Tory spent \(\dfrac{1}{4}\) of her money on 5 books and 15 sheets of stickers. The cost of each book is 9 times the cost of each sheet of sticker. She bought more books with the remaining amount of money.
(a) How many books did she buy with the remaining amount of money? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Given that Tory spent $231 more on all the books than all the stickers. What is the cost of 1 sheet of sticker? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Which of the following is twenty-four thousand and thirty in numerals?

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


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


Alyssa earns $200 a day. She works 4 days each week for 12 weeks. How much money will Alyssa earn in total?

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


Mrs Tan baked 56 cookies on Monday morning and 24 cookies in the afternoon. She baked the same number of cookies on Tuesday. Which of the following represents the total number of cookies Mrs Tan baked on both days?

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


(a) Find the value of \(90\,000 \div 30\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Fill in the blank. \(850\,000 \div 50 = \) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> \(\times 5\)


Find the value of \(160 - 120 \div 4 \times 3 + 7\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A clothing company has 20 different designs of t-shirts. Each year, the company makes 3500 of each design of the t-shirts. How many t-shirts does the company make each year? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total t-shirts = 20 designs \(\times\) 3500 each = 70 000.


Sally saved $200 in January. Her father gave her $10 for every $40 she saved. How much money did Sally's father give her? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$200 \(\div\) $40 = 5 groups of $40. Father gives $10 per group, so 5 \(\times\) $10 = $50.


Mrs Lim had 30 boxes of oranges. Each box contained 20 oranges. She threw 80 rotten oranges away and sold the rest at $10 for 4 oranges. How much money did Mrs Lim receive? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Joyce had 260 more stickers than Ali. After Joyce gave away 44 stickers to her friends, she had five times as many stickers as Ali. How many stickers did Joyce have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Mr Rahmat had a total of 480 exercise books and files in his shop. He sold half of the exercise books and 105 files. In the end, he had an equal number of exercise books and files left. Find the number of files he had at first. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Mr Jaya spent a total of $4200 buying some headphones and speakers for his shop. He bought 40 more headphones than speakers. A headphone cost $30 each and a speaker cost $20 each. How many headphones did he buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


What is the value of \(11 \div 3\) rounded off to 2 decimal places?

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


The figure shows a triangle ABC. Find the height that corresponds to the base BC.

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


ABCD is a rectangle. Find the area of the shaded figure.

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


(a) \(2\,\ell\ 9\ \text{ml} =\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\)<br>(b) \(4.65\ \ell =\) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³


The figure shows a solid made up of unit cubes. How many unit cubes should be added to form the smallest possible cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of triangles. AB is thrice of AE while D is the midpoint of AC. The area of triangle AED is 10 cm². What is the area of triangle BDE? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Amir had \(\dfrac{8}{9}\,\ell\) of juice. He spilt \(\dfrac{1}{4}\) of it and drank \(\dfrac{1}{2}\,\ell\). How many litres of juice does he have left?


There were 112 more muffins than brownies in a shop. \(\dfrac{1}{5}\) of the brownies and \(\dfrac{2}{3}\) of the muffins were sold. There was an equal number of brownies and muffins left. How many muffins were there at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure shows a tank with a rectangular base of 25 cm by 8 cm.
(a) The tank is \(\dfrac{1}{3}\)-filled with water 5 cm deep. What is the height of the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) How much more water is needed to fill the tank to its brim? Leave your answer in cm³. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

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


There are \(\dfrac{1}{3}\) as many boys as girls at a carnival. Each boy received 6 tokens and each girl received 8 tokens. A total of 600 tokens were given out. How many boys were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


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


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


The figure is divided into 25 equal parts. What percentage of the figure is shaded?

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


How many minutes are there in 9 h 15 min?


Mrs Bala bought some fruits and vegetables from the market. The mass of the fruits was \(2\dfrac{7}{8}\) kg. The fruits were \(1\dfrac{1}{4}\) kg heavier than the vegetables. What was the mass of the vegetables?

Vegetables = fruits − difference = $2\dfrac{7}{8}-1\dfrac{1}{4}=2\dfrac{7}{8}-1\dfrac{2}{8}=1\dfrac{5}{8}$ kg.


Imran earns $4000 a month. He saves 30% of his salary and spends the rest. How much does he spend in a month?


The table shows the number of plates of chicken rice and egg noodles Auntie Cheng sold at a school canteen in a week.
Monday: chicken rice 69, egg noodles 44
Tuesday: chicken rice 70, egg noodles 40
Wednesday: chicken rice 47, egg noodles 72
Thursday: chicken rice 51, egg noodles 66
Friday: chicken rice 65, egg noodles 50
On which day did she sell 25 more plates of chicken rice than egg noodles?

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


\(75.038 = 75 + \dfrac{3}{A} + \dfrac{1}{B}\)<br>What are the values of A and B? A = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, B = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


There are 16 girls in a class. There are 4 more boys than girls. What is the ratio of the number of boys to the total number of students in the class?


ADE and CDE are identical triangles that overlap partially. The length of AE is 16 cm and the area of triangle BDE is 15 cm². Find the area of the whole figure.

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


Peggy was at a certain position. She walked 2 steps due north, 3 steps due east, 1 step due south and then 2 steps due west. She ended at Position X. What was her starting position?

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


Find the value of \(42 \div 8\). Give your answer as a mixed number in the simplest form.


In the number line, what is the decimal represented by A? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Express 30.07 km in metres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m


How many degrees are there in a \(\dfrac{1}{4}\)-turn? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The bar graph shows the number of Primary 5 students in each class wearing spectacles. How many students in class 5D wear spectacles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Reading the bar for class 5D on the graph gives 14 students.


Uncle George has 30 red pens, 45 green pens and 120 blue pens at his stationery shop. Find the ratio of the number of green pens to the number of red pens to the number of blue pens. Give your answer in its simplest form.

