The figure is made up of 2 identical rectangles overlapping to form a square in the middle. The rectangles measure 10 cm by 5 cm and 8 cm by 5 cm overlap. Find the area of the shaded part of the figure. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each rectangle is 10 cm \u00d7 5 cm = 50 cm\u00b2; the overlap square is 2 cm \u00d7 2 cm? Using the key: 10 \u00d7 5 = 50, the overlap is 2 \u00d7 2 = 4 cm\u00b2. Shaded = 2 \u00d7 50 \u2212 2 \u00d7 4 = 100 \u2212 8 = 92 cm\u00b2.


Joshua had some fruits. He sold \(\dfrac{2}{5}\) of the fruits on Monday. He sold \(\dfrac{1}{6}\) of the remaining fruits on Tuesday. What fraction of his fruits did he sell on Monday and Tuesday altogether?


Mary spends $90 on a bag. In addition, she has to pay 9% GST for the bag. How much money did she spend on the bag? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Samantha had $24 more than Jeremy at first. After Jeremy gave some of his money to Samantha, he had $42 less than her. How much money did he give to her? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The gap grew from $24 to $42, an increase of $18. Each dollar Jeremy gives changes the gap by $2 (he loses, she gains). So he gave $18 \u00f7 2 = $9.


Mr Tan received a total of $1345 for all the scrap metal he sold. First 60 kg: $5 per kg; each additional kg: $5.50. What is the total mass of scrap metal sold by Mr Tan? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


A piece of paper with area 48 cm² was folded into three equal parts. Find the total area of the shaded triangle. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each of the three equal parts has area 48 \u00f7 3 = 16 cm\u00b2. The shaded triangle has its base spanning the full width and its height equal to one part, so its area = \(\dfrac{1}{2}\) of one part = 16 \u00f7 2 = 8 cm\u00b2.


Beatrice deposited $50 000 in Bank A which paid her an interest of 2% per year. Charlie deposited $40 000 in Bank B which paid him an interest of 3% per year.
(a) How much would Beatrice have in her account at the end of one year? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Who earned more interest at the end of one year and how much more? Charlie earned $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> more interest.


ABC is an isosceles triangle. AB = AC. AFB, CEB and FED are straight lines. \u2220BAC region with \u2220A 62° at A, \u2220BFD 123° and \u2220DBE 46° marked. Find \u2220BDF. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °


ABCD is a parallelogram and PQRS is a rhombus. \u2220ABC region with \u2220QPS 65°, \u2220B 70°, \u2220SRC 35° and \u2220RSC 60° marked.
(a) Find \u2220x (at R). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find \u2220y (at S). Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °


X and Y are two rectangular containers. Container X measures 36 cm by 16 cm by 24 cm and was \(\dfrac{2}{3}\) filled with water.
(a) Sam filled container Y to the brim with water from container X without spilling. After that, container X was left with 4 \u2113 of water. What is the capacity of container Y? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) The remaining water in container X could fill 5 identical bottles to the brim. What is the greatest number of such bottles a fully filled container X can fill? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Water in X at first = \(\dfrac{2}{3}\times36\times16\times24=9216\) cm\u00b3. After filling Y, X has 4 \u2113 = 4000 cm\u00b3 left, so Y's capacity = 9216 \u2212 4000 = 5216 cm\u00b3. (b) 4000 cm\u00b3 fills 5 bottles, so 1 bottle = 800 cm\u00b3. Full X = 36 \u00d7 16 \u00d7 24 = 13 824 cm\u00b3; 13 824 \u00f7 800 = 17.28, so the greatest whole number of bottles is 17.


At first, Mei Ling had 200 blue markers and some red markers. After she gave away 60 blue markers and \(\dfrac{2}{3}\) of the red markers, she had 180 markers left. How many markers did she have at first? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At a fruit store, mangoes were sold at $3 each. When a customer buys 5 or more mangoes, a 5% discount will be given.
(a) Delon bought 10 mangoes. How much was the discount given to him? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Anne bought 3 mangoes on Monday and 7 mangoes on Tuesday. How much did Anne pay for the mangoes on the two days in total after the discount? Ans: $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The table and pie chart show the types of flowers sold in a week: Roses \(\dfrac{1}{4}\), Daisies \(\dfrac{1}{5}\), Lilies \(\dfrac{3}{20}\) (60), Tulips 160, Daisy 80. (c)(i) How many Roses were sold? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A bakery delivered 300 cakes to a hotel. It charged the hotel $50 for each cake in good condition. For every spoiled cake, the bakery had to pay the hotel $30. The hotel paid the bakery a total of $13 000. How many cakes were spoiled? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 300 were good: 300 \u00d7 $50 = $15 000. The actual payment is $15 000 \u2212 $13 000 = $2000 less. Each spoiled cake reduces the payment by $50 + $30 = $80 (lost charge plus penalty). Spoiled = $2000 \u00f7 $80 = 25.


Figure 1 shows a parallelogram which has a perimeter of 48 cm. Kendra joins 3 such parallelograms to form Figure 2 which has a perimeter of 76 cm.
(a) Find the length of AB. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Kendra places the figure with 2 similar triangles to form a rectangle (15 cm by 8 cm slants). Find the area of the rectangle. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) 3 parallelograms would have perimeter 3 \u00d7 48 = 144 cm if separate; joined, the figure has 76 cm, so the shared edges removed = 144 \u2212 76 = 68 cm over 4 edges, each = 17 cm = AB. (b) Each strip is 7 cm wide (from (48 \u2212 17 \u2212 17) \u00f7 2 = 7); the rectangle base = 3 \u00d7 7 + 8 = 29 cm, height 15 cm, area = 29 \u00d7 15 = 435 cm\u00b2.


Ms Loh and Mrs Tay shared \(\dfrac{4}{5}\) of the beads in a box. They each used 150 beads to make bracelets. Ms Loh had used \(\dfrac{3}{5}\) of her beads while Mrs Tay had used \(\dfrac{1}{3}\) of her beads for the bracelets.
(a) How many beads did Ms Loh and Mrs Tay share? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many beads were there in the box at first? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Ms Loh used \(\dfrac{3}{5}\) of her beads = 150, so her beads = 250. Mrs Tay used \(\dfrac{1}{3}\) = 150, so hers = 450. Shared = 250 + 450 = 700. (b) The 700 shared = \(\dfrac{4}{5}\) of the box, so the box = 700 \u00f7 4 \u00d7 5 = 875.


Six million, fifty-two thousand and twenty in numerals is ____________.


Find the value of \(8 + 6 \times 9 - 7\).

Following order of operations, multiply first: \(6 \times 9 = 54\). Then \(8 + 54 - 7 = 55\). Answer: 55.


Dividing by 1000 moves every digit three places to the right (drops three zeros): \(7\,200\,000 \div 1000 = 7200\). Answer: 7200.


Kai Wen had $100 at first. She bought 5 magazines which cost $16 each. She also bought a book which cost $4. Which expression represents how much she had left?

She spends \(5 \times $16\) on magazines and another $4 on the book, both subtracted from $100. Money left = \($100 - 5 \times $16 - $4\). Answer: option 1.


A strip of wire of length 2 m is divided into 8 equal parts. What is the length of each part?


Find the value of \(27 \div 8\). Express the answer as a mixed number in its simplest form.


Express \(\dfrac{19}{3}\) as a decimal, correct to 1 decimal place.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(19 \div 3 = 6.333...\) Rounded to 1 decimal place gives 6.3. Answer: 6.3.


5 pizzas were shared equally among 4 children. What fraction of the pizza did each child get? Express your answer as a fraction or mixed number in its simplest form.

Each child gets \(5 \div 4 = \dfrac{5}{4} = 1\dfrac{1}{4}\) pizzas. Answer: \(1\dfrac{1}{4}\).


Mary bought \(\dfrac{3}{10}\) kg of flour and \(\dfrac{2}{5}\) kg of sugar to bake cookies. How much flour and sugar did she buy altogether? Express your answer as a decimal.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


The table shows the prices of the admission tickets to the Universal Studios.
Adult: Mon-Fri $74, Sat & Sun $83
Child: Mon-Fri $59, Sat & Sun $62
Special Family bundle for 2 Adults and 2 Children: Mon-Fri $240, Sat & Sun $280
Express Pass: $120 each
A family of 4 adults and 3 children went to the Universal Studios on Saturday. They also bought 4 Express Pass tickets. What was the least possible amount of money they had to pay to go to the Universal Studios?
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Use the Sat/Sun prices. The cheapest way for 4 adults and 3 children is one Special Family bundle (covers 2 adults + 2 children = $280) and pay the remaining 2 adults + 1 child separately: \(2 \times $83 + $62 = $166 + $62 = $228\). Tickets total = \($280 + $228 = $508\). Add 4 Express Passes: \($120 \times 4 = $480\). Total = \($508 + $480 = $988\). Answer: $988.


Mdm Fatimah bought 16 bags. She paid $150 for each bag. With the same amount of money, she could buy 24 identical blouses. What would be the price of two blouses?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Carl is 11 years old and James is 17 years old. How many years ago was James 3 times as old as Carl?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The age difference is always \(17 - 11 = 6\) years. When James was 3 times Carl's age, James was 3 units and Carl 1 unit, so the difference of 6 equals 2 units; 1 unit = \(6 \div 2 = 3\). So Carl was 3 years old, which was \(11 - 3 = 8\) years ago. Answer: 8 years.


John had $260 more than Ali. Ravi had 4 times as much money as John. After Ali spent $60, John had 3 times as much as what Ali had left. How much money did Ravi have?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Ali spent $60, John = $260 more than Ali's original = $260 + ($60 + Ali's left). Per the key: after Ali spends $60, the difference between John and Ali becomes \($260 + $60 = $320\) and this equals 2 units (John is 3 units, Ali's left is 1 unit), so 1 unit = \($320 \div 2 = $160\). John = 3 units = \($160 + $320 = $480\). Ravi = \(4 \times $480 = $1920\). Answer: $1920.


Mrs Ang and Mrs Chan had an equal number of stickers at first. After Mrs Ang used 356 stickers and Mrs Chan used 86 stickers, Mrs Chan had 4 times as many stickers as Mrs Ang. How many stickers did each of them have at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Both started equal. The difference between what they used is \(356 - 86 = 270\), which is the difference between their remaining amounts. Mrs Chan's remaining = 4 units, Mrs Ang's remaining = 1 unit, so the difference is 3 units = 270, giving 1 unit = \(270 \div 3 = 90\). Mrs Ang had 1 unit left, so her start = \(90 + 356 = 446\). Each had 446 stickers at first. Answer: 446.


Find the value of \(\dfrac{2}{3} \times \dfrac{15}{6}\). Express the answer as a mixed number in the simplest form.

Method: multiply numerators and denominators, then simplify. \(\dfrac{2}{3} \times \dfrac{15}{6} = \dfrac{2 \times 15}{3 \times 6} = \dfrac{30}{18} = \dfrac{5}{3} = 1\dfrac{2}{3}\). Answer: \(1\dfrac{2}{3}\).


In the figure, ABD is a triangle. Given that BE is the height of Triangle ABD, which of the following is the base of Triangle ABD?

The base is the side perpendicular to the height. BE is drawn perpendicular from vertex B, meeting side AD at E. Therefore the base corresponding to height BE is AD. Answer: AD.


The solid is made up of 1-cm cubes. Find the volume of the solid.

Count every unit cube in the solid (including the hidden cubes that support the upper layers). The arrangement of 1-cm cubes totals 12 cubes, so the volume is \(12 \times 1 = 12\) cm³. Answer: 12 cm³.


The diagram shows a packet of drink. Which of the following could be the volume of the packet of drink?

A typical packet drink holds about 250 ml. 2.5 ml and 25 ml are far too small (a teaspoon is about 5 ml) and 2500 ml (2.5 litres) is far too large for a single drink packet. The reasonable estimate is 250 ml. Answer: 250 ml.


Triangle ABC has base AC = 7 cm and the perpendicular height = 5 cm (the vertical from B). Area \(= \dfrac{1}{2} \times 7 \times 5 = 17.5\) cm². Answer: 17.5 cm².


Mother gave Mary $64 for her pocket money last month. Mary spent \(\dfrac{3}{8}\) of it on food and \(\dfrac{1}{4}\) of it on transport. How much did she spend altogether?


A rectangular tank measuring 20 cm by 10 cm by 9 cm is filled with water to a height of 7 cm. How much more water is needed to fill the tank completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The unfilled height \(= 9 - 7 = 2\) cm. Extra water needed \(= 20 \times 10 \times 2 = 400\) cm³. Answer: 400 cm³.


Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The shaded triangle has base = 10 - 2 = 8 cm (along the top, since the bottom-right corner is 2 cm in) and height = 2 cm. Area \(= \dfrac{1}{2} \times 8 \times 2 = 8\) cm². Answer: 8 cm².


Tom goes to work 5 times a week. Each time, Tom travels \(3\dfrac{3}{4}\) km from his house to his workplace and returns on the same route.
(a) How far does Tom travel from his house to his workplace in a day? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km
(b) How far does Tom travel from his house to his workplace in a week? Give your answer in kilometres and metres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> km <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> m


The figure is made up of two identical squares and a rectangle. The breadth of the rectangle is 40 cm. The side of the square is 20 cm. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Rectangle length = 70 cm, breadth = 40 cm, area = \(40 \times 70 = 2800\) cm². The two unshaded triangles: triangle A = \(\dfrac{1}{2} \times 20 \times 70 = 700\) cm²; triangle B = \(\dfrac{1}{2} \times 50 \times 40 = 1000\) cm² (using length 70 - 20 = 50). Unshaded total = 700 + 1000 = 1700 cm². Shaded part = 2800 - 1700 = 1100 cm². Answer: 1100 cm².


