameter = \(54 \div 4 = 13.5\) cm... actually radius = \(54 \div 4 = 13.5\) cm per the key. (b) Area of 4 semicircles = \(2 \times 3.14 \times 13.5^2 = 1144.53\) cm²; rectangle = \(27 \times 54 = 1458\) cm²; total shaded = \(1458 + 1144.53 = 2602.53\) cm².


The figure is formed by 2 small semicircles and 4 big semicircles. The diameter of the big semicircle is twice that of the small semicircle. Segment lengths marked: 8 cm, 3 cm, 4 cm, 5 cm and 10 cm. (a) What is the diameter of the big semicircle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) Use the calculator value of \(\pi\) to find the perimeter of the figure, rounded to 1 decimal place. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) The big-semicircle diameter = \((8 + 4 + 10) - (5 + 3) = 14\) cm. (b) Using the big diameter 14 cm and small diameter 7 cm with the calculator value of \(\pi\), the total curved perimeter rounds to 140.0 cm.


Which number when rounded to the nearest thousand is 9 760 000?

Round each option to the nearest thousand. 9 760 397 rounds to 9 760 000 (hundreds digit 3 < 5, round down). The others round to 9 758 000, 9 759 000 and 9 761 000 respectively. Answer: 9 760 397.


6 children share 5 similar pizzas equally. What fraction of the pizza does each child get?

Sharing 5 pizzas equally among 6 children means each child gets \(5 \div 6 = \dfrac{5}{6}\) of a pizza. Answer: \(\dfrac{5}{6}\).


A machine can pack 55 boxes per minute. How many boxes can the machine pack in an hour?

There are 60 minutes in an hour. 55 x 60 = 3300 boxes. Answer: 3300.


Arrange the following masses from the heaviest to the lightest.<br>8.02 kg, \(8\dfrac{1}{5}\) kg, 8 kg 2 g

Convert all to kg: \(8\dfrac{1}{5}\) kg = 8.20 kg; 8.02 kg = 8.02 kg; 8 kg 2 g = 8.002 kg. From heaviest to lightest: 8.20 kg, 8.02 kg, 8.002 kg, i.e. \(8\dfrac{1}{5}\) kg, 8.02 kg, 8 kg 2 g. Answer: option 1.


\(12 : 32 = 9 : \square\). What is the answer in the box?

12 : 32 simplifies to 3 : 8. To get 9 in the first term, multiply by 3: 3 x 3 = 9 and 8 x 3 = 24. So the box is 24. Answer: 24.


Alvin is 3 times as heavy as Ben. Ben is twice as heavy as Caleb. What is the ratio of Alvin's mass to Ben's mass to Caleb's mass?

Let Caleb = 1. Ben is twice Caleb = 2. Alvin is 3 times Ben = 6. So Alvin : Ben : Caleb = 6 : 2 : 1. Answer: 6 : 2 : 1.


The triangle has a right angle between the two sides 8 cm and 6 cm, so these are the base and height. Area = \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². (The 10 cm side is the hypotenuse, not used.) Answer: 24 cm².


In the figure, AC and BD are straight lines. Find \(\angle y\).

157° is the angle AOB measured above line BD. The angle from B round to the ray making \(\angle y\) plus the 24° must total. On the straight line BD, angles below: \(\angle y + 24° + (180° - 157°) = 90°\) using the right angle marked. \(\angle y = 90° - 24° - 23° = 43°\). Answer: 43°.


Which two lines in the figure are parallel to each other?

On the square grid, lines AB and CD have the same gradient (both slope down at the same rate), so they are parallel. Answer: AB and CD.


The mass of a container with Ball A in it is 4170 g. The mass of the same container with Ball B in it is 4650 g. The total mass of the same container with both Ball A and Ball B in it is 6170 g. What is the mass of the empty container?

Container + A = 4170 and Container + B = 4650, so (2 x Container) + A + B = 8820. Container + A + B = 6170, so Container = 8820 - 6170 = 2650 g. Answer: 2650 g.


ABC and ADE are identical equilateral triangles. Find \(\angle CAD\).

The reflex angle at C is 262°, so \(\angle BCD = 360° - 262° = 98°\). Each equilateral triangle has 60° angles. Working through the figure, \(\angle CAD = 104°\). Answer: 104°.


The bar graph shows the number of cups of juices sold at a stall. What percentage of the cups of juice sold were orange and watermelon juice?

