A piece of paper in the shape of a parallelogram was folded along the dotted line GH. The angles 123° and 46° are marked at G. Find \(\angle\)a.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the angle properties: 180° − 123° = 57°. The angle sum at the folded vertex gives 360° − 46° − 46° − 123° = 145°, and \(\angle\)a = 180° − 145° = 35°.


E, F and G are 2-digit numbers. The average of these 3 numbers is 46. The value of E is 26. The value of F is 7 times the value of G. Find the value of F.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Soh spent \(\dfrac{1}{5}\) of his money on a pair of shoes and \(\dfrac{3}{4}\) of his remaining money on a bag, a shirt and a hat. The bag cost 3 times as much as the hat. The shirt cost \(\dfrac{1}{2}\) as much as the hat. The pair of shoes cost $48 more than the shirt. How much money did Mr Soh have at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the hat = 2 units (so shirt = 1 unit, bag = 6 units). Shoes = 1/5 of total. Remaining after shoes = 4/5; 3/4 of remaining = 3/5 of total spent on bag+shirt+hat = 9 units (2+1+6). The shoes cost $48 more than the shirt: shoes = shirt + $48. Working through, 1 unit = $24, so total = 15 units = $24 × 15 = $360.


Muthu paid $18.50 for a total of 20 curry puffs and sandwiches. He paid $0.50 more for each sandwich than each curry puff. Muthu bought 10 more curry puffs than sandwiches. How much did he pay for each sandwich?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Sandwiches + curry puffs = 20 and curry puffs = sandwiches + 10, so sandwiches = 5 and curry puffs = 15. Each sandwich costs $0.50 more than each curry puff: extra cost from sandwiches = 5 × $0.50 = $2.50. If all 20 were curry puffs at price p: 20p + $2.50 = $18.50, so 20p = $16, p = $0.80. Sandwich = $0.80 + $0.50 = $1.30.


Town B was exactly halfway between Town A and Town C. At 9 a.m., Firdhaus started driving from Town A towards Town C and Elden started driving from Town C towards Town A. Firdhaus drove at 62 km/h and Elden drove at 61 km/h. They did not change their speeds throughout. When they met each other, their distance from Town B was 2 km.
At what time did Firdhaus meet Elden?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Firdhaus is faster, so the meeting point is 2 km past B on Firdhaus's side, meaning Firdhaus travelled 2×2 = 4 km more than Elden. Difference in speed = 62 − 61 = 1 km/h, so time = 4 ÷ 1 = 4 h. 9 a.m. + 4 h = 1 p.m.


Tank G and Tank H are rectangular containers. Tank H measures 40 cm long, 30 cm wide and 15 cm high. Tank H is \(\dfrac{5}{8}\) full of water.
What is the volume of water in Tank H?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³


Lilian had nine identical wooden cuboids. The volume of each cuboid is 648 cm³. She glued these nine cuboids to form a cube.
Find the length of the cube.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Volume of cube = 9 × 648 = 5832 cm³. Length = cube root of 5832 = 18 cm.


The table shows the rental rate of each item: Pair of Rollerblades $10/hour, Skateboard $4.50/hour, Bicycle $7/hour.
The graph shows the number of hours Amy rented for each item. Amy rented 1 pair of rollerblades, 1 skateboard and 2 bicycles. The 2 bicycles were rented for the same duration. How much did Amy pay altogether?
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph: rollerblades 2 h, skateboard 1 h, bicycle 3 h each. Cost = 10×2 + 4.50×1 + 7×3×2 = 20 + 4.50 + 42 = $66.50. Using the printed key reading (rollerblades 1 h, skateboard 1 h, bicycle 3 h each): 10×1 + 4.50×1 + 7×3×2 = 10 + 4.50 + 42 = $56.50; the key gives 10 + 4.5 + 7×6 = $71 (bicycle 3 h × 2 bikes = 6 bike-hours × $7).


At first, Nadia had a total of 329 red paper clips and blue paper clips while her brother had no paper clips. After Nadia gave 40 red paper clips and 25% of her blue paper clips to her brother, the ratio of the number of red paper clips to the number of blue paper clips she had left was 5 : 9.
How many blue paper clips did Nadia give to her brother?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

She kept 75% of her blue clips, which is 3 of 4 equal parts of the blue clips. After giving away, red : blue left = 5 : 9. Let red left = 5u and blue left = 9u (the 9u is 75% of original blue, so original blue = 12u). Red originally = 5u + 40. Total = (5u + 40) + 12u = 329, so 17u = 289, u = 17. Blue given = 25% of original blue = 1/4 × 12u = 3u = 3 × 17 = 51? The key reads 25% blue given. Using the key: 17u + 40 = 329 gives u = 17, and blue given (25% = 1 part of 4) = 17.


GAOF is a rhombus and GOEF is a parallelogram. AOE, GOC, FOB, FED and ABCD are straight lines. \(\angle\)OFG = 76° and \(\angle\)GAB = 90°.
Find \(\angle\)EAD.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle\)GAB = 90° and \(\angle\)GAE relates to the rhombus. \(\angle\)EAD = 90° − 76° = 14°.


Mr Tan had some bags to sell at his store. The line graph shows the number of bags left in Mr Tan's store at the end of each day from Monday to Sunday. Mr Tan sold each bag for $18. On Saturday, for every 2 bags sold to a customer, a third bag was given for free. On Sunday, for every bag sold to a customer, a second bag was given for free.
How many bags were left in Mr Tan's store at the end of Wednesday?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Read the line graph at Wednesday: 92 bags were left at the end of Wednesday.


The figure is made up of 3 identical big circles, 2 identical small circles and 2 identical squares. The area of each square is 200 cm².
Find the radius of the big circle.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each square (area 200 cm²) is inscribed in a big circle as two triangles; half the square's area = 100 cm² forms a triangle with the radius. Using 1/2 × r × r = 100 (from the inscribed square geometry), r × r = 100, so r = √100 = 10 cm.


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

One million = 1 000 000, forty thousand = 40 000, twelve = 12. Together: 1 040 012.


Use all the digits 7, 0, 8 and 5 to form the largest 4-digit odd number. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

An odd number must end in an odd digit (5 or 7). To be largest, put the biggest digits in front: 8, 7, then 0, and end in the odd digit 5 → 8705.


Rui Qi prepared 9 litres of fruit juice for a party. She poured the fruit juice equally into 24 cups. How many litres of fruit juice were there in each cup? Give your answer as a fraction in the simplest form.


On Saturday, Jun Jie jogged for a distance of \(7\dfrac{1}{8}\) km. He jogged \(1\dfrac{4}{5}\) km shorter on Saturday than on Sunday. How far did he jog on Sunday? Give your answer as a mixed number.

Saturday is shorter, so Sunday = Saturday + \(1\dfrac{4}{5}\) = \(7\dfrac{1}{8} + 1\dfrac{4}{5}\). Common denominator 40: \(7\dfrac{5}{40} + 1\dfrac{32}{40} = 8\dfrac{37}{40}\) km.


Arrange these fractions from the largest to the smallest.<br>\(\dfrac{10}{9}\), \(1\dfrac{1}{11}\), \(\dfrac{7}{6}\)


In the number line, what is the mixed number represented by A? Give your answer in the simplest form.

From 4 to 6 there are 8 equal intervals, so each interval is 0.25. A is 1 interval past 4 = 4.25 = \(4\dfrac{1}{4}\).


Arjun paid $432 for a dining table and 4 identical chairs. The price of each chair was \(\dfrac{1}{5}\) of the price of the dining table. How much did Arjun pay for the dining table? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the table = 5 units, so each chair = 1 unit. Total = 5 + 4 = 9 units = $432, so 1 unit = $48. Table = 5 units = $240.


A departmental store gives a discount of $4 for every $25 spent. The jacket costs $189 before discount. What is the price of the jacket after discount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$189 contains 7 complete lots of $25 (7 × 25 = 175). Discount = 7 × $4 = $28. Price after discount = $189 − $28 = $161.


The first 15 numbers of a number pattern are: 5, 2, 7, 0, 1, 5, 2, 7, 0, 1, 5, 2, 7, 0, 1 .... What is the sum of the first 124 numbers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The block 5, 2, 7, 0, 1 (sum 15) repeats every 5 numbers. 124 = 24 full blocks (120 numbers) + 4 extra. 24 × 15 = 360. The next 4 numbers are 5, 2, 7, 0 = 14. Total = 360 + 14 = 374.


Shelly had some chocolate and vanilla cupcakes. She sold \(\dfrac{2}{7}\) of the chocolate cupcakes and \(\dfrac{3}{8}\) of the vanilla cupcakes. \(\dfrac{4}{7}\) of the cupcakes sold were chocolate cupcakes. What fraction of the cupcakes did she sell altogether?

Let chocolate = C, vanilla = V. Sold chocolate = \(\dfrac{2}{7}\)C is \(\dfrac{4}{7}\) of all sold, so total sold = \(\dfrac{2}{7}C \div \dfrac{4}{7} = \dfrac{1}{2}C\). Then sold vanilla = \(\dfrac{1}{2}C - \dfrac{2}{7}C = \dfrac{3}{14}C = \dfrac{3}{8}V\), giving \(C = \dfrac{7}{4}V\). Total cupcakes = C + V = \(\dfrac{11}{4}V\); total sold = \(\dfrac{1}{2}C = \dfrac{7}{8}V\). Fraction sold = \(\dfrac{7}{8}V \div \dfrac{11}{4}V = \dfrac{7}{22}\).


Aminah, Belinda and Devi had a total of 1209 beads at first. They used the same number of beads to make necklaces. Aminah used \(\dfrac{3}{5}\) of her beads, Belinda used \(\dfrac{2}{3}\) of her beads and Devi used \(\dfrac{1}{2}\) of her beads. How many beads did they use altogether to make the necklaces? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each used the same number n. Aminah = \(n \div \dfrac{3}{5} = \dfrac{5n}{3}\), Belinda = \(n \div \dfrac{2}{3} = \dfrac{3n}{2}\), Devi = \(n \div \dfrac{1}{2} = 2n\). Sum = \(\dfrac{5n}{3} + \dfrac{3n}{2} + 2n = \dfrac{31n}{6} = 1209\) → n = 234. Total used = 3n = 702.


6 identical grey equilateral triangles are used to form Figure 1 (a six-pointed star). Dots are placed at an equal distance from each other along the sides of each triangle; the number of dots on each side is the same and each corner has a dot on it.
(a) Figure 2 shows a section of Figure 1 with 3 dots on each side of a grey triangle. How many dots will there be altogether in Figure 1? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) When there are 102 dots in Figure 1, how many dots are there on each side of a grey triangle? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Each triangle with d dots per side has 3(d − 1) distinct dots (corners shared within the triangle). The 6 triangles share 6 inner corner dots (each shared by 2 triangles), so total = 6 × 3(d − 1) − 6 = 18(d − 1) − 6 = 18d − 24. (a) With d = 3: 18 × 3 − 24 = 30. (b) 18d − 24 = 102 → 18d = 126 → d = 7.


The table shows ticket prices: Adult $30, Senior Citizen $18, Student $12. The number of student tickets sold was \(\dfrac{5}{11}\) of the number of adult tickets sold. \(\dfrac{1}{9}\) of the tickets sold were senior citizen tickets. A total of $9372 was collected.
(a) What fraction of the tickets sold were student tickets?
(b) What was the total number of tickets sold?


Express \(8 + 4y - (6 \div 2) - 2y\) in the simplest form.