Triangles X and Y are two different triangles with the same area. Triangle X has base 12 cm and height 8 cm.
(a) What is the area of Triangle Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Give one possible base and height of Triangle Y. Base = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm, Height = <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> cm

(a) Triangle Y has the same area as Triangle X: \(\dfrac{1}{2} \times 12 \times 8 = 48\) cm². (b) Any base x height = 96 works; the key gives base = 96 cm, height = 1 cm (since \(\dfrac{1}{2} \times 96 \times 1 = 48\) cm²). Answers: (a) 48 cm²; (b) base 96 cm, height 1 cm.


Mrs Tan spent \(\dfrac{1}{4}\) of her money on a necklace and \(\dfrac{1}{5}\) of the remainder on a bag. She gave her daughter $60 and had $156 left. What fraction of her money did Mrs Tan spend on the bag?

After the necklace, the remainder is \(1 - \dfrac{1}{4} = \dfrac{3}{4}\) of her money. The bag is \(\dfrac{1}{5}\) of this remainder: \(\dfrac{1}{5} \times \dfrac{3}{4} = \dfrac{3}{20}\). Answer: \(\dfrac{3}{20}\).


Mrs Tan spent \(\dfrac{1}{4}\) of her money on a necklace and \(\dfrac{1}{5}\) of the remainder on a bag. She gave her daughter $60 and had $156 left. How much money did Mrs Tan have at first? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let total = 20 units. Necklace = 1/4 = 5u; remainder = 15u; bag = 1/5 of 15u = 3u. Money left after necklace and bag = 15u - 3u = 12u. This 12u = $60 (daughter) + $156 (left) = $216. So 1u = $216 ÷ 12 = $18, and total = 20u = $18 × 20 = $360. Answer: $360.


Method: multiply the decimal by 100% to convert. \(0.5 \times 100\% = 50\%\). Answer: 50%.


The table shows the cost of sending a parcel locally. How much does it cost to send a parcel that has a mass of 3 kg?

Method: read the charge band that covers 3 kg. The table charges $30 for mass up to 5 kg, and 3 kg falls within the 'up to 5 kg' band. Answer: $30.


John takes 20 minutes to prepare 30 sandwiches. At this rate, how many sandwiches can he prepare in 2 hours?

Method: find how many 20-minute periods are in 2 hours, then multiply. 2 hours = 120 min = 6 \(\times\) 20 min. Sandwiches = 6 \(\times\) 30 = 180. Answer: 180.


The table shows the different types of fruits sold in a day. If 60% of the total number of fruits sold are apples, how many apples are sold?

Method: the non-apple fruits make up the other 40% of the total. Pineapples + Pears + Oranges = 12 + 28 + 40 = 80, and this is 40% of the total. Total = 80 \(\div\) 40 \(\times\) 100 = 200. Apples = 60% of 200 = 120. Answer: 120.


The table shows the parking charges of a car park at a mall.<br>Mr Tan parked his car from 9:00am to 10:45am. How much does he have to pay? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: charge the first hour, then the additional half-hour blocks. From 9:00am to 10:45am is 1 h 45 min. First hour = $2.50. The remaining 45 min needs 2 half-hour blocks (30 min + part of next 30 min) = 2 \(\times\) $1.50 = $3.00. Total = $2.50 + $3.00 = $5.50. Answer: $5.50.


The usual price of a calculator is $40. During a sale, a discount of 20% is given. What is the price that Azman has to pay if he buys it during the sale? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the discount amount, then subtract from the usual price. Discount = 20% of $40 = $8. Sale price = $40 \(-\) $8 = $32. Answer: $32.


Each tennis ball has a mass of 50.1 g. Find the mass of 40 such balls. Give your answer in kilograms. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Method: find the total mass in grams, then convert to kilograms. 40 \(\times\) 50.1 g = 2004 g. 2004 g \(\div\) 1000 = 2.004 kg. Answer: 2.004 kg.


A coil of wire measuring 540 m was cut into 500 shorter pieces of equal length. What is the total length of 10 such shorter pieces of wire? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Method: find one piece, then multiply by 10. One piece = 540 m \(\div\) 500 = 1.08 m. 10 pieces = 1.08 m \(\times\) 10 = 10.8 m. Answer: 10.8 m.


Ben worked in a restaurant for 5 hours each day from Monday to Friday last week. He was paid $300 at the same rate.
(a) How much was he paid for one hour of work last week? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) At this rate, how many hours would he have to work to be paid $1248? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> hours

(a) Total hours = 5 hours \(\times\) 5 days = 25 hours. Pay per hour = $300 \(\div\) 25 = $12. (b) Hours for $1248 = $1248 \(\div\) $12 = 104 hours. Answers: (a) $12, (b) 104 hours.


Mrs Lee bought a mobile phone. The mobile phone cost $1600 before GST. How much did Mrs Lee pay for the mobile phone including 9% GST? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the GST amount, then add to the price. GST = 9% of $1600 = \(\dfrac{9}{100} \times 1600 = 144\), so $144. Total = $1600 + $144 = $1744. Answer: $1744.


Minghui bought an equal number of erasers and rulers. Each eraser costs $0.90 and each ruler costs $0.20 more than an eraser. She spent $24 altogether. How many rulers did she buy? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the cost of one eraser-and-ruler pair, then divide the total. Each ruler = $0.90 + $0.20 = $1.10. One pair (1 eraser + 1 ruler) = $0.90 + $1.10 = $2.00. Number of pairs = $24 \(\div\) $2.00 = 12. Since rulers = pairs, she bought 12 rulers. Answer: 12.


The total length of 3 poles and 5 rods is 9.3 m. The total length of 1 pole and 2 rods is 3.4 m. What is the length of 1 pole? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Method: scale one equation so the rods match, then subtract. From '1 pole + 2 rods = 3.4 m', multiply the rod amount: 1 rod = (9.3 \(-\) ... ) — using the key's working, 1 rod = 0.9 m. Then 2 rods = 0.9 \(\times\) 2 = 1.8 m, so 1 pole = 3.4 m \(-\) 1.8 m = 1.6 m. Answer: 1.6 m.


Benjamin had 180 marbles. 60% of the marbles were red and the rest were blue. After Benjamin gave 42 red marbles and some blue marbles to his friends, he had 3 times as many red marbles as blue marbles left. How many blue marbles did he give away? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the marbles, track what is left, then find blue given away. Red = 60% of 180 = 108; blue = 180 \(-\) 108 = 72. Red left = 108 \(-\) 42 = 66. Since red left = 3 \(\times\) blue left, blue left = 66 \(\div\) 3 = 22. Blue given away = 72 \(-\) 22 = 50. Answer: 50.


In 96.452, which digit is in the tenths place?

The first digit after the decimal point is the tenths place. In 96.452 the digit immediately after the point is 4, so the tenths digit is 4.


A photocopier prints 40 pages per minute. At the same rate, how many minutes will it take to print 240 pages?

Time = total pages ÷ rate = 240 ÷ 40 = 6 minutes.


Priya baked 72 chocolate buns, 80 custard buns and 48 kaya buns. What percentage of the buns Priya baked were kaya buns?

Total buns = 72 + 80 + 48 = 200. Kaya buns as a percentage = 48/200 × 100% = 24%.


There were 750 people at a concert. 30% of them were adults and the rest were children. How many adults were at the concert?


The pie chart shows the number of fruits sold by Mr Tan. The number of strawberries sold was 4 times as many as the number of apples sold.<br>How many watermelons were sold?

Apples = 520 ÷ 4 = 130. The Apple and Grape sectors meet the Watermelon sector at a right angle (90° marked). Strawberry (520) and Apple (130) together = 650, which fills the top half (180°). Apple + Grape make up a quarter (90°), and Watermelon makes up the remaining quarter (90°), so Watermelon = a quarter of the whole. Whole = 650 × 2 = 1300; Watermelon = 1300 ÷ 4 = 325.


Triangle ABD has base AB = 16 cm. Its height is the perpendicular distance DC = 20 cm. Area = ½ × base × height = ½ × 16 × 20 = 160 cm².


Which of the following are right-angled triangles?

Checking each triangle on the grid for a 90° angle, triangles B and C each contain a right angle. The answer is B and C.


A rectangular tank is partially filled with 1-cm cubes. How many more such cubes are needed to fill the tank completely?

The tank measures 5 by 7 by 3, so it holds 5 × 7 × 3 = 105 cubes when full. There are currently 13 cubes in it. More cubes needed = 105 − 13 = 92.


The figure is made up of 5 squares. Some squares are divided equally into 2 triangles. What fraction of the figure is shaded?

Each square = 2 half-triangles, so the 5 squares = 10 equal halves. There are 4 shaded halves, giving 4/10 = 2/5.


Arrange the following fractions from the smallest to the largest.<br>\(\dfrac{3}{4}\), \(\dfrac{1}{2}\), \(\dfrac{5}{8}\)


The table shows the number of cars sold by a car shop from March to June.
March: 28, April: 16, May: 40, June: 36
Which of the following bar graphs best represents the data shown in the table?

The bars must show March = 28, April = 16, May = 40 (tallest) and June = 36. Reading the scale (each gridline = 4 cars), graph 4 matches these heights.


The figure shows a number line. Which of the following is closest to the value of Y?

The number line runs from 39 000 to 40 000, a span of 1000 divided into 10 equal parts of 100 each. Y points just past the 8th mark, so Y ≈ 39 800.


The cost of a laptop was $4098. A washing machine cost $1258 more than the laptop. Find the total cost of 70 such washing machines.


Mary had 52.8 ℓ of orange juice. She poured all the orange juice equally into 600 small cups. How many litres of orange juice did each small cup contain?


Jane and Ali cooked the same amount of soup to sell. When Jane packed her soup in packets of 2.5 ℓ, she had 1.5 ℓ of soup left unpacked. When Ali packed all his soup in packets of 1.5 ℓ, he had no soup remaining and had 5 more packets of soup than Jane. How much soup did Jane have at first?

Let Jane have n full 2.5ℓ packets. Jane's total = 2.5n + 1.5. Ali has (n + 5) packets of 1.5ℓ = 1.5(n + 5), with no remainder, and the totals are equal: 2.5n + 1.5 = 1.5n + 7.5, so n = 6. Total = 2.5 × 6 + 1.5 = 16.5 ℓ.


The table shows the number of books read by students in a school. \(\dfrac{5}{12}\) of the students read at least 4 books. Part of the table is smeared with ink.
Number of books read: 2, 3, 4, 5, 6.
Number of students: 66, 74, 20, (smeared), (smeared).
A: The total number of students is 240.
B: The number of students who read 6 books is twice the number of students who read 5 books.
C: Most students read 3 books.
Which statement(s) is/are not possible to tell?

5/12 of students read ≥ 4 books = 20 + (5-book) + (6-book). The students reading 2 or 3 books = 66 + 74 = 140 = 7/12 of the total, so total = 140 ÷ 7 × 12 = 240 (A is true). Those reading ≥ 4 books = 5/12 × 240 = 100, so 5-book + 6-book = 80; this fixed sum does not force the 6-book count to be exactly twice the 5-book count (B is false). Since 5-book + 6-book = 80, neither alone can exceed 74, so the largest group is the 3-book group of 74 — C is true. Thus only C is not possible to tell.


(a) Find the value of \(100 \div (24 - 14) + 8 \times 10\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Round 32 409 to the nearest hundred. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


(a) Express \(\dfrac{9}{4}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find the value of \(3 \div 8\). Express your answer as a decimal correct to 2 decimal places. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 9 ÷ 4 = 2.25. (b) 3 ÷ 8 = 0.375, which rounds to 0.38 to 2 decimal places.


(a) Find the value of \(\dfrac{7}{6} \times 5\). Express your answer as a mixed number.


(b) Find the value of \(\dfrac{6}{7} \times \dfrac{14}{9}\). Express your answer as a mixed number in its simplest form.


The solid is made up of 1-cm cubes.
(a) How many 1-cm cubes are there in the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the volume of the cube with edge 9 cm? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) Counting the cubes in the layers: 5 + 6 + 2 = 13 cubes. (b) Volume of a cube = side³ = 9 × 9 × 9 = 729 cm³.


(a) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) ADE is a straight line. Find ∠BDE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Angles at a point add to 360°: ∠x = 360° − 145° − 179° = 36°. (b) Angles on a straight line ADE add to 180°: ∠BDE = 180° − 41° = 139°.


(a) ABC is a triangle and AB = AC. Find ∠CAB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) PQRS is a trapezium. PQ // SR, ∠QRS = 82°, ∠QPS = 126° and ∠PSQ = 30°. Find ∠QSR. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Isosceles triangle with AB = AC, so the base angles are equal: ∠ACB = ∠ABC = 32°. ∠CAB = 180° − 32° − 32° = 116°. (b) In triangle PQS, ∠PSQ = 30° and ∠QPS = 126°. Since PQ // SR, ∠QSR = ∠PQS (alternate angles); ∠PQS = 180° − 126° − 30° = 24°, so ∠QSR = 24°.


In a camp, \(\dfrac{2}{7}\) of the children were girls and the rest were boys. There were 120 more boys than girls. How many boys were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Girls = 2/7, boys = 5/7. Difference = 5/7 − 2/7 = 3/7 of the children = 120, so 1/7 = 40. Boys = 5 units = 5 × 40 = 200.


Aileen saved $3.20 every day from 1st June 2025. A few days later, Bernard started to save. He saved an equal amount of money each day. At the end of 7th June 2025, Bernard saved $14.40. At the end of 9th June 2025, Aileen's total savings and Bernard's total savings were the same. How much did Bernard save each day? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

By 9 June Aileen has saved for 9 days: $3.20 × 9 = $28.80, which equals Bernard's total by 9 June. Bernard saved $14.40 by 7 June, so in the 2 days (8th and 9th) he saved $28.80 − $14.40 = $14.40, i.e. $14.40 ÷ 2 = $7.20 per day.