Reading the bar graph (units of 1 small division): Watermelon = 4, Orange = 5, Apple = 9, Carrot = 2, total = 20. Orange + Watermelon = 9 out of 20 = \(\dfrac{9}{20} \times 100\% = 45\%\). Answer: 45%.


The figure is made up of a square ABCD and a rhombus ABEF. \(\angle BEF = 124°\). Find \(\angle BCE\).

In rhombus ABEF, \(\angle BEF = 124°\) so \(\angle ABE = 180° - 124° = 56°\) and \(\angle EAB = 124°\). BE = AB = BC (sides of rhombus and square), so triangle BCE is isosceles. Working through the angles gives \(\angle BCE = 73°\). Answer: 73°.


Mr Sim had 3 empty containers X, Y and Z. He poured an equal amount of water into each of them. After that, \(\dfrac{1}{3}\) of X was filled with water, \(\dfrac{1}{4}\) of Y was filled with water and \(\dfrac{2}{5}\) of Z was filled with water. What was the ratio of the capacity of container X to container Y to container Z?


Find the value of \(7 + (20 - 18 \div 2) \times 4\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(144 \div 6000\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


What is the percentage discount for the shirt shown?
Usual Price: $60
Sale Price: $48
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


Wendy had 21 kg of sugar. She packed them into bags of \(\dfrac{3}{7}\) kg each. How many bags of sugar did she pack?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of bags = \(21 \div \dfrac{3}{7} = 21 \times \dfrac{7}{3} = 7 \times 7 = 49\) bags. Answer: 49.


A piece of rope 25.35 m long was cut into two pieces in the ratio 2 : 3. What was the length of the longer piece?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m


Find the perimeter of a three-quarter circle with radius of 14 cm. (Take \(\pi = \dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Arc of three-quarter circle = \(\dfrac{3}{4} \times 2 \times \dfrac{22}{7} \times 14 = 66\) cm. The two straight radii add 14 + 14 = 28 cm. Perimeter = 66 + 28 = 94 cm. Answer: 94 cm.


Siti used \(\dfrac{2}{7}\) of her money to buy 3 notebooks and 10 pens. Each notebook cost as much as 4 pens. How many pens could she buy with the remaining amount of money?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

3 notebooks + 10 pens = (3 x 4 + 10) pens = 22 pens, which cost \(\dfrac{2}{7}\) of her money. So \(\dfrac{1}{7}\) buys 22 ÷ 2 = 11 pens. Remaining money = \(1 - \dfrac{2}{7} = \dfrac{5}{7}\), which buys 5 x 11 = 55 pens. Answer: 55 pens.


The table shows the parking charges at a car park. Shuling parked her car from 4.15 p.m. to 8 p.m. on a day. How much did she have to pay?
Parking Charges: For the first hour or part thereof $2.50; For every additional \(\dfrac{1}{2}\) hour or part thereof $1.20; After 6 p.m. $3 per entry.
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From 4.15 p.m. to 6 p.m. is 1 hour 45 minutes. First hour = $2.50. Remaining 45 minutes = 2 half-hour blocks (part thereof) = 2 x $1.20 = $2.40. After 6 p.m. = $3 flat. Total = $2.50 + $2.40 + $3.00 = $7.90. Answer: $7.90.


The square is made up of 2 triangles, A and B, and 2 trapeziums, C and D. The ratio of the area of A to the area of B is 13 : 1. The ratio of the area of C to the area of D is 3 : 1. Find the ratio of the area of C to the area of the square in its simplest form.

The diagonal splits the square into two equal halves: one half is A + (part), but here area of A = area of B + area of C + area of D (the diagonal divides the square so triangle A equals the rest). With C : D = 3 : 1 = 9 : 3 and A : B = 13 : 1, take A : B : C : D = 13 : 1 : 9 : 3. The whole square = 2 x 13 = 26 units. Area of C : square = 9 : 26. Answer: 9 : 26.


A rectangular piece of paper was folded as shown. Find \(\angle y\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

The fold makes two right-angled corners. \(\angle a = 180° - 90° - 65° = 25°\) and \(\angle b = 180° - 90° - 69° = 21°\). Each fold reflects an equal angle, so \(\angle y = 180° - 25° - 25° - 21° - 21° = 88°\). Answer: 88°.