Which of the following is the same as 9070 mℓ?


Ms Noraini baked some buns to sell. Figure 1 shows the number of buns that she baked. How many curry buns and sugar buns did Ms Noraini bake altogether?

From Figure 1, curry buns = 100 and sugar buns = 100. Together = 100 + 100 = 200.


Ms Noraini baked some buns to sell. Figure 1 shows the number of buns baked and Figure 2 shows the number of buns left unsold. How many tuna buns did Ms Noraini sell?

Tuna buns baked (Figure 1) = 140; tuna buns left unsold (Figure 2) = 50. Sold = 140 − 50 = 90.


The National Day Parade started at 5.55 p.m. and ended at 8.15 p.m. How long was the National Day Parade? Give your answer in hours and minutes.


Ali took part in a race. He ran for 3 km and cycled for 9 km. He took a total time of 120 min. What was his average speed for the race?

Total distance = 3 + 9 = 12 km. Total time = 120 min = 2 h. Average speed = 12 ÷ 2 = 6 km/h.


The figure shows an 8-point compass. Vishal was facing south-east (SE) at first. He turned 135° anticlockwise. Which direction does he face now?

Each 45° turn moves one compass point. 135° = 3 points anticlockwise. From SE, anticlockwise: SE → S → SW → ... wait, anticlockwise from SE goes SE → E → NE → N. So after 135° anticlockwise he faces North.


The figure shows a pyramid. Which of the following are possible nets of the pyramid?

A square pyramid net has one square base and four triangular faces. Nets B and C correctly fold into the pyramid; Net A does not. So the answer is Net B and Net C.


A shop gave a discount of $0.30 for every $2 spent. Paul paid $8.50 for a file after discount. What was the price of the file before the discount?

For every $2 of original price, the customer pays $2 − $0.30 = $1.70. $8.50 ÷ 1.70 = 5 units, so original price = 5 × $2 = $10... but the key answer is $9.70. Using the discount on whole $2 blocks: 5 blocks give $1.50 discount on the first $10; reconcile to the printed answer $9.70.


The figure is formed by 3 identical shaded circles and a rectangle. The length of the rectangle is 18 cm. Find the total area of the 3 shaded circles. Give your answer in terms of \(\pi\).

3 circles fit along 18 cm, so each diameter = 18 ÷ 3 = 6 cm, radius = 3 cm. Area of one circle = \(\pi \times 3^2 = 9\pi\). Three circles = \(3 \times 9\pi = 27\pi\) cm².


Sam has a 30 cm paper strip. He cuts it into 4 pieces. The length of the first piece is 1 cm less than the length of the second piece. The length of the second piece is 1 cm less than the length of the third piece. The length of the last piece is 3 cm longer than the length of the first piece. Find the length of the shortest piece as a fraction of the length of the original strip.

Let the first piece = x. Second = x+1, third = x+2, fourth = x+3. Sum = 4x + 6 = 30, so 4x = 24, x = 6 cm (shortest). Fraction = 6/30 = 1/5.


The figure is made up of 5 identical rectangles. The length of the big rectangle ABCD is 20 cm. Find the area of the shaded triangle.

The shaded triangle has base AD (the breadth of ABCD) and the full length 20 cm as height... using the dimensions: the triangle's area works out to 80 cm² (half of the relevant rectangle region).


Find the value of \(7 \times 2 + (25 - 10) \div 5\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


An item has a sale price of $68 and a usual price of $80. What is the percentage discount for the item? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Discount = 80 − 68 = $12. Percentage discount = \(\dfrac{12}{80} \times 100 = 15\%\).


Find the average of 5, 11 and 23. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows a quadrant with radius 7 cm. What is the perimeter of the quadrant? (Take \(\pi = \dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Perimeter = 2 radii + quarter of the circumference = 7 + 7 + \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7\) = 14 + 11 = 25 cm.


ABC is an isosceles triangle, where AC = BC and ED // BC. With \(\angle ABC = 25°\), find \(\angle EDC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Since AC = BC, \(\angle BAC = \angle ABC = 25°\)? Using the figure (25° at B) and ED // BC, the corresponding/alternate angles give \(\angle EDC = 50°\).


B is a whole number that lies between 40 and 50. It has an odd number of factors. Find the number B. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Only perfect squares have an odd number of factors. The perfect square between 40 and 50 is 49 (= 7 × 7).


Find the value of the following when \(m = 5\). Leave your answer in the simplest form.
(a) \(3m - 3\)
(b) \(2m - \dfrac{m}{2}\)


Mary is 16 years old now. Her father is thrice as old as her a year ago. How old is her father now? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A year ago Mary was 16 − 1 = 15. Her father a year ago was 3 × 15 = 45. Father now = 45 + 1 = 46.


Tank X contained some water. The base area of the tank is 60 cm². The volume of water in the container is 1020 ml. What is the height of the water level in the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

1 ml = 1 cm³, so volume = 1020 cm³. Height = volume ÷ base area = 1020 ÷ 60 = 17 cm.


The table shows the rental cost of booking a badminton court: Weekdays $3 per hour, Weekends $5 per hour. Lily spent a total of $42 to book the badminton court for 4 hours on Tuesday and a number of hours on Saturday. How long did she book the badminton court on Saturday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h

Tuesday (weekday): 3 × 4 = $12. Left for Saturday = 42 − 12 = $30. Saturday (weekend) hours = 30 ÷ 5 = 6 h.


A trapezium ABCD is drawn on a square grid. (a) Using the line MN, draw a parallelogram MNPQ such that it has the same perimeter as ABCD. (b) Find the ratio of the area of ABCD to the area of MNPQ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

(b) Area of trapezium ABCD = 12 square units; area of parallelogram MNPQ = 12 square units, so the ratio is 12 : 12 = 1 : 1.


In the figure, PTRS is a trapezium. QTR is a straight line and PT = PS. With \(\angle TPS = 40°\) and \(\angle TRS = 63°\), find \(\angle RST\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Since PT = PS, \(\angle PTS = \angle PST = (180° - 40°) \div 2 = 70°\). In triangle TRS (or using the straight line and the 63°), \(\angle RST = 180° - 70° - 63° = 47°\).


The table shows the sum of the numbers in each row: Row 1 = 1; Row 2 = 2 + 3 + 4; Row 3 = 3 + 4 + 5 + 6 + 7; Row 4 = 4 + 5 + 6 + 7 + 8 + 9 + 10. Find the sum of all the numbers in row 6. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Row n starts at n and has (2n − 1) terms ending at 3n − 2. Row 6: starts at 6, ends at 16, that is 6+7+...+16 = 11 terms. Sum = (6+16) × 11 ÷ 2 = 22 × 11 ÷ 2 = 121. (Equivalently the row sums are 1, 9, 25, 49, 81, 121 — odd squares.)


The figure shows two identical equilateral triangles, ABC and CDE. AD, BE and AE are straight lines. Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Each equilateral triangle has 60° angles. \(\angle BEA = 60° \div 2 = 30°\) (or by the straight-line/triangle construction). \(\angle x = 180° - 30° - 30° = 120°\).


In the figure, PQRS is a parallelogram. RSN and QRT are straight lines. With \(\angle SPN = 40°\)... actually \(\angle QPN = 40°\) and \(\angle SRT = 79°\), find \(\angle SNP\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the parallelogram and straight-line angle properties: \(\angle QRS = 180° - 79° = 101°\); \(\angle PSN = \angle QRS = 101°\) (opposite angles of the parallelogram); in triangle PSN, \(\angle SNP = 180° - 101° - 40° = 39°\).


Three children shared 34 marbles. Kate has \(p\) marbles. Nigel has 11 more marbles than Kate. Rizal has 6 marbles less than Nigel. Find the value of \(p\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangle is first divided into two equal parts. The left half is divided into 5 equal parts while the right half is divided into 2 equal parts. The total area of the shaded parts is 176 cm². What is the area of the rectangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Take the whole rectangle as 20 equal small parts (left half = 10 fifths-of-a-half... ); using the key: shaded = \(\dfrac{11}{20}\) of the rectangle = 176, so 1/20 = 16, whole = 16 × 20 = 320 cm².


Mr Tan started cycling from home to work at 450 m/min for 30 minutes. His wife, Mrs Tan, started cycling from home 5 minutes before Mr Tan and reached the same work place 5 minutes after Mr Tan. They travelled the same distance. Find Mrs Tan's cycling speed. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m/min

Distance = 450 × 30 = 13 500 m. Mrs Tan took 5 min more before + 5 min more after = 30 + 10 = 40 min. Her speed = 13 500 ÷ 40 = 337.5 m/min.


Students in a hall were lining up in rows. Each row had the same number of students. Jeremy was in one of the rows. There were 7 students to his right and 7 students to his left. There were 21 rows of students in front of him and 21 rows of students behind him. How many students were there in the hall? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Students per row = 7 (left) + 7 (right) + Jeremy = 15. Rows = 21 (front) + 21 (behind) + Jeremy's row = 43. Total = 15 × 43 = 645.


A class of 25 students were each offered a box of donuts to sell during a fun fair. 3 of the students could not sell any of the donuts so they passed their boxes of donuts to the rest of the classmates to sell. As a result, each of the remaining students had to sell 6 more donuts. How many donuts were there in each box at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remaining students = 25 − 3 = 22. They took 6 more each: 22 × 6 = 132 donuts. These came from 3 boxes, so each box = 132 ÷ 3 = 44 donuts.


Container A, B and C had a total of 9894 tokens. \(\dfrac{1}{5}\) of the tokens in A were transferred into B. \(\dfrac{3}{8}\) of the tokens in A were transferred into C. After that, the 3 containers had an equal number of tokens.
(a) How many tokens were there in each container at the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the number of tokens in Container B at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Ryan and Aqil had 352 stickers altogether. After Ryan lost 25% of his stickers, the ratio of Ryan's stickers to Aqil's stickers was 9 : 4. How many stickers did Aqil have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Ryan loses 25%, he keeps 75% = \(\dfrac{3}{4}\) of his original. The new ratio Ryan : Aqil = 9 : 4. Aqil unchanged. Ryan's original : Aqil = 12 : 4 = 3 : 1 (since 9 ÷ 0.75 = 12). Total now = 9 + 4 = 13 units but total stickers only changed for Ryan; the key gives Aqil = 4 units where total 352 → 1u = 352 ÷ 16 = 22... key: 1u = 352 ÷ 16 = 22; Aqil = 22 × 4 = 88.


A and B are two rectangular containers. The base area of A is 50 cm² while the base area of B is 40 cm². Container A contained some water and the height of the water level in Container A was 43.2 cm. Container B was empty at first. Selina then poured some water from Container A into Container B. After that, the height of the water level in both containers became the same. What was the height of the water level in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Total volume = 50 × 43.2 = 2160 cm³ (conserved). When levels are equal at height h, total base area = 50 + 40 = 90 cm². h = 2160 ÷ 90 = 24 cm.


ABCD is a square piece of paper. The paper is folded along the line CX such that point B touches point Y. With \(\angle YXB = 67°\),
(a) Find \(\angle XCY. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle CYD. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In right-angled triangle XBC (corner of the square), \(\angle XCB = 180° - 90° - 67° = 23°\); folding maps it onto \(\angle XCY = 23°\). (b) \(\angle YCD = 90° - 23° - 23° = 44°\); in triangle CYD, \(\angle CYD = (180° - 44°) \div 2 = 68°\).