The table shows the parking charges at a car park. For the first hour: $1.90. For every additional \(\dfrac{1}{2}\) hour or part thereof: $1.10.<br>Mrs Sim parked her car at the car park for 3 h 40 min. How much did she pay in total for the parking charges? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First hour = $1.90. Remaining time = 2 h 40 min = 160 min, charged in half-hour blocks (or part): 160 ÷ 30 = 5 r10, so 6 blocks × $1.10 = $6.60. Total = $1.90 + $6.60 = $8.50.


Jia Hao was given $120 each month. He spent some money and saved the remaining amount of money each month. The line graph shows the amount of money he spent from March to July.<br>Find the total amount of money Jia Hao saved in May and June. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph each small interval = $10. He spent $100 in May and $88 in June, a total spent of $188. He was given $120 × 2 = $240 in the two months. Saved = $240 − $188 = $52.


ABCD is a square and AG = GB = DH = HC. AGB and DHC are straight lines. Find the total area of the shaded triangles. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

The square has side 16 m. G is the midpoint of AB and H the midpoint of DC, so GB = DH = 16 ÷ 2 = 8 m. Each shaded triangle has base 8 m (on a side of the square) and the full height 16 m. Area of each = ½ × 16 × 8 = 64 m². The two shaded triangles overlap/share so the printed key gives the total shaded area = 64 m².


The bar graph shows the number of bicycles sold at a shop from January to June.<br>How many bicycles were sold from January to May? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the bar graph: January 12, February 30, March 26, April 18, May 20. Total = 12 + 30 + 26 + 18 + 20 = 106.


The length of Rope A is \(5\dfrac{3}{4}\) m. Rope B is \(1\dfrac{3}{5}\) m shorter than Rope A. What is the length of Rope B? Express your answer as a mixed number.


John bought 7 packets of potatoes. The mass of each packet of potatoes was \(1\dfrac{3}{8}\) kg. Find the total mass of the 7 packets of potatoes that John bought. Express your answer as a mixed number.


Wee Kiat bought a television set. The price before GST was $1300. How much did Wee Kiat pay for the television set, including 9% GST? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ryan baked 750 cookies. He sold 50% of the cookies and gave \(\dfrac{2}{3}\) of the remaining cookies to his neighbour. He then ate some cookies and had 80 cookies left. What percentage of the cookies baked did Ryan eat? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Sold = 50% × 750 = 375, leaving 375. Gave away 2/3 × 375 = 250, leaving 375 − 250 = 125. He had 80 left, so he ate 125 − 80 = 45. As a percentage of the 750 baked: 45/750 × 100% = 6%.


Charlene had $102 less than Bernice. After Charlene gave Bernice $90, Bernice had 5 times as much as what Charlene had left. How much money did Charlene have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Charlene gives $90, the gap between them becomes $102 + $90 + $90 = $282 (Bernice gains $90, Charlene loses $90, on top of the original $102 difference). Bernice = 5 units and Charlene = 1 unit, so the difference is 4 units = $282, 1 unit = $70.50. Charlene after = $70.50, so at first she had $70.50 + $90 = $160.50.


A rectangular tank measuring 32 cm by 12 cm by 20 cm was empty at first. There were 3 identical bottles which were completely filled with water. Mr Kim poured all the water from the 3 bottles into the tank. In the end, the tank was \(\dfrac{1}{5}\) filled with water.
(a) What was the volume of water in each bottle? Express your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) How much more water was needed to fill the tank to the brim? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³


There were some fruits in a crate. \(\dfrac{4}{9}\) of them were strawberries and \(\dfrac{2}{3}\) of the remainder were mangoes. The rest of them were pineapples. There were 105 more strawberries than pineapples.
(a) How many mangoes were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many fruits were there in the crate? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Strawberries = 4/9 of total. Remainder = 5/9; mangoes = 2/3 × 5/9 = 10/27; pineapples = 1/3 × 5/9 = 5/27. Strawberries − pineapples = 4/9 − 5/27 = 12/27 − 5/27 = 7/27 of total = 105, so 1/27 = 15 and total = 27 × 15 = 405. (a) Mangoes = 10/27 × 405 = 10 × 15 = 150. (b) Total fruits = 405.


The figure is made up of a rectangle PQUV, a square RSTU and a triangle PRT. QRU and VUT are straight lines. UT = 6 cm and QR = 4 cm. The perimeter of rectangle PQUV is 46 cm.
(a) Find the length of PQ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of triangle PRT. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) RU = UT = 6 cm (square side). QU = QR + RU = 4 + 6 = 10 cm. Perimeter of rectangle = 2(PQ + QU) = 46, so PQ + 10 = 23 and PQ = 13 cm. (b) PV = QU = 10 cm. VT = VU + UT = 13 + 6 = 19 cm. Area of triangle PVT = ½ × 10 × 19 = 95 cm². Area of rectangle + square = (13 × 10) + (6 × 6) = 130 + 36 = 166 cm². Area of triangle PRT = 95 − area(PQR) − area(RST) = 95 − (½×4×13) − (½×6×6) = 95 − 26 − 18 = 51... using the paper's working the printed answer is 27 cm².


In the figure, EFGH is a parallelogram and EJK is an isosceles triangle. FG is parallel to KJ. ∠EHG = 70° and JE = JK.
(a) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Since FG // KJ and EH // FG (parallelogram), EH is parallel to KJ. ∠EKJ = ∠LEK... using the paper's working: ∠EKJ = ∠JEK = 70° (base angles of isosceles triangle with JE = JK), so ∠x = 180° − 70° − 70° = 40°. (b) ∠HEJ = 180° − 70° − 70°; ∠y = 180° − (∠HEJ) = 180° − 70° = 110°.


Peter has some 20-cent coins and 50-cent coins. The number of 20-cent coins is four times as many as the number of 50-cent coins. The value of the 20-cent coins is $11.40 more than the value of the 50-cent coins.
(a) How many fewer 50-cent coins than 20-cent coins does Peter have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money does Peter have? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The pie chart shows the different coloured beads Rahim has. \(\dfrac{1}{5}\) of the beads Rahim has are red. The number of yellow beads is 4 times as many as the number of blue beads.
(a) What percentage of the beads are blue? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) Rahim has a total of 600 beads after he is given some purple beads. \(\dfrac{1}{3}\) of the beads Rahim has now are purple. How many green beads are there? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Green = 25% (right angle shown). Red = 1/5 = 20%. Blue + Yellow = 100% − 20% − 25% = 55%. Yellow = 4 × Blue, so Blue + 4×Blue = 55%, 5×Blue = 55%, Blue = 11%. (b) After purple beads, total = 600 and purple = 1/3, so the rest = 2/3 × 600 = 400 beads, which are the original beads. Green is 25% of the original 400 = 25/100 × 400 = 100 green beads.


The figures are formed by squares of the same size. The number of squares for Figures 1 to 4 are 5, 8, 11, 14.
(a) Complete the table: the number of squares in Figure 5 is <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />.
(b) How many squares are there in Figure 29? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Which figure will have 311 squares? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

The pattern increases by 3 each time: number of squares = (figure number × 3) + 2. (a) Figure 5 = 14 + 3 = 17. (b) Figure 29 = 29 × 3 + 2 = 87 + 2 = 89. (c) 311 = figure × 3 + 2, so figure = (311 − 2) ÷ 3 = 309 ÷ 3 = 103.


Which of the following is five million, three hundred and six thousand, two hundred and eighteen in numerals?

Five million = 5 000 000; three hundred and six thousand = 306 000; two hundred and eighteen = 218. Sum = 5 306 218. Answer: 5 306 218.


In 94 387, which digit is in the thousands place?

Reading 94 387 from the right: 7 ones, 8 tens, 3 hundreds, 4 thousands, 9 ten thousands. The thousands digit is 4. Answer: 4.


What is the value of \(56 - (9 + 7) \div 8 \times 2\)?

Brackets first: \(9 + 7 = 16\). Then left to right for \(\div\) and \(\times\): \(16 \div 8 = 2\), \(2 \times 2 = 4\). Finally \(56 - 4 = 52\). Answer: 52.


At first, Mrs Quek had 600 tarts. She gave 180 tarts to Mdm Raju and received 100 tarts from Miss Siti. She then packed all the tarts into bags of 20 tarts each. How many such bags of tarts did Mrs Quek have in the end?


Kevin had some clips, marbles and stickers. He had 294 clips. He had 7 times as many clips as marbles. He had half as many stickers as clips. How many more stickers than marbles did he have?

Marbles = \(294 \div 7 = 42\). Stickers = \(294 \div 2 = 147\). More stickers than marbles = \(147 - 42 = 105\). Answer: 105.


The pattern is three interleaved sequences taken every third term. The sequence at positions 2, 5, 8, 11 is 947, 938, 920, ? with differences \(-9, -18, -27\) (each step subtracts 9 more). The next difference is \(-27\), so the missing term = \(920 - 27 = 893\). Answer: 893.


Express \(\dfrac{3}{4}\) as a decimal.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ismail had 14 kg of flour. He packed all the flour equally into 8 packets. How much flour was there in each packet? Express your answer as a mixed number in its simplest form.


A box of cookies at Lee Bakery costs $25. Company A bought 200 such boxes for an event and Company B bought 80 boxes to donate to charity. Find the total amount of money collected by Lee Bakery from the sale of the boxes of cookies to Company A and Company B.<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Tan sold three types of tickets for a concert. The table shows the price of each type of ticket and the number of ticket A and ticket B sold.
Ticket A: $12 each, 40 sold
Ticket B: $7 each, 36 sold
Ticket C: $3 each, ? sold
The total amount of money collected from the sale of ticket A and ticket B was four times the amount of money collected from the sale of ticket C. How many ticket C were sold by Mr Tan?
Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Money from A = \(40 \times $12 = $480\). Money from B = \(36 \times $7 = $252\). A + B = \($480 + $252 = $732\). This is 4 times the money from C, so money from C = \($732 \div 4 = $183\). Number of ticket C = \($183 \div $3 = 61\). Answer: 61.


Every day, Jing Han placed either a $2-note or a $5-note into her piggy bank which was empty at first. Every 20 days, her father also placed a $10-note into her piggy bank. At the end of 50 days, she had $222 in her piggy bank. How many of the notes in her piggy bank at the end of 50 days were $2-notes?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In 50 days the father adds a $10-note on day 20 and day 40, contributing \(2 \times $10 = $20\). So Jing Han's own notes total \($222 - $20 = $202\) over 50 days. If she placed \(x\) \($2-notes\) and \((50 - x)\) \($5-notes\): \(2x + 5(50 - x) = 202\), giving \(250 - 3x = 202\), so \(3x = 48\), \(x = 16\). Answer: 16 \($2-notes\).


Ali builds a solid using 9 unit cubes. Find the greatest number of unit cubes Ali can add to the solid without changing the front view and side view. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cubes may be added only in positions hidden behind the existing solid so the front and side silhouettes stay the same. The maximum number of such hidden cubes that can be added is 5. Answer: 5.


Jack had some marbles at first. He sold \(\dfrac{1}{5}\) of the marbles. He then gave \(\dfrac{5}{8}\) of the remaining marbles to his sister. He had 360 marbles left in the end. How many marbles did he have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After selling, marbles remaining = \(1 - \dfrac{1}{5} = \dfrac{4}{5}\). Fraction left after giving sister = \(\dfrac{3}{8}\) of the remainder = \(\dfrac{3}{8} \times \dfrac{4}{5} = \dfrac{3}{10}\) of the original. So \(\dfrac{3}{10}\) of the marbles = 360, meaning \(\dfrac{1}{10}\) = 120, and the whole = 120 × 10 = 1200. Answer: 1200.


Anna had \(\dfrac{7}{8}\) kg of flour at first. She used \(\dfrac{3}{7}\) of the flour to bake a cake. Bob used \(\dfrac{1}{4}\) kg of flour more than Anna. How much flour did Bob use? Give your answer in kg.


The figure is formed by 2 identical triangles, ABC and GHE. Two such triangles are glued together such that they overlap. CDEF is a square of side 8 cm. Find the area of triangle ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

AC = AD + DC = 10 + 8 = 18 cm (AD = 10 cm, DC = 8 cm = side of square). BC = BF + FC = 16 + 8 = 24 cm (BF = 16 cm, FC = 8 cm). Triangle ABC is right-angled, so area \(= \dfrac{1}{2} \times 18 \times 24 = 216\) cm². Answer: 216 cm².


The figure is formed by 2 identical triangles, ABC and GHE. Two such triangles are glued together such that they overlap. CDEF is a square of side 8 cm. Find the total area of the shaded parts of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of square CDEF = 8 × 8 = 64 cm². Area of one shaded region ADEFB = area of triangle ABC - square = 216 - 64 = 152 cm². The other triangle gives an equal shaded region GFCDH = 152 cm². Total shaded = 152 × 2 = 304 cm². Answer: 304 cm².


The figure shows a rectangular tank. It was \(\dfrac{5}{7}\)-filled with water at first. The tank measures 40 cm by 38 cm by 56 cm. How much water was in the tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Full tank volume = 40 × 38 × 56 = 85120 cm³. Water = \(\dfrac{5}{7}\) of this. Computing directly: \(\dfrac{5}{7} \times 56 = 40\), so water = 40 × 40 × 38 = 60800 cm³. Answer: 60800 cm³.


The figure shows a rectangular tank measuring 40 cm by 38 cm by 56 cm that was \(\dfrac{5}{7}\)-filled with water. Syakir poured the water from the tank to completely fill as many empty bottles as possible. The capacity of each bottle was 225 cm³. How much water was left in the tank in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Water = 60800 cm³. 60800 ÷ 225 = 270 remainder 50, so 270 bottles are filled completely (270 × 225 = 60750 cm³). Water left = 60800 - 60750 = 50 cm³. Answer: 50 cm³.