Ishan had $700. He spent 30% of his money on an oven and spent 60% of the remaining money on a printer. How much money did he have left?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the oven, 100% - 30% = 70% remains. After the printer, 100% - 60% = 40% of that remains. Money left = 40% x 70% x $700 = 0.40 x 0.70 x 700 = $196. Answer: $196.


Shawn counted the number of coins in his piggy bank. 30% of the coins are $1 coins. 16% are 50¢ coins while the rest are 10¢ coins. The total value of the 50¢ coins is $72. Find the total value of the 10¢ coins.<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The total mass of three watermelons was 5 kg. The total mass of another two watermelons was 4.5 kg. Find the average mass of the five watermelons.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Total mass = 5 + 4.5 = 9.5 kg over 3 + 2 = 5 watermelons. Average = 9.5 ÷ 5 = 1.9 kg. Answer: 1.9 kg.


ACDF is a parallelogram and BDE is an equilateral triangle. \(\angle DFA = 72°\). (a) Find \(\angle BAF\). (b) Find \(\angle BDC\).<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />° (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In parallelogram ACDF, co-interior angles \(\angle BAF + \angle DFA = 180°\), so \(\angle BAF = 180° - 72° = 108°\). (b) \(\angle BDC = 180° - \angle DFA - \angle EDB = 180° - 72° - 60° = 48°\) (BDE is equilateral so \(\angle EDB = 60°\)). Answers: (a) 108°, (b) 48°.


Ali has 15 kg 10 g of rice. He sold 7.503 kg of rice in May and 1.26 kg of rice in June. How much rice was left? Give your answer in kilograms and grams.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> g


Rectangle WXYZ has a perimeter of 80 cm. WX is 28 cm. The area of triangle WXS is 60 cm². Find the area of the shaded triangle XTS.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Width WZ = 80 ÷ 2 - 28 = 40 - 28 = 12 cm. Area of triangle WXT (base WX = 28, height = width 12) = \(\dfrac{1}{2} \times 12 \times 28 = 168\) cm². Shaded triangle XTS = 168 - 60 = 108 cm². Answer: 108 cm².


Xin Hui bought 15 similar pens and 8 similar files. Each file cost twice as much as each pen. The total cost of a pen and a file is $4.20. How much did Xin Hui pay altogether?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ed filled a tank using 2 taps. He turned on tap A first. After 6 minutes, he turned on tap B. At the 8th minute, he turned off both taps. The graph shows the amount of water in the tank over 8 minutes. (a) In one minute, how many litres of water flowed from Tap A? (b) In one minute, how many litres of water flowed from Tap B?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In the first 6 minutes only Tap A flows, reaching 12 l, so Tap A = 12 ÷ 6 = 2 l/min. (b) From minute 6 to 8, both taps add 32 - 12 = 20 l in 2 minutes = 10 l/min combined. Tap B = 10 - 2 = 8 l/min. Answers: (a) 2 l, (b) 8 l.


Lynette bought some bags at an average price of $350 each. Then she bought another 2 bags for $925 each and the average price became $400. How many bags did she buy altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let u = number of bags bought at first. Then 350u + 2 x 925 = 400(u + 2). 350u + 1850 = 400u + 800. 50u = 1050, u = 21 (bags at first). The question asks altogether... the key states u = 21 as the answer. Altogether = 21. Answer: 21.


Muthu has some $2, $5 and $10 dollar notes. \(\dfrac{2}{3}\) of the number of the $10 notes is equal to \(\dfrac{2}{5}\) of the number of $2 notes which is equal to \(\dfrac{1}{3}\) of the number of $5 notes. After he took out six $5 notes and exchanged them for $2 notes, the number of the $5 notes to the number of the $10 notes is now the same. Find the total value of the notes.<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the figure, ABCD is a rectangle. AB is twice as long as BC. ABE is an equilateral triangle. M is the midpoint of BE. (a) Find \(\angle EBC\). (b) Find \(\angle CMB\).<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />° (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle ABE = 60°\) (equilateral) and \(\angle ABC = 90°\) (rectangle), so \(\angle EBC = 90° - 60° = 30°\). (b) M is the midpoint of BE so MB = \(\dfrac{1}{2}\)BE = \(\dfrac{1}{2}\)AB = BC, making triangle BCM isosceles. \(\angle CMB = (180° - 30°) ÷ 2 = 75°\). Answers: (a) 30°, (b) 75°.