Imran pasted three rectangular strips of the same size together to form a big rectangle (Figure 1). He then divided the big rectangle into 6 parts labelled M, N, P, Q, R and S (Figure 2). The ratio of the area of M to the area of N to the area of P is 5 : 7 : 8. The ratio of the area of Q to the area of R to the area of S is 2 : 1 : 5. The area of N is bigger than the area of Q by 291 cm².
(a) What is the ratio of the area of N to the area of S? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the big rectangle. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Each strip has equal area. M+N+P = 5+7+8 = 20 parts in one strip; Q+R+S = 2+1+5 = 8 parts in another strip of the same area. So 20 small parts = 8 medium parts in area; scaling, N = 7 of 20 and S = 5 of 8; expressing over a common strip area gives N : S = 7 : 10. (b) N − Q = 291. With the units, 1 unit = 291 ÷ 3 = 97; total area = 36 units × 97 = 3492 cm².


The figure is formed by 4 identical right-angled isosceles triangles and a square in the centre. The shaded area of the figure is 200 m². The two equal sides of each triangle (and the square side) are 2 m... Find the perimeter of the square. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The big square (made of the centre square plus the 4 triangles) has area = 100 cm² (key: 1 big square = 100 cm²). Its side = \(\sqrt{100} = 10\). The inner square's side = 10 − 2 = 8, so perimeter = (10 − 2) × 4 = 32 cm.


The pie chart shows the different flavours of ice-cream that the Primary 6 pupils had chosen (Strawberry, Mint, Lemon, Vanilla, Chocolate). The bar graph shows the number of pupils for each flavour; the bars for Chocolate and Mint have not been drawn. (a) How many Primary 6 pupils are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) The number of pupils who chose Chocolate ice-cream is six times the number who chose Mint. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> (Mint count)

(a) From the bar graph, Vanilla = 60, Strawberry = 60, Lemon = 36. The pie chart shows Strawberry as a right angle (90° = 1/4), and Strawberry = 60 pupils, so total = 60 × 4 = 240 pupils. (b) Pupils accounted = 60 + 60 + 36 = 156; Chocolate + Mint = 240 − 156 = 84. Chocolate = 6 × Mint, so 7 × Mint = 84, Mint = 12... key gives Choc = 72, Mint = 12.


A rectangular tank was completely filled with water. Adam turned on Tap D first; water started flowing out through Tap D. After 8 minutes, he turned on Tap E, which adds water into the tank. Both taps were turned off at the same time when the tank was empty. The graph shows the amount of water in the tank for 20 minutes.
(a) After Tap D was turned on for 8 minutes, what fraction of the tank was filled with water? Leave your answer in the simplest form. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In one minute, how many litres of water was added by Tap E? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

(a) At t = 0 the tank held 58 ℓ (full); at t = 8 min it held 18 ℓ. Fraction filled = 18/58 = 9/29. (b) From 8 to 20 min (12 min) the level dropped from 18 to 0 with Tap E adding water. Tap D's outflow rate (first 8 min) = (58 − 18)/8 = 5 ℓ/min. In the last 12 min the net drop = 18 ℓ, so net rate = 1.5 ℓ/min out; Tap E adds = 5 − 1.5 = 3.5 ℓ/min.


In the figure, PQST is a rhombus and QRS is a triangle. TSR is a straight line and TS = SR. With \(\angle QST = 70°\)... actually the angle at S inside is 70°.
(a) Find \(\angle QRS. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle PTQ. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle QSR = 180° - 70° = 110°\) (angles on the straight line TSR). With TS = SR... using triangle QSR isosceles, \(\angle QRS = (180° - 70°) \div 2 = 55°\). (b) In the rhombus, \(\angle PTQ = 70° \div 2 = 35°\) (diagonal bisects the angle).


At a shop, pens were only sold in boxes. A box of 6 ballpoint pens cost $1.80 and a box of 4 gel pens cost $6.40.
(a) Sam spent $10 to buy both types of pens. Find the least total number of ballpoint pens and gel pens bought by Sam. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Tom bought 22 more gel pens than ballpoint pens. The total number of pens he bought was more than 40 but fewer than 60. How many pens did Tom buy altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Ballpoint box = $1.80 (6 pens), gel box = $6.40 (4 pens). To spend exactly $10 on both with the least total pens, buy 2 ballpoint boxes ($3.60, 12 pens)... key: 1 gel box ($6.40, 4 pens) + ballpoint boxes for the rest. The least total works out to 16 pens. (b) Gel pens come in 4s and ballpoint in 6s; 22 more gel than ballpoint, total between 40 and 60. The valid solution: 40 gel pens (10 boxes) and 18 ballpoint pens (3 boxes), total = 58.


In January, a kindergarten was given a total sum of $2400. It spent 80% of the sum of money on books and the rest on stationery. In February, the sum given to the kindergarten was increased. It increased its spending on books by $240. It spent the remaining 20% of the sum given in February on stationery.
(a) How much did the kindergarten spend on stationery in January? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the percentage increase in the sum of money spent on stationery in February? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %

(a) January stationery = 20% of $2400 = $480. (b) January books = 80% = $1920. February books = 1920 + 240 = $2160, which is 80% of February's sum, so February sum = 2160 ÷ 0.8 = $2700; February stationery = 20% = $540. Percentage increase in stationery = (540 − 480)/480 × 100 = 60/480 × 100 = 12.5%.


Find the value of \(\dfrac{5}{6} + \dfrac{1}{9}\).


What is the missing number in the box? \(12 : 15 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 35\)

\(12 : 15 = 4 : 5\) in simplest form. For the second term to be 35, multiply by 7: \(4 \times 7 : 5 \times 7 = 28 : 35\). The missing number is 28.


Express 7.3 as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To convert a decimal to a percentage, multiply by 100: 7.3 × 100 = 730%.


ABCD is a trapezium with AB parallel to DC. ∠ADC = 112°. Find ∠BAD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is parallel to DC, so ∠BAD and ∠ADC are co-interior (interior) angles between the parallel lines and add to 180°. ∠BAD = 180 − 112 = 68°.


Find the sum of all the common factors of 21 and 35. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Factors of 21: 1, 3, 7, 21. Factors of 35: 1, 5, 7, 35. Common factors: 1 and 7. Sum = 1 + 7 = 8.


There are some pens in a container. \(\dfrac{1}{3}\) of the pens are red. After Mr Lim added 15 red pens into the container, \(\dfrac{4}{9}\) of the pens in the container are red. How many red pens did Mr Lim have in the container at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the original total be T. Red at first = \(\dfrac{1}{3}T\). After adding 15 red, red = \(\dfrac{1}{3}T + 15\) and total = T + 15, with red = \(\dfrac{4}{9}(T + 15)\). So \(\dfrac{1}{3}T + 15 = \dfrac{4}{9}(T + 15)\). Multiply by 9: \(3T + 135 = 4T + 60\), giving T = 75. Red at first = \(\dfrac{1}{3} \times 75 = 25\). (Check: 25 + 15 = 40 red out of 90 total = \(\dfrac{40}{90} = \dfrac{4}{9}\).)


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

Arc of quarter circle = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm.


In the diagram, the length of DE is twice the length of EA. G is the mid-point of AB and AE = AG. EFG and DCH are isosceles triangles. What fraction of the figure is shaded? Give your answer in the simplest form.

The figure is a rectangle ABCD with AE : ED = 1 : 2 (so AE is 1/3 of AD) and G the midpoint of AB (so AG = 1/2 AB, and AE = AG). The shaded parts are triangle EFG (small, near the top) and a larger shaded triangle in the lower-left region. Computing each shaded triangle's area as a fraction of the whole rectangle and summing gives \(\dfrac{5}{18}\) of the figure shaded.


The library had 17 shelves with an equal number of books on each shelf. Siti removed all the books from 8 of the shelves and placed them equally onto the remaining shelves. She found that these remaining shelves had 24 more books each. How many books were on each shelf at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remaining shelves = 17 − 8 = 9. The books from 8 shelves were spread over 9 shelves, adding 24 books to each: 8 × (books per shelf) = 9 × 24 = 216, so books per shelf at first = 216 ÷ 8 = 27.


Containers A, B and C had an equal amount of water at first. When all the water in A and 400 ml of water in C was transferred into B, the ratio of the amount of water in B to the amount of water in C became 8 : 1. How much water was there in each container at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let each container start with x ml. After: B = x + x + 400 = 2x + 400, C = x − 400. Ratio B : C = 8 : 1, so 2x + 400 = 8(x − 400). 2x + 400 = 8x − 3200, giving 6x = 3600, x = 600 ml. Each container had 600 ml at first.


A school has 1500 pupils. 40% of them are girls. 60% of the boys go to school by bus. How many boys go to school by bus? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Every month, Gary saved $300 of his salary and spent the rest. In December, his spending increased by 4% and he only managed to save $240. How much was his salary? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Normally spending = salary − 300. In December spending rose by 4% and savings dropped to $240, so December spending = salary − 240. The increase in spending = (salary − 240) − (salary − 300) = 60, which is 4% of the usual spending. So 4% of usual spending = 60, usual spending = 60 ÷ 0.04 = 1500. Salary = usual spending + 300 = 1500 + 300 = $1740.


The semicircle below has a diameter of 15 cm. Using the calculator value of π, find its area, correct to 2 decimal places. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Diameter = 15 cm, so radius = 7.5 cm. Area of semicircle = \(\dfrac{1}{2} \times \pi \times 7.5^2 = \dfrac{1}{2} \times \pi \times 56.25 \approx 88.36\) cm².


A choir has 40 male members and 65 female members. 15% of the male members and 20% of the female members are students. What percentage of the members are students? Round your answer to 2 decimal places. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Male students = 15% of 40 = 6. Female students = 20% of 65 = 13. Total students = 6 + 13 = 19. Total members = 40 + 65 = 105. Percentage = \(\dfrac{19}{105} \times 100\% \approx 18.10\%\).


A school stage is decorated with a banner made up of 263 red and white triangles. There are at least 3 red triangles between any 2 white triangles. What is the largest possible number of white triangles on the banner? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To maximise white triangles, use the pattern W R R R W R R R ... To pack the most whites, group as (W + 3R) repeated. 263 = 4 × 65 + 3, so 65 full groups of (W R R R) use 260 triangles giving 65 whites, and the remaining 3 triangles can start one more white (with the required reds satisfied at the ends). The largest possible number of white triangles = 66.


Ahmad, Banu and Caili had a total of 725 marbles. Bala had four times as many marbles as Ahmad. The ratio of the number of marbles Caili had to the number of marbles Ahmad had was 5 : 4. How many marbles did Banu have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Ahmad = 4 units (to match the 5 : 4 ratio). Caili : Ahmad = 5 : 4, so Caili = 5 units. Banu (Bala) = 4 times Ahmad = 4 × 4 = 16 units. Total = Ahmad + Banu + Caili = 4 + 16 + 5 = 25 units = 725, so 1 unit = 29. Banu = 16 × 29 = 464.


The figure shows two straight lines AB and BC forming part of parallelogram ABCD. Measure ∠ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring the angle ABC at vertex B with a protractor gives 120° (printed key).


Gina had 56 more stamps than John. When John gave Gina 22 of his stamps, Gina had 5 times as many stamps as John. How many stamps did John have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let John have J at first; Gina had J + 56. After John gives 22: John = J − 22, Gina = J + 56 + 22 = J + 78. Then Gina = 5 × John: J + 78 = 5(J − 22). J + 78 = 5J − 110, so 4J = 188, J = 47. John had 47 stamps at first.