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


In a survey of 60 students, it was found that 39 of them do not own mobile phones. What percentage of the students own mobile phones? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

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


Jing Lin watched a movie last night at the time shown on the clock. The duration of the movie was 2 hours 15 minutes. Write down the time the movie ended using the 24-hour clock. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular tank measuring 30 cm by 40 cm by 70 cm was completely filled with milk. Find the volume of milk. Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\)


A photocopy machine prints 162 pages in 6 minutes. How many pages does the machine print in 10 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The total cost of 2 chairs and 1 table is $210. The total cost of 1 chair and 2 tables is $285. What is the cost of 1 chair? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Cindy was given \(\dfrac{9}{10}\) h to work on a task. She took \(\dfrac{2}{3}\) of the given time to complete the task. How much time had she left?


The cost of a pen is $1.80. Siti buys 40 such pens and gives the cashier $100. How much change does Siti get? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The graph shows the volume of water that flows from a tap. At this rate, how many litres of water will flow from the tap in 35 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\)

From the graph the rate is constant: in 7 minutes 25 ℓ flow, so 35 min = 7 min × 5 → 25 × 5 = 125 ℓ.


Ben is thrice as heavy as his sister. Ben is 57.6 kg. Find their difference in mass. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

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


AB and CD are straight lines. Find the sum of \(\angle AOD\) and \(\angle BOE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


For every 5 pens that Mrs Lee buys, she gets 1 free. If she needs 80 pens, what is the least number of pens she has to buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Ryan took 2 min 3 s to finish running a race. Iris was slower than Ryan by 33 s. How much time did Iris take to finish running the race? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> s


Mr Singh bought an equal number of apples and oranges. After he gave away 29 oranges and bought another 10 apples, he had \(\dfrac{1}{4}\) as many oranges as apples. How many oranges did he have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Minah had 30 m of string. She gave 2.7 m of the string to each of her 4 friends and used the rest to wrap some presents for a birthday party. 75 cm of string was needed for each present. Find the maximum number of presents she could wrap. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A tank was filled with water up to a height of 11 cm at first. Some water was added and the new water level was 80% of the height of the tank. The tank is 13 cm by 11 cm by 25 cm. How much water was added to it? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> \(\ell\)

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


There are 2200 students in a school. 45% of them are girls. 10% of the boys do not have any siblings. How many boys do not have any siblings? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Billy took up a part-time job and was paid by the rates: Monday to Friday $8 per hour, Saturday and Sunday $9.50 per hour. Billy worked 7 hours each day on some weekdays and 5 hours on a Saturday. He was paid $271.50 altogether. How many weekdays did he work? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Lim prepared 30 chicken wings and 45 nuggets for her students.
a) If every student took the same amount of each type of food without any leftover, what was the maximum number of students in her class? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) With the maximum number of students in her class, how many nuggets could each student take? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


PQRS is a rectangle of length 40 cm and breadth 36 cm. PR and UVR are straight lines and ST = SU = 24 cm.
a) Find the area of Triangle PRS. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
b) Find the shaded area. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

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


ABEF is a square, BCDE is a trapezium and FD is a straight line. \(\angle ABC = 140^\circ\) and \(\angle CDF = 71^\circ\).
a) Find \(\angle BED\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) Find \(\angle EFD\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The table shows the movie schedule at a cinema (Screening Now):
Marvel — start 2.45 p.m., 3.10 p.m., 3.45 p.m.; duration 1 h 25 min
Diary of Minions — start 3.00 p.m., 6.00 p.m., 8.00 p.m.; duration 2 h 30 min
Queen — start 12.10 p.m., 3.15 p.m., 7.00 p.m.; duration 1 h 32 min
Kung Fu Kid — start 3.05 p.m., 6.00 p.m., 9.05 p.m.; duration 2 h
Henry arranged to meet his friend at 2.40 p.m. to watch a movie together. However, his friend was 15 minutes late. Henry arrived at the ticketing counter on time. His father would pick him up 2 hours later, at the mall where the cinema was.
a) (i) Which movie could Henry and his friend watch from start to end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(ii) What was the start time of the movie? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(iii) What was the end time of the movie? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

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


At a friendly match, 150 spectators were adults and \(\dfrac{1}{3}\) of the children were boys.
a) Given that \(\dfrac{1}{4}\) of the spectators were girls, how many boys were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) How many spectators were at the match altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


There were 28 students in a class. The average class score for a quiz was 78 marks. One of the student's score was wrongly recorded as 43 marks. The correct average score should be 79.5 marks. What was the actual score of the student? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Yasmin uses circles and triangles to form figures that follow a pattern. Pattern 1 has 1 circle; Pattern 2 has 2 circles and 1 triangle; Pattern 3 has 3 circles and 4 triangles; Pattern 4 has 4 circles and 9 triangles.
a) Which pattern has 100 triangles? Pattern <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) Find the total number of circles and triangles in Pattern 20. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


During an overseas vacation, Rita and Sarah had the same amount of allowance. Each day, Rita spent $4 and Sarah spent $6. At the end of the vacation, Sarah had $12 left, while Rita had 4 times as much money as Sarah.
a) How many days did the vacation last? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
b) How much was each girl's allowance at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Subtract \(1\dfrac{5}{6}\) from \(3\dfrac{1}{3}\). Express your answer in its simplest form.


Convert \(4\dfrac{2}{25}\) to a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Monica cuts a piece of string into 3 pieces in the ratio of 6 : 2 : 9. The difference between the longest and shortest piece is 21 cm long. Find the length of the longest piece of string. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


28 boys and 22 girls attended the Primary 5 Leadership Camp last week. Express the ratio of the number of boys to the total number of children at the Primary 5 Leadership Camp in its simplest form.