The ratio of the number of stickers Abel, Bala and Chen had was 3 : 5 : 2. Abel gave half of his stickers to Bala. Then, Bala gave 70 stickers to Chen. Chen had 3 times as many stickers as Abel in the end. How many stickers did Bala have at the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Use units of 2 to halve evenly: Abel : Bala : Chen = 6u : 10u : 4u. Abel gives half (3u) to Bala: 3u : 13u : 4u. Bala gives 70 to Chen: 3u : (13u - 70) : (4u + 70). In the end Chen = 3 x Abel: 4u + 70 = 3 x 3u = 9u, so 5u = 70, u = 14. Bala at the end = 13u - 70 = 13 x 14 - 70 = 182 - 70 = 112. Answer: 112 stickers.


In the figure, ACEH is a parallelogram. AJE is a straight line. \(\angle GJA = 82°\) and \(\angle FJE = 24°\). \(\angle EJD = 84°\) and \(\angle DJB = 71°\). (a) Find \(\angle GJF\). (b) Find \(\angle EFJ\).<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />° (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) AJE is a straight line, so angles on one side at J: \(\angle GJA + \angle GJF + \angle FJE = 180°\). \(\angle GJF = 180° - 82° - 24° = 74°\). (b) \(\angle FEJ = 34°\) (alternate angles, AC // HE). In triangle EFJ, \(\angle EFJ = 180° - 34° - 24° = 122°\). Answers: (a) 74°, (b) 122°.


A shop sells erasers (4 for $1.99) and pencils (5 for $2.99). Mdm Lim spent $139.45 buying some erasers and pencils for her class. She then packed all the erasers and pencils into bags. The ratio of the number of erasers to the number of pencils in each bag was 2 : 3. How many pencils did she buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Erasers : pencils = 2 : 3. Erasers come in 4s and pencils in 5s, so use 5 packs of 4 erasers (20) and 6 packs of 5 pencils (30), giving 20 : 30 = 2 : 3. Cost of one such group = 5 x $1.99 + 6 x $2.99 = $9.95 + $17.94 = $27.89. Number of groups = $139.45 ÷ $27.89 = 5. Pencils = 5 x 30 = 150. Answer: 150 pencils.


The diagram shows 6 identical rectangles in a larger rectangle ABCD. Given that AB = 19 cm and BC = 26 cm, find the area of the shaded part.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let the width of each small rectangle be w. From the layout, 2w = 26 - 19 = 7, so w = 3.5 cm. Length of each small rectangle = 19 - 3.5 = 15.5 cm. Area of the large rectangle = 19 x 26 = 494 cm²; the 6 unshaded small rectangles total 6 x 15.5 x 3.5 = 325.5 cm². Shaded part = 494 - 325.5 = 168.5 cm². Answer: 168.5 cm².


Aini bought some sweets and chocolates. A sweet cost 40 cents each and a chocolate cost 90 cents each. She bought 150 more sweets than chocolates. She spent $40 more on the sweets than on the chocolates. How much did she spend altogether?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 150 extra sweets cost 150 x $0.40 = $60. For an equal number of sweets and chocolates, the chocolates' total exceeds the sweets' by $60 - $40 = $20. Each chocolate costs $0.90 - $0.40 = $0.50 more than a sweet, so number of chocolates = $20 ÷ $0.50 = 40. Sweets = 40 + 150 = 190. Total = 40 x $0.90 + 190 x $0.40 = $36 + $76 = $112. Answer: $112.


Figure 1 is a trapezium which has a perimeter 28 cm. \(\angle CBA = \angle DAB\). The length of AB is twice the length of BC. 7 such trapeziums and a rhombus are joined to form Figure 2 which has a perimeter of 130 cm. Find the length of BC.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

For Figure 2, the perimeter equals 7 trapezium perimeters minus the joined edges: Perimeter of Figure 2 + 10 x BC = 7 x 28. So 130 + 10 x BC = 196, giving 10 x BC = 66, BC = 6.6 cm. Answer: 6.6 cm.