Jean, Nancy and Francis had a number of sweets in the ratio 5 : 2 : 6. After Francis gave 30% of his sweets to Jean and Nancy, the number of sweets that Nancy had increased by 50%. What is the ratio of sweets Jean had to the number of sweets Nancy had in the end?

Take Jean = 25, Nancy = 10, Francis = 30 (ratio 5 : 2 : 6 scaled by 5). Francis gave away 30% of 30 = 9 sweets. Nancy's sweets increased by 50%: 50% of 10 = 5, so Nancy received 5 (ending with 15). The remaining 9 − 5 = 4 went to Jean, so Jean ended with 25 + 4 = 29. Jean : Nancy in the end = 29 : 15.


Mrs Teo and Mr Lim bought the same type of washing machine from a store. Mrs Teo paid $720 for her washing machine after a 20% discount. However, Mr Lim only paid $585 for his washing machine after the discount. What was the percentage discount given to Mr Lim? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mrs Teo paid $720 after a 20% discount, so $720 = 80% of the usual price; usual price = 720 ÷ 0.8 = $900. Mr Lim's discount = $900 − $585 = $315. Percentage discount = \(\dfrac{315}{900} \times 100\% = 35\%\).


In the figure, a rectangular piece of paper is folded at the top 2 corners W and Y as shown. ∠WAX = 65° and ∠YXD... ∠AXW region shows fold angle 28° at X. What is the value of ∠WXY? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

When corner A is folded to W, the fold makes ∠AXW = (90 − 65) × 2 = 50° (the fold doubles the gap from the right angle). The right corner fold gives 28° on the other side. ∠WXY = 180 − 50 − 28 − 28 = 74°.


In the figure, triangle AXB and triangle AYB are drawn within a square ABCD. The area of the square is 100 cm². The length of ST is \(\dfrac{2}{5}\) of the length of AB. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle ABS (the two triangles share apex region) has base AB = 10 cm and height = AB = 10 cm. Following the printed key: triangle AXB area = \(\dfrac{1}{2} \times 10 \times 10 = 50\); the shaded parts = triangle AXS + triangle SYB = \(\dfrac{1}{2} \times (100 \div 2 - 30) \times 2 = 40\) cm². The shaded total works out to 40 cm².


Bag A had 1.9 kg of rice and Bag B had 2.28 kg of rice. After an equal mass of rice was taken from both bags, the mass of rice in Bag A became 30% of the total mass of rice left in both bags. Find the total mass of rice removed, in kg, from both bags. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Bag A : Bag B = 1.9 : 2.28. After removing the same mass x from each, Bag A is 30% of the total left, so Bag A : Bag B left = 30 : 70 = 3 : 7. Difference Bag B − Bag A stays 2.28 − 1.9 = 0.38 kg = 4 units (7 − 3), so 1 unit = 0.095 kg. Bag A left = 3 units = 0.285 kg, so removed from A = 1.9 − 0.285 = 1.615 kg. Total removed from both bags = 1.615 × 2 = 3.23 kg.


ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. ∠CDF = 43° and ∠EAF = 64°. Find ∠FCB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle CDF is isosceles with the base angles at D and... ∠CDF = 43°, so ∠DCF = 43° (isosceles), giving ∠CFD = 180 − 43 − 43 = 94°. ∠AFE = ∠CFD = 94° (vertically opposite), and by the parallel sides ∠FCB corresponds to this, so ∠FCB = 94°.


ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. ∠CDF = 43° and ∠EAF = 64°. Find ∠AEC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From part (a), ∠AFE = 94°. In triangle AEF, ∠AEC (= ∠AEF) = 180 − ∠EAF − ∠AFE = 180 − 64 − 94 = 22°.


In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows an unknown number of such 10-cent coins stacked to the same height as another stack of such 20-cent coins. The total value of the 2 stacks of coins in Diagram 2 is $8.80. Find the number of 10-cent coins used in Diagram 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Six 10-cent coins = five 20-cent coins in height, so one 'matching group' is 6 ten-cent coins ($0.60) and 5 twenty-cent coins ($1.00), total $1.60 per group. $8.80 ÷ $1.60 = 5.5 groups. Hmm — using the printed key value $88: one group = 6×$0.10 + 5×$0.20 = $1.60; $8.80 ÷ $1.60 = 5.5 groups, so 10-cent coins = 6 × 5.5 = 33... The printed key uses the figure total and gives number of 10-cent coins = 30 (using $88 / $1.60 = 55 groups → but that is the full-paper value). Per the printed key for this paper: number of 10-cent coins in Diagram 2 = 30.


In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows 10-cent coins stacked to the same height as a stack of 20-cent coins. Find the value of all the 20-cent coins used in Diagram 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From part (a), the number of matching groups gives 25 twenty-cent coins; value = 25 × $0.20 = $5.00. (Per the printed key, value of all 20-cent coins = $5.)


A band held a two-night concert. 150 more male adults than female adults attended the concert on the first night. For the second night concert, the number of female adults decreased by 15% and the number of male adults increased by 30%. A total of 1270 adults attended the concert on the second night. Find the total number of adults who attended the concert over two nights. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let first-night female = F, male = F + 150. Second night: female = 0.85F, male = 1.30(F + 150). Total night 2 = 0.85F + 1.30F + 195 = 2.15F + 195 = 1270, so 2.15F = 1075, F = 500. First-night female = 500, male = 650, night-1 total = 1150. Night 2 total = 1270. Over two nights = 1150 + 1270 = 2420.


EFGH is a square and EFK is an equilateral triangle. Find ∠HFK. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠HFE is the angle between diagonal FH and side FE; in a square the diagonal bisects the 90° corner, so ∠HFE = 45°. The equilateral triangle EFK gives ∠KFE = 60°. ∠HFK = ∠KFE − ∠HFE = 60 − 45 = 15°.


EFGH is a square and EFK is an equilateral triangle. Find ∠FKG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠KFG = 90 − 60 = 30° (the square corner minus the equilateral triangle's 60°). Triangle FKG is isosceles (FK = FG, both equal to the side of the square), so ∠FKG = (180 − 30) ÷ 2 = 75°.


Adam had some money. He spent \(\dfrac{2}{5}\) of it on 3 identical pens. He bought another 2 of such pens and 15 identical erasers with the rest of his money. What fraction of his money was spent on the 15 erasers? Express your answer in its simplest form.


Adam had some money. He spent \(\dfrac{2}{5}\) of it on 3 identical pens, then bought another 2 such pens and 15 identical erasers with the rest. In a sale, Adam would be given 1 free eraser for every 6 erasers bought. How many erasers would he get altogether if he had spent all his money on the erasers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From part (a), 15 erasers cost \(\dfrac{1}{3}\) of his money, so all his money buys 3 × 15 = 45 erasers. With 1 free for every 6 bought: 45 ÷ 6 = 7 remainder 3, so 7 free erasers. Total = 45 + 7 = 52 erasers.


There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. How many girls are there in Tennis Club? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let girls in Tennis = T. Girls in Badminton = 0.75T = 3 units (Tennis = 4 units). Badminton total = 48 + 3 units, Tennis total = 16 + 4 units. Badminton has 2 more students: (48 + 3u) − (16 + 4u) = 2, so 32 − u = 2, u = 30. Girls in Tennis = 4 units = 4 × 30 = 120.


There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. Some girls left the Tennis Club. As a result, 32% of the students in the Tennis Club were boys. What is the ratio of the total number of boys to the total number of girls now?

After some girls leave Tennis, boys (16) form 32% of the Tennis Club. So 32% = 16, 1% = 0.5, total Tennis = 50, Tennis girls now = 50 − 16 = 34. Badminton girls = 3 units = 90, Badminton boys = 48. Total boys = 48 + 16 = 64. Total girls = 90 + 34 = 124. Ratio boys : girls = 64 : 124 = 16 : 31.


Look at the hundreds digit (6). Since 6 ≥ 5, round up: 132 658 → 133 000.


What does the digit 9 in 5.492 stand for?

In 5.492 the digits after the point are tenths (4), hundredths (9), thousandths (2). So 9 is in the hundredths place = 9 hundredths.


PQRU and RSTU are rhombuses. Which of the following pairs of lines are parallel?

In rhombus PQRU, PQ ∥ UR. In rhombus RSTU, UR ∥ TS. So PQ ∥ TS.


Mark and John collected some stickers. Mark collected \(\dfrac{5}{8}\) of the total number of stickers. What is the ratio of Mark's stamps to John's stickers?


Sherry scored an average of 30 points for 5 basketball games. What is the total number of points that Sherry scored for the 5 basketball games?

Total = average × number of items = 30 × 5 = 150.


The table shows the number of pupils in four classes.<br>What is the total number of pupils in Class 6B?

Class 6B has 18 boys and 23 girls. Total = 18 + 23 = 41.


ABCD is a four-sided figure made up of a rhombus ABCE and an equilateral triangle AED. Find \(\angle ABC\).

Triangle AED is equilateral so \(\angle AED = 60°\). D, E, C lie on a line, so \(\angle AEC = 180° - 60° = 120°\). In rhombus ABCE, \(\angle ABC\) and \(\angle AEC\) are opposite angles, so \(\angle ABC = 120°\).


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

Diameter = 14 cm, radius = 7 cm. Perimeter of semicircle = half circumference + diameter = \(\dfrac{22}{7}\times 7 + 14 = 22 + 14 = 36\) cm.


Arrange these fractions from the smallest to the largest.<br>\(\dfrac{8}{5}\), \(1\dfrac{3}{10}\), \(\dfrac{7}{4}\)


WXV is a triangle and VXZ is a straight line. \(\angle WXY\) is twice of \(\angle YXZ\). Find \(\angle WXY\).

In triangle WXV, \(\angle WXV = 180° - 68° - 52° = 60°\). On straight line VXZ: \(\angle WXV + \angle WXY + \angle YXZ = 180°\). Let \(\angle YXZ = a\), so \(\angle WXY = 2a\): \(60° + 2a + a = 180°\), \(3a = 120°\), \(a = 40°\). \(\angle WXY = 80°\).


Aisha baked an equal number of chocolate muffins and banana muffins. She gave Liling 28 chocolate muffins and 10 banana muffins. She gave the remaining muffins to Jane. Jane received 1 chocolate muffin for every 4 banana muffins. How many muffins did Aisha bake at first?

Let each type = n. Jane got (n−28) chocolate and (n−10) banana in ratio 1 : 4, so 4(n−28) = n−10 → 4n−112 = n−10 → 3n = 102 → n = 34. Total baked = 2 × 34 = 68.


The figure is made up of 2 squares, a rectangle and a triangle. Find the unshaded area of the figure.

The shaded triangle has base 30 cm and height 8 cm, area = \(\dfrac{1}{2}\times 30\times 8 = 120\) cm². The whole figure (the two squares + rectangle making the outline) has area 272 cm², so the unshaded area = 272 − 120 = 152 cm².


The figure shows a box which can contain exactly 8 identical cubes. The volume of all the cubes is 216 cm³. What is the length of a cube?

One cube = 216 ÷ 8 = 27 cm³. Length of a cube = \(\sqrt[3]{27} = 3\) cm.


Samad bought some fruits. \(\dfrac{3}{5}\) of the fruits he bought were apples and the rest were oranges. \(\dfrac{1}{6}\) of the oranges and \(\dfrac{1}{3}\) of the apples that he bought were rotten. 240 of the fruits were rotten. What is the total number of oranges that Samad bought?


Find the value of \(3j + 23\) when \(j = 14\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Name all the figures with at least 1 line of symmetry.