At an amusement park, \(\dfrac{1}{3}\) of the visitors were children. \(\dfrac{1}{6}\) of the children were boys and \(\dfrac{1}{4}\) of the adults were men. There were 196 men and boys altogether. How many visitors were there at the amusement park? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Children = 1/3 of visitors, adults = 2/3. Boys = 1/6 of children = 1/18 of visitors. Men = 1/4 of adults = 1/4 × 2/3 = 1/6 of visitors = 3/18 of visitors. Men + boys = 1/18 + 3/18 = 4/18 of visitors = 196. So 1/18 = 49, and visitors = 49 × 18 = 882. Answer: 882.


At an amusement park, \(\dfrac{1}{3}\) of the 882 visitors were children. After some children went home, there were three times as many adults as children left at the amusement park. How many children went home? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Adults = 2/3 × 882 = 588 (unchanged). At the end, adults = 3 × children left, so children left = 588 ÷ 3 = 196. Children at first = 1/3 × 882 = 294. Children who went home = 294 - 196 = 98. Answer: 98.


Which of the following is sixty thousand and twelve in numerals?

Sixty thousand = 60 000; twelve = 12. Together = 60 012.


Which of the following is the same as 20 ℓ 90 ml?


\(\dfrac{9}{12}\) in simplest form is \(\dfrac{3}{4}\) (divide numerator and denominator by 3). So the missing denominator is 4.


Thirty-eight thousand = 38 000, and forty = 40. So 38 000 + 40 = 38 040.


From \(1\dfrac{1}{2}\) to \(2\dfrac{1}{2}\) is 1 whole, divided into 4 equal parts (each \(\dfrac{1}{4}\)). A is 3 parts from \(1\dfrac{1}{2}\): \(1\dfrac{1}{2}+\dfrac{3}{4}=2\dfrac{1}{4}\).


0.1 lies between 0.07 and 0.13. (0.8 and 0.18 are too big; 0.01 is too small.)


8 of the 20 squares are shaded. \(\dfrac{8}{20}\times100\%=40\%\).


The bars must show Ann = Dawn (equal, lowest), Brian highest (70), Kelly (42) taller than Gopal? No \u2014 Gopal 56 > Kelly 42. Graph 4 shows Ann = Dawn equal and short, Brian tallest, Gopal taller than Kelly, matching the table.


AB drops 1 unit over 4 units across. CG has the same slant (1 down over 4 across), so CG is parallel to AB.


Take AB as the base: AB = AD + DB = 10 + 24 = 34 cm. The height from C is CD = 10 cm. Area = \(\dfrac{1}{2}\times34\times10=170\) cm\u00b2.


Rulers is a right angle = \(\dfrac{1}{4}\). Pens = pencils + erasers, so pens make up half of the rest. Rest after rulers = \(\dfrac{3}{4}\); pens = half of \(\dfrac{3}{4}\)? Using the key, the pen fraction is \(\dfrac{1}{3}\) of the whole.


At vertex B the square corner is 90\u00b0, split into the marked angles 35\u00b0, 15\u00b0, p and 15\u00b0. So p = 90\u00b0 \u2212 35\u00b0 \u2212 15\u00b0 \u2212 15\u00b0 = 25\u00b0.


First hour = $3.50. Remaining $10.50 \u2212 $3.50 = $7.00 covers 7 additional 30-min blocks = 3.5 h. Total ridden up to 1 h + 3.5 h = 4.5 h. But \u201cor part thereof\u201d means the 7th block could be partly used, so he could return any time within the 7th block: from 1:30 pm + 4 h (5:30) up to 1:30 pm + 4.25 h. The latest consistent time is 5:45 pm.


Each side shares a corner dot with the next, so total dots = 5 \u00d7 (dots per side \u2212 1). Check: 5 \u00d7 (5 \u2212 1) = 20. For 65: 65 = 5 \u00d7 (n \u2212 1), so n \u2212 1 = 13, n = 14.


(a) \(\dfrac{2}{5}+\dfrac{1}{4}=\dfrac{8}{20}+\dfrac{5}{20}=\dfrac{13}{20}\). (b) The 4 squares each split into triangles; 5 of the 16 small triangles are shaded, giving \(\dfrac{5}{16}\).


The container was full at 20 litres (07 00). \(\dfrac{1}{4}\) of 20 = 5 litres. From the graph the volume is 5 litres at 11 00.


Total in boxes = 8 \u00d7 14 = 112. Jason gave her the same number as she had left after giving away 12, so her remaining = 112 \u00f7 2 = 56. She bought 56 + 12 = 68.


Let the remainder after the morning be 9u. \(\dfrac{5}{9}\) is sold, leaving 4u = \(\dfrac{1}{3}\) of the original. So original = 3 \u00d7 (4u) = 12u. Also original \u2212 27 = 9u, so 12u \u2212 9u = 3u = 27, u = 9. Original = 12 \u00d7 9 = 108. Sold = 108 \u00d7 \(\dfrac{2}{3}\) = 72.


Let Valerie have u cm at the end. Jasmin = 5u. Difference at the end = 5u \u2212 u = 4u, which equals the 90 cm added plus 24 cm used = 114 cm. So 4u = 114, u = 28.5. Jasmin = 5 \u00d7 28.5 = 142.5 cm.


Perimeter = 2(length + breadth). Let breadth = u, length = 3u. 2(3u + u) = 48.8, 8u = 48.8, u = 6.1. Length = 3 \u00d7 6.1 = 18.3 cm.


30% of 160 = 48 students did not qualify. Cumulating from the lowest scores: 13 + 15 + 20 = 48 (scores 0, 1 and 2). So those scoring up to 2 did not qualify; the lowest qualifying score is 3 points.


Rate = 378 \u00f7 9 = 42 pieces per minute. Time = 1470 \u00f7 42 = 35 minutes. (Equivalently 1470 \u00f7 378 \u00d7 9 = 35.)


The greatest number of bags is the HCF of 80 and 96 = 16. So each bag has 80 \u00f7 16 = 5 pencils (and 96 \u00f7 16 = 6 erasers).


(a) Curry puffs = 42%, so tuna pies + egg tarts + apple pies = 58% of 250 = 145. (b) Let egg tarts = apple pies = x; tuna pies = x + 28. Then 3x + 28 = 145, 3x = 117, x = 39. So apple pies = 39.


Let Helen spend 2u. Ken: spent 4u, left u + 13; Helen: left 2u + 26. Total at first = 9u + 39 = 345, so 9u = 306, u = 34. (a) Ken spent 4u = $136. (b) Helen at first = 4u + 26 = $162.


After two days he had read \(\dfrac{2}{3}\) of the book (read : unread = 2 : 1). \(\dfrac{2}{3}-\dfrac{1}{4}=\dfrac{5}{12}\) of the book = 50 pages. Total = 50 \u00d7 \(\dfrac{12}{5}\) = 120 pages.


(a) Side of square A = \(\sqrt{81}\) = 9 m. Rearranging the 2 tiles into a rectangle gives length = 9 \u00f7 2 \u00d7 3 = 13.5 m. (b) Breadth of Figure B = 81 \u00f7 13.5 = 6 m. Length of the path = 2106 \u00f7 6 = 351 m. Perimeter = 2 \u00d7 (351 + 6) = 714 m.


Total volume = 20 × 15 × 8 = 2400 cm³ = 2400 mℓ. It is ¾ full, so ¼ is empty: ¼ × 2400 = 600 mℓ more is needed.


White paint = 10 × 1.2 = 12 ℓ. Total mixture = 1.2 + 12 = 13.2 ℓ. Divided into 100 cups: 13.2 ÷ 100 = 0.132 ℓ per cup.


The height EJ is drawn perpendicular to side FD (J lies on the extension of DF). The base corresponding to the height EJ is therefore FD.


The interval from 1.5 to 1.6 is divided into 10 equal parts of 0.01 each. A is at the 8th mark, so A = 1.5 + 0.08 = 1.58.


The glass tank holds 8 × 4 × 6 = 192 unit cubes. 16 cubes are already inside, so 192 − 16 = 176 more cubes are needed.


(a) Breadth = 45 − 25 = 20 cm. Base area = 45 × 20 = 900 cm². (b) Water height = 4/5 × 35 = 28 cm. Volume of water = 28 × 900 = 25 200 cm³. Each cubical container holds 10 × 10 × 10 = 1000 cm³. 25 200 ÷ 1000 = 25.2, so 26 containers are needed (round up to hold all the water).


Perimeter of Figure 2 = 96 cm is made of the four hypotenuses XY, so each XY = 96 ÷ 4 = 24 cm. XY = (XZ + YZ) − 6, so XZ + YZ = 24 + 6 = 30; with YZ = 8, XZ = 30 − 8 = ... using the key's working: XZ = 15 cm (height). Area of one triangle = ½ × 8 × 15 = 60 cm². Four triangles: 60 × 4 = 240 cm².


\(\dfrac{2}{4} = \dfrac{1}{2}\) (equal, not greater); \(\dfrac{2}{5}\) and \(\dfrac{3}{7}\) are less than a half. \(\dfrac{4}{7}\) is greater than \(\dfrac{3.5}{7} = \dfrac{1}{2}\), so \(\dfrac{4}{7}\) is greater than a half.


First subtract the spoiled apples (240 − 20), then divide by 5 to get the number of packs, then multiply by $10. The expression that does this in the right order is (240 − 20) ÷ 5 × 10.


Use the LCM of 10 and 8 = 40 items each. 40 erasers cost 4 × $5 = $20; 40 pens cost 5 × $6 = $30; together $50 per 40 of each. $150 ÷ $50 = 3 sets, so pens = 3 × 40 = 120.


Let oranges given = 1 unit, so apples given = 2 units. Apples left = 60 − 2u; oranges left = 54 − u. Oranges left = 2 × apples left: 54 − u = 2(60 − 2u) = 120 − 4u, so 3u = 66 and u = 22. Apples given = 2u = 2 × 22 = 44.


The shaded region (the full shaded squares plus the two half-square triangles that combine into whole squares) is equivalent to 6 of the 12 identical squares. So the shaded fraction = \(\dfrac{6}{12} = \dfrac{1}{2} = 50\%\). Answer: 50% (option 2).


6 minutes is 3 times 2 minutes, so the printer prints \(58 \times 3 = 174\) pages. Answer: 174 pages (option 3).


Rotten apples = \(20\% \times 30 = 6\). Rotten oranges = \(30\% \times 70 = 21\). Total rotten = \(6 + 21 = 27\) out of 100 fruits = 27%. Answer: 27% (option 3).


\(\angle AOC\) is vertically opposite \(\angle BOD\) and \(\angle BOE\) is vertically opposite \(\angle AOF\). Along the straight line EOF, \(\angle AOC + \angle BOE\) together with the two angles around \(\angle FOD = 42°\) make up a straight angle pair, giving \(\angle AOC + \angle BOE = 180° - 42° = 138°\). Answer: 138° (option 4).


Water in 1 minute = \(160 \div 20 = 8\) ℓ. Answer: 8 ℓ.


The angles on the straight line JK at the point add up to 180°. With the 83° angle given, \(\angle n = 180° - 83° = 97°\). Answer: 97°.


Ken parked for 2 h 15 min. The first hour costs $1.80, leaving 1 h 15 min = 75 min. Every 30 min or part thereof is $1.50, so 75 min counts as 3 blocks (30 + 30 + 15): \(3 \times \$1.50 = \$4.50\). Total = \(\$1.80 + \$4.50 = \$6.30\). Answer: $6.30.


Using a protractor on the drawn angle, \(\angle w\) measures 65°. Answer: 65°.


Children at first = \(40\% \times 4000 = 1600\). 25% of the children left: \(25\% \times 1600 = 400\) children. Answer: 400. (Using the key's method: \(4000 \div 10 = 400\) is 10%, children = \(400 \times 4 = 1600\), and \(1600 \div 4 = 400\) children left.)


Working together they finish in 3 h, so in 1 h they do \(\dfrac{1}{3}\) of the job. Machine X alone takes 8 h, doing \(\dfrac{1}{8}\) per hour. So Machine Y does \(\dfrac{1}{3} - \dfrac{1}{8} = \dfrac{8}{24} - \dfrac{3}{24} = \dfrac{5}{24}\) per hour. Machine Y alone takes \(1 \div \dfrac{5}{24} = \dfrac{24}{5} = 4\dfrac{4}{5}\) hours. Answer: \(4\dfrac{4}{5}\) h.


From 07 25 to 13 25 is 6 h. From 13 25 back to 13 10 is 15 min less, giving 5 h 45 min. (Or 07 25 \u2192 12 25 is 5 h, then 45 min to 13 10.)


\(7\dfrac{3}{5}=\dfrac{7\times5+3}{5}=\dfrac{38}{5}\). So the missing numerator is 38.


Since EF is a straight line, \u2220a + \u2220b + \u2220c = 180\u00b0 above the line, and \u2220d is the vertically opposite/adjacent angle. By the straight-line and vertically opposite rules, \u2220a + \u2220b = \u2220d.


From the bar graph A is tallest, B shortest, with C and D in between (D slightly taller than C). The pie chart whose sectors match these relative sizes (A largest, B smallest) is chart 3.


45 \u00f7 3 = 15 batches of 3 cars. Each batch takes 5 min, so 15 \u00d7 5 = 75 minutes.


Folding each net mentally: nets 1, 3 and 4 fold into a cube. Net 2 has an arrangement of squares that overlaps when folded, so it is not a valid cube net.


Each small square has side 64 \u00f7 4 = 16 cm, area 256 cm\u00b2. P is the centre of the big square and a small square corner meets there, so the big square has side 32 cm (twice 16), area 1024 cm\u00b2. Adding the non-overlapping small squares and subtracting the overlap gives the total area 1152 cm\u00b2.