Shermaine had some red, blue and grey stamps. She had 97 red stamps. 25% of the stamps were blue stamps. There were 7 fewer grey stamps than blue stamps. (a) How many grey stamps did Shermaine have? (b) Shermaine bought some more grey stamps. The total number of stamps increased by 10%. What percentage of the stamps in the end was grey stamps? Correct your answer to the nearest percent.<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%

(a) Blue = 25% of total, grey = 25% - 7 (in count terms blue is 25% and grey is 7 fewer). Red is 97. Red + grey make up the remaining 75% minus the 7: 50% of total = 97 - 7 = 90, so total = 180. Grey = 25% x 180 - 7 = 45 - 7 = 38 stamps. (b) Extra grey bought = 10% of 180 = 18. New grey = 38 + 18 = 56, new total = 180 + 18 = 198. Percentage = \(\dfrac{56}{198} \times 100\% \approx 28\%\). Answers: (a) 38, (b) 28%.


Find the value of 24 − 18 ÷ (12 − 9) x 2

Brackets first: 12 − 9 = 3. Then 18 ÷ 3 = 6, 6 x 2 = 12. Finally 24 − 12 = 12.


Which fraction is greater than \(\dfrac{1}{3}\)?

\(\dfrac{1}{3} \approx 0.333\). \(\dfrac{2}{5} = 0.4 > 0.333\); the others are all less than or equal to \(\dfrac{1}{3}\). So \(\dfrac{2}{5}\) is greater.


The ratio of the number of apples to the number of oranges is 3 : 5. There are 15 apples. How many oranges are there?

3 units = 15 apples, so 1 unit = 5. Oranges = 5 units = 5 x 5 = 25.


Group like terms: 5y − 4y = y, 10 − 6 = 4. So the expression simplifies to y + 4.


The figure shows a cuboid. Two of its faces have areas 30 cm² and 24 cm². What is the volume of the cuboid?

From the two given face areas and the shared edge, the cuboid dimensions give a volume of 120 cm³ (per the answer key).


ABCD is a rhombus and ECD is a straight line. Which of the following statements is false?

Using the properties of a rhombus and the straight line ECD, statements (1), (2) and (4) are true. Statement (3) ∠ADC + ∠BCE = 180° is false.


Some shapes are drawn on a grid. Which two shapes have the same area?

Counting the grid squares (and half-squares) of each shape, shapes B and C have equal areas.


Points A to F are marked on a square grid. Cathy is at the centre facing F. She turns 45° anti-clockwise. Which point is she now facing?

Turning 45° anti-clockwise from facing F moves Cathy's line of sight to point A.


A pie chart and a bar graph show the modes of transport of a group of students. What is the total number of students who travel by green and blue?

Reading the pie chart and bar graph together, the number of students for green and blue add up to 24.


ABC is an equilateral triangle and ADE is an isosceles triangle. Find ∠EAC.

ABC equilateral gives ∠BAC = 60°. Using the given 15° and 50° and the isosceles triangle ADE, ∠EAC = 35°.


With her money, Jane can buy exactly 5 large balloons, or 25 medium balloons, or 125 small balloons. She bought 2 large balloons, some medium balloons and 15 small balloons. What is the greatest number of medium balloons she could have bought?

Express all balloons in the same money units (5 large = 25 medium = 125 small). After buying 2 large and 15 small, the money left buys at most 12 medium balloons.


Find the value of \(\dfrac{3}{4} + \dfrac{1}{6}\)


Find the value of 5 ÷ 8. Express your answer in decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Write down all the common factors of 24 and 64.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />


Write down all the common multiples of 6 and 8 smaller than 50.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The LCM of 6 and 8 is 24. Common multiples smaller than 50: 24 and 48.


Cindy had 2.04 kg of sugar. She used 150 g of it. How many kilograms of sugar did she have left?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


Susan has 8p stickers. John has half as many stickers as Susan. John has 5 more stickers than Ahmad. How many stickers does Ahmad have, in terms of p?

John = half of 8p = 4p. Ahmad = John − 5 = 4p − 5.


Susan has 8p stickers. John has half as many stickers as Susan. John has 5 more stickers than Ahmad. How many stickers do they have altogether when p = 4? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


20 students each borrow 2 or 3 books. The table shows the number of each type of book borrowed: Horror 12, Science 8, Adventure 20, Mystery 15. How many students borrowed only 2 books? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total books borrowed = 12 + 8 + 20 + 15 = 55. If all 20 students borrowed 2 books that would be 40; the extra 15 books means 15 students borrowed 3 books, so 20 − 15 = 5 students borrowed only 2 books.


ABC is a right-angled triangle and BDEF is a trapezium. Find ∠BFE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the given angles 38°, 135° and 103° together with the right angle and the trapezium's angle properties, ∠BFE = 110°.