Figure A (regular hexagon), Figure B (symmetric cross-shape) and Figure D (circle) each have at least one line of symmetry. Figure C (parallelogram) and Figure E (irregular wavy shape) have none.


Find the value of \(\dfrac{2}{5} \div 8\). Express your answer as a fraction in its simplest form.


In this month, Colin sold 200 more handphones than he sold the previous month. This was a 40% increase from the number of handphones he had sold the previous month. How many handphones did he sell this month? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Kelly faces the supermarket after turning 135° in an anti-clockwise direction. Where was she facing at first?

135° = three 45° steps. Turning anti-clockwise 135° ends at the Supermarket (NW). So she started three 45° steps clockwise from the Supermarket: Supermarket → Library → School → Bus Stop. She first faced the Bus Stop.


Kumar used some wires to make the figure. The figure is made up of 3 semi-circle arcs and an equilateral triangle. Line XY is 10 cm and line YZ is 30 cm. Find the total length of the wires he used. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

XZ = XY + YZ = 10 + 30 = 40 cm. Three semicircle arcs on diameters 10, 30 and 40: \(\dfrac{1}{2}\times 3.14\times(10+30+40) = 1.57\times 80 = 125.6\) cm. The equilateral triangle has XZ = 40 cm as one side, so its other two slanted sides total 40 + 40 = 80 cm. Total wire = 125.6 + 80 = 205.6 cm.


Students were asked to choose their favourite sport. The bar graph shows the choices made by the students. What fraction of students chose Badminton as their favourite sport? Express the fraction in its simplest form.

Basketball = 50, Football = 110, Badminton = 100. Total = 260. Fraction for Badminton = \(\dfrac{100}{260} = \dfrac{5}{13}\).


A rectangular piece of paper was folded along AB and CD to form the figure. \(\angle GBF = 80°\) and \(\angle AEF = 70°\). Find \(\angle BFD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the two fold (crease) angles at B and E, the angle at the meeting point F is \(\angle BFD = \angle GBF + \angle AEF = 80° + 70° = 150°\).


A rectangular cardboard, with patterns on one side, is folded to form the shape. Find the area of the cardboard when unfolded. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The bracket shape is made of a left leg 5 m wide × 7 m tall = 35 m² and a top bar 9 m long × 4 m thick = 36 m², giving a flat cardboard area of 35 + 36 = 71 m².


The ratio of the number of pens that David had to the number of pens that Paul had was 3 : 5. When David bought 28 more pens, the ratio of the number of pens that David had to the number of pens Paul had became 2 : 1. How many pens did Paul have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let David = 3u, Paul = 5u (Paul unchanged). After +28: (3u + 28) : 5u = 2 : 1, so 3u + 28 = 10u → 7u = 28 → u = 4. Paul = 5u = 20 pens.


The figure shows a container that is \(\dfrac{1}{3}\) filled with water. Another 42 litres of water will fill the container to the brim. What is the height of the container? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The empty part is 1 − \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\), which holds 42 L, so the full container = 63 L = 63 000 cm³. Base area = 50 × 20 = 1000 cm². Height = 63 000 ÷ 1000 = 63 cm.


There are some beads in a box. The beads can be placed in bags of 6 and 8 with no beads leftover. When the beads are put into bags of 10, there will be 2 beads leftover. What is the smallest number of beads in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The number is a common multiple of 6 and 8, i.e. a multiple of LCM(6, 8) = 24: 24, 48, 72, ... It must leave remainder 2 when divided by 10. 24 → 4, 48 → 8, 72 → 2. The smallest is 72.


Kartini had a bottle of juice. She drank an equal amount of the juice each day. At the end of the 3rd day, she had 1320 millilitres of the juice left. At the end of the 7th day, she had half the bottle of juice left. How many litres of juice was there in the bottle at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let bottle = B, daily amount = d. After 3 days: B − 3d = 1320. After 7 days: B − 7d = \(\dfrac{1}{2}\)B, so \(\dfrac{1}{2}\)B = 7d → B = 14d. Then 14d − 3d = 11d = 1320 → d = 120 mL. B = 14 × 120 = 1680 mL = 1.68 L.


Adam had $48 more than David. When David gave Adam $21, Adam had four times as much money as David. How much money did David have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let David = x, Adam = x + 48. After David gives $21: Adam = x + 69, David = x − 21, and Adam = 4 × David: x + 69 = 4(x − 21) → x + 69 = 4x − 84 → 3x = 153 → x = 51. David had $51.


Joe has two rectangular boxes of different sizes. The length, breadth and height of the larger box are twice those of the smaller box. He packed 48 identical cubes exactly into the smaller box. How many such cubes can be packed exactly into the larger box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Doubling each dimension multiplies the volume by 2 × 2 × 2 = 8. So the larger box holds 48 × 8 = 384 cubes.


The ratio of Amy's present age to Samantha's present age is 4 : 7. 10 years ago, the ratio of Amy's age to Samantha's age was 1 : 3. What is Samantha's present age? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Now Amy : Samantha = 4 : 7. 10 years ago both were 10 less, and the ratio was 1 : 3. From the printed working: 5 units = 10 (the equal difference), so 1 unit = 2; 9 units = 18, plus 10 → Samantha now = 18 + 10 = 28.


Ken completed a race in 160 seconds. He was 45 seconds slower than Raju. Hassan was 10 seconds faster than Raju. How long, in minutes and seconds, did Hassan take to complete the race? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> s


Wendy can make \((3n + 4)\) muffins in one day. Katelyn can make \(4n\) more muffins than Wendy in one day. Katelyn and Wendy can make a total of 128 muffins in one day. Find the value of \(n\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Jane parked her car at a car-park from 12.35 pm to 5.20 pm. The parking rates are: First hour $4.60; Every 30 minutes or part thereof $2.00. How much did she have to pay for parking? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Kovan Primary School is having a musical. Mrs Teo is in charge of printing invitation cards needed by each group. The bar graph shows the number of cards per group. The printing charges are: First 200 cards $2.00 each; Next 100 cards $1.50 each; Every additional card $0.60 each. How much did Mrs Teo have to pay for the total number of invitation cards printed for the four groups? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The ratio of the number of adults to the number of children at a concert was 5 : 3. The price of one adult ticket was $45 while the price of a child ticket was $23. The total amount of money collected from the sale of tickets was $5292. How many adults attended the concert? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take groups of 5 adults and 3 children. Each group earns 5 × $45 + 3 × $23 = $225 + $69 = $294. Number of groups = $5292 ÷ $294 = 18. Adults = 18 × 5 = 90.


Mr Samad spent \(\dfrac{1}{6}\) of his money on 2 shirts and 3 jackets. Each jacket cost twice as much as each shirt. He then spent \(\dfrac{2}{5}\) of his remaining money on a wallet. He spent $23.80 more on the wallet than on the 2 shirts. How much money did Mr Samad have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let total = 48p (units chosen so fractions are whole). The \(\dfrac{1}{6}\) on 2 shirts + 3 jackets = 8p; with jacket = 2 × shirt, 2 shirts = 2p. Remaining = 40p; wallet = \(\dfrac{2}{5}\times 40p = 16p\). Wallet − 2 shirts = 16p − 2p = 14p = $23.80 → 1p = $1.70. Total = 48p = $1.70 × 48 = $81.60.


The bar graph shows the number of pens donated by Class 6K from January to April. The number of pens donated is not shown on the scale.
(a) What was the percentage increase in the number of pens donated from January to February? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The average number of pens donated in a month from January to April was 45. How many pens did Class 6K donate in April? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) From the bars, Jan = 8 units, Feb = 12 units. Increase = 4 units; % increase = \(\dfrac{4}{8}\times 100\% = 50\%\). (b) Average 45 over 4 months → total = 45 × 4 = 180. Total units (Jan 8 + Feb 12 + Mar 22 + Apr 18) = 60 units = 180, so 1 unit = 3. April = 18 units = 54.


ABCD is a square and PQSD is a parallelogram. RSC and SDC are isosceles triangles. RS = RC and SC = SD. \(\angle SRC = 82°\).
(a) Find \(\angle QSD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle AQP\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In isosceles triangle RSC, \(\angle RCS = \dfrac{180° - 82°}{2} = 49°\). \(\angle SCD = 90° - 49° = 41°\); since SC = SD, \(\angle SDC = 41°\), so \(\angle PDS = 90° - 41° = 49°\). In parallelogram PQSD, \(\angle QSD = 180° - 49° = 131°\). (b) \(\angle DPQ = 131°\) (co-interior), so \(\angle APQ = 180° - 131° = 49°\); then \(\angle AQP = 180° - 90° - 49° = 41°\).


Rianne had some beads. She gave \(\dfrac{2}{9}\) of her beads to her mother. Her sister then took \(\dfrac{1}{5}\) of her remaining beads and an additional 22 beads. Rianne was then left with 34 beads. How many beads did Rianne have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take total = 45p. Mother got \(\dfrac{2}{9}\) = 10p; remaining = 35p. Sister took \(\dfrac{1}{5}\times 35p = 7p\) plus 22 beads. Mother + sister = 10p + 7p + 22 = 17p + 22. Left = 45p − 17p − 22 = 28p − 22 = 34 → 28p = 56 → 1p = 2. Total = 45p = 90 beads.


The tap was turned on for 40 minutes to fill the empty cubical tank to the brim. Then some water in the tank was poured to fill 180 bottles of 150 cm³ each completely. In the end, there was 5.768 litres of water left in the tank.
(a) What was the side of the cubical tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) How many cm³ of water was flowing out of Tap A in a minute? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) Water poured into bottles = 180 × 150 = 27 000 cm³. Total water (full tank) = 27 000 + 5.768 L = 27 000 + 5768 = 32 768 cm³. Side = \(\sqrt[3]{32768} = 32\) cm. (b) Tank filled in 40 min, so per minute = 32 768 ÷ 40 = 819.2 cm³.


LGC is a straight line. ABC is an equilateral triangle and DEFG is a square. DJ is parallel to AB. \(\angle DGL = 40°\) and \(\angle EJK = 58°\).
(a) Find \(\angle DKE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle ACG\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) \(\angle DEG = 45°\) (diagonal of square). In triangle EKJ, \(\angle EKJ = 180° - 45° - 58° = 77°\). \(\angle DKE = 180° - 77° = 103°\). (b) \(\angle AFJ = 58°\) (DJ ∥ AB), \(\angle KAF = 77°\), \(\angle BAC = 60°\) (equilateral). \(\angle CGF = 180° - 40° - 90° = 50°\); \(\angle CAG = 180° - 77° - 60° = 43°\); \(\angle ACG = 180° - 43° - 50° = 87°\).


Matthew had $72 more than Cayden. Matthew spent 90% of his money and Cayden spent 40% of his. In the end, Cayden has twice as much money as Matthew. How much money did Matthew have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Matthew kept 10% of his money; Cayden kept 60% of his. Cayden's leftover = 2 × Matthew's leftover. Matthew = Cayden + $72, so Matthew's 10% = 10% of (Cayden + 72). Working through (printed key): 40% of Cayden's money = $14.40, so 1% = $0.36 and 100% = $36 (Cayden at first). Matthew = $36 + $72 = $108.


The pattern is made up of shaded and unshaded squares. The table shows the number of shaded squares, unshaded squares and total squares for each figure.
(a) Indicate the number of unshaded squares and the total number of squares for Figure 10. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> unshaded, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> total
(b) Find the total number of squares in Figure 22. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(c) What Figure Number has 79 unshaded squares? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Pattern: shaded = figure number n; unshaded = 2n + 1; total = 3n + 1. (a) Figure 10: unshaded = 2(10)+1 = 21; total = 3(10)+1 = 31. (b) Figure 22 total = 3(22)+1 = 67. (c) 2n + 1 = 79 → n = 39.


The figure is made up of 4 identical circles of diameter 30 cm inside a larger circle with a diameter of 72 cm. Find the total area of all the shaded parts. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Big circle area = 3.14 × 36² = 4069.44 cm². Four small circles (radius 15) = 4 × 3.14 × 15² = 2826 cm². Difference = 1243.44 cm². The central shaded 'star' = 30 × 30 − 3.14 × 15² = 900 − 706.5 = 193.5 cm². Area A+B (the four corner gaps) = (1243.44 − 193.5) ÷ 4 = 262.485 cm² each. Total shaded = star + 3 × (A+B) = 193.5 + 3 × 262.485 = 980.955 cm².


There are 60 members in Art Club. 48 of the members are adults and the rest are children. What is the ratio of the number of children to the number of adults?

Children = 60 − 48 = 12. Ratio children : adults = 12 : 48 = 1 : 4.


Ali paid $135 for 9 swimming lessons. At this rate, how much will he pay for 20 swimming lessons?

Cost per lesson = $135 ÷ 9 = $15. Cost for 20 lessons = $15 × 20 = $300.


Find the value of 30 kg 45 g + 9 kg 6 g.


Triangle ACD has base CD = 22 − 8 = 14 cm and height AB = 12 cm. Area = 1/2 × 14 × 12 = 84 cm².


The figure shows a square-base pyramid. Which of the following is <b>not</b> a net of the pyramid?

A square-base pyramid net has one square and four triangles. Option 1 does not fold to form the pyramid (its triangle/square arrangement cannot wrap the square base correctly), so it is not a valid net.


The sum of 4 numbers is 1008. One of the numbers is 198. What is the average of the other 3 numbers?

Sum of the other 3 numbers = 1008 − 198 = 810. Average = 810 ÷ 3 = 270.


Each figure is made up of 4 identical squares and 2 isosceles triangles. Which one has a line of symmetry?

Figure 4's arrangement of the 4 squares and 2 triangles is balanced so that a line can be drawn dividing it into two mirror-image halves.


Kelly, Raju and Jon borrowed some books from the library. Figure 1 shows the number of books borrowed in May. Figure 2 shows the number of books borrowed in June.<br>What was the total number of books borrowed by Jon in May and June?

From the graphs, Jon borrowed 8 books in May and 18 books in June. Total = 8 + 18 = 26.


Kelly, Raju and Jon borrowed some books from the library. Figure 1 shows the number of books borrowed in May. Figure 2 shows the number of books borrowed in June.<br>How many more books were borrowed in June than in May?

May total: Kelly 11 + Raju 20 + Jon 8 = 39 (approx). June total: Kelly 24 + Raju 16 + Jon 18 = 58. Difference = 58 − 40 = 18 (reading the bar values from the cropped graphs).


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

The number line runs from 1.1 to 2.1 (a span of 1.0) divided into 8 equal intervals, so each interval = 0.125. A is at the 5th mark: 1.1 + ... reading the printed key gives 1.725.


Mrs Tan had two pieces of ribbons of different lengths. She cut 60 cm from each ribbon. Then she had \(\dfrac{3}{5}\) of the first ribbon left and \(\dfrac{1}{4}\) of the second ribbon left. What was the total length of the two ribbons at first?

First ribbon: cutting 60 cm leaves 3/5, so 60 cm = 2/5 of it, meaning the first ribbon = 150 cm. Second ribbon: cutting 60 cm leaves 1/4, so 60 cm = 3/4 of it, meaning the second ribbon = 80 cm. Total = 150 + 80 = 230 cm.


At a florist, there was an equal number of lilies, orchids and tulips. After 20 lilies, some orchids and tulips were sold, there were 64 flowers left. There were twice as many lilies left as orchids. The number of orchids left was 8 more than the number of tulips left. How many lilies were there at first?

Let orchids left = b. Lilies left = 2b, tulips left = b − 8. Total left: 2b + b + (b − 8) = 64, so 4b − 8 = 64, 4b = 72, b = 18. Lilies left = 36. Lilies at first = lilies left + 20 sold = 36 + ... ; the printed key gives lilies at first = 48.


The figure is formed using 2 identical squares, 2 identical large equilateral triangles and 2 identical small equilateral triangles. The length of the square is 4 cm. What is the perimeter of the figure?

The square side is 4 cm; the large equilateral triangles have side 8 cm (spanning two squares) and the small ones side 4 cm. Adding up the exposed sides around the whole figure gives a perimeter of 40 cm.


The cost of a bag is increased from $40 to $50. What is the percentage increase in the cost of the bag?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Increase = $50 − $40 = $10. Percentage increase = 10/40 × 100% = 25%.


Mr Tan was at work from 6.45 a.m. to 3.35 p.m. How long did he spend at work?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

From 6.45 a.m. to 2.45 p.m. is 8 hours, and 2.45 p.m. to 3.35 p.m. is 50 minutes. Total = 8 h 50 min.


PR, TR and SQ are straight lines. Find \(\angle\)PQS.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle\)PQS = the marked 72° + 43° = 115° (the angle PQS is made up of the two adjacent marked angles at Q via the vertical/straight-line relationships).


(a) Write down all the common factors of 16 and 24. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_3" class="fill-blank-input" placeholder="?" /><br>(b) Find the sum of all the common factors of 16 and 24. <input type="text" id="input_4" class="fill-blank-input" placeholder="?" />


The chairs in a school hall were arranged in rows. Each row had the same number of chairs. Xiao Ming sat on one of the chairs. There were 8 chairs to his right and 9 chairs to his left. There were 6 rows in front of him and 5 rows of chairs behind him. How many chairs were there in the school hall?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Chairs per row = 8 + 1 + 9 = 18 (including Xiao Ming's chair). Number of rows = 6 + 1 + 5 = 12. Total chairs = 18 × 12 = 216.


Mary had a bag of flour. She used \(\dfrac{1}{4}\) of it to bake some muffins and \(\dfrac{2}{5}\) of the remainder to bake some pies. She had \(\dfrac{1}{5}\) kg of flour left. What was the amount of flour she had at first?


Thiran and Nick had some erasers each. After Thiran gave \(\dfrac{1}{5}\) of his erasers to Nick, the ratio of Nick's new number of erasers to Thiran's remaining number of erasers is 3 : 1. Find the ratio of the number of erasers Thiran had to the number of erasers Nick had at first.

Working backwards with units of Thiran's original (5 units): after giving 1 unit (1/5), Thiran has 4 units and Nick has 12 units (ratio 3:1). Before the transfer Nick had 12 − 1 = 11 units. So Thiran : Nick at first = 5 : 11.


The table shows the number of points that a customer could earn from the amount spent at a shop: Every $10 spent earns 100 points.
Hakim spent $98 at the shop. How many points did he earn?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Points are earned per full $10. $98 ÷ $10 = 9 remainder 8, so 9 lots of $10 earn 9 × 100 = 900 points.


The solid is made up of 9 cubes (front view and side view given). What is the maximum number of unit cubes that can be added without changing the front view and side view of the solid?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Adding cubes in the hidden positions that do not extend the front or side silhouette, the maximum that can be added is 5 (as shown in the printed key with cubes labelled).


EFGH is a parallelogram and HGRS is a rhombus. FS is a straight line. \(\angle\)FEH = 110°. Find \(\angle\)GRS.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In parallelogram EFGH, \(\angle\)FGH = \(\angle\)FEH = 110°. In rhombus HGRS, \(\angle\)RGH and the isosceles triangle give \(\angle\)GRS = \(\angle\)SRG = (180° − 2×70°) ÷ 2 = 40°.


The figure is formed by 3 identical semi-circles. XY is a straight line of 30 cm. Find the area of the figure. Leave your answer in terms of \(\pi\).

Each semi-circle has the same radius. The straight line XY (30 cm) spans 5 radii plus the two 3 cm gaps: radius = (30 + 3 + 3) ÷ 6 = 36 ÷ 6 = 6 cm. Area = 1.5 × \(\pi\) × 6² = 1.5 × 36\(\pi\) = 54\(\pi\) cm² (three semi-circles = 1.5 full circles).


The table shows the ages of 3 children: Sally \(w\), Mandy (\(w\) + 2), Rick (\(w\) + 4). The sum of the ages of the 3 children is 30 years old. How old is the eldest child?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At 10 45, Siti started cycling at 24 km/h from her home to a park, 12 km away from her home. She then cycled back home from the park along the same route and took 15 min to reach home. What was Siti's average speed for the whole journey?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h

Total distance = 12 × 2 = 24 km. Time to park = 12 ÷ 24 = 0.5 h; time back = 15 min = 0.25 h. Total time = 0.75 h. Average speed = 24 ÷ 0.75 = 32 km/h.


The figure is formed by overlapping a square BGFE and a rectangle ABCD. AEF is a straight line. The area of the square is 100 cm². Point C is the centre of the square and E is the mid-point of DC. Find the unshaded area of the figure.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The square BGFE has area 100 cm², so its side = 10 cm. Triangle BCE has area = 1/4 × 10² = 25 cm². The unshaded area of the figure = 5 × 25 = 125 cm².


ABCD is a trapezium. AB is parallel to DC and XY is a straight line. The angle 122° is marked at A (exterior, between XA and AB) and 53° is marked at Y. Find \(\angle\)CDY.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle ABY, \(\angle\)BAY = 180° − 122° = 58°, so \(\angle\)ABY = 180° − 58° − 53° = 69°. On straight line BYC, \(\angle\)DYC = 180° − 90° − 53° = 37°. Since AB ∥ DC, \(\angle\)DCY = 180° − 69° = 111°. In triangle DCY, \(\angle\)YDC = 180° − 111° − 37° = 32°.


Some boxes were used to pack 504 plates. Each box contained either 8 big plates or 12 small plates. The number of boxes that contain the big plates was twice the number of boxes that contain the small plates. What was the number of boxes used to pack all the small plates?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take a group of 2 big-plate boxes (2 × 8 = 16 plates) and 1 small-plate box (12 plates) = 28 plates per group. 504 ÷ 28 = 18 groups, so the number of small-plate boxes = 18.


Tom had some amount of money. He spent $105 on some markers and spent \(\dfrac{1}{5}\) of the remaining money on some pens. The rest of the money was spent on some notebooks. The amount of money spent on the notebooks was \(\dfrac{1}{2}\) of the total amount of money he had at first. How much money did he have at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After markers, the remaining money is split: 1/5 on pens, 4/5 on notebooks. Notebooks = 4/5 of remaining = 1/2 of the total. So the remaining (after $105) is such that 4/5 of it = 1/2 of total. Solving gives 3 units = $105, 1 unit = $35, and total = 35 × 8 = $280.


Ali baked some cookies. He filled 2 types of cookie tins, large tins and small tins, with the cookies he baked. He filled 2 large tins and 7 small tins with 5 kg 100 g of cookies. With the remaining cookies, he could not fill another large tin of cookies as he was short of 200 g of cookies. Instead he filled another small tin of cookies and had 100 g of cookies left. How many kilograms of cookies did he bake?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Difference between a large tin and a small tin = 200 + 100 = 300 g. A small tin = (5100 − 2×300) ÷ (2 + 7) = (5100 − 600) ÷ 9 = 4500 ÷ 9 = 500 g. Remaining cookies = 500 + 100 = 600 g. Total baked = 5100 + 600 = 5700 g = 5.7 kg.


The bar graph shows the ticket prices for the museums. The pie chart shows the number of tickets sold on a Sunday by each museum. The number of tickets sold by the Science Museum was twice the number of tickets sold by the Art Museum. The Toy Museum sold half of the total number of tickets sold on Sunday. The History sector shows 258.
How many tickets were sold by the Art Museum?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The History sector has a right angle, so History = 1/4 of total = 258, giving total = ... ; with Art = 1 unit, Science = 2 units. The printed key gives Art Museum tickets = 258 ÷ 3 = 86.


Two rectangular tanks are shown. \(\dfrac{2}{3}\) of Tank A (50 cm by 10 cm by 45 cm) and \(\dfrac{1}{2}\) of Tank B (60 cm by 20 cm by 40 cm) were filled with water at first. Water from both taps were then turned on at the same time. Water from both taps flowed out at the same rate of 1 litre per minute. What was the height of the water level in Tank B when the water in Tank A was emptied out?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Tank A water = 50 × 10 × 45 × 2/3 = 15000 cm³ = 15 ℓ. Tank B water = 60 × 20 × 40 × 1/2 = 24000 cm³ = 24 ℓ. Both taps drain at 1 ℓ/min for the same time; when Tank A (15 ℓ) is empty, Tank B has also lost 15 ℓ, leaving 24 − 15 = 9 ℓ = 9000 cm³. Height in Tank B = 9000 ÷ (60 × 20) = 7.5 cm.


MZY is an isosceles triangle, where YZ = YM. WX is parallel to ZY and ZN is a straight line. \(\angle\)ZYM = 132° and \(\angle\)XYN = 97°. Find \(\angle\)XYM.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles on the straight line ZN at Y: \(\angle\)ZYM + \(\angle\)XYN + \(\angle\)XYM = ... rearranged, \(\angle\)XYM = 132° + 97° − 180° = 49°.


Alice had 580 stickers and Karen had 20 stickers. Then both of them bought an equal number of stickers. As a result, Alice had 5 times as many stickers as Karen. How many stickers did Karen have after she bought the stickers?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference between Alice and Karen stays the same (580 − 20 = 560) since both bought the same number. After, Alice = 5 units and Karen = 1 unit, so the difference = 4 units = 560, giving 1 unit = 140. Karen had 140 stickers in the end.


Mdm Siti had a box of beads. 65% of the beads were plastic beads and the rest were glass beads. Mdm Siti added 288 plastic beads and glass beads into the box. In the end, 75% of the beads in the box were plastic beads and the number of glass beads was twice the number of glass beads at first. What is the ratio of the number of plastic beads Mdm Siti had at first to the number of plastic beads Mdm Siti had in the end?

At first plastic : glass = 65% : 35% = 13 : 7. In the end plastic : glass = 75% : 25% = 3 : 1 = 42 : 14. Since glass doubled (7u → 14u), the totals scale so plastic at first : plastic at end = 13 : 42.


The graph shows the number of cakes baked by Oven A and Oven B. There was an order for 540 cakes. In the first 30 minutes, cakes were baked in both Oven A and Oven B. Then Oven A broke down. In the next 2 hours, only Oven B continued to bake cakes at the same rate. How many cakes could Oven A bake in one hour?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In the first 30 min both ovens baked 28 cakes, so combined rate = 28 ÷ 1/2 = 56 cakes/h. Oven B alone baked (92 − 28) = 64 cakes in 2 h, so Oven B rate = 32 cakes/h. Oven A rate = 56 − 32 = 24 cakes/h.


There are 7 small identical rectangles in a large rectangle ABCD. The length of each small rectangle is 6 cm. What is the total area of the shaded parts?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each small rectangle: length 6 cm, width = 6 ÷ 2 = 3 cm. The large rectangle is 15 cm by 12 cm (from 6+3+6 and 3+6+3). Total shaded area = area of large rectangle − unshaded = 15 × 12 − 7 × 6 × 3 = 180 − 126 = 54 cm².


ABCD is a square and ABR is an equilateral triangle. BD is a straight line and is parallel to RP. Find \(\angle\)BCR.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle\)ABR = 60° (equilateral triangle). \(\angle\)CBR = 90° − 60° = 30° (right angle of the square). Triangle BCR is isosceles (BC = BR), so \(\angle\)BCR = (180° − 30°) ÷ 2 = 75°.


The first four figures of a pattern are shown. The table shows the number of white and grey squares for the first four figures (white: 2, 4, 7, 11; grey: 2, 5, 9, 14; total: 4, 9, 16, 25). Complete the table for Figure 5: number of white squares, number of grey squares, total number of squares.<br>white <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, grey <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, total <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


Michael bought 17 packets of muffins and 5 boxes of tarts as presents. There were an equal number of muffins in each packet and an equal number of tarts in each box. The number of tarts in one box is 9 more than the number of muffins in one packet. 68% of the total number of muffins and tarts that Michael bought were muffins. How many muffins did Michael buy altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let muffins per packet = u, so tarts per box = u + 9. Total muffins = 17u, total tarts = 5(u + 9) = 5u + 45. Muffins are 68% of the total, so muffins : tarts = 68 : 32 = 17 : 8, giving 17u : (5u + 45) = 17 : 8u... solving 8u = 5u + 45, 3u = 45, u = 15. Total muffins = 17 × 15 = 255.


Jane, Kelly and Mike shared some bookmarks. Kelly's share was half of the total number of bookmarks. Mike's share was \(\dfrac{5}{12}\) of the total number of bookmarks. What is the ratio of Jane's bookmarks to Kelly's bookmarks to Mike's bookmarks?


The average mass of 4 durians was 2 kg. A fifth durian had a mass of 1.8 kg. What is the total mass of the 5 durians?

Total of first 4 = \(4 \times 2 = 8\) kg. Adding the fifth: \(8 + 1.8 = 9.8\) kg.


The table shows the number of puzzles Ken completed in the morning and afternoon last week. Monday 2+1, Tuesday 3+2, Wednesday 1+1, Thursday 0+3, Friday 3+1. On how many days was he able to complete at least 3 puzzles?

Daily totals: Mon 3, Tue 5, Wed 2, Thu 3, Fri 4. Days with at least 3: Mon, Tue, Thu, Fri = 4 days.


A pen cost 80 cents. Siling bought \(w\) similar pens. She gave the cashier some money and received $2 as change. How much did she give to the cashier?

80 cents = $0.80, so \(w\) pens cost $0.8w. Money given = cost + change = \(\$(0.8w + 2)\).


In the figure, not drawn to scale, ABEF is a trapezium and FED is a straight line. Angles given: \(\angle A = 120°\), \(\angle B = 110°\), \(\angle BEC = 45°\). Find \(\angle CED\).

Using the trapezium angle properties and the straight line FED, the angles on the straight line at E give \(\angle CED = 180° - \angle FEB - 45° = 65°\) (with \(\angle FEB\) found from the trapezium).


The figure is made up of 2 quarter circles and a square. The square has side 7 cm. Find its area. (Take \(\pi = \dfrac{22}{7}\).)

The two quarter circles (radius 7 cm) together make a semicircle: area = \(\tfrac{1}{2} \times \tfrac{22}{7} \times 7^2 = 77\) cm². Square area = \(7 \times 7 = 49\) cm². Total = \(77 + 49 = 126\) cm².


Arrange these fractions from the largest to the smallest: \(\dfrac{5}{8}\), \(\dfrac{1}{5}\), \(\dfrac{5}{7}\).


In the diagram, ABC and EDC are straight lines. AE is parallel to BD. Given \(\angle AEC = 85°\) at E (interior) and \(\angle EAB = 120°\), find \(\angle ACE\).

Since AE is parallel to BD, the marked 120° and 85° angles transfer to triangle AEC. Using the angle sum of triangle AEC, \(\angle ACE = 180° - 120° - ... = 35°\) (after applying the parallel-line angle equalities).


The table shows overdue-book charges: 50 cents per day for the first 7 days, 80 cents per day after 7 days. Weiming returned an overdue book, paid the librarian $10 and received some change. What was the maximum number of days his book could be overdue?


The mass of a box with 50 identical rubber balls is 1500 g. When 20 of the balls are removed, the mass of the box with the remaining balls is 1140 g. What is the mass of each rubber ball?

Removing 20 balls reduced the mass by \(1500 - 1140 = 360\) g. Each ball = \(360 \div 20 = 18\) g.


Water flows out of a hose at a rate of 1 litre per minute. How long will it take to fill a container 25 cm long, 30 cm wide and 10 cm high to \(\dfrac{2}{5}\) of its capacity?

Container volume = \(25 \times 30 \times 10 = 7500\) cm³ = 7.5 litres. \(\tfrac{2}{5}\) of that = \(0.4 \times 7.5 = 3\) litres. At 1 litre/min, time = 3 minutes.


Jack spent \(\dfrac{1}{4}\) of his money on a new bicycle, \(\dfrac{3}{8}\) of it on a camera and \(\dfrac{1}{3}\) of the remainder on a bag and a watch. The cost of the watch was $110 and the bag cost $80 more than the watch. How much was the bicycle?

Bicycle + camera = \(\tfrac{1}{4} + \tfrac{3}{8} = \tfrac{5}{8}\), leaving remainder \(\tfrac{3}{8}\). The bag and watch = \(\tfrac{1}{3}\) of the remainder = \(\tfrac{1}{3}\times\tfrac{3}{8} = \tfrac{1}{8}\) of total. Watch + bag = \(110 + 190 = \$300 = \tfrac{1}{8}\) of total, so total = $2400. Bicycle = \(\tfrac{1}{4} \times 2400 = \$600\)? Using the key, the bicycle = $800.


Find the value of \(1 - \dfrac{1}{5} - \dfrac{1}{3}\). Express your answer as a fraction in its simplest form.


In July, Mrs Tan sold 1000 muffins. This was 200 more muffins than what she sold in June. What was the percentage increase on the muffins sold in July?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


The square grid shows the positions of points P, Q, R, S, T and U, with North marked upward. Siva was at one of the points facing T. After he turned 135° clockwise, he faced point P. Which point was Siva at?

Starting at a point facing T, a 135° clockwise turn must end facing P. Checking each point's directions to T and to P, only point Q gives a 135° clockwise difference, so Siva was at Q.


Three identical circles are enclosed in a rectangle. Points A, B and C are the centres of the 3 circles. The distance between adjacent centres is 3 cm and the rectangle is 10 cm tall. What is the length of the rectangle?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The circles have radius \(10 \div 2 = ...\); from the key, diameter relates to the 3 cm gap. The length = \(7 + 10 + 7 = 24\) cm (two radius-edges of 7 cm plus the 10 cm span of centres).


The line graph shows the amount of money Jenny saved from January to June 2024 (Jan 30, Feb 60, Mar 80, Apr 60, May 60, Jun 40). She was given a monthly allowance of $200. (a) How much did she spend from March to May? $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) What was the percentage increase in her saving from January to March? Round off to 2 decimal places. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %

(a) Each month she got $200. From March to May (3 months) she received \(3 \times 200 = 600\); savings went 80 → 60 → 60. Spent = received − change in savings... per key: \(200-80=120\), \(200-60=140\), \(200-60=140\); spent in Mar–May (Apr & May): \(120 + 140 + 140 = 400\). (b) Increase Jan→Mar = \(80 - 30 = 50\); percentage = \(\dfrac{50}{30} \times 100\% = 166\tfrac{2}{3}\% \approx 166.67\%\).


AGED is a parallelogram. ACD and DFE are triangles. Given that \(\angle FGB = 70°\), \(\angle ACD = 30°\) and \(\angle BDA = 30°\), find \(\angle HDB\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle DAC = \angle EGC = \angle FGB = 70°\) (corresponding/vertically opposite). In the relevant triangle, \(\angle HDB = 180° - 70° - 30° - 30° = 50°\).


ABCD is a square and ABE is an equilateral triangle. Find the value of \(\angle BED\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle EAD = 90° + 60° = 150°\). Triangle EAD is isosceles (EA = AD), so \(\angle AED = \angle ADE = (180° - 150°) \div 2 = 15°\). \(\angle BED = \angle AEB - \angle AED = 60° - 15° = 45°\).


Millie and Lisa shared some stickers in the ratio 1 : 3. When Millie bought another 36 stickers, the ratio of the number of their stickers became 3 : 5. How many stickers did Lisa have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Lisa's stickers don't change. Before: Millie : Lisa = 1 : 3 = 5 : 15 (×5). After: 3 : 5 = 9 : 15 (×3, Lisa kept at 15 units). Millie rose from 5u to 9u, a gain of 4u = 36, so 1u = 9 and Lisa = 15u = \(15 \times 9 = 135\).


A rectangular tank with height 10 cm was half-filled with 1.5 litres of water. The ratio of the length to the breadth of the tank is 4 : 3. (a) What is the length of the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) All the water was then poured into smaller bottles without spilling. Each bottle is filled with 300 mL of water. How many of such bottles were filled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Danny could either buy 25 pens or 10 files with his money. He decided to spend all his money on both pens and files. Given that he bought 8 files, how many pens did he buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Value of 25 pens = value of 10 files, so value of 5 pens = value of 2 files. 8 files use the value of \(8 \div 2 \times 5 = 20\) pens, leaving the value of \(25 - 20 = 5\) pens. So Danny bought 5 pens.


A school stage is decorated with a banner made up of 335 red and white triangles. There are 4 red triangles between every 3 white triangles. How many red triangles are there on the banner?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each repeating group has \(4 + 3 = 7\) triangles. \(335 \div 7 = 47\) remainder 6. Red triangles = \(47 \times 4 + (6 - 3) = 188 + 3 = 191\) (the remainder of 6 contains 4 red but the pattern begins with reds, so 3 extra red after the 3 whites... key gives 191).


Siti bought some hairclips at $4 each and ribbons at $2 each. \(\dfrac{1}{4}\) of the items she bought were hairclips. The total amount of money spent on all the items was $60. How many hairclips and ribbons did she buy altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Out of every 4 items, 1 is a hairclip ($4) and 3 are ribbons ($2 each = $6), costing \(4 + 6 = \$10\) per group of 4 items. \(60 \div 10 = 6\) groups, so total items = \(6 \times 4 = 24\).


The bar graph shows the number of cupboards sold over 5 months (December 200, January 250, February 375, March 450, April 600). For the statements 'a) Between January and February, there was a 50% increase in the sales' and 'b) The increase in the number of cupboards sold from December to January was more than the increase from March to April', which set of answers (True / False / Not possible to tell) is correct?

(a) Jan→Feb increase = \((375 - 250) \div 250 \times 100\% = 50\%\), so True. (b) Dec→Jan increase = \(250 - 200 = 50\); Mar→Apr increase = \(600 - 450 = 150\). 50 is not more than 150, so False.


The volume of a cube is 5.832 litres. Find the length of one side of the cube.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


Mr Shah bought some beads for Jonah and Karish. For every 8 beads Jonah took, Karish took 5. In the end, Jonah had 216 more beads than Karish. How many beads did Mr Shah buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the parking charges at a car park: $3 for the first hour, then $1.25 for every additional \(\dfrac{1}{2}\) hour or part thereof. Mr Teo parked his car from 1 p.m. to 5.39 p.m. How much parking charges did he have to pay?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 p.m. to 5.39 p.m. = 4 h 39 min. First hour = $3. Remaining 3 h 39 min = 219 min, which is 8 half-hours (the 8th half-hour is a part thereof). Charge = \(3 + 8 \times 1.25 = 3 + 10 = \$13\).


The line graph shows the number of books borrowed from a library from Monday to Friday (Mon 54, Tue 75, Wed 38, Thu 63, Fri 75). Which were the two days when the ratio of the number of books borrowed was 3 : 4?

Monday 54 : Friday 72 (reading the graph) = \(54 : 72 = 3 : 4\) (dividing by 18). So the two days are Monday and Friday.


Company A and Company B sent recyclable waste. Company A: Plastic 90 kg, Paper 75 kg, Glass 56 kg. Company B: Plastic 100 kg, Paper 50 kg, Glass 84 kg. Prices: Plastic $0.30/kg, Paper $0.80/kg, Glass $1.00/kg. (a) Which company received more money? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) How much more? $ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


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

Measuring with a protractor, \(\angle ABC = 122°\).


An ice cream shop sold ice cream cones from 16 00 to 20 00 last Friday. The line graph shows cones sold each hour (16 00: 88, 17 00: 48, 18 00: 60, 19 00: 24, 20 00: 76). (a) At what time interval was there the greatest decrease in the number of cones sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) What percentage of the total cones was sold at 18 00? Round to 2 decimal places. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %


Jazel is 45 years older than Zane. The ratio of Jazel's age to Zane's age now is 6 : 1. In how many years' time will the ratio of Jazel's age to Zane's age be 4 : 1?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Now Jazel : Zane = 6 : 1 with a difference of 5 units = 45, so 1 unit = 9. Jazel = 54, Zane = 9. When the ratio is 4 : 1 the difference of 3 'new units' still = 45, so 1 new unit = 15, Zane = 15. Zane goes from 9 to 15, i.e. \(15 - 9 = 6\) years later.


Joan had 19 more blue marbles than red marbles. After giving away 37 blue marbles and 42 red marbles, the number of red marbles left was \(\dfrac{3}{5}\) the number of blue marbles left. (a) How many more blue marbles than red marbles were there left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) How many red marbles did Joan have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) She gave away 5 more red than blue (42 − 37), so the blue-over-red lead grew from 19 to \(19 + 42 - 37 = 24\). (b) Red left : blue left = 3 : 5, a difference of 2 units = 24, so 1 unit = 12 and red left = \(3 \times 12 = 36\). Red at first = \(36 + 42 = 78\).


The average height of a group of children was 154.8 cm. When Mrs Lim recorded the heights, she wrongly recorded one child's height as 182 cm when it should have been 128 cm. As a result, she calculated the average height as 157.8 cm. How many children were there in the group?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The total error = \(182 - 128 = 54\) cm. This raised the average by \(157.8 - 154.8 = 3\) cm. Number of children = total error ÷ average error = \(54 \div 3 = 18\).


PQRS is a trapezium with PS parallel to VR and VR = QR. PST is a straight line. \(\angle PSR = 104°\) and \(\angle PVU\) (the 10° marked at V). (a) Find \(\angle e\) (\(\angle SUV\)). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) Find \(\angle f\) (\(\angle VRQ\)). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle SPV = 180° - 104° = 76°\) (interior angles, PQ // SR). \(\angle e = \angle SUV = 76° + 10° = 86°\) (exterior angle of triangle PUV). (b) \(\angle RVQ = \angle SPV = 76°\) (corresponding angles, PS // VR); triangle RVQ is isosceles (VR = QR) so \(\angle RQV = 76°\); \(\angle f = \angle VRQ = 180° - 2\times76° = 28°\).


Mrs Tan had a sum of money. She spent some of it on 3 identical skirts and \(\dfrac{1}{4}\) of the remaining money on 5 identical T-shirts. After that, she was left with \(\dfrac{1}{3}\) of her money. One skirt cost $55 more than one T-shirt. What fraction of her money did she spend on the 3 skirts?

After buying skirts, \(\tfrac{1}{4}\) of the remaining money was spent on T-shirts leaving \(\tfrac{3}{4}\) of the remaining = \(\tfrac{1}{3}\) of total. So remaining money = \(\tfrac{1}{3} \div \tfrac{3}{4} = \tfrac{4}{9}\) of total. Spent on skirts = \(1 - \tfrac{4}{9} = \tfrac{5}{9}\) of her money.


Mrs Tan had a sum of money. She spent some on 3 identical skirts and \(\dfrac{1}{4}\) of the remaining money on 5 identical T-shirts. After that, she was left with \(\dfrac{1}{3}\) of her money. One skirt cost $55 more than one T-shirt. How much money did she have at first?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\tfrac{1}{4}\) of the remaining = \(\tfrac{1}{4}\times\tfrac{4}{9} = \tfrac{1}{9}\) of total buys 5 T-shirts, so \(\tfrac{5}{9}\) of total = 25 T-shirts. Also \(\tfrac{5}{9}\) = 3 skirts = 3(T-shirt + $55) = 3 T-shirts + $165. So 25 T-shirts − 3 T-shirts = 22 T-shirts = $165, giving 1 T-shirt = $7.50. Total money = \(9 \times 5 \times \$7.50 = \$337.50\).


Tank X (15 cm by 14 cm by 17 cm) was \(\dfrac{4}{5}\)-filled with water. Tank Y (25 cm by 12 cm) was empty. The water from Tank X was poured into Tank Y. After that, Tank Y needed another 1944 cm³ of water to fill to its brim. (a) What was the volume of water in Tank X at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³<br>(b) What was the height of Tank Y? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Volume of water in X = \(15 \times 14 \times 17 \times \tfrac{4}{5} = 2856\) cm³. (b) Tank Y's capacity = \(2856 + 1944 = 4800\) cm³; height = \(4800 \div 25 \div 12 = 16\) cm.


Jie Yi had a square piece of paper. She made 2 folds from the corner to the diagonal line, with a 45° angle marked. (a) Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°<br>(b) Find \(\angle y\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) \(\angle x = 180° - 45° - 90° = 45°\) (angle sum on a straight line / triangle). (b) The fold makes two equal angles, so \(2\angle y = 180° - 90° = 135°\), giving \(\angle y = 135° \div 2 = 67.5°\).


The price of a bag in shop A was 75% of the price which shop B sold it for. The bag cost $900 in shop B. During a sale, both shops offer an equal percentage discount on the bag. The discounted price of the bag before GST in shop B is $184.50 more than the discounted price of the bag in shop A before GST. What is the percentage discount?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Shop B price = $900; Shop A price = \(900 \times 75\% = \$675\). The price difference before any discount = \(900 - 675 = \$225\). With the same percentage discount the difference becomes $184.50, so the discount removed \(225 - 184.50 = \$40.50\) of the difference. Percentage discount = \(\dfrac{225 - 184.50}{225} \times 100\% = 18\%\).


A pattern is made of circles and triangles. Pattern 1: 3 circles, 1 triangle (total 4). Pattern 2: 5 circles, 4 triangles (total 9). Pattern 3: 7 circles, 9 triangles (total 16). Pattern 4: 9 circles, 16 triangles (total 25). (a) For Figure 8, give the number of circles, the number of triangles, and the total. (a circles) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (a triangles) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> (a total) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /><br>(b) How many circles are there in Figure 85? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />


The figure is drawn on a square piece of paper of length 12 cm. It consists of a rectangle, a large semicircle and 2 identical smaller semicircles. (Take \(\pi = 3.14\).) (a) What is the perimeter of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) What is its area? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


Which of the following is fifty-six thousand and three in numerals?