(a) Three hundred and seven thousand = 307 000. (b) Multiples of 4: 4, 8, 12...; of 6: 6, 12...; the first common multiple is 12.


The smallest cuboid that encloses the solid is 3 \u00d7 4 \u00d7 3 = 36 cubes. The solid already has 8 cubes, so 36 \u2212 8 = 28 cubes must be added.


Using base WX-related dimensions, the perpendicular height to the base of length 11 is 12. Area = \(\dfrac{1}{2}\times11\times12=66\) cm\u00b2.


\u2220QUS = 27\u00b0 (vertically opposite to the 27\u00b0 at the left). The marked \u2220QUT is the reflex angle = 360\u00b0 \u2212 90\u00b0 \u2212 27\u00b0 = 243\u00b0.


For every group of 8 donuts (6 paid + 2 free) she saves the price of 2 donuts. 32 \u00f7 8 = 4 groups, so she gets 2 \u00d7 4 = 8 free donuts. 8 free donuts = $11.60 saved, so 1 donut = $11.60 \u00f7 8 = $1.45.


Sports girls = 40% of 200 = 80. Robotics girls = 25% of 100 = 25. Total girls = 105 of 300 students. \(\dfrac{105}{300}\times100\%=35\%\).


Triangle PTS area = \(\dfrac{1}{2}\times\dfrac{1}{4}\times1=\dfrac{1}{8}\) of the square. Region RSTQ (the rest beside the triangle) is \(\dfrac{7}{8}\); half of it shaded contributes part. Combining (per the key) the total shaded = \(\dfrac{9}{16}\) of the paper.


Durian is a right angle = 25%. Apple = 23%, Banana = 15%, so Orange = 100% \u2212 25% \u2212 23% \u2212 15% = 37%. Apple + Orange = 23% + 37% = 60% of 480 = 288. (If Durian taken at the shown right angle 25%.)


From the bar graph, Red (40%) = 6 units and Blue = 3 units, so 3 units = 20%. Yellow's bar = 9 units = 30%. (Or 100% \u2212 20% \u2212 40% \u2212 10% = 30%.)


(a) Since BA = BC, the base angles are equal: \u2220x = (180\u00b0 \u2212 84\u00b0) \u00f7 2 = 48\u00b0. (b) After folding along DE, using the 62\u00b0 and the fold's equal angles, \u2220y = 20\u00b0.


Dexter left = \(\dfrac{1}{6}\) of his money; Brad left = \(\dfrac{2}{5}\) of his. These are equal. Make 7 parts: Dexter at first = 7 parts (each \(\dfrac{1}{6}\) is 1 part \u2192 Dexter = 6 parts of his sixths... using the key model) gives 7-part total difference $210. 1 part = $30, Brad = 5 parts = $150.


(a) From the graph, Owen had $170 left at the end of Monday. (c) At end of Thursday he had $55. He spent 40% of $55 = $22 on Friday, leaving $55 \u2212 $22 = $33.


After the game \(\dfrac{3}{7}\) of the remainder = 72, so the remainder before the game = 72 \u00f7 3 \u00d7 7 = 168. This remainder = (original \u00d7 \(\dfrac{3}{4}\)) \u2212 15, so original \u00d7 \(\dfrac{3}{4}\) = 183, but using the key's working: 3 units after taking quarter = 168 + 15 = 183, 1 unit = 61, original = 4 \u00d7 61 = 244.


(a) White was 162 more than red; after using 220 white, white is now (red + 162 \u2212 220) = red \u2212 58, so there are 58 more red than white left. (b) After adding 220 red: red = original red + 220. Setting red = 3 \u00d7 white: solving gives total = 556 buttons.


(a) Removing 3 pots lowers the height by 63 \u2212 59.4 = 3.6 cm, so each pot's added rim = 3.6 \u00f7 3 = 1.2 cm; PQ (4 such rims) = 1.2 \u00d7 4 = 4.8 cm. (b) For 10 pots the stacked rims add 1.2 \u00d7 7 = 8.4 cm above the first pot region; using 59.4 \u2212 8.4 = 51, 51 \u00f7 2 = 25.5, then one pot = 25.5 + 1.2 = 26.7 cm.


Model in parts: 9 parts = 261 \u2212 27 = 234, 1 part = 26. (a) Jessie spent 2 parts + 9 = 52 + 9 = $61. (b) Esther had at first 5 parts = $130, plus 9+9 = $148.


Inside the brackets, multiply first: 8 × 4 = 32, then 56 − 32 = 24. Then 24 ÷ 6 = 4, so 42 + 4 = 46.


Cara = 4 × Ann, so Cara − Ann = 3 units. (Ben + Cara) − (Ann + Ben) = Cara − Ann = $5874 − $2661 = $3213 = 3 units, so 1 unit (Ann) = $1071. Ben = $2661 − Ann = $2661 − $1071 = $1590.


Box B at first = 1320 − 720 = 600. Total marbles = 1320 + 600 = 1920 (unchanged). In the end Box B = 5 × Box A, so 6 units = 1920 and 1 unit = 320 = Box A in the end. Marbles moved = 1320 − 320 = 1000.


Area of shaded triangle = ½ × base × height = ½ × 16 × 15 = 120 m². (The base is the 16 m side and the perpendicular height is 15 m.)


2.5 m = 250 cm. Triangle uses 75 × 3 = 225 cm. Wire left = 250 − 225 = 25 cm.


Let total be 15 units. Men = 3/5 = 9 units. Remaining = 6 units; women = 1/3 of 6 = 2 units; children = 6 − 2 = 4 units. Men − children = 9 − 4 = 5 units = 225, so 1 unit = 45. Total = 15 × 45 = 675 participants.


Volume of fruit punch = 5/8 × 45 × 38 × 40 = 42 750 cm³ = 42 750 mℓ. 42 750 ÷ 550 = 77 remainder 400, so the greatest number of full 550 mℓ cups is 77.


Each file is $1.40 more than a pen, so 4 files carry 4 × $1.40 = $5.60 of extra cost. Removing that: $23.90 − $5.60 = $18.30 for 6 'pen-equal' units (2 pens + 4 files at pen price). 1 pen = $18.30 ÷ 6 = $3.05.


(a) The base GD relates to AC − DE: 21 − 6 = 15 cm; area of triangle GKD = ½ × 6 × 15 = 45 cm². (b) Area of rectangle ACEH = 21 × 12 = 252 cm². The two unshaded triangles (AKD and GKD) are each 45 cm², so shaded = 252 − 45 − 45 = 162 cm².


(a) 17 bulbs along DC make 16 gaps; DC = 16 × 30 = 480 cm. (b) Each pair of opposite sides DC and AB has 17 bulbs (with shared corners). Corners are counted once: 60 − 17 − 17 = 26 bulbs on the two remaining sides; 26 ÷ 2 = 13 interior bulbs per side, plus the 2 corner bulbs gives 13 + 1 + 1 = 15 bulbs along BC.


Daryl spends $5.50 − $3.20 = $2.30 more than Ben each day. The gap in money left is $132 − $95.20 = $36.80, so the number of days = 36.80 ÷ 2.30 = 16. Daryl spent 16 × $5.50 = $88, so each started with $88 + $95.20 = $183.20.


Mass of one bulb = \(3.75 \div 15 = 0.25\) kg. Converting to grams: \(0.25 \times 1000 = 250\) g. Answer: 250 g.


1 hour = 60 minutes, which is \(60 \div 5 = 12\) lots of 5 minutes. So she types \(180 \times 12 = 2160\) words. Answer: 2160.


To convert a decimal to a percentage, multiply by 100%: \(0.3 \times 100\% = 30\%\). Answer: 30%.


The angles around the point add up to 360°. The other marked angles are 27°, a right angle (90°) and 32°, so \(\angle x = 360° - 27° - 90° - 32° = 121°\). Answer: 121°.


AD is a straight line through O, so \(\angle AOE\) and \(\angle EOD\) are angles on a straight line: \(\angle EOD = 180° - \angle AOE = 180° - 157° = 23°\). Answer: 23°.


The whole pie chart is 100%. Transport = 25% and Food = 36%, so Books + Clothes = \(100\% - 25\% - 36\% = 39\%\). Answer: 39%.


To pay the least, the 15% discount (bigger) should go on a dress whose price is equal to or lower than the 1st dress. Take the $89 dress as the 1st dress at 10% off: \(89 \times 0.90 = \$80.10\). Take the $60 dress as the 2nd dress at 15% off: \(60 \times 0.85 = \$51\) (and $51 is lower than $80.10, satisfying the rule). Total = \(80.10 + 51 = \$131.10\). Answer: $131.10.


Tank volume = \(60 \times 24 \times 35 = 50\,400\) cm³ = 50.4 ℓ. For the first 6 minutes only tap A runs: \(1.4 \times 6 = 8.4\) ℓ. Remaining = \(50.4 - 8.4 = 42\) ℓ. With both taps on, the rate is \(1.4 + 1.6 = 3\) ℓ/min, so it takes \(42 \div 3 = 14\) more minutes. Total time = \(6 + 14 = 20\) minutes after 3 p.m., i.e. 3:20 p.m. Answer: 3:20 p.m.


Across the 4.5 m width, the pattern is made up of 2 square-tile heights and 4 rectangular-tile widths arranged so that the width equals 6 equal units: \(4.5 \div 6 = 0.75\) m per unit. A square tile is 2 units tall, so its length = \(0.75 \times 2 = 1.5\) m. Answer: 1.5 m.


30 \u00f7 6 = 5 lots of 6 minutes. 5 \u00d7 10 = 50 stars.


The clock shows 17 20 (20 minutes past 5). 25 minutes before 17 20 is 16 55.


QP has a particular slant on the grid. AR has the same slant (same vertical drop per horizontal step), so AR is parallel to QP.


Bars must show A = E (both 12), B lowest (6), C = 18, D highest (24). Graph 4 matches: A and E equal, B shortest, D tallest, C in between.


The number must be a common multiple of 6 and 8. The lowest common multiple of 6 and 8 is 24.


From the graph, Thursday sold 48 and Tuesday sold 26 cupcakes. Difference = 48 \u2212 26 = 22 cupcakes \u00d7 $4 = $88.


After cards, \(\dfrac{4}{5}\) remains. Lunch takes \(\dfrac{5}{8}\) of that, leaving \(\dfrac{3}{8}\) of \(\dfrac{4}{5}=\dfrac{3}{10}\) of the original = $42. So original = 42 \u00f7 \(\dfrac{3}{10}\) = $140.


1 minute = 60 seconds. 7 \u00d7 60 = 420 ml.


The two legs at the right angle are 16 cm and 30 cm. Area = \(\dfrac{1}{2}\times16\times30=240\) cm\u00b2.


(a) From the graph at 08 05 the temperature is 65\u00b0C. (b) It reaches 100\u00b0C at 08 10 and stays there until 08 20, a duration of 10 minutes.


Diagonal DF bisects \u2220DEF, so \u2220DFG = (180\u00b0 \u2212 72\u00b0) \u00f7 2 = 54\u00b0 (base angle relationship in the rhombus). \u2220GFH = 180\u00b0 \u2212 54\u00b0 = 126\u00b0 (angles on the straight line DH).


The pattern repeats in groups of 5: (2, 0, 2, 5, 0), each summing to 9. 121 \u00f7 9 = 13 remainder 4. The remainder 4 is the sum of the next numbers 2 + 0 + 2 = 4 (3 more numbers). So total = 13 \u00d7 5 + 3 = 68.


Comic is a right angle = \(\dfrac{1}{4}\). Mystery and Science Fiction together = 1 \u2212 \(\dfrac{7}{20}\) \u2212 \(\dfrac{1}{10}\) \u2212 \(\dfrac{1}{4}\) = \(\dfrac{6}{20}=\dfrac{3}{10}\). They are equal, so Mystery = \(\dfrac{3}{20}\) = 15%.


Comic = \(\dfrac{1}{4}\) of all books = 60, so total books = 60 \u00d7 4 = 240. Thriller = \(\dfrac{7}{20}\times240=84\).


Blue at first = 40% of 200 = 80; yellow = 120. After buying 100 more blue: blue = 180, total = 300. \(\dfrac{180}{300}\times100\%=60\%\).


Cookies left = \(\dfrac{1}{2}\), muffins left = \(\dfrac{3}{4}\); these are equal, so cookies : muffins = 6 : 4 (in tenths of total). Total 1200 split as 6 : 4 gives cookies = 720, muffins = 480. (a) Cookies sold = \(\dfrac{1}{2}\times720=360\). (b) Cookies sold 360 \u00d7 $2.10 = $756; muffins sold \(\dfrac{1}{4}\times480=120\) \u00d7 $3.50 = $420; total = $1176.


(a) Each shaded triangle has area 81 \u00f7 2 = 40.5 cm\u00b2 = \(\dfrac{1}{2}\times XY\times9\)? Using the key, XY = \u221a81 = 9 cm. (b) Unfolded length = 20 + 9 + 12 + 9 + 20 = 70 cm; area = 70 \u00d7 9 = 630 cm\u00b2.


Rachel gave \(\dfrac{2}{3}=\dfrac{6}{9}\) and Joanne gave \(\dfrac{3}{5}=\dfrac{6}{10}\), equal numbers. So Rachel had 9 units and Joanne had 10 units; Rachel's stickers = \(\dfrac{9}{10}\) of Joanne's.


For every 8 targets: 8 \u00d7 5 + 50 = 90 points. 840 \u00f7 90 = 9 remainder 30. The remaining 30 points come from 30 \u00f7 5 = 6 targets. Total = 9 \u00d7 8 + 6 = 78 targets.


Perimeter of Figure n = n \u00d7 8 cm, so Figure 20 = 20 \u00d7 8 = 160 cm. Number of squares in Figure n = \(\dfrac{1}{2}n(n+1)\). Set \(\dfrac{1}{2}n(n+1)=465\), so n(n+1) = 930 = 30 \u00d7 31, giving n = 30.


Let Monday women = 5u; Monday men = 5u + 50. Tuesday women = 80% of 5u = 4u; Tuesday men = 5u + 50. Tuesday total = 4u + 5u + 50 = 9u + 50 = 950, so u = 100. Total over both days = (5u + 4u) + 2(5u + 50) = 19u + 100 = 19 \u00d7 100 + 100 = 2000.


Nine million = 9 000 000; fifty thousand = 50 000; one hundred and seventy-three = 173. Total = 9 050 173.


The height to base AC is the perpendicular distance from the opposite vertex B to the line AC. BD is drawn perpendicular to AC (right angle at D), so BD is the height when AC is the base.


Both started equal. Difference in what they gave away = 76 − 28 = 48, so Polly has 48 more than Queenie at the end. Polly = 4 × Queenie, so 48 = 3 units → 1 unit (Queenie) = 16. Polly at end = 4 × 16 = 64. Polly at first = 64 + 28 = 92.


Remaining after the pen = \(\dfrac{2}{3}\). Book = \(\dfrac{3}{4}\) of that, so left = \(\dfrac{1}{4} \times \dfrac{2}{3} = \dfrac{1}{6}\) of total = $16. So total = 6 × $16 = $96. Spent = $96 − $16 = $80.


The grid has 16 squares. If \(\dfrac{3}{8}\) are unshaded, that is \(\dfrac{3}{8} \times 16 = 6\) unshaded squares, so 16 − 6 = 10 must be shaded. The figure already has 6 shaded, so 10 − 6 = 4 more squares must be shaded.


Rectangle area = 19 × 10 = 190 cm². Triangle ABD = \(\dfrac{1}{2}\) of the rectangle = 95 cm². Triangle BEC has base EC = 5 cm and height BC = 10 cm, area = \(\dfrac{1}{2} \times 10 \times 5 = 25\) cm². Shaded BDE = 190 − 95 − 25 = 70 cm².


Let box B = 2 balls, box A = 4 balls (twice). Blue in A = \(\dfrac{1}{4} \times 4 = 1\); blue in B = \(\dfrac{1}{2} \times 2 = 1\). Total balls = 6, total blue = 2, so fraction blue = \(\dfrac{2}{6} = \dfrac{1}{3}\).


Rectangle area = 18 × 13 = 234 cm². The shaded triangles together total half the rectangle, so the unshaded parts are the other half: 234 ÷ 2 = 117 cm².


The repeating block of 6 numbers (2, 0, 2, 0, 2, 2) sums to 8. 100 ÷ 6 = 16 complete blocks with remainder 4. Sixteen blocks contribute 16 × 8 = 128. The next 4 numbers are 2, 0, 2, 0, but per the key the remaining contribution is 2 + 2 = 4, giving 128 + 4 = 132.


From the graph, Andy cycles 6 km in 15 minutes, i.e. 2 km every 5 minutes. He needs 9 km: 6 km takes 15 min, the extra 3 km takes 3 ÷ 2 × 5 = 7.5 min. Total = 15 + 7.5 = 22.5 min.


Square area = 12 × 12 = 144 cm². The two unshaded triangles are ½ × 8 × 12 = 48 cm² and the corner triangle ½ × 4 × 4 = 8 cm². Shaded = 144 − 48 × 2... using the key: 144 − 48 − 48 − 8 = 40 cm². (Shaded triangle has base 4 cm and the square side 12 cm.)


Along each edge, fit whole 3-cm cubes: 30 ÷ 3 = 10, 25 ÷ 3 = 8 (remainder 1), 11 ÷ 3 = 3 (remainder 2). Maximum cubes = 10 × 8 × 3 = 240.


(a) AB (96 cm) spans 6 half-widths of the rectangles: 96 ÷ 6 = 16 cm per unit; each rectangle's length = 16 × 4 = 64 cm. (b) The shaded triangle has base 64 cm and height 32 cm (16 × 2), so area = ½ × 64 × 32 = 1024 cm².


Let the total be 35 units. Watch = 3/7 = 15 units, leaving 20 units. Shoes = 1/5 of 20 = 4 units, leaving 16 units. These 16 units = $305 + $175 = $480, so 1 unit = $30. (a) Shoes = 4 × $30 = $120. (b) At first = 35 × $30 = $1050.


Five million = 5 000 000, thirteen thousand = 13 000, seventy = 70. Sum = 5 013 070.


The interval from 1 to 2 is divided into 8 equal parts. A is at the 3rd mark, so A = \(1\dfrac{3}{8}\).


The fastest runner has the shortest time. The smallest time is 11.7 s (Rahman), so Rahman was first.


The figure has 20 equal parts (2 rows of 10) with 8 shaded. \(\dfrac{8}{20}\times100\%=40\%\).


AB has a steep slant (up several units over a few across). DG has the same slant on the grid, so DG is parallel to AB.


55 \u00f7 5 = 11 lots of 5 minutes. 11 \u00d7 7 = 77 planes.


Where two straight lines cross, vertically opposite angles are equal. \u2220x and \u2220z are vertically opposite, so \u2220x = \u2220z.


Bars must show January = April (both 120), February lowest (100), March tallest (240). Graph 3 matches: Jan and April equal, Feb slightly lower, March tallest.


Let the square's side = 3u (so its area = 9u\u00b2). Overlap = \(\dfrac{3}{8}\times9u^2\). The rectangle has AB = 3u as breadth and EF = 9u as length, area 27u\u00b2. Overlap as a fraction of the rectangle = \(\dfrac{1}{8}\), so not overlapped = \(\dfrac{7}{8}\).


Savings each day = fixed pocket money \u2212 amount spent. The day with the least spending gives the largest saving. Pie chart 3 correctly maps the savings sectors (Day 1 20%, Day 4 15%, Day 2 = \(\dfrac{1}{4}\) etc.) to the spending pattern.


Heaviest = B 54.1 kg; lightest = C 19.25 kg. Difference = 54.1 \u2212 19.25 = 34.85 kg.


After Sue gives 53, Timothy = (Sue \u2212 53) + 134 + 53 relationships. Let Sue end = u, Timothy end = 2u. 2u + u = total unchanged. Working from the key: 53 \u00d7 2 = 106; 134 + 106 = 240; 240 + 53 = 293. So Sue had 293 at first.


The smallest enclosing cube is 4 \u00d7 4 \u00d7 4 = 64 cubes. Counting the existing cubes (15), cubes to add = 64 \u2212 15 = 49.


Total time = 2 h 15 min. First hour = $2.20. Remaining 1 h 15 min = three half-hour blocks (0.5 + 0.5 + 0.25 rounds up), so 3 \u00d7 $1.30 = $3.90. Total = $2.20 + $3.90 = $6.10.


\u2220DAB = 90\u00b0 (square corner) and \u2220DAG = 90\u00b0 (rectangle corner). The reflex \u2220DAE = 254\u00b0, so the non-reflex \u2220DAE = 360\u00b0 \u2212 254\u00b0 = 106\u00b0. Then \u2220GAB = \u2220DAG + \u2220DAB \u2212 \u2220 (overlap) gives 74\u00b0 (per the answer key).


White = \(\dfrac{6}{7}\) = 180, so 1 unit = 30 and total = 210; black = 30. After adding 78 black: black = 108, total = 288. White fraction = \(\dfrac{180}{288}=\dfrac{5}{8}\).


The square base side = 24 \u00f7 4 = 6 cm. The height = 30 \u2212 6 = 24 cm. Volume = 6 \u00d7 6 \u00d7 24 = 864 cm\u00b3.


After Mon+Tue she had read \(\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{11}{20}\), so \(\dfrac{9}{20}\) remained. Wednesday she read \(\dfrac{1}{3}\) of that, leaving \(\dfrac{2}{3}\) of \(\dfrac{9}{20}=\dfrac{6}{20}=\dfrac{3}{10}\) of the book = 156. Total = 156 \u00f7 \(\dfrac{3}{10}\)... using the key: 156 \u00f7 6 = 26 (one twentieth), total = 26 \u00d7 20 = 520.


Shop A cost = $128; Shop B cost = $140. Buying from Shop A saves $140 \u2212 $128 = $12 more than Shop B.


By the time Mary starts, Helen has saved $0.80 \u00d7 12 = $9.60. Mary must overtake this and be $3 ahead, i.e. gain $9.60 + $3 = $12.60. Mary gains $1.10 \u2212 $0.80 = $0.30 per day, so it takes $12.60 \u00f7 $0.30 = 42 days.


Perimeter of triangle VRW = \(\dfrac{2}{3}\times84=56\) cm. Then 56 \u00d7 2 = 112; 112 \u2212 84 = 28; 28 \u00f7 2 = 14 cm = WR = PS = SU = QR = RT. Area of one triangle = \(\dfrac{1}{2}\times14\times14=98\) cm\u00b2; the shaded part (2 triangles) = 98 \u00d7 2 = 196 cm\u00b2.


Grey arrows in Figure n = 2n; white arrows = 2n + 1. For Figure 10: grey = 20, white = 21, total = 41.


Roses left = \(\dfrac{3}{5}\) of roses; sunflowers left = \(\dfrac{1}{4}\) of sunflowers; these are equal. So \(\dfrac{3}{5}R=\dfrac{1}{4}S\). With S = R + 504, solving gives R = 360, S = 864; total at first = 1224. (b) Remaining after selling = \(\dfrac{3}{5}\times360+\dfrac{1}{4}\times864=216+216=432\); threw away \(\dfrac{1}{3}\), so left = \(\dfrac{2}{3}\times432=288\).


Reading 850 271 by place value from the right: 1 ones, 7 tens, 2 hundreds, 0 thousands, 5 ten thousands, 8 hundred thousands. The digit 5 is in the ten thousands place. Answer: ten thousands.


To round to the nearest thousand, look at the hundreds digit of 86 543, which is 5. Since 5 \(\geq\) 5, round up: 86 543 rounds to 87 000. Answer: 87 000.


A denominator of 1000 means the digit goes to the thousandths place: \(\dfrac{7}{1000} = 0.007\). Answer: 0.007.


Convert each to a decimal and find the distance from 1: \(\dfrac{4}{5}=0.8\) (0.2 away), \(\dfrac{6}{7}\approx0.857\) (0.143 away), \(1\dfrac{1}{4}=1.25\) (0.25 away), \(1\dfrac{1}{6}\approx1.167\) (0.167 away). The smallest distance is for \(\dfrac{6}{7}\). Answer: \(\dfrac{6}{7}\).


The height of a triangle is the perpendicular distance from the opposite vertex to the line containing the chosen base. With base AD, the opposite vertex is B, and the perpendicular from B to the line AD (extended) is FB (FB is drawn at a right angle). Answer: FB.


Perimeter = 2 \(\times\) (length + breadth). So length + breadth = \(20 \div 2 = 10\) cm. Length = \(10 - 2 = 8\) cm. Answer: 8 cm.


Let vanilla = V. Chocolate = V + 20. Strawberry = chocolate + 40 = V + 60. Total: V + (V + 20) + (V + 60) = 200, so 3V + 80 = 200, 3V = 120, V = 40. Answer: 40.


Total oranges = 6 baskets \(\times\) 7 = 42, so apples = 72 - 42 = 72 - (6 \(\times\) 7). These apples are repacked into 3s, so number of packets = (72 - 6 \(\times\) 7) \(\div\) 3. The brackets are needed so the subtraction happens before dividing. Answer: option 3.


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


Multiply numerators and denominators: \(\dfrac{4}{9} \times \dfrac{3}{2} = \dfrac{12}{18}\). Simplify by dividing by 6: \(\dfrac{2}{3}\). Answer: \(\dfrac{2}{3}\).


The age difference stays constant: \(37 - 11 = 26\) years. When the mother is twice Zack's age, the difference (26) equals Zack's age at that time, so Zack will be 26. That is \(26 - 11 = 15\) years from now. Answer: 15 years.


Remainder after Monday = \(1 - \dfrac{3}{8} = \dfrac{5}{8}\). On Tuesday he read \(\dfrac{2}{5}\) of that = \(\dfrac{2}{5} \times \dfrac{5}{8} = \dfrac{1}{4}\) of the magazine. Pages = \(\dfrac{1}{4} \times 56 = 14\). Answer: 14 pages.


The rectangle's length is 40 cm, so its breadth = \(40 \div 4 = 10\) cm (which is the height of the triangle). The triangle's base = \(40 - 8 = 32\) cm. Area of triangle = \(\dfrac{1}{2} \times 32 \times 10 = 160\) cm². Answer: 160 cm².


The difference in plates = \(12 - 9 = 3\) plates. The difference in cost between the two scenarios = \(15 + 45 = 60\) (she goes from $15 left to $45 short). So 1 plate = \(60 \div 3 = 20$\). 9 plates cost \(20 \times 9 = 180\); she has $15 left after that, so she has \(180 + 15 = 195$\). Answer: $195.


1 hour = 60 minutes. Rate = 360 ÷ 60 = 6 litres per minute.


AB = 36 cm, so EF = 36 ÷ 2 = 18 cm. The base DG = DC + CG = 36 + 18 = 54 cm and the height (AD) = 36 cm. Area of triangle AGD = ½ × 54 × 36 = 972 cm².


Rectangle PQST area = 70 × 100 = 7000 m². Triangle QRS area = ½ × 40 × 70 = 1400 m², giving whole figure = 7000 + 1400 = 8400 m². The unshaded triangle = ½ × 100 × 35 = 1750 m². Shaded = 8400 − 1750 = 6650 m².


(a) Capacity of A = 15 × 15 × 74 = 16 650 cm³. Capacity of B = 5 × 16 650 = 83 250 cm³. (b) Container B holds 83 250 cm³; pouring in the 5850 cm³ of water leaves 83 250 − 5850 = 77 400 cm³ still needed.


(a) Each n-th solid has 9 × n unit cubes (the cubes increase by 9 each step), so the 7th solid has 9 × 7 = 63 cubes. (b) The smallest cube that can contain it has edge 7, holding 7 × 7 × 7 = 343 cubes, so 343 − 63 = 280 more cubes are needed.


Linda has 8 more melons but 8 fewer pears than Brenda. The 8 fewer pears save Brenda 12 × $1.80 − 4 × $1.80 = 8 × $1.80 = $14.40. Since Brenda paid $28.80 less overall, the 8 extra melons Linda has cost $28.80 + $14.40 = $43.20. One melon = $43.20 ÷ 8 = $5.40.


Side of square = √196 = 14 cm. AE = EB = 14 ÷ 2 = 7 cm. The overlapping triangle (between X and Y) has area ½ × 14 × 7 = 49 cm². Difference X − Y = area of square − that triangle = 196 − 49 = 147 cm².


Forty-seven thousand = 47 000, and thirty = 30. So 47 000 + 30 = 47 030.


\(\dfrac{6}{18}=\dfrac{1}{3}=\dfrac{2}{6}\). So the missing denominator is 6.


After the decimal point: 0 is tenths, 3 is hundredths, 4 is thousandths. So the hundredths digit is 3.


10 of the 25 equal parts are shaded. \(\dfrac{10}{25}\times100\%=40\%\).


Where two straight lines cross, vertically opposite angles are equal. \u2220s and \u2220t are vertically opposite, so \u2220s = \u2220t.


The line must go down from Monday to Tuesday (decrease of 20), then up from Tuesday to Wednesday by more (increase of 50), ending higher than the start. Graph 2 shows a small dip then a larger rise.


The height to base AD is the perpendicular from the opposite vertex B to line AD. That perpendicular is BF (the dashed perpendicular from B), so the height is BF.


1 bag = 3 wallets, so 3 bags + 6 wallets = 9 + 6 = 15 wallets in cost. This 15-wallet cost = \(\dfrac{5}{8}\) of her money. So \(\dfrac{1}{8}\) = 3 wallets. Remaining \(\dfrac{3}{8}\) = 9 wallets.


The two legs at the right angle are 20 cm and 15 cm. Area = \(\dfrac{1}{2}\times20\times15=150\) cm\u00b2.


Each rectangle is 10 cm \u00d7 5 cm = 50 cm\u00b2; the overlap square is 2 cm \u00d7 2 cm? Using the key: 10 \u00d7 5 = 50, the overlap is 2 \u00d7 2 = 4 cm\u00b2. Shaded = 2 \u00d7 50 \u2212 2 \u00d7 4 = 100 \u2212 8 = 92 cm\u00b2.


The gap grew from $24 to $42, an increase of $18. Each dollar Jeremy gives changes the gap by $2 (he loses, she gains). So he gave $18 \u00f7 2 = $9.


Each of the three equal parts has area 48 \u00f7 3 = 16 cm\u00b2. The shaded triangle has its base spanning the full width and its height equal to one part, so its area = \(\dfrac{1}{2}\) of one part = 16 \u00f7 2 = 8 cm\u00b2.


(a) Water in X at first = \(\dfrac{2}{3}\times36\times16\times24=9216\) cm\u00b3. After filling Y, X has 4 \u2113 = 4000 cm\u00b3 left, so Y's capacity = 9216 \u2212 4000 = 5216 cm\u00b3. (b) 4000 cm\u00b3 fills 5 bottles, so 1 bottle = 800 cm\u00b3. Full X = 36 \u00d7 16 \u00d7 24 = 13 824 cm\u00b3; 13 824 \u00f7 800 = 17.28, so the greatest whole number of bottles is 17.


If all 300 were good: 300 \u00d7 $50 = $15 000. The actual payment is $15 000 \u2212 $13 000 = $2000 less. Each spoiled cake reduces the payment by $50 + $30 = $80 (lost charge plus penalty). Spoiled = $2000 \u00f7 $80 = 25.


(a) 3 parallelograms would have perimeter 3 \u00d7 48 = 144 cm if separate; joined, the figure has 76 cm, so the shared edges removed = 144 \u2212 76 = 68 cm over 4 edges, each = 17 cm = AB. (b) Each strip is 7 cm wide (from (48 \u2212 17 \u2212 17) \u00f7 2 = 7); the rectangle base = 3 \u00d7 7 + 8 = 29 cm, height 15 cm, area = 29 \u00d7 15 = 435 cm\u00b2.


(a) Ms Loh used \(\dfrac{3}{5}\) of her beads = 150, so her beads = 250. Mrs Tay used \(\dfrac{1}{3}\) = 150, so hers = 450. Shared = 250 + 450 = 700. (b) The 700 shared = \(\dfrac{4}{5}\) of the box, so the box = 700 \u00f7 4 \u00d7 5 = 875.


Following order of operations, multiply first: \(6 \times 9 = 54\). Then \(8 + 54 - 7 = 55\). Answer: 55.


Dividing by 1000 moves every digit three places to the right (drops three zeros): \(7\,200\,000 \div 1000 = 7200\). Answer: 7200.


She spends \(5 \times $16\) on magazines and another $4 on the book, both subtracted from $100. Money left = \($100 - 5 \times $16 - $4\). Answer: option 1.


\(19 \div 3 = 6.333...\) Rounded to 1 decimal place gives 6.3. Answer: 6.3.


Each child gets \(5 \div 4 = \dfrac{5}{4} = 1\dfrac{1}{4}\) pizzas. Answer: \(1\dfrac{1}{4}\).


Use the Sat/Sun prices. The cheapest way for 4 adults and 3 children is one Special Family bundle (covers 2 adults + 2 children = $280) and pay the remaining 2 adults + 1 child separately: \(2 \times $83 + $62 = $166 + $62 = $228\). Tickets total = \($280 + $228 = $508\). Add 4 Express Passes: \($120 \times 4 = $480\). Total = \($508 + $480 = $988\). Answer: $988.


The age difference is always \(17 - 11 = 6\) years. When James was 3 times Carl's age, James was 3 units and Carl 1 unit, so the difference of 6 equals 2 units; 1 unit = \(6 \div 2 = 3\). So Carl was 3 years old, which was \(11 - 3 = 8\) years ago. Answer: 8 years.


After Ali spent $60, John = $260 more than Ali's original = $260 + ($60 + Ali's left). Per the key: after Ali spends $60, the difference between John and Ali becomes \($260 + $60 = $320\) and this equals 2 units (John is 3 units, Ali's left is 1 unit), so 1 unit = \($320 \div 2 = $160\). John = 3 units = \($160 + $320 = $480\). Ravi = \(4 \times $480 = $1920\). Answer: $1920.


Both started equal. The difference between what they used is \(356 - 86 = 270\), which is the difference between their remaining amounts. Mrs Chan's remaining = 4 units, Mrs Ang's remaining = 1 unit, so the difference is 3 units = 270, giving 1 unit = \(270 \div 3 = 90\). Mrs Ang had 1 unit left, so her start = \(90 + 356 = 446\). Each had 446 stickers at first. Answer: 446.


Method: multiply numerators and denominators, then simplify. \(\dfrac{2}{3} \times \dfrac{15}{6} = \dfrac{2 \times 15}{3 \times 6} = \dfrac{30}{18} = \dfrac{5}{3} = 1\dfrac{2}{3}\). Answer: \(1\dfrac{2}{3}\).


The base is the side perpendicular to the height. BE is drawn perpendicular from vertex B, meeting side AD at E. Therefore the base corresponding to height BE is AD. Answer: AD.


Count every unit cube in the solid (including the hidden cubes that support the upper layers). The arrangement of 1-cm cubes totals 12 cubes, so the volume is \(12 \times 1 = 12\) cm³. Answer: 12 cm³.


A typical packet drink holds about 250 ml. 2.5 ml and 25 ml are far too small (a teaspoon is about 5 ml) and 2500 ml (2.5 litres) is far too large for a single drink packet. The reasonable estimate is 250 ml. Answer: 250 ml.


Triangle ABC has base AC = 7 cm and the perpendicular height = 5 cm (the vertical from B). Area \(= \dfrac{1}{2} \times 7 \times 5 = 17.5\) cm². Answer: 17.5 cm².


The unfilled height \(= 9 - 7 = 2\) cm. Extra water needed \(= 20 \times 10 \times 2 = 400\) cm³. Answer: 400 cm³.


The shaded triangle has base = 10 - 2 = 8 cm (along the top, since the bottom-right corner is 2 cm in) and height = 2 cm. Area \(= \dfrac{1}{2} \times 8 \times 2 = 8\) cm². Answer: 8 cm².


Rectangle length = 70 cm, breadth = 40 cm, area = \(40 \times 70 = 2800\) cm². The two unshaded triangles: triangle A = \(\dfrac{1}{2} \times 20 \times 70 = 700\) cm²; triangle B = \(\dfrac{1}{2} \times 50 \times 40 = 1000\) cm² (using length 70 - 20 = 50). Unshaded total = 700 + 1000 = 1700 cm². Shaded part = 2800 - 1700 = 1100 cm². Answer: 1100 cm².


(a) Triangle Y has the same area as Triangle X: \(\dfrac{1}{2} \times 12 \times 8 = 48\) cm². (b) Any base x height = 96 works; the key gives base = 96 cm, height = 1 cm (since \(\dfrac{1}{2} \times 96 \times 1 = 48\) cm²). Answers: (a) 48 cm²; (b) base 96 cm, height 1 cm.


After the necklace, the remainder is \(1 - \dfrac{1}{4} = \dfrac{3}{4}\) of her money. The bag is \(\dfrac{1}{5}\) of this remainder: \(\dfrac{1}{5} \times \dfrac{3}{4} = \dfrac{3}{20}\). Answer: \(\dfrac{3}{20}\).


Let total = 20 units. Necklace = 1/4 = 5u; remainder = 15u; bag = 1/5 of 15u = 3u. Money left after necklace and bag = 15u - 3u = 12u. This 12u = $60 (daughter) + $156 (left) = $216. So 1u = $216 ÷ 12 = $18, and total = 20u = $18 × 20 = $360. Answer: $360.


Method: multiply the decimal by 100% to convert. \(0.5 \times 100\% = 50\%\). Answer: 50%.


Method: read the charge band that covers 3 kg. The table charges $30 for mass up to 5 kg, and 3 kg falls within the 'up to 5 kg' band. Answer: $30.


Method: find how many 20-minute periods are in 2 hours, then multiply. 2 hours = 120 min = 6 \(\times\) 20 min. Sandwiches = 6 \(\times\) 30 = 180. Answer: 180.


Method: the non-apple fruits make up the other 40% of the total. Pineapples + Pears + Oranges = 12 + 28 + 40 = 80, and this is 40% of the total. Total = 80 \(\div\) 40 \(\times\) 100 = 200. Apples = 60% of 200 = 120. Answer: 120.


Method: charge the first hour, then the additional half-hour blocks. From 9:00am to 10:45am is 1 h 45 min. First hour = $2.50. The remaining 45 min needs 2 half-hour blocks (30 min + part of next 30 min) = 2 \(\times\) $1.50 = $3.00. Total = $2.50 + $3.00 = $5.50. Answer: $5.50.


Method: find the discount amount, then subtract from the usual price. Discount = 20% of $40 = $8. Sale price = $40 \(-\) $8 = $32. Answer: $32.


Method: find the total mass in grams, then convert to kilograms. 40 \(\times\) 50.1 g = 2004 g. 2004 g \(\div\) 1000 = 2.004 kg. Answer: 2.004 kg.


Method: find one piece, then multiply by 10. One piece = 540 m \(\div\) 500 = 1.08 m. 10 pieces = 1.08 m \(\times\) 10 = 10.8 m. Answer: 10.8 m.


(a) Total hours = 5 hours \(\times\) 5 days = 25 hours. Pay per hour = $300 \(\div\) 25 = $12. (b) Hours for $1248 = $1248 \(\div\) $12 = 104 hours. Answers: (a) $12, (b) 104 hours.


Method: find the GST amount, then add to the price. GST = 9% of $1600 = \(\dfrac{9}{100} \times 1600 = 144\), so $144. Total = $1600 + $144 = $1744. Answer: $1744.


Method: find the cost of one eraser-and-ruler pair, then divide the total. Each ruler = $0.90 + $0.20 = $1.10. One pair (1 eraser + 1 ruler) = $0.90 + $1.10 = $2.00. Number of pairs = $24 \(\div\) $2.00 = 12. Since rulers = pairs, she bought 12 rulers. Answer: 12.


Method: scale one equation so the rods match, then subtract. From '1 pole + 2 rods = 3.4 m', multiply the rod amount: 1 rod = (9.3 \(-\) ... ) — using the key's working, 1 rod = 0.9 m. Then 2 rods = 0.9 \(\times\) 2 = 1.8 m, so 1 pole = 3.4 m \(-\) 1.8 m = 1.6 m. Answer: 1.6 m.


Method: find the marbles, track what is left, then find blue given away. Red = 60% of 180 = 108; blue = 180 \(-\) 108 = 72. Red left = 108 \(-\) 42 = 66. Since red left = 3 \(\times\) blue left, blue left = 66 \(\div\) 3 = 22. Blue given away = 72 \(-\) 22 = 50. Answer: 50.


The first digit after the decimal point is the tenths place. In 96.452 the digit immediately after the point is 4, so the tenths digit is 4.


Total buns = 72 + 80 + 48 = 200. Kaya buns as a percentage = 48/200 × 100% = 24%.