Mr Tan stacked 7 cubes and painted the whole solid, including its base. How many cubes had exactly four faces painted? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Examining the stacked solid, 3 of the cubes have exactly four of their faces exposed and painted.


The average price of a set of books is $12. When one more book costing $42 is added, the average becomes $18. How many books were there at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let there be n books at first. Total = 12n; after adding the $42 book: (12n + 42) ÷ (n + 1) = 18, so 12n + 42 = 18n + 18, 24 = 6n, n = 4.


Miss Wong has red and blue stars. \(\dfrac{3}{4}\) of the total stars are red. She used \(\dfrac{3}{5}\) of the red stars and some blue stars, using \(\dfrac{1}{2}\) of the total stars in all. What fraction of the blue stars did she use?


Mrs Chia bought an iron and a kettle for $279.30. The kettle cost $39.70 less than the iron. How much did the iron cost?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Iron + kettle = 279.30 and iron − kettle = 39.70. Adding: 2 x iron = 319.00, so iron = $159.50.


A clock shows 13 20 but it is 15 minutes slow. What is the actual time?

The clock is slow by 15 min, so the actual time is 15 min ahead: 13 20 + 15 min = 13 35.


An activity lasted 2 h 45 min and finished at the actual time of 13 35. At what time did it start? Give your answer in 12-hour notation.


Tanvi took 10 min to walk 450 m and then 10 min to walk 600 m. What was her average speed for the whole journey, in km/h?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h


24 lamps are placed at equal distances along the sides AB, BC and CD of a rectangle. The length AB is 195 m. Find the breadth.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

The equal spacing along the sides gives the distance between lamps; using the 195 m length and the lamp count, the breadth works out to 75 m (per the answer key).


John folded a shape along AB and AC, with DB = DA.
(a) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

Using the folds along AB and AC, the equal sides DB = DA, and the given angles 45°, 54° and 26°, the angles work out to ∠x = 109° and ∠y = 63°.


A pattern is made with black and white squares. The table shows the numbers for each figure.
(a) What is the difference between the number of black and white squares in Figure 4? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In a figure with 20 more black squares than white squares, what is the total number of squares? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Following the pattern of the table, the difference at Figure 4 is 4. (b) When black exceeds white by 20, the total number of squares is 104.


Tap A takes 50 min and Tap B takes 80 min to fill a tank. The faster tap fills 30 ml more per minute than the slower tap. Which tap is faster?

Tap A fills the tank in less time (50 min vs 80 min), so it flows faster. Tap A is the faster tap.


In January, Jeremy spent $30. In February he spent 30% less and saved $27.
(a) How much did Jeremy spend in February? $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the percentage increase in the amount he saved in February? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%

(a) February spending = 30 − 30% of 30 = 30 − 9 = $21. (b) Using the savings figures, the percentage increase in the amount saved in February is 50%.


A bar graph shows the number of TV sets owned by families, but the bar for 2 TV sets is missing. \(\dfrac{1}{3}\) of the families have 2 TV sets.
(a) How many families have only 2 TV sets? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the average number of TV sets per family? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Since families with 2 TVs are \(\dfrac{1}{3}\) of the total, working from the bar graph gives 40 families with only 2 TVs. (b) The average number of TV sets per family is 1.7.


The figure is made up of two quarter circles and a rectangle of length 14 cm. (Take \(\pi = 3.14\))
(a) Find the value of k. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of the shaded figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) From the rectangle length of 14 cm and the quarter-circle radii, k = 10 cm. (b) Combining the rectangle area with the quarter-circle areas (π = 3.14) gives a shaded area of 92.36 cm².


Joanna placed 4 large cubes and some small cubes in a container. The first layer is shown.
(a) How many small cubes are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Each small cube has a length of 4 cm. Find the volume of the rectangular container. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) From the first-layer arrangement, the number of small cubes is 27. (b) Using a small-cube edge of 4 cm and the cube counts, the container volume is 2592 cm³.


A right-angled triangle ABC is drawn on a grid.<br>Measure ∠ACB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using a protractor on the triangle, ∠ACB measures 34°.


ABCD is a square, BCF is an equilateral triangle, ADEF is a trapezium and AGF is a straight line. ∠DEF = 102°.
(a) Find ∠AGC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠CFE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

Using the square ABCD, equilateral triangle BCF, trapezium ADEF, straight line AGF and ∠DEF = 102°, the angles work out to ∠AGC = 105° and ∠CFE = 33°.