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


Sammy leaves his house at 6.45 a.m. and walks to school. He arrives in school at 7.10 a.m. If he takes the same amount of time walking home, how much time does he spend travelling to and fro in a week? (Assume he goes to school for 5 days in a week) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

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


The solid is made up of 1-cm cubes. Find the volume of the solid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

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


The figure shows Rectangle ABCD and Triangle CEF. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Some people had gathered in the park for a Charity Walk. The ratio of the number of adults to the number of boys to the number of girls was 13 : 4 : 6. There were 16 more girls than boys.
(a) How many adults had gathered for the walk? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many people were there? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


The area of Triangle A is twice the area of Triangle B. Find the total area of the two triangles. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


A rectangular container measuring 25 cm by 10 cm by 18 cm was filled to the brim with iced tea. Jane drank some of it and the depth of the liquid in the container became 15 cm.
(a) How much iced tea did Jane drink? Give your answer in millilitres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml
(b) Jane then poured some of the remaining iced tea into 5 mugs to serve her guests. She had 2 ℓ 500 ml of iced tea left. How much iced tea was poured into each mug? Give your answer in litres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

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


\(\dfrac{1}{3}\) of the fruits in a basket are oranges. \(\dfrac{1}{3}\) of the remainder are pears and the rest are apples. There are 84 apples in the basket.
(a) How many oranges are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After selling some oranges, \(\dfrac{3}{10}\) of the fruits left in the basket are oranges. How many oranges are sold? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Sharon was shopping for snacks for her goodie bags. She spent $42.50 altogether. She filled her goodie bags with chocolates and sweets. Each chocolate cost $2.70 and each sweet cost $1.40 less than the chocolate. There were 5 more sweets than chocolates. How many sweets were there altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


In 9.358, which digit is in the hundredths place?

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


Dan participated in a school activity from 9 a.m. to 2 p.m. How many hours did the school activity last?

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


AB and CD are straight lines. Which of the following is true?

Angles x and z are vertically opposite angles formed by the two straight lines AB and CD, so \(\angle x = \angle z\).


What percentage of the shapes are squares \(\square\)?

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


Which of the following is the related height of base AC?

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


Bala has $1200. He saves $800 and spends the remainder. What fraction of his savings is his spending?


At a bookshop, the number of books sold in July is equal to the total number of books sold from April to June. Which of the following bar graphs represents the information correctly?

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


Carl gave away \(\dfrac{1}{4}\) of his stamps and sold 60% of the remainder. What percentage of his stamps was left?

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


What is the value of \((2 + 2 \times 2) + 2 + 2 \times 2\)?

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


Arrange the following from the lightest to the heaviest:<br>7.45 kg, \(7\dfrac{4}{5}\) kg, 7 kg 405 g

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


At a shop, the ratio of the number of erasers to the number of pens to the number of rulers is 5 : 3 : 2. There is a total of 280 erasers and rulers. How many pens are there?

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


A rectangular tile is printed with square patterns. The perimeter of the tile is 32 cm. What is the area of each square pattern?

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


A piece of paper in the shape of an equilateral triangle is folded along the dotted line. Find \(\angle y\).

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


Find the value of 902 ÷ 5. Express your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Express \(1\dfrac{3}{20}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The line graph shows the amount of rice left in a container at the start of each day from Day 1 to Day 4. What is the amount of rice left in the container at the start of Day 2? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Reading the line graph at Day 2, the amount of rice left is 10 kg.


Find the value of A in the ratio.<br>\(63 : 56 : A = 81 : 72 : 27\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The table shows the time taken by 6 runners to complete a race. Who was last in the race?

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


The figure shows five roads drawn on a map in a square grid.
(a) Name two roads that are parallel to each other.
(b) Name two roads that are perpendicular to each other.

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


In the scale, round the value marked X to 1 decimal place. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A repeated pattern is formed using the numbers 0, 1 and 2. The first 13 numbers are: 0, 1, 2, 0, 2, 1, 0, 1, 2, 0, 2, 1, 0. What is the average value of the first 30 numbers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


6 jugs of water can fill \(\dfrac{5}{8}\) of a pail. Another 3 jugs and 6 cups of water are needed to fill the pail completely. How many cups of water can the pail hold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Ali had $36 of 20-cent and 50-cent coins in his piggy bank. There were twice as many 20-cent as 50-cent coins. Find the number of 50-cent coins in the piggy bank. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a rectangle and an equilateral triangle. Find \(\angle a\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Peter was given a sum of money. He gave \(\dfrac{4}{5}\) of the money to his parents and shared the rest of the money equally with his 2 sisters. His parents received $240. How much did Peter receive in the end? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Ms Lim bought 25 boxes of doughnuts for a party. Each large box contained 12 doughnuts and each small box contained 8 doughnuts. There were 252 doughnuts in all. How many large boxes of doughnuts did she buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A tank contains 12.8 ℓ of water. Mr Lee pours 3 pails of water, each containing 0.65 ℓ of water, into the tank without spilling. Find the volume of water in the tank in the end. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


A florist takes 25 minutes to make a bouquet of flowers. How much time is needed to make 9 such bouquets of flowers? Give your answer in hours.


Mr Tan bought some sweets for his students. When he gave each student 3 sweets, he would have 4 sweets left over. When he gave each student 4 sweets, he would need another 2 sweets. How many sweets did Mr Tan buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Bala had 3 m of wire. He cut some of the wire to bend into the shape shown. All the sides are equal in length. He then had 0.4 m of wire left. Find the length of 1 side of the shape. Give your answer in metres, correct to 2 decimal places. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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


Ali needs 1 full tank of spring water for a gardening job. Given that 1 ℓ of spring water costs $0.65, how much money does Ali need for a tank measuring 100 cm by 80 cm by 50 cm? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Shelby took a bank loan of $90 000 to buy a car. The bank charged him an interest rate of 5% per year. He paid the loan in equal monthly payments over 2 years. What was the amount he had to pay each month? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is not drawn to scale and all lines meet at right angles. Find the perimeter of the figure. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


One afternoon, Ahmad, Benny and Charlie took turns to play games on 2 computers from 2 p.m. to 4.30 p.m. At any time, 2 of them played on the computers while the other one watched. Benny and Charlie had the same amount of playing time while Ahmad had 30 minutes more playing time than Benny. Find the ratio of the amount of time Ahmad played to the amount of time Benny played to the amount of time Charlie played. Give your answer in the simplest form.

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


The heights of 4 children were shown in the table: Aini 1.42 m, Ben 1.43 m, Caleb 1.37 m, Devi 1.26 m.
(a) Find the average height of the 4 children. Give your answer in metres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The average height of the children became 1.34 m when another child, Eric, joined the group. Find Eric's height. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Mr Lee started making shirts at 9 a.m. He took 1 h 40 min to make each shirt. After making each shirt, he took a 15-min break. He finished making all the shirts at 6.20 p.m. How many shirts did he make? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A rectangular tank measures 30 cm by 12 cm by 18 cm.
(a) Given \(\dfrac{1}{3}\) of the tank was filled with water, find the volume of water in the tank. Give your answer in litres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) Find the volume of water needed to fill the tank to half its height. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ


Four identical triangles are used to form the figure. The triangles have legs 1 cm and 3 cm. Find the unshaded area of the figure. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The first four figures of a bead pattern are shown. The table gives, for each figure, the number of grey beads, white beads and total beads: Fig 1 (1, 0, 1); Fig 2 (1, 2, 3); Fig 3 (4, 2, 6); Fig 4 (4, 6, 10); Fig 5 (<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> grey, 6 white, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> total). Find the number of grey beads and the total number of beads for Figure 5.

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


In the bead pattern, the number of white beads in each figure follows the sequence 0, 2, 2, 6, 6, ... where the white beads added grow by even numbers. Find the number of white beads for Figure 21. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Using the bead pattern, decide each statement. (i) The number of white beads in each figure is always even. (ii) In Figure 32, the ratio of the number of grey beads to the number of white beads is 16 : 17. For each statement, is it True, False or Not possible to tell?

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


Caili and Deepa had different amounts of money. After Caili spent 25% of her money and Deepa spent \(\dfrac{2}{3}\) of her money, both of them had the same amount of money left. The total amount of money left was $480. Find the total amount of money they had at first. Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Which set of operations makes the equation \(30 \bigcirc 8 \heartsuit 2 = 120\) true?

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


John folded 18 paper planes in \(\dfrac{2}{3}\) h. He spends the same amount of time folding each paper plane. How long does he take to fold one paper plane?


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

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


Jane is m years old. Her father is 3 times as old as she is. Her mother is 4 years younger than her father. How old was her mother?

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


The mass of a bag of flour was 8.4 kg. All the flour was put into 3 bags. The first bag was twice as heavy as the second bag. The second bag was 3 times as heavy as the third bag. What was the mass of the second bag of flour?

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


Different beads were used to make a necklace with a repeated pattern. The first 12 beads are shown. What will the 74th bead be?

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


ABCD is a square and BEF is an isosceles triangle. DF = DE and BE = BF. Find \(\angle BFC\).

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


Write the greatest 4-digit even number without using the digits 2, 4, 8 and 9. The digits cannot be repeated.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Find the value of \(8 \div \dfrac{3}{7}\). Leave your answer as a mixed number.


What is the value of \(\dfrac{6a + 12}{5}\) when \(a = 3\)?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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

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


In the figure, ABCD is a rectangle. DE = 2 cm. Find the area of the triangle ACE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The figure shows two circles. The diameter of the small circle is \(\dfrac{1}{4}\) the diameter of the big circle. The diameter of the big circle is 56 cm. Find the area of the shaded part. (Take \(\pi = \dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The price of a muffin was $w. Miss Tan bought 28 muffins. She was given a discount of $1.50 for every 3 muffins. How much did she pay for the muffins altogether?

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


In the figure, BCDE is a trapezium. AFC and BFE are straight lines and AB = BC. \(\angle ABF = 28°\) and \(\angle FCD = 60°\). Find \(\angle BCF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


Hugo had 18 fewer stickers than Jamie. Jamie gave some stickers to Hugo. In the end, Hugo had 24 more than Jamie. How many stickers did Jamie give to Hugo?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Three identical containers filled with different marbles were weighed. Container 1 (X and Y) = 235 g; Container 2 (Y, Y, Z) = 0.62 kg; Container 3 (Z, Z, Y, Y, Y) = 1.04 kg. What was the mass of Marble X? Give your answer in grams.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

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


A group of boys planned to make 18 cards each for Teachers' Day. 4 more boys joined in to help them. As a result, each boy needed to make 12 cards only. How many cards were made altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Kieran formed a figure. Its outline consists of 1 big semi-circle and 4 identical small semi-circles. The radius of each small semi-circle is 10 cm. Find the area of the shaded part. Take \(\pi = 3.14\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Rishu had some money left after spending $104 on a pair of shoes, a cap and a book. He could not buy another similar pair of shoes with his remaining money as he was short of $29. He decided to buy another similar book instead and had $11 left in the end. (a) How much more did the pair of shoes cost than the book? (b) A pair of shoes cost 4 times as much as the cap. How much did Rishu have at first?<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) After spending $104, the remaining money was $29 short of a pair of shoes. Buying a book instead left $11. So shoes - book = (remaining + 29) - (remaining - 11) = 29 + 11 = $40. Answer (a): $40. (b) Shoes = 4 x cap. With shoes - book = $40 and the total first spend $104 = shoes + cap + book, working through gives Rishu started with $139. Answer (b): $139.


Which of the following decimals is the smallest?

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


Which of the following shows \(\dfrac{1}{5}\) of the figure shaded?

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


Which of the following is likely to be the length of a road bicycle?

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


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

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


Rachel watched a movie at the cinema. The movie lasted for 2 h 15 min. After the movie ended, she walked 25 minutes to reach her home. She reached home at 6.55 p.m. What time did the movie start?

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


The square grid shows the position of 5 landmarks A, B, C, D and E.<br>Julian is standing at a location north-west of landmark D and east of landmark A. In what direction is landmark C from Julian?

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


The pie chart shows the number of pupils taking part in various activities at a sports carnival. The same information is shown in a bar graph, but the names of the activities are not shown on the bar graph.<br>How many pupils took part in Soccer and Hockey altogether?

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


In the figure, FCEB and DGE are straight lines. ABC is an equilateral triangle. DGE = FCE and ∠CFD = 70°. Find ∠CGE.

ABC equilateral so ∠ACB = 60°, giving ∠FCD relationships; with ∠CFD = 70° and the parallel/equal-angle conditions, ∠CGE works out to 100° (the printed key gives option 3).


Alex and Ben had $190 altogether at first. After Alex gave Ben $30, Alex had $40 more than Ben. How much did Ben have at first?

After the transfer the total is still $190. Alex = (190 + 40)/2 = $115, Ben = $75 after. Before, Ben had $75 - $30 = $45.


Arrange these volumes from the smallest to the largest.<br>2.25 ℓ, \(2\dfrac{2}{5}\) ℓ, 2 ℓ 225 ml


The figure is made up of one big semicircle and a small semicircle. The large semicircle with centre O has a radius of 6 cm. Find the perimeter of the figure. Leave your answer in terms of π.

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


Joe baked some chocolate muffins and vanilla muffins. He sold an equal number of chocolate muffins and vanilla muffins. He had \(\dfrac{3}{4}\) of the vanilla muffins and \(\dfrac{3}{7}\) of the chocolate muffins left. What fraction of the muffins were sold?

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


Write three million, six thousand and forty in numerals.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Find the value of \(\dfrac{3}{8} \div 18\). Give your answer as a fraction in the simplest form.


Write down all the common factors of 12 and 20.


At a fruit stall, there were 320 fruits altogether. \(\dfrac{1}{4}\) of them were apples, \(\dfrac{5}{8}\) of them were oranges and the rest were mangoes. The fruit stall sold \(\dfrac{3}{4}\) of the mangoes. How many mangoes did the fruit stall sell?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the charges to post a parcel.
First 500 g: $2; Every additional 1 kg: $1.50.
Mr Yap posted a parcel that weighed 4.8 kg. How much did Mr Yap pay to post the parcel?
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First 500 g = $2. Remaining 4.3 kg needs 5 lots of 'additional 1 kg' (round up): the key takes 4.8 kg = 0.5 kg + 4 kg + 0.3 kg, charging $2 + 4 x $1.50 + $1.50 = $9.50.


ABCD is a rectangle with AB = 12 cm and BC = 5 cm. Point E lies on DC. DE = 4 cm. Find the area of the shaded triangle.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


Jack wanted to buy 16 chicken wings but found that he needed another $5. He bought 7 chicken wings and had $5.80 left. What was the cost of a chicken wing?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Jack bought 7 instead of 16, i.e. 9 fewer wings, leaving him $5.00 (the shortfall he no longer needed) + $5.80 = $10.80. So 9 wings cost $10.80, and one wing = $10.80 ÷ 9 = $1.20.


7 unit cubes are glued together to form the solid shown (Top, Front and Side views indicated).
(a) Draw the side view of the solid on the grid.
(b) What is the least number of unit cubes to be added to the above solid to form a cube?
Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a square and a rectangle overlapping each other. The ratio of the shaded area to the area of the square is 1 : 5. The ratio of shaded area to the area of rectangle is 2 : 9. What is the ratio of the area of square to the area of the rectangle?

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


At a party, 87 blue balloons and green balloons line one side of the wall. There are at least 3 blue balloons between any 2 green balloons. What is the greatest possible number of green balloons along the wall?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Richard took 90 minutes to run 5.4 km. What was his average speed in km/h?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h

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


A fruit seller packed 860 apples into bags of 4 for sale. The price of each bag of apples was $7.80. How much money did the fruit seller collect after selling all the bags of apples?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Figure 1 is a triangle with a perimeter of 39 cm. Each side of the triangle has a different design. Figure 2 is made up of 5 such triangles. The perimeter of Figure 2 is 87 cm. What is the length of ST in Figure 1?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


A shop sells only three types of toys. The table shows the number of each type of toy sold: Robot = p, Car = 2p + 8, Doll = 45.
(a) Find the total number of toys sold by the shop in terms of p. Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The shop sold a total of 140 toys. How many toy robots did the shop sell? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The table shows the number of visitors to a zoo from Thursday to Saturday (Sunday not shown): Thursday 3100, Friday 2690, Saturday 3580, Sunday ?
(a) The number of visitors to the zoo on Sunday increased by 15% when compared to Saturday. What was the number of visitors to the zoo on Sunday? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the percentage decrease in the number of visitors to the zoo on Friday compared to Thursday? Round your answer to 1 decimal place. Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%


In the figure, ABC is an isosceles triangle. AB = BC. BDEC is a parallelogram. AFE is a straight line. ∠CBD = 142°. Find ∠FED.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


A baker sold some boxes of tarts and cookies. Each box of tarts cost $3.10 and each box of cookies cost $5.50. The baker sold twice as many boxes of tarts as boxes of cookies. He earned a total of $1439.10. How many boxes of tarts did he sell?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The first four figures of a pattern are shown. The table shows the number of dots and lines: Figure 1 = 4 dots/4 lines, Figure 2 = 7/8, Figure 3 = 10/12, Figure 4 = 13/16.
(a) A figure in the pattern has 232 lines. What is the Figure number? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the total number of dots in Figure 70. Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Walter and Cedric started cycling from the same place in opposite directions along a straight path. Walter's speed was 40 m/min faster than Cedric's speed. Both of them did not change their speeds throughout. They were 23.38 km apart after they finished cycling. Walter cycled 2.8 km more than Cedric. What was Cedric's speed in m/min?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m/min

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


The figure shows two stacks of identical paper cups. There are 4 paper cups in the shorter stack and 6 paper cups in the taller stack. The height of the shorter stack is 21 cm and the height of the taller stack is 25 cm.
(a) Find the height of a paper cup. Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Mrs Ong wants to pack the paper cups as a single stack into a box of length 50 cm. What is the most number of paper cups she can pack into the box? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


At first, \(\dfrac{1}{11}\) of a tank was filled with water. A tap was turned on at 10 00 for more water to flow into the tank. It was turned off at 10 20. The graph shows the volume of water in the tank over the 20 minutes.
(a) How many litres of water flowed from the tap in 1 minute? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) At 10 30, the tap was turned on again to fill the tank to the brim at the same rate as before. At what time will the tank be filled to the brim with water? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Susan designed a logo. The logo is made up of 3 identical small quarter circles, 3 identical big quarter circles and 2 different squares. O is the centre of the small and big quarter circles. The radius of the small quarter circle is 10 cm. (Take π = 3.14)
(a) What is the radius of a big quarter circle? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the total area of the logo that is shaded? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


A and B are two rectangular containers. The base area of container A is twice the base area of container B. Container A was filled with water to a height of 10 cm and container B was empty. Container A has base 36 cm by 12 cm; container B has height 32 cm.
(a) What was the volume of the water in container A? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml
(b) All the water from container A was poured into container B without spilling. How much more water was needed to fill container B to the brim? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

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


Mr Lee and Mrs Sim bought muffins at the price shown: chocolate muffins 3 for $5, vanilla muffins 4 for $6.
(a) Mr Lee bought an equal number of chocolate muffins and vanilla muffins. He spent $24 more on the chocolate muffins. How many muffins did he buy altogether? Ans: (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mrs Sim spent an equal amount of money on the chocolate muffins and vanilla muffins. What fraction of the muffins she bought were chocolate? Leave your answer in the simplest form. Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Chloe spent \(\dfrac{5}{8}\) of her money on 3 identical books and 7 identical files. The cost of each book was 3 times the cost of each file. She bought some more files with \(\dfrac{5}{6}\) of her remaining money and had $4 left. How much did Chloe spend on the files altogether?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Three boys Lionel, Michael and Nate had the same number of coins. Lionel and Michael each had a mix of fifty-cent and ten-cent coins. Lionel had 10 ten-cent coins, while Michael had 15 ten-cent coins. Nate had only fifty-cent coins.
(a) How much more money did Lionel have than Michael? Ans: (a) $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Michael used all his fifty-cent coins to buy some food. He then had $8 less in coins than Nate. How many fifty-cent coins did Nate have? Ans: (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


A bowl of noodles at a restaurant cost $27.25 including 9% GST. Find the cost of the bowl of noodles before GST.<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The radius of a circle is 21 cm. Find the circumference of the circle. Leave your answer in terms of \(\pi\).

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


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

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


ABCD is a square. CEFD is a rhombus. AGF and CGD are straight lines. \(\angle GDF = 44°\). Find \(\angle GFD\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The diameter of a circle is 70 cm. Find the area of the circle. Take \(\pi = \dfrac{22}{7}\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


The figure is made up of a quarter circle and a square. Find the area of the figure. Take \(\pi = 3.14\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


EFGH is a parallelogram and EBHA is a trapezium. AE is parallel to HB. \(\angle AEH = 34°\), \(\angle EBH = 76°\), \(\angle FGH = 108°\). Find \(\angle BEF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


The figure shows a rectangle and two identical semicircles. The length of the rectangle is 24 m and the radius of each semicircle is 4 m. Find the perimeter of the figure. Take \(\pi = 3.14\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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


The number of people who visited a museum in February increased by 20% when compared to January. The number of people who visited the same museum in March decreased by 5% when compared to February. 855 people visited the museum in March. How many people visited the museum in January?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


There were 140 pupils in an Art Club at first. 40% of the pupils were girls and the rest were boys. After some girls left the Art Club, 20% of the number of pupils who remained in the Art Club were girls. (a) How many pupils in the Art Club were boys? (b) How many girls left the Art Club?<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) Boys = 60% of 140 = \(\dfrac{60}{100} \times 140 = 84\) boys. (b) The number of boys stays 84. After some girls left, girls are 20% and boys are 80% of those remaining: 80% (= 80u) = 84, so 1u = 84 ÷ 80 = 1.05; girls remaining = 20u = 1.05 x 20 = 21. Girls at first = 140 - 84 = 56. Girls who left = 56 - 21 = 35. Answers: (a) 84 boys, (b) 35 girls.


Which of the following is likely to be the length of a bed for an adult?

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


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


3 children shared a pack of stickers. The ratio of Elly's stickers to Flynn's stickers to Gerry's stickers is 4 : 2 : 5. Gerry received 180 stickers. How many stickers did Elly and Flynn receive?

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


A furniture store sold 50 sofas in January. In February, it sold 40 sofas. What is the percentage decrease in the number of sofas sold in February?

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


The semicircle has a radius of 10 cm. What is the perimeter of the shaded figure? Take \(\pi = 3.14\).

Perimeter = half circumference + diameter = \(\pi r + 2r = 3.14\times 10 + 20 = 31.4 + 20 = 51.4\) cm.


A solid cuboid of height 10 cm has a square base of side 4 cm. What is its volume?

Volume = base area × height = (4 × 4) × 10 = 16 × 10 = 160 cm³.


The figure shows trapezium PQRS. Which of the following is a property of a trapezium?

A trapezium is defined by having exactly one pair of parallel sides.


In the figure, which two lines are parallel?

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


The bar graph shows the number of burgers and pies sold in a shop over four days. Find the total number of burgers sold on Thursday and Sunday.

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


The number of pies sold on Monday was an 80% increase from the number sold on Thursday. How many pies were sold on Monday?

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


A wire of length 7.2 m was cut into 3 pieces. The first piece was 3 times as long as the second piece. The second piece was twice as long as the third piece. How long was the second piece of wire?

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


Figure 1 is a right-angled triangle with a base to height ratio of 2 : 3. Figure 2 is made up of four such right-angled triangles. The length of AB is 60 cm. What is the area of figure 1?

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


After a discount of 25%, an adult would have to pay $60 for a buffet lunch. A senior citizen gets an additional discount of $12. Find the percentage discount given to the senior citizen for the buffet lunch.

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


Oman and Pauline made figurines over 2 days. On Monday, Oman made 19 figurines more than Pauline. On Tuesday, Oman made another 20 figurines and Pauline made another 15. At the end of the two days, Oman made \(\dfrac{3}{5}\) of the total number of figurines. What was the total number of figurines Pauline made on Monday and Tuesday?


Express \(9\dfrac{7}{25}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The opening hours of a clinic are: Daily opening hours 9.15 a.m. to 5 p.m., Lunch break 1 p.m. to 2 p.m. How long is the clinic open each day? Give your answer in hours and minutes. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

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


AB and PQ are straight lines. Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The line graph shows the number of fish in a pond from January to June. Between which 1-month period was the increase in the number of fishes in the pond the greatest?

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


Use all the digits 3, 5, 6, 7 to form
(a) the smallest 4-digit multiple of 5 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) the number closest to 7000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Chelsea used 12 identical cubes of side 1 cm to form a solid. She then added some cubes to form a cuboid 4 cm by 6 cm by 6 cm. How many cubes did she add? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is a rectangle with a length of 9 cm and breadth of 6 cm. Find the total shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shaded triangles together have bases that add up to the full 9 cm width and a height equal to the 6 cm breadth, so the shaded area = \(\dfrac{1}{2}\times 9\times 6 = 27\) cm² (half the rectangle).


Shival paid $945 for a golf club and 4 boxes of golf balls. The price of a box of golf balls was \(\dfrac{2}{7}\) of the price of the golf club. How much did Shival pay for the golf club? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A choir has 50 male and 70 female members. 16% of the male members and 10% of the female members are university students and the remaining members are not university students. What percentage of the members are not university students? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A square and a rectangle overlap to form the figure. The ratio of the area of the shaded part to the area of the square to the area of the rectangle is 1 : 2 : 7. The area of the rectangle is 84 cm². Find the area of the whole figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Mr Huang has balloons of 3 different colours. \(\dfrac{1}{4}\) of the balloons are purple. The ratio of the number of yellow balloons to the number of orange balloons is 2 : 3. What is the ratio of the number of purple balloons to the number of orange balloons?

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


Pauline, Queenie and Roger had 162 marbles altogether. Pauline gave \(\dfrac{3}{10}\) of her marbles to Queenie and \(\dfrac{1}{4}\) of her marbles to Roger. After that, all 3 children had the same number of marbles.
(a) What is the ratio of the marbles Pauline had to Queenie had to Roger had at first?
(b) How many more marbles did Pauline have than Queenie at first?


The figure shows a cylinder with a diameter of 35 cm placed a distance away from a wall. The distance between the centre of the cylinder to the wall is 12.1 m. How many complete turns will the cylinder have to make before it touches the wall? Take \(\pi = \dfrac{22}{7}\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Mrs Chan bought 150 pencils and 100 erasers for her students. She divided the pencils equally among them and had 17 pencils left. She also divided the erasers equally among the students and had 5 erasers left. How many students were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Five boys shared the cost of a present equally. When calculating the amount for each share, one of the boys made a mistake by dividing the cost of the present by 4 instead of 5. Each boy paid $3.75 more than his original share. What is the cost of the present? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of three identical squares. Given that EH = 72 cm, CY = 16 cm and BX = YF, find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The average cost of 3 different backpacks is $12.70. The total cost of 2 of the backpacks is $16.80. What is the cost of the third backpack? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Malik had 140 postcards and Neil had 112 postcards. After Malik gave some of his postcards to Neil, the ratio of the number of postcards that Malik had to the number of postcards that Neil had was 1 : 2. How many postcards did Neil have in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABC bookshop sells only 1 type of calculator. The bar graph shows the number of calculators sold from January to June; the bar for June has not been drawn.
(a) Given that the average number of calculators sold from January to June was 55, find the number of calculators sold in June. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The total amount collected from all the calculators sold in March was $1046.40 including 9% GST. Find the cost of 1 calculator excluding GST. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


ABPJ is a parallelogram and LDEG is a trapezium. AMPF is a straight line. LD is parallel to JP. \(\angle ABP = 65°\), \(\angle MPH = 48°\) and \(\angle LHJ = 54°\).
(a) Find \(\angle JAP\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle GLD\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Candice takes 30 days to knit 5 identical scarves while Shelley takes 20 days to knit 4 identical sweaters. Given that Candice and Shelley started knitting on the same day, how many sweaters would Shelley have knitted completely by the time Candice has knitted 24 scarves? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Pauline, Queenie and Roger had 162 marbles altogether. Pauline gave \(\dfrac{3}{10}\) of her marbles to Queenie and \(\dfrac{1}{4}\) of her marbles to Roger. After that, all 3 children had the same number of marbles. How many more marbles did Pauline have than Queenie at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Andrea baked some cookies. She gave \(\dfrac{1}{5}\) of them to her relatives and 56 of them to her friends. She was left with \(\dfrac{2}{3}\) of the cookies. She packed these cookies into 34 bags. Some bags contained 4 cookies while the rest contained 12 cookies.
(a) How many cookies were packed into the 34 bags? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many bags contained 4 cookies each? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


At a musical, a total of 1932 photocards were given out. Each adult received 5 photocards, each boy received 4 photocards and each girl received 3 photocards. \(\dfrac{1}{4}\) of the people who attended the musical were adults, \(\dfrac{4}{9}\) of the remainder were boys and the rest were girls. How many adults were there at the musical? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A shop had 60 t-shirts and some blouses at first. After 10 t-shirts and 20% of the blouses were sold, the shop had a total of 118 t-shirts and blouses left. How many blouses did the shop have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a big semicircle, two small semicircles and a square. The base of the big semicircle is 120 cm.
(a) Find the radius of the small semicircle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the unshaded parts of the figure. (Take \(\pi = 3.14\)) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Jason has several identical square-based wooden blocks. He glued 5 such blocks to form a cuboid as shown in Figure 1. He stacked 4 such cuboids to form a larger cuboid shown in Figure 2. Given that the length of the cuboid is 80 cm, find the volume of the cuboid shown in Figure 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


ABXD and WXYZ are overlapping rectangles and XMNO is a square. \(\angle YXO = \angle OXW\) and \(\angle MXD = 20°\).
(a) Find \(\angle t\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle p\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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


Bessie stacked 8 identical cups: the stack of 8 cups is 82 cm tall and a stack of 4 cups is 44 cm tall. Find the height of the 6 stacked cups. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Nelson used some sticks to form figures in a pattern. The number of sticks per figure is 3, 7, 10, 14, 17, ... for Figures 1, 2, 3, 4, 5.
(a) How many sticks are there in Figure 6 and Figure 7? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (Figure 6), <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> (Figure 7)
(b) How many sticks are there in Figure 107? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(c) Nelson used 2327 sticks to form a figure. Which Figure number did he form? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

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


Onions are sold at 25¢ per 100 g at a supermarket. What is the price of 3.6 kg of onions?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Siti had a bottle of honey. She used an equal amount of honey each day. At the end of the 6th day, \(\dfrac{2}{3}\) of the honey was left. At the end of the 8th day, 400 ml of honey was left. What was the amount of honey in the bottle at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

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


The diagram shows a rectangle containing two identical quarter circles, each with a radius of 37 cm. The rectangle has a perimeter of 236 cm. What is the perimeter of the unshaded part? Give your answer correct to 2 decimal places.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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


At first, Kathy had a total of 78 yellow and red balloons. After 19 red balloons burst and she increased the number of yellow balloons by 75%, Kathy then had a total of 83 balloons. How many red balloons did Kathy have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The pie chart represents the sports played by some Primary 6 students. Each student played only one sport. The number of students who played each sport is shown: Basketball 47, Badminton 38, Table Tennis 75, Soccer unknown. How many students played soccer?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Jacob earned $240 from selling a carton of 12 bottles of vitamins. He sold 36 000 bottles of vitamins in total. (a) How many cartons of vitamins did he sell? (b) How much did he earn from selling all the bottles of vitamins?<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) Each carton holds 12 bottles, so cartons = 36 000 ÷ 12 = 3000 cartons. (b) Earnings = 3000 cartons x $240 = $720 000. Answers: (a) 3000, (b) $720 000.


ABCD is a rhombus and DEFG is a square. Find \(\angle FHD\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

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


5 identical tables cost as much as 7 identical chairs. Each table cost $54 more than the chair. Find the total cost of the 5 tables and 7 chairs.<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


A hawker had some eggs. He used \(\dfrac{1}{3}\) of them on Saturday and \(\dfrac{7}{12}\) of the rest on Sunday. After that, he bought 104 eggs and then had as many eggs as he had at first. How many eggs did he have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


Eagle Express Delivery Company charges $3.50 for parcels delivered on time and $1.80 for parcels delivered late. In May, the company could earn $382.50 more had all the parcels been delivered on time. (a) How many parcels were delivered late in May? (b) For every parcel that was delivered late in May, 9 parcels were delivered on time. How much did the company earn in May?<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="?" />


Jacinta had two identical bottles each completely filled with mixtures of oil and water. The capacity of each bottle is 600 ml. The ratio of the amount of oil to the amount of water in the first bottle was 5 : 3 and in the second bottle it was 3 : 1. Jacinta emptied both bottles into an empty pail. What was the ratio of the amount of oil to the amount of water in the pail?


25% of Lina's money was spent on 5 files and 10 erasers. The cost of each file was 4 times the cost of each eraser. Lina bought more files with 40% of her remaining money. How many files did she buy altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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


The figure is made up of a square, a quarter circle and a semicircle. The area of the square is 196 cm². Find the total area of the shaded parts. (Take \(\pi = \dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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


At the start of a party, there were 70 children. Each boy was given 5 candies and each girl was given 3 candies. A total of 260 candies were given out. (a) Find the number of boys at the party. (b) Halfway through the party, 3 boys and some girls joined in. After that, the ratio of the number of boys to the number of girls at the party became 1 : 2. Find the number of girls who joined the party later.<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) Assume all 70 were girls: 70 x 3 = 210 candies. Extra candies = 260 - 210 = 50, and each boy adds 5 - 3 = 2 extra, so boys = 50 ÷ 2 = 25. (b) Girls at start = 70 - 25 = 45. Boys become 25 + 3 = 28. New ratio boys : girls = 1 : 2, so girls now = 28 x 2 = 56. Girls who joined = 56 - 45 = 11. Answers: (a) 25, (b) 11.


The pie chart shows the number of $10, $20, $50 and $100 concert tickets sold by an event organiser. The $100 sector is 5%, the $50 sector is \(\dfrac{3}{20}\), and AB is a straight line. What fraction of the tickets sold were $20 tickets?


In the figure, ABC is a triangle, BGF and ADJ are 2 identical quarter circles, CDEF and EGHJ are squares. The area of square CDEF is 25 cm² and the radius of the quarter circles is 18 cm. Find the total area of the unshaded parts of the figure. (Take \(\pi = 3.14\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

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