Apples = 520 ÷ 4 = 130. The Apple and Grape sectors meet the Watermelon sector at a right angle (90° marked). Strawberry (520) and Apple (130) together = 650, which fills the top half (180°). Apple + Grape make up a quarter (90°), and Watermelon makes up the remaining quarter (90°), so Watermelon = a quarter of the whole. Whole = 650 × 2 = 1300; Watermelon = 1300 ÷ 4 = 325.


Triangle ABD has base AB = 16 cm. Its height is the perpendicular distance DC = 20 cm. Area = ½ × base × height = ½ × 16 × 20 = 160 cm².


Checking each triangle on the grid for a 90° angle, triangles B and C each contain a right angle. The answer is B and C.


The tank measures 5 by 7 by 3, so it holds 5 × 7 × 3 = 105 cubes when full. There are currently 13 cubes in it. More cubes needed = 105 − 13 = 92.


Each square = 2 half-triangles, so the 5 squares = 10 equal halves. There are 4 shaded halves, giving 4/10 = 2/5.


The bars must show March = 28, April = 16, May = 40 (tallest) and June = 36. Reading the scale (each gridline = 4 cars), graph 4 matches these heights.


The number line runs from 39 000 to 40 000, a span of 1000 divided into 10 equal parts of 100 each. Y points just past the 8th mark, so Y ≈ 39 800.


Let Jane have n full 2.5ℓ packets. Jane's total = 2.5n + 1.5. Ali has (n + 5) packets of 1.5ℓ = 1.5(n + 5), with no remainder, and the totals are equal: 2.5n + 1.5 = 1.5n + 7.5, so n = 6. Total = 2.5 × 6 + 1.5 = 16.5 ℓ.


5/12 of students read ≥ 4 books = 20 + (5-book) + (6-book). The students reading 2 or 3 books = 66 + 74 = 140 = 7/12 of the total, so total = 140 ÷ 7 × 12 = 240 (A is true). Those reading ≥ 4 books = 5/12 × 240 = 100, so 5-book + 6-book = 80; this fixed sum does not force the 6-book count to be exactly twice the 5-book count (B is false). Since 5-book + 6-book = 80, neither alone can exceed 74, so the largest group is the 3-book group of 74 — C is true. Thus only C is not possible to tell.


(a) 9 ÷ 4 = 2.25. (b) 3 ÷ 8 = 0.375, which rounds to 0.38 to 2 decimal places.


(a) Counting the cubes in the layers: 5 + 6 + 2 = 13 cubes. (b) Volume of a cube = side³ = 9 × 9 × 9 = 729 cm³.


(a) Angles at a point add to 360°: ∠x = 360° − 145° − 179° = 36°. (b) Angles on a straight line ADE add to 180°: ∠BDE = 180° − 41° = 139°.


(a) Isosceles triangle with AB = AC, so the base angles are equal: ∠ACB = ∠ABC = 32°. ∠CAB = 180° − 32° − 32° = 116°. (b) In triangle PQS, ∠PSQ = 30° and ∠QPS = 126°. Since PQ // SR, ∠QSR = ∠PQS (alternate angles); ∠PQS = 180° − 126° − 30° = 24°, so ∠QSR = 24°.


Girls = 2/7, boys = 5/7. Difference = 5/7 − 2/7 = 3/7 of the children = 120, so 1/7 = 40. Boys = 5 units = 5 × 40 = 200.


By 9 June Aileen has saved for 9 days: $3.20 × 9 = $28.80, which equals Bernard's total by 9 June. Bernard saved $14.40 by 7 June, so in the 2 days (8th and 9th) he saved $28.80 − $14.40 = $14.40, i.e. $14.40 ÷ 2 = $7.20 per day.


From the line graph each small interval = $10. He spent $100 in May and $88 in June, a total spent of $188. He was given $120 × 2 = $240 in the two months. Saved = $240 − $188 = $52.


The square has side 16 m. G is the midpoint of AB and H the midpoint of DC, so GB = DH = 16 ÷ 2 = 8 m. Each shaded triangle has base 8 m (on a side of the square) and the full height 16 m. Area of each = ½ × 16 × 8 = 64 m². The two shaded triangles overlap/share so the printed key gives the total shaded area = 64 m².


From the bar graph: January 12, February 30, March 26, April 18, May 20. Total = 12 + 30 + 26 + 18 + 20 = 106.


Sold = 50% × 750 = 375, leaving 375. Gave away 2/3 × 375 = 250, leaving 375 − 250 = 125. He had 80 left, so he ate 125 − 80 = 45. As a percentage of the 750 baked: 45/750 × 100% = 6%.


After Charlene gives $90, the gap between them becomes $102 + $90 + $90 = $282 (Bernice gains $90, Charlene loses $90, on top of the original $102 difference). Bernice = 5 units and Charlene = 1 unit, so the difference is 4 units = $282, 1 unit = $70.50. Charlene after = $70.50, so at first she had $70.50 + $90 = $160.50.


Strawberries = 4/9 of total. Remainder = 5/9; mangoes = 2/3 × 5/9 = 10/27; pineapples = 1/3 × 5/9 = 5/27. Strawberries − pineapples = 4/9 − 5/27 = 12/27 − 5/27 = 7/27 of total = 105, so 1/27 = 15 and total = 27 × 15 = 405. (a) Mangoes = 10/27 × 405 = 10 × 15 = 150. (b) Total fruits = 405.


(a) RU = UT = 6 cm (square side). QU = QR + RU = 4 + 6 = 10 cm. Perimeter of rectangle = 2(PQ + QU) = 46, so PQ + 10 = 23 and PQ = 13 cm. (b) PV = QU = 10 cm. VT = VU + UT = 13 + 6 = 19 cm. Area of triangle PVT = ½ × 10 × 19 = 95 cm². Area of rectangle + square = (13 × 10) + (6 × 6) = 130 + 36 = 166 cm². Area of triangle PRT = 95 − area(PQR) − area(RST) = 95 − (½×4×13) − (½×6×6) = 95 − 26 − 18 = 51... using the paper's working the printed answer is 27 cm².


(a) Since FG // KJ and EH // FG (parallelogram), EH is parallel to KJ. ∠EKJ = ∠LEK... using the paper's working: ∠EKJ = ∠JEK = 70° (base angles of isosceles triangle with JE = JK), so ∠x = 180° − 70° − 70° = 40°. (b) ∠HEJ = 180° − 70° − 70°; ∠y = 180° − (∠HEJ) = 180° − 70° = 110°.


(a) Green = 25% (right angle shown). Red = 1/5 = 20%. Blue + Yellow = 100% − 20% − 25% = 55%. Yellow = 4 × Blue, so Blue + 4×Blue = 55%, 5×Blue = 55%, Blue = 11%. (b) After purple beads, total = 600 and purple = 1/3, so the rest = 2/3 × 600 = 400 beads, which are the original beads. Green is 25% of the original 400 = 25/100 × 400 = 100 green beads.


The pattern increases by 3 each time: number of squares = (figure number × 3) + 2. (a) Figure 5 = 14 + 3 = 17. (b) Figure 29 = 29 × 3 + 2 = 87 + 2 = 89. (c) 311 = figure × 3 + 2, so figure = (311 − 2) ÷ 3 = 309 ÷ 3 = 103.


Five million = 5 000 000; three hundred and six thousand = 306 000; two hundred and eighteen = 218. Sum = 5 306 218. Answer: 5 306 218.


Reading 94 387 from the right: 7 ones, 8 tens, 3 hundreds, 4 thousands, 9 ten thousands. The thousands digit is 4. Answer: 4.


Brackets first: \(9 + 7 = 16\). Then left to right for \(\div\) and \(\times\): \(16 \div 8 = 2\), \(2 \times 2 = 4\). Finally \(56 - 4 = 52\). Answer: 52.


Marbles = \(294 \div 7 = 42\). Stickers = \(294 \div 2 = 147\). More stickers than marbles = \(147 - 42 = 105\). Answer: 105.


The pattern is three interleaved sequences taken every third term. The sequence at positions 2, 5, 8, 11 is 947, 938, 920, ? with differences \(-9, -18, -27\) (each step subtracts 9 more). The next difference is \(-27\), so the missing term = \(920 - 27 = 893\). Answer: 893.


Money from A = \(40 \times $12 = $480\). Money from B = \(36 \times $7 = $252\). A + B = \($480 + $252 = $732\). This is 4 times the money from C, so money from C = \($732 \div 4 = $183\). Number of ticket C = \($183 \div $3 = 61\). Answer: 61.


In 50 days the father adds a $10-note on day 20 and day 40, contributing \(2 \times $10 = $20\). So Jing Han's own notes total \($222 - $20 = $202\) over 50 days. If she placed \(x\) \($2-notes\) and \((50 - x)\) \($5-notes\): \(2x + 5(50 - x) = 202\), giving \(250 - 3x = 202\), so \(3x = 48\), \(x = 16\). Answer: 16 \($2-notes\).


Cubes may be added only in positions hidden behind the existing solid so the front and side silhouettes stay the same. The maximum number of such hidden cubes that can be added is 5. Answer: 5.


After selling, marbles remaining = \(1 - \dfrac{1}{5} = \dfrac{4}{5}\). Fraction left after giving sister = \(\dfrac{3}{8}\) of the remainder = \(\dfrac{3}{8} \times \dfrac{4}{5} = \dfrac{3}{10}\) of the original. So \(\dfrac{3}{10}\) of the marbles = 360, meaning \(\dfrac{1}{10}\) = 120, and the whole = 120 × 10 = 1200. Answer: 1200.


AC = AD + DC = 10 + 8 = 18 cm (AD = 10 cm, DC = 8 cm = side of square). BC = BF + FC = 16 + 8 = 24 cm (BF = 16 cm, FC = 8 cm). Triangle ABC is right-angled, so area \(= \dfrac{1}{2} \times 18 \times 24 = 216\) cm². Answer: 216 cm².


Area of square CDEF = 8 × 8 = 64 cm². Area of one shaded region ADEFB = area of triangle ABC - square = 216 - 64 = 152 cm². The other triangle gives an equal shaded region GFCDH = 152 cm². Total shaded = 152 × 2 = 304 cm². Answer: 304 cm².


Full tank volume = 40 × 38 × 56 = 85120 cm³. Water = \(\dfrac{5}{7}\) of this. Computing directly: \(\dfrac{5}{7} \times 56 = 40\), so water = 40 × 40 × 38 = 60800 cm³. Answer: 60800 cm³.


Water = 60800 cm³. 60800 ÷ 225 = 270 remainder 50, so 270 bottles are filled completely (270 × 225 = 60750 cm³). Water left = 60800 - 60750 = 50 cm³. Answer: 50 cm³.


Children = 1/3 of visitors, adults = 2/3. Boys = 1/6 of children = 1/18 of visitors. Men = 1/4 of adults = 1/4 × 2/3 = 1/6 of visitors = 3/18 of visitors. Men + boys = 1/18 + 3/18 = 4/18 of visitors = 196. So 1/18 = 49, and visitors = 49 × 18 = 882. Answer: 882.


Adults = 2/3 × 882 = 588 (unchanged). At the end, adults = 3 × children left, so children left = 588 ÷ 3 = 196. Children at first = 1/3 × 882 = 294. Children who went home = 294 - 196 = 98. Answer: 98.


Sixty thousand = 60 000; twelve = 12. Together = 60 012.


The figure is divided into 20 equal parts. What percentage of the figure is shaded?

8 of the 20 equal parts are shaded. 8/20 = 40/100 = 40%.


Given that AE is the height of triangle ABC, find its base.

The base is the side perpendicular to the given height. AE is perpendicular to BC, so the base corresponding to height AE is BC.


A solid cuboid of height 10 cm has a square base of side 6 cm. What is its volume?

Volume = base area × height = (6 × 6) × 10 = 36 × 10 = 360 cm³.


In the figure, ST and PR are straight lines. Which of the following is true?

PR is a straight line, so the angles ∠a and ∠b lying on PR above the line add up to 180° (angles on a straight line). Thus ∠a + ∠b = 180°.


Which of the following is an isosceles triangle?

Triangle 4 has angles 30° and 120°, so its third angle = 180° − 30° − 120° = 30°. Two equal angles (30° and 30°) means two equal sides, so it is isosceles.


The table shows the number of boys and girls in 4 classes.
Class A: 23 boys, 25 girls. Class B: 18 boys, 23 girls. Class C: 20 boys, 22 girls. Class D: 21 boys, 20 girls.
Which class has more boys than girls?

Compare boys with girls in each class: A 23<25, B 18<23, C 20<22, D 21>20. Only Class D has more boys than girls.


What is the value of \(40 + (20 - 8) \div 2 \times 3\)?

Brackets: 20 − 8 = 12. Then ÷ and × left to right: 12 ÷ 2 = 6, 6 × 3 = 18. Add: 40 + 18 = 58.


John is 16 years old and his sister is 4 years older. In how many years will their total age be 70 years?

Sister = 16 + 4 = 20. Current total = 16 + 20 = 36. Each year both ages rise by 1, so the total rises by 2 each year. Needed increase = 70 − 36 = 34, so years = 34 ÷ 2 = 17.


Dinesh ran 3.55 km. He ran 40 m more than Siti. What was the distance that Siti ran?

Siti ran 40 m less than Dinesh: 3.55 km = 3550 m. 3550 − 40 = 3510 m = 3 km 510 m.


Which of the following is a net of the cube shown?

A valid cube net must fold into a cube with exactly 6 faces and no overlap. Net 1 folds correctly to form the cube.


The usual price of a blouse is $80. Fatimah bought the blouse at a 20% discount. How much did she pay for the blouse?


Suzy folds 4 paper hearts in 3 minutes. At this rate, how long will she take to fold 24 paper hearts?

24 ÷ 4 = 6 sets of 4 hearts. Each set takes 3 minutes, so 6 × 3 = 18 minutes.