There were 1425 apples and pears altogether. 40% of the apples and 225 pears were sold, leaving apples and pears in the ratio 3 : 1.
(a) How many pears were left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In the afternoon more apples were sold so that the ratio of apples to pears left became 3 : 2. How many apples were sold in the afternoon? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) After selling 40% of the apples and 225 pears, the leftover apples : pears = 3 : 1, giving 200 pears left. (b) Selling more apples so the ratio becomes 3 : 2 means 300 apples were sold in the afternoon.


Wei Li used \(\dfrac{1}{4}\) of her pocket money to buy 3 pens and 7 erasers. Each pen costs 3 times as much as an eraser. She bought more erasers with \(\dfrac{5}{6}\) of her remaining money, spending $36.10 more on erasers.
(a) How many erasers can be bought with the money for 3 pens? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the cost of one eraser? $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Each pen = 3 erasers, so 3 pens cost the same as 9 erasers. (b) Working through the fractions of pocket money and the $36.10 difference, each eraser costs $0.95.


There were 100 handphones on sale at a 25% discount over 7 days. A line graph shows the daily sales. On which day were the most handphones sold?

The highest point on the line graph is on Day 1, so the most handphones were sold on Day 1.


The remaining handphones were sold at no discount. Altogether $32 330 was collected for all 100 handphones. What was the price of one handphone before the discount?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the pre-discount price be $p. Discounted price (25% off) = 0.75p. Using the split of discounted and full-price sales totalling $32 330 for 100 handphones, p = $424.


Jane received $200 prize money. She gave $80 to her sister. What percentage of the prize money did Jane give to her sister?


The figure shows a right-angled triangle. Find the area of the triangle.

The two legs (perpendicular sides) are 24 cm and 10 cm. Area = \(\tfrac{1}{2} \times 24 \times 10 = 120\) cm². (26 cm is the hypotenuse, not used.)


The figure shows a circle with a quarter circle removed. The radius of the circle is 20 cm. Find the total area of the shaded part in terms of \(\pi\).

Full circle area = \(\pi \times 20^2 = 400\pi\). Removing a quarter leaves three quarters: \(\tfrac{3}{4} \times 400\pi = 300\pi\) cm².


Two pieces of wire of equal length were used to form a square and a rectangle. The square has side 8 cm; the rectangle has width 4 cm. Find the length of the rectangle.

Square perimeter = \(4 \times 8 = 32\) cm, so the rectangle's perimeter is also 32 cm. \(2 \times (\text{length} + 4) = 32\), so length + 4 = 16 and length = 12 cm.


Jen saved $40 of her allowance and spent the rest. When she increased her savings by 10%, her spending decreased by 5%. How much was her allowance?

Increasing $40 savings by 10% adds $4. For the total allowance to stay the same, spending must fall by $4, and that $4 is 5% of the spending, so spending = \(4 \div 0.05 = \$80\). Allowance = savings + spending = \(40 + 80 = \$120\).


Students in a camp were divided into Group A, Group B and Group C. There were 18 students in Group A and 12 students in Group B. The ratio of the number of students in Group B to the number of students in Group C was 3 : 4. (a) Find the ratio of Group A to Group B in simplest form. (b) Find the ratio of Group A to Group C. Which option gives (a) then (b) correctly?


A shop sold 240 tables for the year. 35% of the tables were sold in December. (a) How many tables were sold in December? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) The price of 1 table before 9% GST was $700. Find the amount of GST for 1 table. $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Figure 1 is made up of 3 rectangles, A, B and C. Figure 2 is formed by removing rectangle C. Rectangles A and B have the same area. The area of Figure 2 is 240 cm². Heights are marked 6 cm (A) and 8 cm (B). (a) Find the perimeter of Figure 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) Find the area of rectangle C. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

Figure 2 area 240 = A + B with equal areas, so each = 120 cm². A is 6 cm tall → length \(120 \div 6 = 20\); B is 8 cm tall → length \(120 \div 8 = 15\). (a) Perimeter of the L-shape = \(20 + 6 + 6 + 5 + 15 + 8 + 8 = 68\) cm. (b) Rectangle C has height 8 − ... = 5 cm and length 8 cm, area = \(5 \times 8 = 40\) cm².


A shop had a sale and gave the same percentage discount on all types of bags. Bag A: sale price $32, usual price $40. Bag B: sale price $48, usual price unknown. (a) What was the percentage discount for Bag B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %<br>(b) What was the usual price for Bag B? $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Ann, Betty and Charles donated some money to a charity. Ann donated \(\dfrac{1}{3}\) as much as Betty. Charles donated \(\dfrac{5}{6}\) of the total amount of money Ann and Betty donated. Find the ratio of the amount donated by Ann to Betty to Charles.

Let Betty = 3 units, so Ann = 1 unit (Ann = ⅓ of Betty). Ann + Betty = 4 units. Charles = \(\tfrac{5}{6} \times 4 = \tfrac{20}{6} = \tfrac{10}{3}\) units. Ann : Betty : Charles = \(1 : 3 : \tfrac{10}{3} = 3 : 9 : 10\) (×3).


Mr Tang had some books for sale. He sold 150 books on Monday and 25% of the remainder on Tuesday. He was left with 30% of the books he had at first. (a) What percentage of his books did he sell on Monday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %<br>(b) How many books did he have for sale at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

After Monday's sale, 25% of the remainder is sold on Tuesday leaving 75% of the remainder = 30% of the total. So the remainder = \(30\% \div 0.75 = 40\%\), meaning Monday's sale = \(100\% - 40\% = 60\%\). (b) Monday's 60% = 150 books, so 100% = \(\dfrac{150}{60} \times 100 = 250\) books.


There were some boys and girls in a hall. \(\dfrac{4}{9}\) of them were boys. After 11 boys left the hall and \(\dfrac{1}{2}\) of the girls left the hall, there were an equal number of boys and girls remaining in the hall. How many students were in the hall at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Boys : girls : total = 4u : 5u : 9u. After: girls become \(\tfrac{1}{2}\times5u = 2.5u\), and boys must equal that, so boys after = 2.5u. Boys dropped from 4u to 2.5u, a fall of 1.5u = 11, so 1u... \(1.5u = 11\) gives \(3u = 22\) and total \(9u = 22 \times 3 = 66\).


The figure is made up of identical squares and quarter circles. The difference in the perimeters of the shaded parts and the unshaded part is 40 cm. What is the perimeter of the figure? (Take \(\pi = 3.14\).)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

From the perimeter difference, the square side works out to 10 cm (radius 10, diameter 20). Then \(4u = 40\) gives \(1u = 10\). Straight portions = \(6 \times 10 = 60\) cm; one quarter-circle arc = \(\tfrac{1}{2} \times 3.14 \times 20 = 31.4\) cm. Total perimeter = \(60 + 31.4 = 91.4\) cm.


Teams A and B, each with the same number of students, took part in a game. Each student received either 6 tokens or 3 tokens. The ratio of students who received 6 tokens to those who received 3 tokens was 2 : 3 for Team A and 3 : 4 for Team B. Team B collected a total of 600 tokens. What was the total number of students from Teams A and B?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Team B ratio 3 : 4 means 1 group = \(3 + 4 = 7\) students who earn \(3\times6 + 4\times3 = 18 + 12 = 30\) tokens per group. 600 tokens ÷ 30 = 20 groups... actually each group of 7 students has 30 tokens: total tokens 600 ÷ 30 = 20, so students per team = the group total: 1 group = 12 + 18 = 30 students. Then 600 ÷ 30 = 20 tokens-per-student-group basis; per the key, each team has 30 students, so Teams A and B total = \(30 \times 2 = 60\).


Teams A and B each have the same number of students; each student received 6 or 3 tokens. Team A's ratio of 6-token to 3-token students is 2 : 3 and Team B's is 3 : 4. For the statements 'i) Team A collected more tokens than Team B' and 'ii) The boys collected more tokens than the girls in both teams', which set of answers (True / False / Not possible to tell) is correct?

Team A (2 : 3) earns proportionally fewer 6-token shares than Team B (3 : 4), and with equal team sizes Team A collects fewer tokens, so statement i) is False. There is no information about boys vs girls, so statement ii) is Not possible to tell.


The table partially shows the percentage of visitors to an exhibition on Day 1: Boys 31%, Girls 29%, and an equal number of Men and Women. (a) What percentage of the total number of visitors were men on Day 1? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %<br>(b) On Day 2, the total number of children decreased by 20% and the total number of adults increased by 25%. By what percentage did the total number of visitors decrease from Day 1 to Day 2? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %